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ially moving with a velocity of 15 m/s in the +x-direction. Mass B is twice as massive and is initially moving with a velocity of 10 m/s in the –x-direction. The two objects collide, and after the collision, mass A moves with a speed of 15 m/s in the –x-direction. (a) What is the final velocity of mass B after the collision? (b) Calculate the change in kinetic energy as a result of the collision, assuming mass A is 5.0 kg. 8.6 Collisions of Point Masses in Two Dimensions 51. Two cars of equal mass approach an intersection. Car A is moving east at a speed of 45 m/s. Car B is moving south at a speed of 35 m/s. They collide inelastically and stick together after the collision, moving as one object. Which of the following statements is true about the center-of-mass velocity of this system? a. The center-of-mass velocity will decrease after the collision as a result of lost energy (but not drop to zero). b. The center-of-mass velocity will remain the same after the collision since momentum is conserved. c. The center-of-mass velocity will drop to zero since the two objects stick together. d. The magnitude of the center-of-mass velocity will remain the same, but the direction of the velocity will change. 52. Car A has a mass of 2000 kg and approaches an intersection with a velocity of 38 m/s directed to the east. Car B has a mass of 3500 kg and approaches the intersection with a velocity of 53 m/s directed 63° north of east. The two cars collide and stick together after the collision. Will the center-of-mass velocity change as a result of the collision? Explain why or why not. Calculate the center-of-mass velocity before and after the collision. 356 Chapter 8 \| Linear Momentum and Collisions This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 9 \| Statics and Torque 357 9 STATICS AND TORQUE Figure 9.1 On a short time scale, rocks like these in Australia's Kings Canyon are static, or motionless relative to the Earth. (credit: freeaussiestock.com) Chapter Outline 9.1. The First Condition for Equilibrium 9.2. The Second Condition for Equilibrium 9.3. Stability 9.4. Applications of Statics, Including Problem-Solving Strategies 9.5. Simple Machines 9.6. Forces and Torques in Muscles and Joints Connection for AP® Courses What might desks, bridges, buildings, trees, and mountains have in common? What do these objects have in common with a car moving at a constant velocity? While it may be apparent that the objects in the first group are all motionless relative to Earth, they also share something with the moving car and all objects moving at a constant velocity. All of these objects, stationary and moving, share an acceleration of zero. How can this be? Consider Newton's second law, F = ma. When acceleration is zero, as is the case for both stationary objects and objects moving at a constant velocity, the net external force must also be zero (Big Idea 3). Forces are acting on both stationary objects and on objects moving at a constant velocity, but the forces are balanced. That is, they are in equilibrium. In equilibrium, the net force is zero. The first two sections of this chapter will focus on the two conditions necessary for equilibrium. They will not only help you to distinguish between stationary bridges and cars moving at constant velocity, but will introduce a second equilibrium condition, this time involving rotation. As you explore the second equilibrium condition, you will learn about torque, in support of both Enduring Understanding 3.F and Essential Knowledge 3.F.1. Much like a force, torque provides the capability for acceleration; however, with careful attention, torques may also be balanced and equilibrium can be reached. The remainder of this chapter will discuss a variety of interesting equilibrium applications. From the art of balancing, to simple machines, to the muscles in your body, the world around you relies upon the principles of equilibrium to remain stable. This chapter will help you to see just how closely related these events truly are. 358 Chapter 9 \| Statics and Torque The content in this chapter supports: Big Idea 3 The interactions of an object with other objects can be described by forces. Enduring Understanding 3.F A force exerted on an object can cause a torque on that object. Essential Knowledge 3.F.1 Only the force component perpendicular to the line connecting the axis of rotation and the point of application of the force results in a torque about that axis. 9.1 The First Condition for Equilibrium Learning Objectives By the end of this section, you will be able to: • State the first condition of equilibrium. • Explain static equilibrium. • Explain dynamic equilibrium. The first condition necessary to achieve equilibrium is the one already mentioned: the net external force on the system must be zero. Expressed as an equation, this is simply net F = 0 Note that if net is zero, then the net external force in any direction is zero. For example, the net external forces along the typical x- and y-axes are zero. This is written as net = 0 and = 0 Figure 9.2 and Figure 9.3 illustrate situations where net = 0 for both static equilibrium (motionless), and dynamic equilibrium (constant velocity). (9.1) (9.2) Figure 9.2 This motionless person is in static equilibrium. The forces acting on him add up to zero. Both forces are vertical in this case. Figure 9.3 This car is in dynamic equilibrium because it is moving at constant velocity. There are horizontal and vertical forces, but the net external force in any direction is zero. The applied force app between the tires and the road is balanced by air friction, and the weight of the car is supported by the normal forces, here shown to be equal for all four tires. However, it is not sufficient for the net external force of a system to be zero for a system to be in equilibrium. Consider the two situations illustrated in Figure 9.4 and Figure 9.5 where forces are applied to an ice hockey stick lying flat on ice. The net external force is zero in both situations shown in the figure; but in one case, equilibrium is achieved, whereas in the other, it is not. In Figure 9.4, the ice hockey stick remains motionless. But in Figure 9.5, with the same forces applied in different places, This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 9 \| Statics and Torque 359 the stick experiences accelerated rotation. Therefore, we know that the point at which a force is applied is another factor in determining whether or not equilibrium is achieved. This will be explored further in the next section. Figure 9.4 An ice hockey stick lying flat on ice with two equal and opposite horizontal forces applied to it. Friction is negligible, and the gravitational force is balanced by the support of the ice (a normal force). Thus, net = 0 . Equilibrium is achieved, which is static equilibrium in this case. Figure 9.5 The same forces are applied at other points and the stick rotates—in fact, it experiences an accelerated rotation. Here net = 0 but the system is not at equilibrium. Hence, the net = 0 is a necessary—but not sufficient—condition for achieving equilibrium. PhET Explorations: Torque Investigate how torque causes an object to rotate. Discover the relationships between angular acceleration, moment of inertia, angular momentum and torque. Figure 9.6 Torque (http://cnx.org/content/m55176/1.2/torque_en.jar) 9.2 The Second Condition for Equilibrium Learning Objectives By the end of this section, you will be able to: • State the second condition that is necessary to achieve equilibrium. • Explain torque and the factors on which it depends. • Describe the role of torque in rotational mechanics. The information presented in this section supports the following AP® learning objectives and science practices: • 3.F.1.1 The student is able to use representations of the relationship between force and torque. (S.P. 1.4) • 3.F.1.2 The student is able to compare the torques on an object caused by various forces. (S.P. 1.4) • 3.F.1.3 The student is able to estimate the torque on an object caused by various forces in comparison to other situations. (S.P. 2.3) 360 Torque Chapter 9 \| Statics and Torque The second condition necessary to achieve equilibrium involves avoiding accelerated rotation (maintaining a constant angular velocity. A rotating body or system can be in equilibrium if its rate of rotation is constant and remains unchanged by the forces acting on it. To understand what factors affect rotation, let us think about what happens when you open an ordinary door by rotating it on its hinges. Several familiar factors determine how effective you are in opening the door. See Figure 9.7. First of all, the larger the force, the more effective it is in opening the door—obviously, the harder you push, the more rapidly the door opens. Also, the point at which you push is crucial. If you apply your force too close to the hinges, the door will open slowly, if at all. Most people have been embarrassed by making this mistake and bumping up against a door when it did not open as quickly as expected. Finally, the direction in which you push is also important. The most effective direction is perpendicular to the door—we push in this direction almost instinctively. Figure 9.7 Torque is the turning or twisting effectiveness of a force, illustrated here for door rotation on its hinges (as viewed from overhead). Torque has both magnitude and direction. (a) Counterclockwise torque is produced by this force, which means that the door will rotate in a counterclockwise due to F . Note that ⊥ is the perpendicular distance of the pivot from the line of action of the force. (b) A smaller counterclockwise torque is produced by a smaller force F′ acting at the same distance from the hinges (the pivot point). (c) The same force as in (a) produces a smaller counterclockwise torque when applied at a smaller distance from the hinges. (d) The same force as i
n (a), but acting in the opposite direction, produces a clockwise torque. (e) A smaller counterclockwise torque is produced by the same magnitude force acting at the same point but in a different direction. Here, is less than 90º . (f) Torque is zero here since the force just pulls on the hinges, producing no rotation. In this case, = 0º . The magnitude, direction, and point of application of the force are incorporated into the definition of the physical quantity called torque. Torque is the rotational equivalent of a force. It is a measure of the effectiveness of a force in changing or accelerating a rotation (changing the angular velocity over a period of time). In equation form, the magnitude of torque is defined to be = sin (9.3) where (the Greek letter tau) is the symbol for torque, is the distance from the pivot point to the point where the force is applied, is the magnitude of the force, and is the angle between the force and the vector directed from the point of application to the pivot point, as seen in Figure 9.7 and Figure 9.8. An alternative expression for torque is given in terms of the perpendicular lever arm ⊥ as shown in Figure 9.7 and Figure 9.8, which is defined as so that ⊥ = sin = ⊥ . (9.4) (9.5) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 9 \| Statics and Torque 361 Figure 9.8 A force applied to an object can produce a torque, which depends on the location of the pivot point. (a) The three factors , , and for pivot point A on a body are shown here— is the distance from the chosen pivot point to the point where the force is applied, and is the angle between F and the vector directed from the point of application to the pivot point. If the object can rotate around point A, it will rotate counterclockwise. This means that torque is counterclockwise relative to pivot A. (b) In this case, point B is the pivot point. The torque from the applied force will cause a clockwise rotation around point B, and so it is a clockwise torque relative to B. The perpendicular lever arm ⊥ is the shortest distance from the pivot point to the line along which F acts; it is shown as a dashed line in Figure 9.7 and Figure 9.8. Note that the line segment that defines the distance ⊥ is perpendicular to F , as its name implies. It is sometimes easier to find or visualize ⊥ than to find both and . In such cases, it may be more convenient to use τ = r ⊥ F rather than = sin for torque, but both are equally valid. The SI unit of torque is newtons times meters, usually written as N · m . For example, if you push perpendicular to the door with a force of 40 N at a distance of 0.800 m from the hinges, you exert a torque of 32 N·m(0.800 m×40 N×sin 90º) relative to the hinges. If you reduce the force to 20 N, the torque is reduced to 16 N·m , and so on. The torque is always calculated with reference to some chosen pivot point. For the same applied force, a different choice for the location of the pivot will give you a different value for the torque, since both and depend on the location of the pivot. Any point in any object can be chosen to calculate the torque about that point. The object may not actually pivot about the chosen “pivot point.” Note that for rotation in a plane, torque has two possible directions. Torque is either clockwise or counterclockwise relative to the chosen pivot point, as illustrated for points B and A, respectively, in Figure 9.8. If the object can rotate about point A, it will rotate counterclockwise, which means that the torque for the force is shown as counterclockwise relative to A. But if the object can rotate about point B, it will rotate clockwise, which means the torque for the force shown is clockwise relative to B. Also, the magnitude of the torque is greater when the lever arm is longer. Making Connections: Pivoting Block A solid block of length d is pinned to a wall on its right end. Three forces act on the block as shown below: FA, FB, and FC. While all three forces are of equal magnitude, and all three are equal distances away from the pivot point, all three forces will create a different torque upon the object. FA is vectored perpendicular to its distance from the pivot point; as a result, the magnitude of its torque can be found by the equation τ=FAd. Vector FB is parallel to the line connecting the point of application of force and the pivot point. As a result, it does not provide an ability to rotate the object and, subsequently, its torque is zero. FC, however, is directed at an angle ϴto the line connecting the point of application of force and the pivot point. In this instance, only the component perpendicular to this line is exerting a torque. This component, labeled F⊥, can be found using the equation F⊥=FCsinθ. The component of the force parallel to this line, labeled F∥, does not provide an ability to rotate the object and, as a result, does not provide a torque. Therefore, the resulting torque created by FC is τ=F⊥d. 362 Chapter 9 \| Statics and Torque Figure 9.9 Forces on a block pinned to a wall. A solid block of length d is pinned to a wall on its right end. Three forces act on the block: FA, FB, and FC. Now, the second condition necessary to achieve equilibrium is that the net external torque on a system must be zero. An external torque is one that is created by an external force. You can choose the point around which the torque is calculated. The point can be the physical pivot point of a system or any other point in space—but it must be the same point for all torques. If the second condition (net external torque on a system is zero) is satisfied for one choice of pivot point, it will also hold true for any other choice of pivot point in or out of the system of interest. (This is true only in an inertial frame of reference.) The second condition necessary to achieve equilibrium is stated in equation form as net = 0 (9.6) where net means total. Torques, which are in opposite directions are assigned opposite signs. A common convention is to call counterclockwise (ccw) torques positive and clockwise (cw) torques negative. When two children balance a seesaw as shown in Figure 9.10, they satisfy the two conditions for equilibrium. Most people have perfect intuition about seesaws, knowing that the lighter child must sit farther from the pivot and that a heavier child can keep a lighter one off the ground indefinitely. Figure 9.10 Two children balancing a seesaw satisfy both conditions for equilibrium. The lighter child sits farther from the pivot to create a torque equal in magnitude to that of the heavier child. Example 9.1 She Saw Torques On A Seesaw The two children shown in Figure 9.10 are balanced on a seesaw of negligible mass. (This assumption is made to keep the example simple—more involved examples will follow.) The first child has a mass of 26.0 kg and sits 1.60 m from the pivot.(a) If the second child has a mass of 32.0 kg, how far is she from the pivot? (b) What is p , the supporting force exerted by the pivot? Strategy Both conditions for equilibrium must be satisfied. In part (a), we are asked for a distance; thus, the second condition (regarding torques) must be used, since the first (regarding only forces) has no distances in it. To apply the second condition for equilibrium, we first identify the system of interest to be the seesaw plus the two children. We take the supporting pivot to be the point about which the torques are calculated. We then identify all external forces acting on the system. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 9 \| Statics and Torque 363 Solution (a) The three external forces acting on the system are the weights of the two children and the supporting force of the pivot. Let us examine the torque produced by each. Torque is defined to be = sin . (9.7) Here = 90º , so that sin = 1 for all three forces. That means ⊥ = for all three. The torques exerted by the three forces are first, second, and third, 1 = 11 2 = – 22 p = pp = 0 ⋅ p = 0. (9.8) (9.9) (9.10) Note that a minus sign has been inserted into the second equation because this torque is clockwise and is therefore negative by convention. Since p acts directly on the pivot point, the distance p is zero. A force acting on the pivot cannot cause a rotation, just as pushing directly on the hinges of a door will not cause it to rotate. Now, the second condition for equilibrium is that the sum of the torques on both children is zero. Therefore or 2 = – 1, 2 2 = 11. Weight is mass times the acceleration due to gravity. Entering for , we get Solve this for the unknown . The quantities on the right side of the equation are known; thus, 2 is 2 = (1.60 m) 26.0 kg 32.0 kg = 1.30 m. As expected, the heavier child must sit closer to the pivot (1.30 m versus 1.60 m) to balance the seesaw. Solution (b) This part asks for a force p . The easiest way to find it is to use the first condition for equilibrium, which is The forces are all vertical, so that we are dealing with a one-dimensional problem along the vertical axis; hence, the condition can be written as net = 0 net F = 0. (9.11) (9.12) (9.13) (9.14) (9.15) (9.16) (9.17) where we again call the vertical axis the y-axis. Choosing upward to be the positive direction, and using plus and minus signs to indicate the directions of the forces, we see that p – 1 – 2 = 0. This equation yields what might have been guessed at the beginning: p = 1 + 2. So, the pivot supplies a supporting force equal to the total weight of the system: p = 1 + 2. (9.18) (9.19) (9.20) 364 Chapter 9 \| Statics and Torque Entering known values gives p = 26.0 kg 9.80 m/s2 + 32.0 kg 9.80 m/s2 (9.21) Discussion = 568 N. The two results make intuitive sense. The heavier child sits closer to the pivot. The pivot supports the weight of the two children. Part (b) can also be solved using the second condition for equilibrium, since both distances are known, but only if the pivot point is chosen to be somew
here other than the location of the seesaw's actual pivot! Several aspects of the preceding example have broad implications. First, the choice of the pivot as the point around which torques are calculated simplified the problem. Since p is exerted on the pivot point, its lever arm is zero. Hence, the torque exerted by the supporting force p is zero relative to that pivot point. The second condition for equilibrium holds for any choice of pivot point, and so we choose the pivot point to simplify the solution of the problem. Second, the acceleration due to gravity canceled in this problem, and we were left with a ratio of masses. This will not always be the case. Always enter the correct forces—do not jump ahead to enter some ratio of masses. Third, the weight of each child is distributed over an area of the seesaw, yet we treated the weights as if each force were exerted at a single point. This is not an approximation—the distances 1 and 2 are the distances to points directly below the center of gravity of each child. As we shall see in the next section, the mass and weight of a system can act as if they are located at a single point. Finally, note that the concept of torque has an importance beyond static equilibrium. Torque plays the same role in rotational motion that force plays in linear motion. We will examine this in the next chapter. Take-Home Experiment Take a piece of modeling clay and put it on a table, then mash a cylinder down into it so that a ruler can balance on the round side of the cylinder while everything remains still. Put a penny 8 cm away from the pivot. Where would you need to put two pennies to balance? Three pennies? 9.3 Stability Learning Objectives By the end of this section, you will be able to: • State the types of equilibrium. • Describe stable and unstable equilibriums. • Describe neutral equilibrium. The information presented in this section supports the following AP® learning objectives and science practices: • 3.F.1.1 The student is able to use representations of the relationship between force and torque. (S.P. 1.4) • 3.F.1.2 The student is able to compare the torques on an object caused by various forces. (S.P. 1.4) • 3.F.1.3 The student is able to estimate the torque on an object caused by various forces in comparison to other situations. (S.P. 2.3) • 3.F.1.4 The student is able to design an experiment and analyze data testing a question about torques in a balanced rigid system. (S.P. 4.1, 4.2, 5.1) • 3.F.1.5 The student is able to calculate torques on a two-dimensional system in static equilibrium, by examining a representation or model (such as a diagram or physical construction). (S.P. 1.4, 2.2) It is one thing to have a system in equilibrium; it is quite another for it to be stable. The toy doll perched on the man's hand in Figure 9.11, for example, is not in stable equilibrium. There are three types of equilibrium: stable, unstable, and neutral. Figures throughout this module illustrate various examples. Figure 9.11 presents a balanced system, such as the toy doll on the man's hand, which has its center of gravity (cg) directly over the pivot, so that the torque of the total weight is zero. This is equivalent to having the torques of the individual parts balanced about the pivot point, in this case the hand. The cgs of the arms, legs, head, and torso are labeled with smaller type. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 9 \| Statics and Torque 365 Figure 9.11 A man balances a toy doll on one hand. A system is said to be in stable equilibrium if, when displaced from equilibrium, it experiences a net force or torque in a direction opposite to the direction of the displacement. For example, a marble at the bottom of a bowl will experience a restoring force when displaced from its equilibrium position. This force moves it back toward the equilibrium position. Most systems are in stable equilibrium, especially for small displacements. For another example of stable equilibrium, see the pencil in Figure 9.12. Figure 9.12 This pencil is in the condition of equilibrium. The net force on the pencil is zero and the total torque about any pivot is zero. A system is in unstable equilibrium if, when displaced, it experiences a net force or torque in the same direction as the displacement from equilibrium. A system in unstable equilibrium accelerates away from its equilibrium position if displaced even slightly. An obvious example is a ball resting on top of a hill. Once displaced, it accelerates away from the crest. See the next several figures for examples of unstable equilibrium. Figure 9.13 If the pencil is displaced slightly to the side (counterclockwise), it is no longer in equilibrium. Its weight produces a clockwise torque that returns the pencil to its equilibrium position. 366 Chapter 9 \| Statics and Torque Figure 9.14 If the pencil is displaced too far, the torque caused by its weight changes direction to counterclockwise and causes the displacement to increase. Figure 9.15 This figure shows unstable equilibrium, although both conditions for equilibrium are satisfied. Figure 9.16 If the pencil is displaced even slightly, a torque is created by its weight that is in the same direction as the displacement, causing the displacement to increase. A system is in neutral equilibrium if its equilibrium is independent of displacements from its original position. A marble on a flat horizontal surface is an example. Combinations of these situations are possible. For example, a marble on a saddle is stable for displacements toward the front or back of the saddle and unstable for displacements to the side. Figure 9.17 shows another example of neutral equilibrium. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 9 \| Statics and Torque 367 Figure 9.17 (a) Here we see neutral equilibrium. The cg of a sphere on a flat surface lies directly above the point of support, independent of the position on the surface. The sphere is therefore in equilibrium in any location, and if displaced, it will remain put. (b) Because it has a circular cross section, the pencil is in neutral equilibrium for displacements perpendicular to its length. When we consider how far a system in stable equilibrium can be displaced before it becomes unstable, we find that some systems in stable equilibrium are more stable than others. The pencil in Figure 9.12 and the person in Figure 9.18(a) are in stable equilibrium, but become unstable for relatively small displacements to the side. The critical point is reached when the cg is no longer above the base of support. Additionally, since the cg of a person's body is above the pivots in the hips, displacements must be quickly controlled. This control is a central nervous system function that is developed when we learn to hold our bodies erect as infants. For increased stability while standing, the feet should be spread apart, giving a larger base of support. Stability is also increased by lowering one's center of gravity by bending the knees, as when a football player prepares to receive a ball or braces themselves for a tackle. A cane, a crutch, or a walker increases the stability of the user, even more as the base of support widens. Usually, the cg of a female is lower (closer to the ground) than a male. Young children have their center of gravity between their shoulders, which increases the challenge of learning to walk. Figure 9.18 (a) The center of gravity of an adult is above the hip joints (one of the main pivots in the body) and lies between two narrowly-separated feet. Like a pencil standing on its eraser, this person is in stable equilibrium in relation to sideways displacements, but relatively small displacements take his cg outside the base of support and make him unstable. Humans are less stable relative to forward and backward displacements because the feet are not very long. Muscles are used extensively to balance the body in the front-to-back direction. (b) While bending in the manner shown, stability is increased by lowering the center of gravity. Stability is also increased if the base is expanded by placing the feet farther apart. Animals such as chickens have easier systems to control. Figure 9.19 shows that the cg of a chicken lies below its hip joints and between its widely separated and broad feet. Even relatively large displacements of the chicken's cg are stable and result in 368 Chapter 9 \| Statics and Torque restoring forces and torques that return the cg to its equilibrium position with little effort on the chicken's part. Not all birds are like chickens, of course. Some birds, such as the flamingo, have balance systems that are almost as sophisticated as that of humans. Figure 9.19 shows that the cg of a chicken is below the hip joints and lies above a broad base of support formed by widelyseparated and large feet. Hence, the chicken is in very stable equilibrium, since a relatively large displacement is needed to render it unstable. The body of the chicken is supported from above by the hips and acts as a pendulum between the hips. Therefore, the chicken is stable for front-to-back displacements as well as for side-to-side displacements. Figure 9.19 The center of gravity of a chicken is below the hip joints. The chicken is in stable equilibrium. The body of the chicken is supported from above by the hips and acts as a pendulum between them. Engineers and architects strive to achieve extremely stable equilibriums for buildings and other systems that must withstand wind, earthquakes, and other forces that displace them from equilibrium. Although the examples in this section emphasize gravitational forces, the basic conditions for equilibrium are the same for all types of forces. The net external force must be zero, and the net torque must also be zero. Take-Home Experiment Stand straight with your heels, back, and head against a wall. Bend forward from your waist, keeping
your heels and bottom against the wall, to touch your toes. Can you do this without toppling over? Explain why and what you need to do to be able to touch your toes without losing your balance. Is it easier for a woman to do this? 9.4 Applications of Statics, Including Problem-Solving Strategies Learning Objectives By the end of this section, you will be able to: • Discuss the applications of statics in real life. • State and discuss various problem-solving strategies in statics. The information presented in this section supports the following AP® learning objectives and science practices: • 3.F.1.1 The student is able to use representations of the relationship between force and torque. (S.P. 1.4) • 3.F.1.2 The student is able to compare the torques on an object caused by various forces. (S.P. 1.4) • 3.F.1.3 The student is able to estimate the torque on an object caused by various forces in comparison to other situations. (S.P. 2.3) • 3.F.1.4 The student is able to design an experiment and analyze data testing a question about torques in a balanced rigid system. (S.P. 4.1, 4.2, 5.1) • 3.F.1.5 The student is able to calculate torques on a two-dimensional system in static equilibrium, by examining a representation or model (such as a diagram or physical construction). (S.P. 1.4, 2.2) Statics can be applied to a variety of situations, ranging from raising a drawbridge to bad posture and back strain. We begin with a discussion of problem-solving strategies specifically used for statics. Since statics is a special case of Newton's laws, both the general problem-solving strategies and the special strategies for Newton's laws, discussed in Problem-Solving Strategies, still apply. Problem-Solving Strategy: Static Equilibrium Situations 1. The first step is to determine whether or not the system is in static equilibrium. This condition is always the case when the acceleration of the system is zero and accelerated rotation does not occur. 2. It is particularly important to draw a free body diagram for the system of interest. Carefully label all forces, and note their relative magnitudes, directions, and points of application whenever these are known. 3. Solve the problem by applying either or both of the conditions for equilibrium (represented by the equations net = 0 and net = 0 , depending on the list of known and unknown factors. If the second condition is involved, This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 9 \| Statics and Torque 369 choose the pivot point to simplify the solution. Any pivot point can be chosen, but the most useful ones cause torques by unknown forces to be zero. (Torque is zero if the force is applied at the pivot (then = 0 ), or along a line through the pivot point (then = 0 )). Always choose a convenient coordinate system for projecting forces. 4. Check the solution to see if it is reasonable by examining the magnitude, direction, and units of the answer. The importance of this last step never diminishes, although in unfamiliar applications, it is usually more difficult to judge reasonableness. These judgments become progressively easier with experience. Now let us apply this problem-solving strategy for the pole vaulter shown in the three figures below. The pole is uniform and has a mass of 5.00 kg. In Figure 9.20, the pole's cg lies halfway between the vaulter's hands. It seems reasonable that the force exerted by each hand is equal to half the weight of the pole, or 24.5 N. This obviously satisfies the first condition for equilibrium (net = 0) . The second condition (net = 0) is also satisfied, as we can see by choosing the cg to be the pivot point. The weight exerts no torque about a pivot point located at the cg, since it is applied at that point and its lever arm is zero. The equal forces exerted by the hands are equidistant from the chosen pivot, and so they exert equal and opposite torques. Similar arguments hold for other systems where supporting forces are exerted symmetrically about the cg. For example, the four legs of a uniform table each support one-fourth of its weight. In Figure 9.20, a pole vaulter holding a pole with its cg halfway between his hands is shown. Each hand exerts a force equal to half the weight of the pole, = = / 2 . (b) The pole vaulter moves the pole to his left, and the forces that the hands exert are no longer equal. See Figure 9.20. If the pole is held with its cg to the left of the person, then he must push down with his right hand and up with his left. The forces he exerts are larger here because they are in opposite directions and the cg is at a long distance from either hand. Similar observations can be made using a meter stick held at different locations along its length. Figure 9.20 A pole vaulter holds a pole horizontally with both hands. Figure 9.21 A pole vaulter is holding a pole horizontally with both hands. The center of gravity is near his right hand. 370 Chapter 9 \| Statics and Torque Figure 9.22 A pole vaulter is holding a pole horizontally with both hands. The center of gravity is to the left side of the vaulter. If the pole vaulter holds the pole as shown in Figure 9.20, the situation is not as simple. The total force he exerts is still equal to the weight of the pole, but it is not evenly divided between his hands. (If = , then the torques about the cg would not be equal since the lever arms are different.) Logically, the right hand should support more weight, since it is closer to the cg. In fact, if the right hand is moved directly under the cg, it will support all the weight. This situation is exactly analogous to two people carrying a load; the one closer to the cg carries more of its weight. Finding the forces and is straightforward, as the next example shows. If the pole vaulter holds the pole from near the end of the pole (Figure 9.22), the direction of the force applied by the right hand of the vaulter reverses its direction. Example 9.2 What Force Is Needed to Support a Weight Held Near Its CG? For the situation shown in Figure 9.20, calculate: (a) , the force exerted by the right hand, and (b) , the force exerted by the left hand. The hands are 0.900 m apart, and the cg of the pole is 0.600 m from the left hand. Strategy Figure 9.20 includes a free body diagram for the pole, the system of interest. There is not enough information to use the first condition for equilibrium (net = 0 ), since two of the three forces are unknown and the hand forces cannot be assumed to be equal in this case. There is enough information to use the second condition for equilibrium (net = 0) if the pivot point is chosen to be at either hand, thereby making the torque from that hand zero. We choose to locate the pivot at the left hand in this part of the problem, to eliminate the torque from the left hand. Solution for (a) There are now only two nonzero torques, those from the gravitational force ( w ) and from the push or pull of the right hand ( ). Stating the second condition in terms of clockwise and counterclockwise torques, or the algebraic sum of the torques is zero. Here this is net cw = –net ccw. = –τw (9.22) (9.23) since the weight of the pole creates a counterclockwise torque and the right hand counters with a clockwise toque. Using the definition of torque, = sin , noting that = 90º , and substituting known values, we obtain (0.900 m) = (0.600 m)(). Thus, Solution for (b) = (0.667) = 32.7 N. 5.00 kg 9.80 m/s2 (9.24) (9.25) The first condition for equilibrium is based on the free body diagram in the figure. This implies that by Newton's second law: + – = 0 (9.26) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 9 \| Statics and Torque From this we can conclude: Solving for , we obtain + = = = − = − 32.7 N 371 (9.27) (9.28) = 5.00 kg 9.80 m/s2 − 32.7 N Discussion FL is seen to be exactly half of , as we might have guessed, since is applied twice as far from the cg as . = 16.3 N If the pole vaulter holds the pole as he might at the start of a run, shown in Figure 9.22, the forces change again. Both are considerably greater, and one force reverses direction. Take-Home Experiment This is an experiment to perform while standing in a bus or a train. Stand facing sideways. How do you move your body to readjust the distribution of your mass as the bus accelerates and decelerates? Now stand facing forward. How do you move your body to readjust the distribution of your mass as the bus accelerates and decelerates? Why is it easier and safer to stand facing sideways rather than forward? Note: For your safety (and those around you), make sure you are holding onto something while you carry out this activity! PhET Explorations: Balancing Act Play with objects on a teeter totter to learn about balance. Test what you've learned by trying the Balance Challenge game. Figure 9.23 Balancing Act (http://phet.colorado.edu/en/simulation/balancing-act) 9.5 Simple Machines By the end of this section, you will be able to: • Describe different simple machines. • Calculate the mechanical advantage. Learning Objectives The information presented in this section supports the following AP® learning objectives and science practices: • 3.F.1.1 The student is able to use representations of the relationship between force and torque. (S.P. 1.4) • 3.F.1.2 The student is able to compare the torques on an object caused by various forces. (S.P. 1.4) • 3.F.1.3 The student is able to estimate the torque on an object caused by various forces in comparison to other situations. (S.P. 2.3) • 3.F.1.5 The student is able to calculate torques on a two-dimensional system in static equilibrium, by examining a representation or model (such as a diagram or physical construction). (S.P. 1.4, 2.2) Simple machines are devices that can be used to multiply or augment a force that we apply – often at the expense of a distance through which we apply the force. The word for “machine” comes from the Greek word meaning “to help make thi
ngs easier.” Levers, gears, pulleys, wedges, and screws are some examples of machines. Energy is still conserved for these devices because a machine cannot do more work than the energy put into it. However, machines can reduce the input force that is needed to perform the job. The ratio of output to input force magnitudes for any simple machine is called its mechanical advantage (MA). MA = o i (9.29) 380 Chapter 9 \| Statics and Torque Figure 9.31 This figure shows that large forces are exerted by the back muscles and experienced in the vertebrae when a person lifts with their back, since these muscles have small effective perpendicular lever arms. The data shown here are analyzed in the preceding example, Example 9.5. What are the benefits of having most skeletal muscles attached so close to joints? One advantage is speed because small muscle contractions can produce large movements of limbs in a short period of time. Other advantages are flexibility and agility, made possible by the large numbers of joints and the ranges over which they function. For example, it is difficult to imagine a system with biceps muscles attached at the wrist that would be capable of the broad range of movement we vertebrates possess. There are some interesting complexities in real systems of muscles, bones, and joints. For instance, the pivot point in many joints changes location as the joint is flexed, so that the perpendicular lever arms and the mechanical advantage of the system change, too. Thus the force the biceps muscle must exert to hold up a book varies as the forearm is flexed. Similar mechanisms operate in the legs, which explain, for example, why there is less leg strain when a bicycle seat is set at the proper height. The methods employed in this section give a reasonable description of real systems provided enough is known about the dimensions of the system. There are many other interesting examples of force and torque in the body—a few of these are the subject of endof-chapter problems. Glossary center of gravity: the point where the total weight of the body is assumed to be concentrated dynamic equilibrium: velocity are zero a state of equilibrium in which the net external force and torque on a system moving with constant mechanical advantage: the ratio of output to input forces for any simple machine neutral equilibrium: a state of equilibrium that is independent of a system's displacements from its original position perpendicular lever arm: the shortest distance from the pivot point to the line along which F lies SI units of torque: newton times meters, usually written as N·m stable equilibrium: displacement a system, when displaced, experiences a net force or torque in a direction opposite to the direction of the static equilibrium: a state of equilibrium in which the net external force and torque acting on a system is zero static equilibrium: equilibrium in which the acceleration of the system is zero and accelerated rotation does not occur torque: turning or twisting effectiveness of a force unstable equilibrium: from equilibrium a system, when displaced, experiences a net force or torque in the same direction as the displacement Section Summary 9.1 The First Condition for Equilibrium • Statics is the study of forces in equilibrium. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 9 \| Statics and Torque 381 • Two conditions must be met to achieve equilibrium, which is defined to be motion without linear or rotational acceleration. • The first condition necessary to achieve equilibrium is that the net external force on the system must be zero, so that net F = 0 . 9.2 The Second Condition for Equilibrium • The second condition assures those torques are also balanced. Torque is the rotational equivalent of a force in producing a rotation and is defined to be = sin where is torque, is the distance from the pivot point to the point where the force is applied, is the magnitude of the force, and is the angle between F and the vector directed from the point where the force acts to the pivot point. The perpendicular lever arm ⊥ is defined to be so that ⊥ = sin = ⊥ . • The perpendicular lever arm ⊥ is the shortest distance from the pivot point to the line along which acts. The SI unit for torque is newton-meter (N·m) . The second condition necessary to achieve equilibrium is that the net external torque on a system must be zero: By convention, counterclockwise torques are positive, and clockwise torques are negative. net = 0 9.3 Stability • A system is said to be in stable equilibrium if, when displaced from equilibrium, it experiences a net force or torque in a direction opposite the direction of the displacement. • A system is in unstable equilibrium if, when displaced from equilibrium, it experiences a net force or torque in the same direction as the displacement from equilibrium. • A system is in neutral equilibrium if its equilibrium is independent of displacements from its original position. 9.4 Applications of Statics, Including Problem-Solving Strategies • Statics can be applied to a variety of situations, ranging from raising a drawbridge to bad posture and back strain. We have discussed the problem-solving strategies specifically useful for statics. Statics is a special case of Newton's laws, both the general problem-solving strategies and the special strategies for Newton's laws, discussed in Problem-Solving Strategies, still apply. 9.5 Simple Machines • Simple machines are devices that can be used to multiply or augment a force that we apply – often at the expense of a distance through which we have to apply the force. • The ratio of output to input forces for any simple machine is called its mechanical advantage • A few simple machines are the lever, nail puller, wheelbarrow, crank, etc. 9.6 Forces and Torques in Muscles and Joints • Statics plays an important part in understanding everyday strains in our muscles and bones. • Many lever systems in the body have a mechanical advantage of significantly less than one, as many of our muscles are attached close to joints. • Someone with good posture stands or sits in such as way that their center of gravity lies directly above the pivot point in their hips, thereby avoiding back strain and damage to disks. Conceptual Questions 9.1 The First Condition for Equilibrium 1. What can you say about the velocity of a moving body that is in dynamic equilibrium? Draw a sketch of such a body using clearly labeled arrows to represent all external forces on the body. 2. Under what conditions can a rotating body be in equilibrium? Give an example. 9.2 The Second Condition for Equilibrium 3. What three factors affect the torque created by a force relative to a specific pivot point? 4. A wrecking ball is being used to knock down a building. One tall unsupported concrete wall remains standing. If the wrecking ball hits the wall near the top, is the wall more likely to fall over by rotating at its base or by falling straight down? Explain your answer. How is it most likely to fall if it is struck with the same force at its base? Note that this depends on how firmly the wall is attached at its base. 382 Chapter 9 \| Statics and Torque 5. Mechanics sometimes put a length of pipe over the handle of a wrench when trying to remove a very tight bolt. How does this help? (It is also hazardous since it can break the bolt.) 9.3 Stability 6. A round pencil lying on its side as in Figure 9.14 is in neutral equilibrium relative to displacements perpendicular to its length. What is its stability relative to displacements parallel to its length? 7. Explain the need for tall towers on a suspension bridge to ensure stable equilibrium. 9.4 Applications of Statics, Including Problem-Solving Strategies 8. When visiting some countries, you may see a person balancing a load on the head. Explain why the center of mass of the load needs to be directly above the person's neck vertebrae. 9.5 Simple Machines 9. Scissors are like a double-lever system. Which of the simple machines in Figure 9.24 and Figure 9.25 is most analogous to scissors? 10. Suppose you pull a nail at a constant rate using a nail puller as shown in Figure 9.24. Is the nail puller in equilibrium? What if you pull the nail with some acceleration – is the nail puller in equilibrium then? In which case is the force applied to the nail puller larger and why? 11. Why are the forces exerted on the outside world by the limbs of our bodies usually much smaller than the forces exerted by muscles inside the body? 12. Explain why the forces in our joints are several times larger than the forces we exert on the outside world with our limbs. Can these forces be even greater than muscle forces (see previous Question)? 9.6 Forces and Torques in Muscles and Joints 13. Why are the forces exerted on the outside world by the limbs of our bodies usually much smaller than the forces exerted by muscles inside the body? 14. Explain why the forces in our joints are several times larger than the forces we exert on the outside world with our limbs. Can these forces be even greater than muscle forces? 15. Certain types of dinosaurs were bipedal (walked on two legs). What is a good reason that these creatures invariably had long tails if they had long necks? 16. Swimmers and athletes during competition need to go through certain postures at the beginning of the race. Consider the balance of the person and why start-offs are so important for races. 17. If the maximum force the biceps muscle can exert is 1000 N, can we pick up an object that weighs 1000 N? Explain your answer. 18. Suppose the biceps muscle was attached through tendons to the upper arm close to the elbow and the forearm near the wrist. What would be the advantages and disadvantages of this type of construction for the motion of the arm? 19. Explain one of the reasons why pregnant women often suffer from back strain late in their pregnancy. This content is available fo
r free at http://cnx.org/content/col11844/1.13 Chapter 9 \| Statics and Torque 383 Problems & Exercises 9.2 The Second Condition for Equilibrium 1. (a) When opening a door, you push on it perpendicularly with a force of 55.0 N at a distance of 0.850m from the hinges. What torque are you exerting relative to the hinges? (b) Does it matter if you push at the same height as the hinges? 2. When tightening a bolt, you push perpendicularly on a wrench with a force of 165 N at a distance of 0.140 m from the center of the bolt. (a) How much torque are you exerting in newton × meters (relative to the center of the bolt)? (b) Convert this torque to footpounds. 3. Two children push on opposite sides of a door during play. Both push horizontally and perpendicular to the door. One child pushes with a force of 17.5 N at a distance of 0.600 m from the hinges, and the second child pushes at a distance of 0.450 m. What force must the second child exert to keep the door from moving? Assume friction is negligible. 4. Use the second condition for equilibrium (net τ = 0) to calculate p in Example 9.1, employing any data given or solved for in part (a) of the example. 5. Repeat the seesaw problem in Example 9.1 with the center of mass of the seesaw 0.160 m to the left of the pivot (on the side of the lighter child) and assuming a mass of 12.0 kg for the seesaw. The other data given in the example remain unchanged. Explicitly show how you follow the steps in the Problem-Solving Strategy for static equilibrium. 9.3 Stability 6. Suppose a horse leans against a wall as in Figure 9.32. Calculate the force exerted on the wall assuming that force is horizontal while using the data in the schematic representation of the situation. Note that the force exerted on the wall is equal in magnitude and opposite in direction to the force exerted on the horse, keeping it in equilibrium. The total mass of the horse and rider is 500 kg. Take the data to be accurate to three digits. 8. (a) Calculate the magnitude and direction of the force on each foot of the horse in Figure 9.32 (two are on the ground), assuming the center of mass of the horse is midway between the feet. The total mass of the horse and rider is 500kg. (b) What is the minimum coefficient of friction between the hooves and ground? Note that the force exerted by the wall is horizontal. 9. A person carries a plank of wood 2 m long with one hand pushing down on it at one end with a force F1 and the other hand holding it up at 50 cm from the end of the plank with force F2 . If the plank has a mass of 20 kg and its center of gravity is at the middle of the plank, what are the magnitudes of the forces F1 and F2 ? 10. A 17.0-m-high and 11.0-m-long wall under construction and its bracing are shown in Figure 9.33. The wall is in stable equilibrium without the bracing but can pivot at its base. Calculate the force exerted by each of the 10 braces if a strong wind exerts a horizontal force of 650 N on each square meter of the wall. Assume that the net force from the wind acts at a height halfway up the wall and that all braces exert equal forces parallel to their lengths. Neglect the thickness of the wall. Figure 9.33 11. (a) What force must be exerted by the wind to support a 2.50-kg chicken in the position shown in Figure 9.34? (b) What is the ratio of this force to the chicken's weight? (c) Does this support the contention that the chicken has a relatively stable construction? Figure 9.32 7. Two children of mass 20 kg and 30 kg sit balanced on a seesaw with the pivot point located at the center of the seesaw. If the children are separated by a distance of 3 m, at what distance from the pivot point is the small child sitting in order to maintain the balance? Figure 9.34 12. Suppose the weight of the drawbridge in Figure 9.35 is supported entirely by its hinges and the opposite shore, so that its cables are slack. (a) What fraction of the weight is supported by the opposite shore if the point of support is directly beneath the cable attachments? (b) What is the direction and magnitude of the force the hinges exert on the bridge under these circumstances? The mass of the bridge is 2500 kg. 384 Chapter 9 \| Statics and Torque Figure 9.35 A small drawbridge, showing the forces on the hinges ( F ), its weight ( w ), and the tension in its wires ( T ). 13. Suppose a 900-kg car is on the bridge in Figure 9.35 with its center of mass halfway between the hinges and the cable attachments. (The bridge is supported by the cables and hinges only.) (a) Find the force in the cables. (b) Find the direction and magnitude of the force exerted by the hinges on the bridge. 14. A sandwich board advertising sign is constructed as shown in Figure 9.36. The sign's mass is 8.00 kg. (a) Calculate the tension in the chain assuming no friction between the legs and the sidewalk. (b) What force is exerted by each side on the hinge? Figure 9.36 A sandwich board advertising sign demonstrates tension. 15. (a) What minimum coefficient of friction is needed between the legs and the ground to keep the sign in Figure 9.36 in the position shown if the chain breaks? (b) What force is exerted by each side on the hinge? 16. A gymnast is attempting to perform splits. From the information given in Figure 9.37, calculate the magnitude and direction of the force exerted on each foot by the floor. This content is available for free at http://cnx.org/content/col11844/1.13 Figure 9.37 A gymnast performs full split. The center of gravity and the various distances from it are shown. 9.4 Applications of Statics, Including ProblemSolving Strategies 17. To get up on the roof, a person (mass 70.0 kg) places a 6.00-m aluminum ladder (mass 10.0 kg) against the house on a concrete pad with the base of the ladder 2.00 m from the house. The ladder rests against a plastic rain gutter, which we can assume to be frictionless. The center of mass of the ladder is 2 m from the bottom. The person is standing 3 m from the bottom. What are the magnitudes of the forces on the ladder at the top and bottom? 18. In Figure 9.22, the cg of the pole held by the pole vaulter is 2.00 m from the left hand, and the hands are 0.700 m apart. Calculate the force exerted by (a) his right hand and (b) his left hand. (c) If each hand supports half the weight of the pole in Figure 9.20, show that the second condition for equilibrium (net = 0) is satisfied for a pivot other than the one located at the center of gravity of the pole. Explicitly show how you follow the steps in the Problem-Solving Strategy for static equilibrium described above. 9.5 Simple Machines 19. What is the mechanical advantage of a nail puller—similar to the one shown in Figure 9.24 —where you exert a force 45 cm from the pivot and the nail is 1.8 cm on the other side? What minimum force must you exert to apply a force of 1250 N to the nail? 20. Suppose you needed to raise a 250-kg mower a distance of 6.0 cm above the ground to change a tire. If you had a 2.0-m long lever, where would you place the fulcrum if your force was limited to 300 N? 21. a) What is the mechanical advantage of a wheelbarrow, such as the one in Figure 9.25, if the center of gravity of the wheelbarrow and its load has a perpendicular lever arm of 5.50 cm, while the hands have a perpendicular lever arm of 1.02 m? (b) What upward force should you exert to support the wheelbarrow and its load if their combined mass is 55.0 kg? (c) What force does the wheel exert on the ground? 22. A typical car has an axle with 1.10 cm radius driving a tire with a radius of 27.5 cm . What is its mechanical advantage assuming the very simplified model in Figure 9.26(b)? 23. What force does the nail puller in Exercise 9.19 exert on the supporting surface? The nail puller has a mass of 2.10 kg. Chapter 9 \| Statics and Torque 385 equivalent lever system. Calculate the force exerted by the upper leg muscle to lift the mass at a constant speed. Explicitly show how you follow the steps in the ProblemSolving Strategy for static equilibrium in Applications of Statistics, Including Problem-Solving Strategies. 24. If you used an ideal pulley of the type shown in Figure 9.27(a) to support a car engine of mass 115 kg , (a) What would be the tension in the rope? (b) What force must the ceiling supply, assuming you pull straight down on the rope? Neglect the pulley system's mass. 25. Repeat Exercise 9.24 for the pulley shown in Figure 9.27(c), assuming you pull straight up on the rope. The pulley system's mass is 7.00 kg . 9.6 Forces and Torques in Muscles and Joints 26. Verify that the force in the elbow joint in Example 9.4 is 407 N, as stated in the text. 27. Two muscles in the back of the leg pull on the Achilles tendon as shown in Figure 9.38. What total force do they exert? Figure 9.40 A mass is connected by pulleys and wires to the ankle in this exercise device. 30. A person working at a drafting board may hold her head as shown in Figure 9.41, requiring muscle action to support the head. The three major acting forces are shown. Calculate the direction and magnitude of the force supplied by the upper vertebrae FV to hold the head stationary, assuming that this force acts along a line through the center of mass as do the weight and muscle force. Figure 9.38 The Achilles tendon of the posterior leg serves to attach plantaris, gastrocnemius, and soleus muscles to calcaneus bone. 28. The upper leg muscle (quadriceps) exerts a force of 1250 N, which is carried by a tendon over the kneecap (the patella) at the angles shown in Figure 9.39. Find the direction and magnitude of the force exerted by the kneecap on the upper leg bone (the femur). Figure 9.39 The knee joint works like a hinge to bend and straighten the lower leg. It permits a person to sit, stand, and pivot. 29. A device for exercising the upper leg muscle is shown in Figure 9.40, together with a schematic representation of an Figure 9.41 31. We analyzed the biceps muscle example with the angle between forearm
and upper arm set at 90º . Using the same numbers as in Example 9.4, find the force exerted by the biceps muscle when the angle is 120º and the forearm is in a downward position. 32. Even when the head is held erect, as in Figure 9.42, its center of mass is not directly over the principal point of support (the atlanto-occipital joint). The muscles at the back of the neck should therefore exert a force to keep the head 386 Chapter 9 \| Statics and Torque erect. That is why your head falls forward when you fall asleep in the class. (a) Calculate the force exerted by these muscles using the information in the figure. (b) What is the force exerted by the pivot on the head? Figure 9.44 A child being lifted by a father's lower leg. 35. Unlike most of the other muscles in our bodies, the masseter muscle in the jaw, as illustrated in Figure 9.45, is attached relatively far from the joint, enabling large forces to be exerted by the back teeth. (a) Using the information in the figure, calculate the force exerted by the lower teeth on the bullet. (b) Calculate the force on the joint. Figure 9.42 The center of mass of the head lies in front of its major point of support, requiring muscle action to hold the head erect. A simplified lever system is shown. 33. A 75-kg man stands on his toes by exerting an upward force through the Achilles tendon, as in Figure 9.43. (a) What is the force in the Achilles tendon if he stands on one foot? (b) Calculate the force at the pivot of the simplified lever system shown—that force is representative of forces in the ankle joint. Figure 9.45 A person clenching a bullet between his teeth. 36. Integrated Concepts Suppose we replace the 4.0-kg book in Exercise 9.31 of the biceps muscle with an elastic exercise rope that obeys Hooke's Law. Assume its force constant = 600 N/m . (a) How much is the rope stretched (past equilibrium) to provide the same force B as in this example? Assume the rope is held in the hand at the same location as the book. (b) What force is on the biceps muscle if the exercise rope is pulled straight up so that the forearm makes an angle of 25º with the horizontal? Assume the biceps muscle is still perpendicular to the forearm. 37. (a) What force should the woman in Figure 9.46 exert on the floor with each hand to do a push-up? Assume that she moves up at a constant speed. (b) The triceps muscle at the back of her upper arm has an effective lever arm of 1.75 cm, and she exerts force on the floor at a horizontal distance of 20.0 cm from the elbow joint. Calculate the magnitude of the force in each triceps muscle, and compare it to her weight. (c) Figure 9.43 The muscles in the back of the leg pull the Achilles tendon when one stands on one's toes. A simplified lever system is shown. 34. A father lifts his child as shown in Figure 9.44. What force should the upper leg muscle exert to lift the child at a constant speed? This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 9 \| Statics and Torque 387 How much work does she do if her center of mass rises 0.240 m? (d) What is her useful power output if she does 25 pushups in one minute? Figure 9.46 A woman doing pushups. 38. You have just planted a sturdy 2-m-tall palm tree in your front lawn for your mother's birthday. Your brother kicks a 500 g ball, which hits the top of the tree at a speed of 5 m/s and stays in contact with it for 10 ms. The ball falls to the ground near the base of the tree and the recoil of the tree is minimal. (a) What is the force on the tree? (b) The length of the sturdy section of the root is only 20 cm. Furthermore, the soil around the roots is loose and we can assume that an effective force is applied at the tip of the 20 cm length. What is the effective force exerted by the end of the tip of the root to keep the tree from toppling? Assume the tree will be uprooted rather than bend. (c) What could you have done to ensure that the tree does not uproot easily? 39. Unreasonable Results Suppose two children are using a uniform seesaw that is 3.00 m long and has its center of mass over the pivot. The first child has a mass of 30.0 kg and sits 1.40 m from the pivot. (a) Calculate where the second 18.0 kg child must sit to balance the seesaw. (b) What is unreasonable about the result? (c) Which premise is unreasonable, or which premises are inconsistent? 40. Construct Your Own Problem Consider a method for measuring the mass of a person's arm in anatomical studies. The subject lies on her back, extends her relaxed arm to the side and two scales are placed below the arm. One is placed under the elbow and the other under the back of her hand. Construct a problem in which you calculate the mass of the arm and find its center of mass based on the scale readings and the distances of the scales from the shoulder joint. You must include a free body diagram of the arm to direct the analysis. Consider changing the position of the scale under the hand to provide more information, if needed. You may wish to consult references to obtain reasonable mass values. Test Prep for AP® Courses 9.2 The Second Condition for Equilibrium 1. Which of the following is not an example of an object undergoing a torque? a. A car is rounding a bend at a constant speed. b. A merry-go-round increases from rest to a constant rotational speed. c. A pendulum swings back and forth. d. A bowling ball rolls down a bowling alley. 2. Five forces of equal magnitude, labeled A–E, are applied to the object shown below. If the object is anchored at point P, which force will provide the greatest torque? Figure 9.47 Five forces acting on an object. a. Force A 388 b. Force B c. Force C d. Force D e. Force E 9.3 Stability 3. Using the concept of torque, explain why a traffic cone placed on its base is in stable equilibrium, while a traffic cone placed on its tip is in unstable equilibrium. 9.4 Applications of Statics, Including ProblemSolving Strategies 4. A child sits on the end of a playground see-saw. Which of the following values is the most appropriate estimate of the torque created by the child? a. 6 N•m b. 60 N•m c. 600 N•m d. 6000 N•m 5. A group of students is stacking a set of identical books, each one overhanging the one below it by 1 inch. They would like to estimate how many books they could place on top of each other before the stack tipped. What information below would they need to know to make this calculation? Figure 9.48 3 overlapping stacked books. I. The mass of each book II. The width of each book III. The depth of each book a. b. c. d. e. I only I and II only I and III only II only I, II, and III 6. A 10 N board of uniform density is 5 meters long. It is supported on the left by a string bearing a 3 N upward force. In order to prevent the string from breaking, a person must place an upward force of 7 N at a position along the bottom surface of the board. At what distance from its left edge would they need to place this force in order for the board to be in static equilibrium? Chapter 9 \| Statics and Torque a. If a 1000 kg car comes to rest at a point 5 meters from the left pier, how much force will the bridge provide to the left and right piers? b. How will FL and FR change as the car drives to the right side of the bridge? 8. An object of unknown mass is provided to a student. Without using a scale, design an experimental procedure detailing how the magnitude of this mass could be experimentally found. Your explanation must include the concept of torque and all steps should be provided in an orderly sequence. You may include a labeled diagram of your setup to help in your description. Include enough detail so that another student could carry out your procedure. 9.5 Simple Machines 9. As a young student, you likely learned that simple machines are capable of increasing the ability to lift and move objects. Now, as an educated AP Physics student, you are aware that this capability is governed by the relationship between force and torque. In the space below, explain why torque is integral to the increase in force created by a simple machine. You may use an example or diagram to assist in your explanation. Be sure to cite the mechanical advantage in your explanation as well. 10. Figure 9.24(a) shows a wheelbarrow being lifted by an applied force Fi. If the wheelbarrow is filled with twenty bricks massing 3 kg each, estimate the value of the applied force Fi. Provide an explanation behind the total weight w and any reasoning toward your final answer. Additionally, provide a range of values over which you feel the force could exist. 9.6 Forces and Torques in Muscles and Joints 11. When you use your hand to raise a 20 lb dumbbell in a curling motion, the force on your bicep muscle is not equal to 20 lb. a. Compare the size of the force placed on your bicep muscle to the force of the 20 lb dumbbell lifted by your hand. Using the concept of torque, which force is greater and explain why the two forces are not identical. b. Does the force placed on your bicep muscle change as you curl the weight closer toward your body? (In other words, is the force on your muscle different when your forearm is 90° to your upper arm than when it is 45° to your upper arm?) Explain your answer using torque. m m a. b. 3 7 5 2 25 7 30 7 e. 5 m d. c. m m 7. A bridge is supported by two piers located 20 meters apart. Both the left and right piers provide an upward force on the bridge, labeled FL and FR respectively. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 10 \| Rotational Motion and Angular Momentum 389 10 ROTATIONAL MOTION AND ANGULAR MOMENTUM Figure 10.1 The mention of a tornado conjures up images of raw destructive power. Tornadoes blow houses away as if they were made of paper and have been known to pierce tree trunks with pieces of straw. They descend from clouds in funnel-like shapes that spin violently, particularly at the bottom where they are most narrow, producing winds as high as 500
km/h. (credit: Daphne Zaras, U.S. National Oceanic and Atmospheric Administration) Chapter Outline 10.1. Angular Acceleration 10.2. Kinematics of Rotational Motion 10.3. Dynamics of Rotational Motion: Rotational Inertia 10.4. Rotational Kinetic Energy: Work and Energy Revisited 10.5. Angular Momentum and Its Conservation 10.6. Collisions of Extended Bodies in Two Dimensions 10.7. Gyroscopic Effects: Vector Aspects of Angular Momentum Connection for AP® Courses Why do tornados spin? And why do tornados spin so rapidly? The answer is that the air masses that produce tornados are themselves rotating, and when the radii of the air masses decrease, their rate of rotation increases. An ice skater increases her spin in an exactly analogous manner, as seen in Figure 10.2. The skater starts her rotation with outstretched limbs and increases her rate of spin by pulling them in toward her body. The same physics describes the exhilarating spin of a skater and the wrenching force of a tornado. We will find that this is another example of the importance of conservation laws and their role in determining how changes happen in a system, supporting Big Idea 5. The idea that a change of a conserved quantity is always equal to the transfer of that quantity between interacting systems (Enduring Understanding 5.A) is presented for both energy and angular momentum (Enduring Understanding 5.E). The conservation of angular momentum in relation to the external net torque (Essential Knowledge 5.E.1) parallels that of linear momentum conservation in relation to the external net force. The concept of rotational inertia is introduced, a concept that takes into account not only the mass of an object or a system, but also the distribution of mass within the object or system. Therefore, changes in the rotational inertia of a system could lead to changes in the motion (Essential Knowledge 5.E.2) of the system. We shall see that all important aspects of rotational motion either have already been defined for linear motion or have exact analogues in linear motion. Clearly, therefore, force, energy, and power are associated with rotational motion. This supports Big Idea 3, that interactions are described by forces. The ability of forces to cause torques (Enduring Understanding 3.F) is extended to the interactions between objects that result in nonzero net torque. This nonzero net torque in turn causes changes in the rotational motion of an object (Essential Knowledge 3.F.2) and results in changes of the angular momentum of an object (Essential Knowledge 3.F.3). 390 Chapter 10 \| Rotational Motion and Angular Momentum Similarly, Big Idea 4, that interactions between systems cause changes in those systems, is supported by the empirical observation that when torques are exerted on rigid bodies these torques cause changes in the angular momentum of the system (Enduring Understanding 4.D). Again, there is a clear analogy between linear and rotational motion in this interaction. Both the angular kinematics variables (angular displacement, angular velocity, and angular acceleration) and the dynamics variables (torque and angular momentum) are vectors with direction depending on whether the rotation is clockwise or counterclockwise with respect to an axis of rotation (Essential Knowledge 4.D.1). The angular momentum of the system can change due to interactions (Essential Knowledge 4.D.2). This change is defined as the product of the average torque and the time interval during which torque is exerted (Essential Knowledge 4.D.3), analogous to the impulse-momentum theorem for linear motion. The concepts in this chapter support: Big Idea 3. The interactions of an object with other objects can be described by forces. Enduring Understanding 3.F. A force exerted on an object can cause a torque on that object. Extended Knowledge 3.F.2. The presence of a net torque along any axis will cause a rigid system to change its rotational motion or an object to change its rotational motion about that axis. Extended Knowledge 3.F.3. A torque exerted on an object can change the angular momentum of an object. Big Idea 4. Interactions between systems can result in changes in those systems. Enduring Understanding 4.D. A net torque exerted on a system by other objects or systems will change the angular momentum of the system. Extended Knowledge 4.D.1. Torque, angular velocity, angular acceleration, and angular momentum are vectors and can be characterized as positive or negative depending upon whether they give rise to or correspond to counterclockwise or clockwise rotation with respect to an axis. Extended Knowledge 4.D.2. The angular momentum of a system may change due to interactions with other objects or systems. Extended Knowledge 4.D.3. The change in angular momentum is given by the product of the average torque and the time interval during which the torque is exerted. Big Idea 5. Changes that occur as a result of interactions are constrained by conservation laws. Enduring Understanding 5.A. Certain quantities are conserved, in the sense that the changes of those quantities in a given system are always equal to the transfer of that quantity to or from the system by all possible interactions with other systems. Extended Knowledge 5.A.2. For all systems under all circumstances, energy, charge, linear momentum, and angular momentum are conserved. Enduring Understanding 5.E. The angular momentum of a system is conserved. Extended Knowledge 5.E.1. If the net external torque exerted on the system is zero, the angular momentum of the system does not change. Extended Knowledge 5.E.2. The angular momentum of a system is determined by the locations and velocities of the objects that make up the system. The rotational inertia of an object or system depends upon the distribution of mass within the object or system. Changes in the radius of a system or in the distribution of mass within the system result in changes in the system's rotational inertia, and hence in its angular velocity and linear speed for a given angular momentum. Examples should include elliptical orbits in an Earth-satellite system. Mathematical expressions for the moments of inertia will be provided where needed. Students will not be expected to know the parallel axis theorem. Figure 10.2 This figure skater increases her rate of spin by pulling her arms and her extended leg closer to her axis of rotation. (credit: Luu, Wikimedia Commons) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 10 \| Rotational Motion and Angular Momentum 391 10.1 Angular Acceleration Learning Objectives By the end of this section, you will be able to: • Describe uniform circular motion. • Explain nonuniform circular motion. • Calculate angular acceleration of an object. • Observe the link between linear and angular acceleration. Uniform Circular Motion and Gravitation discussed only uniform circular motion, which is motion in a circle at constant speed and, hence, constant angular velocity. Recall that angular velocity was defined as the time rate of change of angle : = Δ Δ , (10.1) where is the angle of rotation as seen in Figure 10.3. The relationship between angular velocity and linear velocity was also defined in Rotation Angle and Angular Velocity as = (10.2) or = , where is the radius of curvature, also seen in Figure 10.3. According to the sign convention, the counter clockwise direction is considered as positive direction and clockwise direction as negative (10.3) Figure 10.3 This figure shows uniform circular motion and some of its defined quantities. Angular velocity is not constant when a skater pulls in her arms, when a child starts up a merry-go-round from rest, or when a computer's hard disk slows to a halt when switched off. In all these cases, there is an angular acceleration, in which changes. The faster the change occurs, the greater the angular acceleration. Angular acceleration is defined as the rate of change of angular velocity. In equation form, angular acceleration is expressed as follows: where Δ is the change in angular velocity and Δ is the change in time. The units of angular acceleration are (rad/s)/s , or rad/s2 . If increases, then is positive. If decreases, then is negative. = Δ Δ , (10.4) Example 10.1 Calculating the Angular Acceleration and Deceleration of a Bike Wheel Suppose a teenager puts her bicycle on its back and starts the rear wheel spinning from rest to a final angular velocity of 250 rpm in 5.00 s. (a) Calculate the angular acceleration in rad/s2 . (b) If she now slams on the brakes, causing an angular acceleration of – 87.3 rad/s2 , how long does it take the wheel to stop? Strategy for (a) The angular acceleration can be found directly from its definition in = Δ Δ are given. We see that Δ is 250 rpm and Δ is 5.00 s. Solution for (a) Entering known information into the definition of angular acceleration, we get because the final angular velocity and time 392 Chapter 10 \| Rotational Motion and Angular Momentum = Δ Δ 250 rpm 5.00 s = . (10.5) Because Δ is in revolutions per minute (rpm) and we want the standard units of rad/s2 for angular acceleration, we need to convert Δ from rpm to rad/s: Δ = 250 rev min = 26.2rad s . ⋅ 2π rad rev ⋅ 1 min 60 sec Entering this quantity into the expression for , we get = Δ Δ = 26.2 rad/s 5.00 s = 5.24 rad/s2. (10.6) (10.7) Strategy for (b) In this part, we know the angular acceleration and the initial angular velocity. We can find the stoppage time by using the definition of angular acceleration and solving for Δ , yielding Δ = Δ . (10.8) Solution for (b) Here the angular velocity decreases from 26.2 rad/s (250 rpm) to zero, so that Δ is – 26.2 rad/s , and is given to be – 87.3 rad/s2 . Thus, Δ = – 26.2 rad/s – 87.3 rad/s2 = 0.300 s. (10.9) Discussion Note that the angular acceleration as the girl spins the wheel is small and positive; it takes 5 s to produce an appreciable angular velocity. When she hits the brake, the angular ac
celeration is large and negative. The angular velocity quickly goes to zero. In both cases, the relationships are analogous to what happens with linear motion. For example, there is a large deceleration when you crash into a brick wall—the velocity change is large in a short time interval. If the bicycle in the preceding example had been on its wheels instead of upside-down, it would first have accelerated along the ground and then come to a stop. This connection between circular motion and linear motion needs to be explored. For example, it would be useful to know how linear and angular acceleration are related. In circular motion, linear acceleration is tangent to the circle at the point of interest, as seen in Figure 10.4. Thus, linear acceleration is called tangential acceleration t . Figure 10.4 In circular motion, linear acceleration , occurs as the magnitude of the velocity changes: is tangent to the motion. In the context of circular motion, linear acceleration is also called tangential acceleration t . This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 10 \| Rotational Motion and Angular Momentum 393 Linear or tangential acceleration refers to changes in the magnitude of velocity but not its direction. We know from Uniform Circular Motion and Gravitation that in circular motion centripetal acceleration, c , refers to changes in the direction of the velocity but not its magnitude. An object undergoing circular motion experiences centripetal acceleration, as seen in Figure 10.5. Thus, t and c are perpendicular and independent of one another. Tangential acceleration t is directly related to the angular acceleration and is linked to an increase or decrease in the velocity, but not its direction. Figure 10.5 Centripetal acceleration c occurs as the direction of velocity changes; it is perpendicular to the circular motion. Centripetal and tangential acceleration are thus perpendicular to each other. Now we can find the exact relationship between linear acceleration t and angular acceleration . Because linear acceleration is proportional to a change in the magnitude of the velocity, it is defined (as it was in One-Dimensional Kinematics) to be t = Δ Δ (10.10) . For circular motion, note that = , so that t = Δ() Δ . The radius is constant for circular motion, and so Δ() = (Δ) . Thus, By definition, = Δ Δ . Thus, or 10.11) (10.12) (10.13) (10.14) These equations mean that linear acceleration and angular acceleration are directly proportional. The greater the angular acceleration is, the larger the linear (tangential) acceleration is, and vice versa. For example, the greater the angular acceleration of a car's drive wheels, the greater the acceleration of the car. The radius also matters. For example, the smaller a wheel, the smaller its linear acceleration for a given angular acceleration . Example 10.2 Calculating the Angular Acceleration of a Motorcycle Wheel A powerful motorcycle can accelerate from 0 to 30.0 m/s (about 108 km/h) in 4.20 s. What is the angular acceleration of its 0.320-m-radius wheels? (See Figure 10.6.) 394 Chapter 10 \| Rotational Motion and Angular Momentum Figure 10.6 The linear acceleration of a motorcycle is accompanied by an angular acceleration of its wheels. Strategy We are given information about the linear velocities of the motorcycle. Thus, we can find its linear acceleration t . Then, the expression = t can be used to find the angular acceleration. Solution The linear acceleration is t = Δ Δ = 30.0 m/s 4.20 s = 7.14 m/s2. We also know the radius of the wheels. Entering the values for t and into = t , we get = t = 7.14 m/s2 0.320 m = 22.3 rad/s2. (10.15) (10.16) Discussion Units of radians are dimensionless and appear in any relationship between angular and linear quantities. So far, we have defined three rotational quantities— , , and . These quantities are analogous to the translational quantities , , and . Table 10.1 displays rotational quantities, the analogous translational quantities, and the relationships between them. Table 10.1 Rotational and Translational Quantities Rotational Translational Relationship = = = Making Connections: Take-Home Experiment Sit down with your feet on the ground on a chair that rotates. Lift one of your legs such that it is unbent (straightened out). Using the other leg, begin to rotate yourself by pushing on the ground. Stop using your leg to push the ground but allow the chair to rotate. From the origin where you began, sketch the angle, angular velocity, and angular acceleration of your leg as a function of time in the form of three separate graphs. Estimate the magnitudes of these quantities. Check Your Understanding Angular acceleration is a vector, having both magnitude and direction. How do we denote its magnitude and direction? Illustrate with an example. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 10 \| Rotational Motion and Angular Momentum 395 Solution The magnitude of angular acceleration is and its most common units are rad/s2 . The direction of angular acceleration along a fixed axis is denoted by a + or a – sign, just as the direction of linear acceleration in one dimension is denoted by a + or a – sign. For example, consider a gymnast doing a forward flip. Her angular momentum would be parallel to the mat and to her left. The magnitude of her angular acceleration would be proportional to her angular velocity (spin rate) and her moment of inertia about her spin axis. PhET Explorations: Ladybug Revolution Join the ladybug in an exploration of rotational motion. Rotate the merry-go-round to change its angle, or choose a constant angular velocity or angular acceleration. Explore how circular motion relates to the bug's x,y position, velocity, and acceleration using vectors or graphs. Figure 10.7 Ladybug Revolution (http://cnx.org/content/m55183/1.2/rotation_en.jar) 10.2 Kinematics of Rotational Motion Learning Objectives By the end of this section, you will be able to: • Observe the kinematics of rotational motion. • Derive rotational kinematic equations. • Evaluate problem solving strategies for rotational kinematics. Just by using our intuition, we can begin to see how rotational quantities like , , and are related to one another. For example, if a motorcycle wheel has a large angular acceleration for a fairly long time, it ends up spinning rapidly and rotates through many revolutions. In more technical terms, if the wheel's angular acceleration is large for a long period of time , then the final angular velocity and angle of rotation are large. The wheel's rotational motion is exactly analogous to the fact that the motorcycle's large translational acceleration produces a large final velocity, and the distance traveled will also be large. Kinematics is the description of motion. The kinematics of rotational motion describes the relationships among rotation angle, angular velocity, angular acceleration, and time. Let us start by finding an equation relating , , and . To determine this equation, we recall a familiar kinematic equation for translational, or straight-line, motion: = 0 + (constant ) (10.17) Note that in rotational motion = t , and we shall use the symbol for tangential or linear acceleration from now on. As in linear kinematics, we assume is constant, which means that angular acceleration is also a constant, because = . Now, let us substitute = and = into the linear equation above: The radius cancels in the equation, yielding = 0 + (constant ) = 0 + . (10.18) (10.19) where 0 is the initial angular velocity. This last equation is a kinematic relationship among , , and —that is, it describes their relationship without reference to forces or masses that may affect rotation. It is also precisely analogous in form to its translational counterpart. Making Connections Kinematics for rotational motion is completely analogous to translational kinematics, first presented in One-Dimensional Kinematics. Kinematics is concerned with the description of motion without regard to force or mass. We will find that translational kinematic quantities, such as displacement, velocity, and acceleration have direct analogs in rotational motion. Starting with the four kinematic equations we developed in One-Dimensional Kinematics, we can derive the following four rotational kinematic equations (presented together with their translational counterparts): 396 Chapter 10 \| Rotational Motion and Angular Momentum Table 10.2 Rotational Kinematic Equations Rotational Translational ¯ = = 0 + = = 0 + (constant , ) = constant , ) 2 + 2 (constant , ) In these equations, the subscript 0 denotes initial values ( 0 , 0 , and 0 are initial values), and the average angular velocity - and average velocity - are defined as follows: ¯ = 0 + 2 and ¯ = 0 + 2 . (10.20) The equations given above in Table 10.2 can be used to solve any rotational or translational kinematics problem in which and are constant. Problem-Solving Strategy for Rotational Kinematics 1. Examine the situation to determine that rotational kinematics (rotational motion) is involved. Rotation must be involved, but without the need to consider forces or masses that affect the motion. 2. Identify exactly what needs to be determined in the problem (identify the unknowns). A sketch of the situation is useful. 3. Make a list of what is given or can be inferred from the problem as stated (identify the knowns). 4. Solve the appropriate equation or equations for the quantity to be determined (the unknown). It can be useful to think in terms of a translational analog because by now you are familiar with such motion. 5. Substitute the known values along with their units into the appropriate equation, and obtain numerical solutions complete with units. Be sure to use units of radians for angles. 6. Check your answer to see if it is reasonable: Does your answer make sense? Example 10.3 Calculating the Acceleration of a Fishing Reel A deep-sea
fisherman hooks a big fish that swims away from the boat pulling the fishing line from his fishing reel. The whole system is initially at rest and the fishing line unwinds from the reel at a radius of 4.50 cm from its axis of rotation. The reel is given an angular acceleration of 110 rad/s2 for 2.00 s as seen in Figure 10.8. (a) What is the final angular velocity of the reel? (b) At what speed is fishing line leaving the reel after 2.00 s elapses? (c) How many revolutions does the reel make? (d) How many meters of fishing line come off the reel in this time? Strategy In each part of this example, the strategy is the same as it was for solving problems in linear kinematics. In particular, known values are identified and a relationship is then sought that can be used to solve for the unknown. Solution for (a) Here and are given and needs to be determined. The most straightforward equation to use is = 0 + because the unknown is already on one side and all other terms are known. That equation states that We are also given that 0 = 0 (it starts from rest), so that = 0 + . = 0 + 110 rad/s2 (2.00s) = 220 rad/s. Solution for (b) Now that is known, the speed can most easily be found using the relationship = , This content is available for free at http://cnx.org/content/col11844/1.13 (10.21) (10.22) (10.23) Chapter 10 \| Rotational Motion and Angular Momentum 397 where the radius of the reel is given to be 4.50 cm; thus, = (0.0450 m)(220 rad/s) = 9.90 m/s. (10.24) Note again that radians must always be used in any calculation relating linear and angular quantities. Also, because radians are dimensionless, we have m×rad = m . Solution for (c) Here, we are asked to find the number of revolutions. Because 1 rev = 2π rad , we can find the number of revolutions by finding in radians. We are given and , and we know 0 is zero, so that can be obtained using = 0.500) 110 rad/s2 (2.00 s)2 = 220 rad. Converting radians to revolutions gives = (220 rad) 1 rev 2π rad = 35.0 rev. Solution for (d) The number of meters of fishing line is , which can be obtained through its relationship with : = = (0.0450 m)(220 rad) = 9.90 m. Discussion (10.25) (10.26) (10.27) This example illustrates that relationships among rotational quantities are highly analogous to those among linear quantities. We also see in this example how linear and rotational quantities are connected. The answers to the questions are realistic. After unwinding for two seconds, the reel is found to spin at 220 rad/s, which is 2100 rpm. (No wonder reels sometimes make high-pitched sounds.) The amount of fishing line played out is 9.90 m, about right for when the big fish bites. Figure 10.8 Fishing line coming off a rotating reel moves linearly. Example 10.3 and Example 10.4 consider relationships between rotational and linear quantities associated with a fishing reel. Example 10.4 Calculating the Duration When the Fishing Reel Slows Down and Stops Now let us consider what happens if the fisherman applies a brake to the spinning reel, achieving an angular acceleration of – 300 rad/s2 . How long does it take the reel to come to a stop? Strategy We are asked to find the time for the reel to come to a stop. The initial and final conditions are different from those in the previous problem, which involved the same fishing reel. Now we see that the initial angular velocity is 0 = 220 rad/s and the final angular velocity is zero. The angular acceleration is given to be = −300 rad/s2 . Examining the available equations, we see all quantities but t are known in = 0 + , making it easiest to use this equation. Solution The equation states 398 Chapter 10 \| Rotational Motion and Angular Momentum = 0 + . We solve the equation algebraically for t, and then substitute the known values as usual, yielding = − 0 = 0 − 220 rad/s −300 rad/s2 = 0.733 s. (10.28) (10.29) Discussion Note that care must be taken with the signs that indicate the directions of various quantities. Also, note that the time to stop the reel is fairly small because the acceleration is rather large. Fishing lines sometimes snap because of the accelerations involved, and fishermen often let the fish swim for a while before applying brakes on the reel. A tired fish will be slower, requiring a smaller acceleration. Example 10.5 Calculating the Slow Acceleration of Trains and Their Wheels Large freight trains accelerate very slowly. Suppose one such train accelerates from rest, giving its 0.350-m-radius wheels an angular acceleration of 0.250 rad/s2 . After the wheels have made 200 revolutions (assume no slippage): (a) How far has the train moved down the track? (b) What are the final angular velocity of the wheels and the linear velocity of the train? Strategy In part (a), we are asked to find , and in (b) we are asked to find and . We are given the number of revolutions , the radius of the wheels , and the angular acceleration . Solution for (a) The distance is very easily found from the relationship between distance and rotation angle: Solving this equation for yields = . = Before using this equation, we must convert the number of revolutions into radians, because we are dealing with a relationship between linear and rotational quantities: = (200 rev)2π rad 1 rev = 1257 rad. Now we can substitute the known values into = to find the distance the train moved down the track: = = (0.350 m)(1257 rad) = 440 m. (10.30) (10.31) (10.32) (10.33) Solution for (b) We cannot use any equation that incorporates to find , because the equation would have at least two unknown values. 2 + 2 will work, because we know the values for all variables except : The equation 2 = 0 Taking the square root of this equation and entering the known values gives (0.250 rad/s2)(1257 rad) 1 / 2 = 25.1 rad/s. We can find the linear velocity of the train, , through its relationship to : = = (0.350 m)(25.1 rad/s) = 8.77 m/s. Discussion The distance traveled is fairly large and the final velocity is fairly slow (just under 32 km/h). (10.34) (10.35) (10.36) There is translational motion even for something spinning in place, as the following example illustrates. Figure 10.9 shows a fly on the edge of a rotating microwave oven plate. The example below calculates the total distance it travels. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 10 \| Rotational Motion and Angular Momentum 399 Figure 10.9 The image shows a microwave plate. The fly makes revolutions while the food is heated (along with the fly). Example 10.6 Calculating the Distance Traveled by a Fly on the Edge of a Microwave Oven Plate A person decides to use a microwave oven to reheat some lunch. In the process, a fly accidentally flies into the microwave and lands on the outer edge of the rotating plate and remains there. If the plate has a radius of 0.15 m and rotates at 6.0 rpm, calculate the total distance traveled by the fly during a 2.0-min cooking period. (Ignore the start-up and slow-down times.) Strategy ¯ First, find the total number of revolutions , and then the linear distance traveled. = because - is given to be 6.0 rpm. can be used to find Solution ¯ Entering known values into = gives = = 6.0 rpm (2.0 min) = 12 rev. As always, it is necessary to convert revolutions to radians before calculating a linear quantity like from an angular quantity like : = (12 rev) 2 rad 1 rev = 75.4 rad. Now, using the relationship between and , we can determine the distance traveled: = = (0.15 m)(75.4 rad) = 11 m. Discussion (10.37) (10.38) (10.39) Quite a trip (if it survives)! Note that this distance is the total distance traveled by the fly. Displacement is actually zero for complete revolutions because they bring the fly back to its original position. The distinction between total distance traveled and displacement was first noted in One-Dimensional Kinematics. Check Your Understanding Rotational kinematics has many useful relationships, often expressed in equation form. Are these relationships laws of physics or are they simply descriptive? (Hint: the same question applies to linear kinematics.) Solution Rotational kinematics (just like linear kinematics) is descriptive and does not represent laws of nature. With kinematics, we can describe many things to great precision but kinematics does not consider causes. For example, a large angular acceleration describes a very rapid change in angular velocity without any consideration of its cause. 400 Chapter 10 \| Rotational Motion and Angular Momentum 10.3 Dynamics of Rotational Motion: Rotational Inertia Learning Objectives By the end of this section, you will be able to: • Understand the relationship between force, mass, and acceleration. • Study the turning effect of force. • Study the analogy between force and torque, mass and moment of inertia, and linear acceleration and angular acceleration. The information presented in this section supports the following AP® learning objectives and science practices: • 4.D.1.1 The student is able to describe a representation and use it to analyze a situation in which several forces exerted on a rotating system of rigidly connected objects change the angular velocity and angular momentum of the system. (S.P. 1.2, 1.4) • 4.D.1.2 The student is able to plan data collection strategies designed to establish that torque, angular velocity, angular acceleration, and angular momentum can be predicted accurately when the variables are treated as being clockwise or counterclockwise with respect to a well-defined axis of rotation, and refine the research question based on the examination of data. (S.P. 3.2, 4.1, 5.1, 5.3) • 5.E.2.1 The student is able to describe or calculate the angular momentum and rotational inertia of a system in terms of the locations and velocities of objects that make up the system. Students are expected to do qualitative reasoning with compound objects. Students are expected to do calculations with a fixed set of extended objects and point masses. (S.P. 2.2) If you have ever sp
un a bike wheel or pushed a merry-go-round, you know that force is needed to change angular velocity as seen in Figure 10.10. In fact, your intuition is reliable in predicting many of the factors that are involved. For example, we know that a door opens slowly if we push too close to its hinges. Furthermore, we know that the more massive the door, the more slowly it opens. The first example implies that the farther the force is applied from the pivot, the greater the angular acceleration; another implication is that angular acceleration is inversely proportional to mass. These relationships should seem very similar to the familiar relationships among force, mass, and acceleration embodied in Newton's second law of motion. There are, in fact, precise rotational analogs to both force and mass. Figure 10.10 Force is required to spin the bike wheel. The greater the force, the greater the angular acceleration produced. The more massive the wheel, the smaller the angular acceleration. If you push on a spoke closer to the axle, the angular acceleration will be smaller. To develop the precise relationship among force, mass, radius, and angular acceleration, consider what happens if we exert a force on a point mass that is at a distance from a pivot point, as shown in Figure 10.11. Because the force is perpendicular to , an acceleration = is obtained in the direction of . We can rearrange this equation such that = and then look for ways to relate this expression to expressions for rotational quantities. We note that = , and we substitute this expression into = , yielding Recall that torque is the turning effectiveness of a force. In this case, because F is perpendicular to , torque is simply = . So, if we multiply both sides of the equation above by , we get torque on the left-hand side. That is, = . or = 2 = 2α. (10.40) (10.41) (10.42) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 10 \| Rotational Motion and Angular Momentum 401 This last equation is the rotational analog of Newton's second law ( = ), where torque is analogous to force, angular acceleration is analogous to translational acceleration, and 2 is analogous to mass (or inertia). The quantity 2 is called the rotational inertia or moment of inertia of a point mass a distance from the center of rotation. Figure 10.11 An object is supported by a horizontal frictionless table and is attached to a pivot point by a cord that supplies centripetal force. A force is applied to the object perpendicular to the radius , causing it to accelerate about the pivot point. The force is kept perpendicular to . Making Connections: Rotational Motion Dynamics Dynamics for rotational motion is completely analogous to linear or translational dynamics. Dynamics is concerned with force and mass and their effects on motion. For rotational motion, we will find direct analogs to force and mass that behave just as we would expect from our earlier experiences. Rotational Inertia and Moment of Inertia Before we can consider the rotation of anything other than a point mass like the one in Figure 10.11, we must extend the idea of rotational inertia to all types of objects. To expand our concept of rotational inertia, we define the moment of inertia of an object to be the sum of 2 for all the point masses of which it is composed. That is, = ∑ 2 . Here is analogous to in translational motion. Because of the distance , the moment of inertia for any object depends on the chosen axis. Actually, calculating is beyond the scope of this text except for one simple case—that of a hoop, which has all its mass at the same distance from its axis. A hoop's moment of inertia around its axis is therefore 2 , where is its total mass and its radius. (We use and for an entire object to distinguish them from and for point masses.) In all other cases, we must consult Figure 10.12 (note that the table is piece of artwork that has shapes as well as formulae) for formulas for that have been derived from integration over the continuous body. Note that has units of mass multiplied by distance squared ( kg ⋅ m2 ), as we might expect from its definition. The general relationship among torque, moment of inertia, and angular acceleration is or net τ = = net τ , (10.43) (10.44) where net is the total torque from all forces relative to a chosen axis. For simplicity, we will only consider torques exerted by forces in the plane of the rotation. Such torques are either positive or negative and add like ordinary numbers. The relationship in = , = net τ is the rotational analog to Newton's second law and is very generally applicable. This equation is actually valid for any torque, applied to any object, relative to any axis. As we might expect, the larger the torque is, the larger the angular acceleration is. For example, the harder a child pushes on a merry-go-round, the faster it accelerates. Furthermore, the more massive a merry-go-round, the slower it accelerates for the same torque. The basic relationship between moment of inertia and angular acceleration is that the larger the moment of inertia, the smaller is the angular acceleration. But there is an additional twist. The moment of inertia depends not only on the mass of an object, but also on its distribution of mass relative to the axis around which it rotates. For example, it will be much easier to accelerate a merry-go-round full of children if they stand close to its axis than if they all stand at the outer edge. The mass is the same in both cases; but the moment of inertia is much larger when the children are at the edge. Take-Home Experiment Cut out a circle that has about a 10 cm radius from stiff cardboard. Near the edge of the circle, write numbers 1 to 12 like hours on a clock face. Position the circle so that it can rotate freely about a horizontal axis through its center, like a wheel. (You could loosely nail the circle to a wall.) Hold the circle stationary and with the number 12 positioned at the top, attach a 402 Chapter 10 \| Rotational Motion and Angular Momentum lump of blue putty (sticky material used for fixing posters to walls) at the number 3. How large does the lump need to be to just rotate the circle? Describe how you can change the moment of inertia of the circle. How does this change affect the amount of blue putty needed at the number 3 to just rotate the circle? Change the circle's moment of inertia and then try rotating the circle by using different amounts of blue putty. Repeat this process several times. In what direction did the circle rotate when you added putty at the number 3 (clockwise or counterclockwise)? In which of these directions was the resulting angular velocity? Was the angular velocity constant? What can we say about the direction (clockwise or counterclockwise) of the angular acceleration? How could you change the placement of the putty to create angular velocity in the opposite direction? Problem-Solving Strategy for Rotational Dynamics 1. Examine the situation to determine that torque and mass are involved in the rotation. Draw a careful sketch of the situation. 2. Determine the system of interest. 3. Draw a free body diagram. That is, draw and label all external forces acting on the system of interest. 4. Apply , the rotational equivalent of Newton's second law, to solve the problem. Care must be taken to use the correct moment of inertia and to consider the torque about the point of rotation. 5. As always, check the solution to see if it is reasonable. Making Connections In statics, the net torque is zero, and there is no angular acceleration. In rotational motion, net torque is the cause of angular acceleration, exactly as in Newton's second law of motion for rotation. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 10 \| Rotational Motion and Angular Momentum 403 Figure 10.12 Some rotational inertias. Example 10.7 Calculating the Effect of Mass Distribution on a Merry-Go-Round Consider the father pushing a playground merry-go-round in Figure 10.13. He exerts a force of 250 N at the edge of the 50.0-kg merry-go-round, which has a 1.50 m radius. Calculate the angular acceleration produced (a) when no one is on the merry-go-round and (b) when an 18.0-kg child sits 1.25 m away from the center. Consider the merry-go-round itself to be a uniform disk with negligible retarding friction. Figure 10.13 A father pushes a playground merry-go-round at its edge and perpendicular to its radius to achieve maximum torque. Strategy Angular acceleration is given directly by the expression = net τ : 404 Chapter 10 \| Rotational Motion and Angular Momentum To solve for , we must first calculate the torque (which is the same in both cases) and moment of inertia (which is greater in the second case). To find the torque, we note that the applied force is perpendicular to the radius and friction is negligible, so that = . (10.45) τ = sin θ = (1.50 m)(250 N) = 375 N ⋅ m. Solution for (a) The moment of inertia of a solid disk about this axis is given in Figure 10.12 to be 1 22, where = 50.0 kg and = 1.50 m , so that Now, after we substitute the known values, we find the angular acceleration to be = (0.500)(50.0 kg)(1.50 m)2 = 56.25 kg ⋅ m2. = = 375 N ⋅ m 56.25 kg ⋅ m2 = 6.67rad s2 . Solution for (b) (10.46) (10.47) (10.48) (10.49) We expect the angular acceleration for the system to be less in this part, because the moment of inertia is greater when the child is on the merry-go-round. To find the total moment of inertia , we first find the child's moment of inertia c by considering the child to be equivalent to a point mass at a distance of 1.25 m from the axis. Then, c = 2 = (18.0 kg)(1.25 m)2 = 28.13 kg ⋅ m2. (10.50) The total moment of inertia is the sum of moments of inertia of the merry-go-round and the child (about the same axis). To justify this sum to yourself, examine the definition of : = 28.13 kg ⋅ m2 + 56.25 kg ⋅ m2 = 84.38 kg ⋅ m2. Substituting known values into
the equation for gives = τ = 375 N ⋅ m 84.38 kg ⋅ m2 = 4.44rad s2 . (10.51) (10.52) Discussion The angular acceleration is less when the child is on the merry-go-round than when the merry-go-round is empty, as expected. The angular accelerations found are quite large, partly due to the fact that friction was considered to be negligible. If, for example, the father kept pushing perpendicularly for 2.00 s, he would give the merry-go-round an angular velocity of 13.3 rad/s when it is empty but only 8.89 rad/s when the child is on it. In terms of revolutions per second, these angular velocities are 2.12 rev/s and 1.41 rev/s, respectively. The father would end up running at about 50 km/h in the first case. Summer Olympics, here he comes! Confirmation of these numbers is left as an exercise for the reader. Making Connections: Multiple Forces on One System A large potter's wheel has a diameter of 60.0 cm and a mass of 8.0 kg. It is powered by a 20.0 N motor acting on the outer edge. There is also a brake capable of exerting a 15.0 N force at a radius of 12.0 cm from the axis of rotation, on the underside. What is the angular acceleration when the motor is in use? The torque is found by = sin = (0.300 m)(20.0 N) = 6.00 N·m . The moment of inertia is calculated as = 1 2 2 = 1 2 8.0 kg (0.300 m)2 = 0.36 kg ⋅ m2 . Thus, the angular acceleration would be = = 6.00 N ⋅ m 0.36 kg ⋅ m2 = 17 rad/s2 . Note that the friction is always acting in a direction opposite to the rotation that is currently happening in this system. If the potter makes a mistake and has both the brake and motor on simultaneously, the friction force of the brake will exert a torque opposite that of the motor. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 10 \| Rotational Motion and Angular Momentum 405 The torque from the brake is = sin = (0.120 m)(15.0 N) = 1.80 N⋅m . Thus, the net torque is 6.00 N⋅m − 1.80 Ν⋅m = 4.20 Ν⋅m . And the angular acceleration is = = 4.20 N ⋅ m 0.36 kg ⋅ m2 = 12 rad/s2 . Check Your Understanding Torque is the analog of force and moment of inertia is the analog of mass. Force and mass are physical quantities that depend on only one factor. For example, mass is related solely to the numbers of atoms of various types in an object. Are torque and moment of inertia similarly simple? Solution No. Torque depends on three factors: force magnitude, force direction, and point of application. Moment of inertia depends on both mass and its distribution relative to the axis of rotation. So, while the analogies are precise, these rotational quantities depend on more factors. 10.4 Rotational Kinetic Energy: Work and Energy Revisited Learning Objectives By the end of this section, you will be able to: • Derive the equation for rotational work. • Calculate rotational kinetic energy. • Demonstrate the law of conservation of energy. The information presented in this section supports the following AP® learning objectives and science practices: • 3.F.2.1 The student is able to make predictions about the change in the angular velocity about an axis for an object when forces exerted on the object cause a torque about that axis. (S.P. 6.4) • 3.F.2.2 The student is able to plan data collection and analysis strategies designed to test the relationship between a torque exerted on an object and the change in angular velocity of that object about an axis. (S.P. 4.1, 4.2, 5.1) In this module, we will learn about work and energy associated with rotational motion. Figure 10.14 shows a worker using an electric grindstone propelled by a motor. Sparks are flying, and noise and vibration are created as layers of steel are pared from the pole. The stone continues to turn even after the motor is turned off, but it is eventually brought to a stop by friction. Clearly, the motor had to work to get the stone spinning. This work went into heat, light, sound, vibration, and considerable rotational kinetic energy. Figure 10.14 The motor works in spinning the grindstone, giving it rotational kinetic energy. That energy is then converted to heat, light, sound, and vibration. (credit: U.S. Navy photo by Mass Communication Specialist Seaman Zachary David Bell) Work must be done to rotate objects such as grindstones or merry-go-rounds. Work was defined in Uniform Circular Motion and Gravitation for translational motion, and we can build on that knowledge when considering work done in rotational motion. The simplest rotational situation is one in which the net force is exerted perpendicular to the radius of a disk (as shown in Figure 10.15) and remains perpendicular as the disk starts to rotate. The force is parallel to the displacement, and so the net work done is the product of the force times the arc length traveled: net = (net )Δ. (10.53) To get torque and other rotational quantities into the equation, we multiply and divide the right-hand side of the equation by , and gather terms: 406 Chapter 10 \| Rotational Motion and Angular Momentum We recognize that net = net τ and Δ / = , so that net = (net τ). net = ( net )Δ . (10.54) (10.55) This equation is the expression for rotational work. It is very similar to the familiar definition of translational work as force multiplied by distance. Here, torque is analogous to force, and angle is analogous to distance. The equation net = (net τ) is valid in general, even though it was derived for a special case. To get an expression for rotational kinetic energy, we must again perform some algebraic manipulations. The first step is to note that net τ = , so that net = . (10.56) Figure 10.15 The net force on this disk is kept perpendicular to its radius as the force causes the disk to rotate. The net work done is thus (net )Δ . The net work goes into rotational kinetic energy. Making Connections qWork and energy in rotational motion are completely analogous to work and energy in translational motion, first presented in Uniform Circular Motion and Gravitation. Now, we solve one of the rotational kinematics equations for . We start with the equation Next, we solve for : 2 = 0 2 + 2. = 2 − 0 2 2 . Substituting this into the equation for net and gathering terms yields net = 1 22 − 1 20 (10.57) (10.58) (10.59) 2. This equation is the work-energy theorem for rotational motion only. As you may recall, net work changes the kinetic energy of a system. Through an analogy with translational motion, we define the term 2 to be rotational kinetic energy KErot for 1 2 an object with a moment of inertia and an angular velocity : KErot = 1 22. (10.60) The expression for rotational kinetic energy is exactly analogous to translational kinetic energy, with being analogous to and to . Rotational kinetic energy has important effects. Flywheels, for example, can be used to store large amounts of rotational kinetic energy in a vehicle, as seen in Figure 10.16. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 10 \| Rotational Motion and Angular Momentum 407 Figure 10.16 Experimental vehicles, such as this bus, have been constructed in which rotational kinetic energy is stored in a large flywheel. When the bus goes down a hill, its transmission converts its gravitational potential energy into KErot . It can also convert translational kinetic energy, when the bus stops, into KErot . The flywheel's energy can then be used to accelerate, to go up another hill, or to keep the bus from going against friction. Example 10.8 Calculating the Work and Energy for Spinning a Grindstone Consider a person who spins a large grindstone by placing her hand on its edge and exerting a force through part of a revolution as shown in Figure 10.17. In this example, we verify that the work done by the torque she exerts equals the change in rotational energy. (a) How much work is done if she exerts a force of 200 N through a rotation of 1.00 rad(57.3º) ? The force is kept perpendicular to the grindstone's 0.320-m radius at the point of application, and the effects of friction are negligible. (b) What is the final angular velocity if the grindstone has a mass of 85.0 kg? (c) What is the final rotational kinetic energy? (It should equal the work.) Strategy To find the work, we can use the equation net = (net τ) . We have enough information to calculate the torque and are given the rotation angle. In the second part, we can find the final angular velocity using one of the kinematic relationships. In the last part, we can calculate the rotational kinetic energy from its expression in KErot = 1 22 . Solution for (a) The net work is expressed in the equation net = (net τ), where net is the applied force multiplied by the radius () because there is no retarding friction, and the force is perpendicular to . The angle is given. Substituting the given values in the equation above yields net = = (0.320 m)(200 N)(1.00 rad) = 64.0 N ⋅ m. Noting that 1 N · m = 1 J , net = 64.0 J. (10.61) (10.62) (10.63) Figure 10.17 A large grindstone is given a spin by a person grasping its outer edge. Solution for (b) To find from the given information requires more than one step. We start with the kinematic relationship in the equation 2 = 0 2 + 2. (10.64) 408 Chapter 10 \| Rotational Motion and Angular Momentum Note that 0 = 0 because we start from rest. Taking the square root of the resulting equation gives Now we need to find . One possibility is = (2)1 / 2. = net τ , where the torque is The formula for the moment of inertia for a disk is found in Figure 10.12: net τ = = (0.320 m)(200 N) = 64.0 N ⋅ m. = 1 22 = 0.5 85.0 kg (0.320 m)2 = 4.352 kg ⋅ m2. Substituting the values of torque and moment of inertia into the expression for , we obtain = 64.0 N ⋅ m 4.352 kg ⋅ m2 = 14.7rad s2 . Now, substitute this value and the given value for into the above expression for : = (2)1 / 2 = 2 14.7rad s2 1 / 2 (1.00 rad) = 5.42rad s . Solution for (c) The final rotational kinetic energy is Both and were found above. Thus, KErot = 1 22. KErot = (0.5) 4.352 kg ⋅
m2 (5.42 rad/s)2 = 64.0 J. (10.65) (10.66) (10.67) (10.68) (10.69) (10.70) (10.71) (10.72) Discussion The final rotational kinetic energy equals the work done by the torque, which confirms that the work done went into rotational kinetic energy. We could, in fact, have used an expression for energy instead of a kinematic relation to solve part (b). We will do this in later examples. Helicopter pilots are quite familiar with rotational kinetic energy. They know, for example, that a point of no return will be reached if they allow their blades to slow below a critical angular velocity during flight. The blades lose lift, and it is impossible to immediately get the blades spinning fast enough to regain it. Rotational kinetic energy must be supplied to the blades to get them to rotate faster, and enough energy cannot be supplied in time to avoid a crash. Because of weight limitations, helicopter engines are too small to supply both the energy needed for lift and to replenish the rotational kinetic energy of the blades once they have slowed down. The rotational kinetic energy is put into them before takeoff and must not be allowed to drop below this crucial level. One possible way to avoid a crash is to use the gravitational potential energy of the helicopter to replenish the rotational kinetic energy of the blades by losing altitude and aligning the blades so that the helicopter is spun up in the descent. Of course, if the helicopter's altitude is too low, then there is insufficient time for the blade to regain lift before reaching the ground. Take-Home Experiment Rotational motion can be observed in wrenches, clocks, wheels or spools on axels, and seesaws. Choose an object or system that exhibits rotational motion and plan an experiment to test how torque affects angular velocity. How will you create and measure different amounts of torque? How will you measure angular velocity? Remember that net = , ∝ 2, and = . Problem-Solving Strategy for Rotational Energy 1. Determine that energy or work is involved in the rotation. 2. Determine the system of interest. A sketch usually helps. 3. Analyze the situation to determine the types of work and energy involved. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 10 \| Rotational Motion and Angular Momentum 409 4. For closed systems, mechanical energy is conserved. That is, KEi + PEi = KEf + PEf. Note that KEi and KEf may each include translational and rotational contributions. 5. For open systems, mechanical energy may not be conserved, and other forms of energy (referred to previously as ), such as heat transfer, may enter or leave the system. Determine what they are, and calculate them as necessary. 6. Eliminate terms wherever possible to simplify the algebra. 7. Check the answer to see if it is reasonable. Example 10.9 Calculating Helicopter Energies A typical small rescue helicopter, similar to the one in Figure 10.18, has four blades, each is 4.00 m long and has a mass of 50.0 kg. The blades can be approximated as thin rods that rotate about one end of an axis perpendicular to their length. The helicopter has a total loaded mass of 1000 kg. (a) Calculate the rotational kinetic energy in the blades when they rotate at 300 rpm. (b) Calculate the translational kinetic energy of the helicopter when it flies at 20.0 m/s, and compare it with the rotational energy in the blades. (c) To what height could the helicopter be raised if all of the rotational kinetic energy could be used to lift it? Strategy Rotational and translational kinetic energies can be calculated from their definitions. The last part of the problem relates to the idea that energy can change form, in this case from rotational kinetic energy to gravitational potential energy. Solution for (a) The rotational kinetic energy is 22. We must convert the angular velocity to radians per second and calculate the moment of inertia before we can find KErot . The angular velocity is KErot = 1 (10.73) = 300 rev 1.00 min ⋅ 2π rad 1 rev ⋅ 1.00 min 60.0 s = 31.4rad s . (10.74) The moment of inertia of one blade will be that of a thin rod rotated about its end, found in Figure 10.12. The total is four times this moment of inertia, because there are four blades. Thus, = 42 3 = 4× (4.00 m)2 50.0 kg 3 = 1067 kg ⋅ m2. Entering and into the expression for rotational kinetic energy gives KErot = 0.5(1067 kg ⋅ m2)(31.4 rad/s)2 = 5.26×105 J (10.75) (10.76) Solution for (b) Translational kinetic energy was defined in Uniform Circular Motion and Gravitation. Entering the given values of mass and velocity, we obtain KEtrans = 1 22 = (0.5) 1000 kg (20.0 m/s)2 = 2.00×105 J. To compare kinetic energies, we take the ratio of translational kinetic energy to rotational kinetic energy. This ratio is 2.00×105 J 5.26×105 J = 0.380. (10.77) (10.78) Solution for (c) At the maximum height, all rotational kinetic energy will have been converted to gravitational energy. To find this height, we equate those two energies: or KErot = PEgrav 1 22 = . (10.79) (10.80) 410 Chapter 10 \| Rotational Motion and Angular Momentum We now solve for and substitute known values into the resulting equation 2 1 2 = = 5.26×105 J 1000 kg 9.80 m/s2 = 53.7 m. (10.81) Discussion The ratio of translational energy to rotational kinetic energy is only 0.380. This ratio tells us that most of the kinetic energy of the helicopter is in its spinning blades—something you probably would not suspect. The 53.7 m height to which the helicopter could be raised with the rotational kinetic energy is also impressive, again emphasizing the amount of rotational kinetic energy in the blades. Figure 10.18 The first image shows how helicopters store large amounts of rotational kinetic energy in their blades. This energy must be put into the blades before takeoff and maintained until the end of the flight. The engines do not have enough power to simultaneously provide lift and put significant rotational energy into the blades. The second image shows a helicopter from the Auckland Westpac Rescue Helicopter Service. Over 50,000 lives have been saved since its operations beginning in 1973. Here, a water rescue operation is shown. (credit: 111 Emergency, Flickr) Making Connections Conservation of energy includes rotational motion, because rotational kinetic energy is another form of KE . Uniform Circular Motion and Gravitation has a detailed treatment of conservation of energy. How Thick Is the Soup? Or Why Don't All Objects Roll Downhill at the Same Rate? One of the quality controls in a tomato soup factory consists of rolling filled cans down a ramp. If they roll too fast, the soup is too thin. Why should cans of identical size and mass roll down an incline at different rates? And why should the thickest soup roll the slowest? The easiest way to answer these questions is to consider energy. Suppose each can starts down the ramp from rest. Each can starting from rest means each starts with the same gravitational potential energy PEgrav , which is converted entirely to KE , provided each rolls without slipping. KE , however, can take the form of KEtrans or KErot , and total KE is the sum of the two. If a can rolls down a ramp, it puts part of its energy into rotation, leaving less for translation. Thus, the can goes slower than it would if it slid down. Furthermore, the thin soup does not rotate, whereas the thick soup does, because it sticks to the can. The thick soup thus puts more of the can's original gravitational potential energy into rotation than the thin soup, and the can rolls more slowly, as seen in Figure 10.19. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 10 \| Rotational Motion and Angular Momentum 411 Figure 10.19 Three cans of soup with identical masses race down an incline. The first can has a low friction coating and does not roll but just slides down the incline. It wins because it converts its entire PE into translational KE. The second and third cans both roll down the incline without slipping. The second can contains thin soup and comes in second because part of its initial PE goes into rotating the can (but not the thin soup). The third can contains thick soup. It comes in third because the soup rotates along with the can, taking even more of the initial PE for rotational KE, leaving less for translational KE. Assuming no losses due to friction, there is only one force doing work—gravity. Therefore the total work done is the change in kinetic energy. As the cans start moving, the potential energy is changing into kinetic energy. Conservation of energy gives PEi = KEf. (10.82) More specifically, or PEgrav = KEtrans + KErot (10.83) 22 + 1 So, the initial is divided between translational kinetic energy and rotational kinetic energy; and the greater is, the less energy goes into translation. If the can slides down without friction, then = 0 and all the energy goes into translation; thus, the can goes faster. = 1 22. (10.84) Take-Home Experiment Locate several cans each containing different types of food. First, predict which can will win the race down an inclined plane and explain why. See if your prediction is correct. You could also do this experiment by collecting several empty cylindrical containers of the same size and filling them with different materials such as wet or dry sand. Example 10.10 Calculating the Speed of a Cylinder Rolling Down an Incline Calculate the final speed of a solid cylinder that rolls down a 2.00-m-high incline. The cylinder starts from rest, has a mass of 0.750 kg, and has a radius of 4.00 cm. Strategy We can solve for the final velocity using conservation of energy, but we must first express rotational quantities in terms of translational quantities to end up with as the only unknown. Solution Conservation of energy for this situation is written as described above: 22 + 1 Before we can solve for , we must get an expression for from Figure 10.12. Because and are related (note her
e that the cylinder is rolling without slipping), we must also substitute the relationship = / into the expression. These substitutions yield = 1 22. (10.85) = 1 22 + 1 2 1 22 2 2 . Interestingly, the cylinder's radius and mass cancel, yielding 42 = 3 22 + 1 Solving algebraically, the equation for the final velocity gives = 1 42. (10.86) (10.87) 412 Chapter 10 \| Rotational Motion and Angular Momentum = 4 3 1 / 2 . Substituting known values into the resulting expression yields = 9.80 m/s2 4 3 (2.00 m) 1 / 2 = 5.11 m/s. (10.88) (10.89) Discussion Because and cancel, the result = 1 / 2 4 3 is valid for any solid cylinder, implying that all solid cylinders will roll down an incline at the same rate independent of their masses and sizes. (Rolling cylinders down inclines is what Galileo actually did to show that objects fall at the same rate independent of mass.) Note that if the cylinder slid without friction down the incline without rolling, then the entire gravitational potential energy would go into translational kinetic energy. Thus, 1 22 = and = (2)1 / 2 , which is 22% greater than (4 / 3)1 / 2 . That is, the cylinder would go faster at the bottom. Check Your Understanding Analogy of Rotational and Translational Kinetic Energy Is rotational kinetic energy completely analogous to translational kinetic energy? What, if any, are their differences? Give an example of each type of kinetic energy. Solution Yes, rotational and translational kinetic energy are exact analogs. They both are the energy of motion involved with the coordinated (non-random) movement of mass relative to some reference frame. The only difference between rotational and translational kinetic energy is that translational is straight line motion while rotational is not. An example of both kinetic and translational kinetic energy is found in a bike tire while being ridden down a bike path. The rotational motion of the tire means it has rotational kinetic energy while the movement of the bike along the path means the tire also has translational kinetic energy. If you were to lift the front wheel of the bike and spin it while the bike is stationary, then the wheel would have only rotational kinetic energy relative to the Earth. PhET Explorations: My Solar System Build your own system of heavenly bodies and watch the gravitational ballet. With this orbit simulator, you can set initial positions, velocities, and masses of 2, 3, or 4 bodies, and then see them orbit each other. Figure 10.20 My Solar System (http://cnx.org/content/m55188/1.4/my-solar-system_en.jar) 10.5 Angular Momentum and Its Conservation Learning Objectives By the end of this section, you will be able to: • Understand the analogy between angular momentum and linear momentum. • Observe the relationship between torque and angular momentum. • Apply the law of conservation of angular momentum. The information presented in this section supports the following AP® learning objectives and science practices: • 4.D.2.1 The student is able to describe a model of a rotational system and use that model to analyze a situation in which angular momentum changes due to interaction with other objects or systems. (S.P. 1.2, 1.4) • 4.D.2.2 The student is able to plan a data collection and analysis strategy to determine the change in angular momentum of a system and relate it to interactions with other objects and systems. (S.P. 2.2) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 10 \| Rotational Motion and Angular Momentum 413 • 4.D.3.1 The student is able to use appropriate mathematical routines to calculate values for initial or final angular momentum, or change in angular momentum of a system, or average torque or time during which the torque is exerted in analyzing a situation involving torque and angular momentum. (S.P. 2.2) • 4.D.3.2 The student is able to plan a data collection strategy designed to test the relationship between the change in angular momentum of a system and the product of the average torque applied to the system and the time interval during which the torque is exerted. (S.P. 4.1, 4.2) • 5.E.1.1 The student is able to make qualitative predictions about the angular momentum of a system for a situation in which there is no net external torque. (S.P. 6.4, 7.2) • 5.E.1.2 The student is able to make calculations of quantities related to the angular momentum of a system when the net external torque on the system is zero. (S.P. 2.1, 2.2) • 5.E.2.1 The student is able to describe or calculate the angular momentum and rotational inertia of a system in terms of the locations and velocities of objects that make up the system. Students are expected to do qualitative reasoning with compound objects. Students are expected to do calculations with a fixed set of extended objects and point masses. (S.P. 2.2) Why does Earth keep on spinning? What started it spinning to begin with? And how does an ice skater manage to spin faster and faster simply by pulling her arms in? Why does she not have to exert a torque to spin faster? Questions like these have answers based in angular momentum, the rotational analog to linear momentum. By now the pattern is clear—every rotational phenomenon has a direct translational analog. It seems quite reasonable, then, to define angular momentum as = . (10.90) This equation is an analog to the definition of linear momentum as = . Units for linear momentum are kg ⋅ m/s while units for angular momentum are kg ⋅ m2/s . As we would expect, an object that has a large moment of inertia , such as Earth, has a very large angular momentum. An object that has a large angular velocity , such as a centrifuge, also has a rather large angular momentum. Making Connections Angular momentum is completely analogous to linear momentum, first presented in Uniform Circular Motion and Gravitation. It has the same implications in terms of carrying rotation forward, and it is conserved when the net external torque is zero. Angular momentum, like linear momentum, is also a property of the atoms and subatomic particles. Example 10.11 Calculating Angular Momentum of the Earth Strategy No information is given in the statement of the problem; so we must look up pertinent data before we can calculate = . First, according to Figure 10.12, the formula for the moment of inertia of a sphere is so that = 22 5 = = 22 5 . (10.91) (10.92) Earth's mass is 5.979×1024 kg and its radius is 6.376×106 m . The Earth's angular velocity is, of course, exactly one revolution per day, but we must covert to radians per second to do the calculation in SI units. Solution Substituting known information into the expression for and converting to radians per second gives = 0.4 5.979×1024 kg = 9.72×1037 kg ⋅ m2 ⋅ rev/d. 6.376×106 m 2 1 rev d Substituting 2π rad for 1 rev and 8.64×104 s for 1 day gives = 2π rad/rev 9.72×1037 kg ⋅ m2 8.64×104 s/d = 7.07×1033 kg ⋅ m2/s. (1 rev/d) (10.93) (10.94) 414 Chapter 10 \| Rotational Motion and Angular Momentum Discussion This number is large, demonstrating that Earth, as expected, has a tremendous angular momentum. The answer is approximate, because we have assumed a constant density for Earth in order to estimate its moment of inertia. When you push a merry-go-round, spin a bike wheel, or open a door, you exert a torque. If the torque you exert is greater than opposing torques, then the rotation accelerates, and angular momentum increases. The greater the net torque, the more rapid the increase in . The relationship between torque and angular momentum is net = Δ Δ . (10.95) This expression is exactly analogous to the relationship between force and linear momentum, = Δ / Δ . The equation net = Δ Δ is very fundamental and broadly applicable. It is, in fact, the rotational form of Newton's second law. Example 10.12 Calculating the Torque Putting Angular Momentum Into a Lazy Susan Figure 10.21 shows a Lazy Susan food tray being rotated by a person in quest of sustenance. Suppose the person exerts a 2.50 N force perpendicular to the lazy Susan's 0.260-m radius for 0.150 s. (a) What is the final angular momentum of the lazy Susan if it starts from rest, assuming friction is negligible? (b) What is the final angular velocity of the lazy Susan, given that its mass is 4.00 kg and assuming its moment of inertia is that of a disk? Figure 10.21 A partygoer exerts a torque on a lazy Susan to make it rotate. The equation net = Δ Δ gives the relationship between torque and the angular momentum produced. Strategy We can find the angular momentum by solving net = Δ Δ for Δ , and using the given information to calculate the torque. The final angular momentum equals the change in angular momentum, because the lazy Susan starts from rest. That is, Δ = . To find the final velocity, we must calculate from the definition of in = . Solution for (a) Solving net = Δ Δ for Δ gives Because the force is perpendicular to , we see that net = , so that Δ = (net τ)Δt. = rFΔ = (0.260 m)(2.50 N)(0.150 s) = 9.75×10−2 kg ⋅ m2 / s. Solution for (b) The final angular velocity can be calculated from the definition of angular momentum, = . Solving for and substituting the formula for the moment of inertia of a disk into the resulting equation gives And substituting known values into the preceding equation yields = = 1 22 . (10.96) (10.97) (10.98) (10.99) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 10 \| Rotational Motion and Angular Momentum = Discussion 9.75×10−2 kg ⋅ m2/s 4.00 kg (0.260 m) (0.500) = 0.721 rad/s. 415 (10.100) Note that the imparted angular momentum does not depend on any property of the object but only on torque and time. The final angular velocity is equivalent to one revolution in 8.71 s (determination of the time period is left as an exercise for the reader), which is about right for a lazy Susan. Take-Home Experiment Plan an experiment to analyze changes to a system's angular momentum. Choose a system capable of rotational motion such
as a lazy Susan or a merry-go-round. Predict how the angular momentum of this system will change when you add an object to the lazy Susan or jump onto the merry-go-round. What variables can you control? What are you measuring? In other words, what are your independent and dependent variables? Are there any independent variables that it would be useful to keep constant (angular velocity, perhaps)? Collect data in order to calculate or estimate the angular momentum of your system when in motion. What do you observe? Collect data in order to calculate the change in angular momentum as a result of the interaction you performed. Using your data, how does the angular momentum vary with the size and location of an object added to the rotating system? Example 10.13 Calculating the Torque in a Kick The person whose leg is shown in Figure 10.22 kicks his leg by exerting a 2000-N force with his upper leg muscle. The effective perpendicular lever arm is 2.20 cm. Given the moment of inertia of the lower leg is 1.25 kg ⋅ m2 , (a) find the angular acceleration of the leg. (b) Neglecting the gravitational force, what is the rotational kinetic energy of the leg after it has rotated through 57.3º (1.00 rad)? Figure 10.22 The muscle in the upper leg gives the lower leg an angular acceleration and imparts rotational kinetic energy to it by exerting a torque about the knee. F is a vector that is perpendicular to . This example examines the situation. Strategy The angular acceleration can be found using the rotational analog to Newton's second law, or = net / . The moment of inertia is given and the torque can be found easily from the given force and perpendicular lever arm. Once the angular acceleration is known, the final angular velocity and rotational kinetic energy can be calculated. Solution to (a) From the rotational analog to Newton's second law, the angular acceleration is = net . (10.101) Because the force and the perpendicular lever arm are given and the leg is vertical so that its weight does not create a torque, the net torque is thus net = ⊥ = (0.0220 m)(2000 N) = 44.0 N ⋅ m. (10.102) Substituting this value for the torque and the given value for the moment of inertia into the expression for gives 416 Chapter 10 \| Rotational Motion and Angular Momentum = 44.0 N ⋅ m 1.25 kg ⋅ m2 = 35.2 rad/s2. Solution to (b) The final angular velocity can be calculated from the kinematic expression or 2 = 0 2 + 2 2 = 2 because the initial angular velocity is zero. The kinetic energy of rotation is (10.103) (10.104) (10.105) 22 so it is most convenient to use the value of 2 just found and the given value for the moment of inertia. The kinetic energy is then KErot = 1 (10.106) KErot = 0.5 1.25 kg ⋅ m2 70.4 rad2 / s2 . (10.107) Discussion = 44.0 J These values are reasonable for a person kicking his leg starting from the position shown. The weight of the leg can be neglected in part (a) because it exerts no torque when the center of gravity of the lower leg is directly beneath the pivot in the knee. In part (b), the force exerted by the upper leg is so large that its torque is much greater than that created by the weight of the lower leg as it rotates. The rotational kinetic energy given to the lower leg is enough that it could give a ball a significant velocity by transferring some of this energy in a kick. Making Connections: Conservation Laws Angular momentum, like energy and linear momentum, is conserved. This universally applicable law is another sign of underlying unity in physical laws. Angular momentum is conserved when net external torque is zero, just as linear momentum is conserved when the net external force is zero. Conservation of Angular Momentum We can now understand why Earth keeps on spinning. As we saw in the previous example, Δ = (net )Δ . This equation means that, to change angular momentum, a torque must act over some period of time. Because Earth has a large angular momentum, a large torque acting over a long time is needed to change its rate of spin. So what external torques are there? Tidal friction exerts torque that is slowing Earth's rotation, but tens of millions of years must pass before the change is very significant. Recent research indicates the length of the day was 18 h some 900 million years ago. Only the tides exert significant retarding torques on Earth, and so it will continue to spin, although ever more slowly, for many billions of years. What we have here is, in fact, another conservation law. If the net torque is zero, then angular momentum is constant or conserved. We can see this rigorously by considering net = Δ Δ for the situation in which the net torque is zero. In that case, implying that net = 0 Δ Δ = 0. If the change in angular momentum Δ is zero, then the angular momentum is constant; thus, or = constant (net = 0) = ′(net = 0). (10.108) (10.109) (10.110) (10.111) These expressions are the law of conservation of angular momentum. Conservation laws are as scarce as they are important. An example of conservation of angular momentum is seen in Figure 10.23, in which an ice skater is executing a spin. The net torque on her is very close to zero, because there is relatively little friction between her skates and the ice and because the friction is exerted very close to the pivot point. (Both and are small, and so is negligibly small.) Consequently, she can This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 10 \| Rotational Motion and Angular Momentum 417 spin for quite some time. She can do something else, too. She can increase her rate of spin by pulling her arms and legs in. Why does pulling her arms and legs in increase her rate of spin? The answer is that her angular momentum is constant, so that = ′. (10.112) Expressing this equation in terms of the moment of inertia, = ′′, (10.113) where the primed quantities refer to conditions after she has pulled in her arms and reduced her moment of inertia. Because ′ is smaller, the angular velocity ′ must increase to keep the angular momentum constant. The change can be dramatic, as the following example shows. Figure 10.23 (a) An ice skater is spinning on the tip of her skate with her arms extended. Her angular momentum is conserved because the net torque on her is negligibly small. In the next image, her rate of spin increases greatly when she pulls in her arms, decreasing her moment of inertia. The work she does to pull in her arms results in an increase in rotational kinetic energy. Example 10.14 Calculating the Angular Momentum of a Spinning Skater Suppose an ice skater, such as the one in Figure 10.23, is spinning at 0.800 rev/ s with her arms extended. She has a moment of inertia of 2.34 kg ⋅ m2 with her arms extended and of 0.363 kg ⋅ m2 with her arms close to her body. (These moments of inertia are based on reasonable assumptions about a 60.0-kg skater.) (a) What is her angular velocity in revolutions per second after she pulls in her arms? (b) What is her rotational kinetic energy before and after she does this? Strategy In the first part of the problem, we are looking for the skater's angular velocity ′ after she has pulled in her arms. To find this quantity, we use the conservation of angular momentum and note that the moments of inertia and initial angular velocity are given. To find the initial and final kinetic energies, we use the definition of rotational kinetic energy given by KErot = 1 (10.114) 22. Solution for (a) Because torque is negligible (as discussed above), the conservation of angular momentum given in = ′′ is applicable. Thus, or = ′ = ′′ Solving for ′ and substituting known values into the resulting equation gives ′ = ′ = = 5.16 rev/s. 2.34 kg ⋅ m2 0.363 kg ⋅ m2 (0.800 rev/s) Solution for (b) Rotational kinetic energy is given by KErot = 1 22. The initial value is found by substituting known values into the equation and converting the angular velocity to rad/s: (10.115) (10.116) (10.117) (10.118) 418 Chapter 10 \| Rotational Motion and Angular Momentum KErot = (0.5) = 29.6 J. 2.34 kg ⋅ m2 (0.800 rev/s)(2π rad/rev) 2 (10.119) The final rotational kinetic energy is KErot ′ = 1 2′′2. (10.120) Substituting known values into this equation gives rot′ = (0.5) = 191 J. 0.363 kg ⋅ m2 (5.16 rev/s)(2π rad/rev) 2 (10.121) Discussion In both parts, there is an impressive increase. First, the final angular velocity is large, although most world-class skaters can achieve spin rates about this great. Second, the final kinetic energy is much greater than the initial kinetic energy. The increase in rotational kinetic energy comes from work done by the skater in pulling in her arms. This work is internal work that depletes some of the skater's food energy. There are several other examples of objects that increase their rate of spin because something reduced their moment of inertia. Tornadoes are one example. Storm systems that create tornadoes are slowly rotating. When the radius of rotation narrows, even in a local region, angular velocity increases, sometimes to the furious level of a tornado. Earth is another example. Our planet was born from a huge cloud of gas and dust, the rotation of which came from turbulence in an even larger cloud. Gravitational forces caused the cloud to contract, and the rotation rate increased as a result. (See Figure 10.24.) Figure 10.24 The Solar System coalesced from a cloud of gas and dust that was originally rotating. The orbital motions and spins of the planets are in the same direction as the original spin and conserve the angular momentum of the parent cloud. In case of human motion, one would not expect angular momentum to be conserved when a body interacts with the environment as its foot pushes off the ground. Astronauts floating in space aboard the International Space Station have no angular momentum relative to the inside of the ship if they are motionless. Their bodies will continue to have this zero value no matter how they twist about as long as they do not give
themselves a push off the side of the vessel. Check Your Undestanding Is angular momentum completely analogous to linear momentum? What, if any, are their differences? Solution Yes, angular and linear momentums are completely analogous. While they are exact analogs they have different units and are not directly inter-convertible like forms of energy are. 10.6 Collisions of Extended Bodies in Two Dimensions By the end of this section, you will be able to: Learning Objectives This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 10 \| Rotational Motion and Angular Momentum 425 Figure 10.31 As seen in figure (a), the forces on a spinning gyroscope are its weight and the supporting force from the stand. These forces create a horizontal torque on the gyroscope, which create a change in angular momentum ΔL that is also horizontal. In figure (b), ΔL and L add to produce a new angular momentum with the same magnitude, but different direction, so that the gyroscope precesses in the direction shown instead of falling over. Check Your Understanding Rotational kinetic energy is associated with angular momentum? Does that mean that rotational kinetic energy is a vector? Solution No, energy is always a scalar whether motion is involved or not. No form of energy has a direction in space and you can see that rotational kinetic energy does not depend on the direction of motion just as linear kinetic energy is independent of the direction of motion. Glossary angular acceleration: the rate of change of angular velocity with time angular momentum: the product of moment of inertia and angular velocity change in angular velocity: the difference between final and initial values of angular velocity kinematics of rotational motion: describes the relationships among rotation angle, angular velocity, angular acceleration, and time law of conservation of angular momentum: angular momentum is conserved, i.e., the initial angular momentum is equal to the final angular momentum when no external torque is applied to the system moment of inertia: mass times the square of perpendicular distance from the rotation axis; for a point mass, it is = 2 and, because any object can be built up from a collection of point masses, this relationship is the basis for all other moments of inertia right-hand rule: direction of angular velocity ω and angular momentum L in which the thumb of your right hand points when you curl your fingers in the direction of the disk's rotation rotational inertia: resistance to change of rotation. The more rotational inertia an object has, the harder it is to rotate rotational kinetic energy: the kinetic energy due to the rotation of an object. This is part of its total kinetic energy tangential acceleration: the acceleration in a direction tangent to the circle at the point of interest in circular motion torque: the turning effectiveness of a force work-energy theorem: if one or more external forces act upon a rigid object, causing its kinetic energy to change from KE1 to KE2 , then the work done by the net force is equal to the change in kinetic energy 426 Chapter 10 \| Rotational Motion and Angular Momentum Section Summary 10.1 Angular Acceleration • Uniform circular motion is the motion with a constant angular velocity = Δ Δ . • In non-uniform circular motion, the velocity changes with time and the rate of change of angular velocity (i.e. angular acceleration) is = Δ Δ . • Linear or tangential acceleration refers to changes in the magnitude of velocity but not its direction, given as t = Δ Δ • For circular motion, note that = , so that . • The radius r is constant for circular motion, and so Δ() = Δ . Thus, t = Δ() Δ . • By definition, Δ / Δ = . Thus, or t = Δ Δ . t = = t . 10.2 Kinematics of Rotational Motion • Kinematics is the description of motion. • The kinematics of rotational motion describes the relationships among rotation angle, angular velocity, angular acceleration, and time. • Starting with the four kinematic equations we developed in the One-Dimensional Kinematics, we can derive the four • rotational kinematic equations (presented together with their translational counterparts) seen in Table 10.2. In these equations, the subscript 0 denotes initial values ( 0 and 0 are initial values), and the average angular velocity - and average velocity - are defined as follows: ¯ = 0 + 2 and ¯ = 0 + 2 . 10.3 Dynamics of Rotational Motion: Rotational Inertia • The farther the force is applied from the pivot, the greater is the angular acceleration; angular acceleration is inversely • proportional to mass. If we exert a force on a point mass that is at a distance from a pivot point and because the force is perpendicular to , an acceleration is obtained in the direction of . We can rearrange this equation such that , and then look for ways to relate this expression to expressions for rotational quantities. We note that , and we substitute this expression into , yielding • Torque is the turning effectiveness of a force. In this case, because is perpendicular to , torque is simply = . If we multiply both sides of the equation above by , we get torque on the left-hand side. That is, or = 2 • The moment of inertia of an object is the sum of 2 for all the point masses of which it is composed. That is, = 2. • The general relationship among torque, moment of inertia, and angular acceleration is = ∑ 2. or = = net τ ⋅ This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 10 \| Rotational Motion and Angular Momentum 427 10.4 Rotational Kinetic Energy: Work and Energy Revisited • The rotational kinetic energy KErot for an object with a moment of inertia and an angular velocity is given by KErot = 1 • Helicopters store large amounts of rotational kinetic energy in their blades. This energy must be put into the blades before takeoff and maintained until the end of the flight. The engines do not have enough power to simultaneously provide lift and put significant rotational energy into the blades. 22. • Work and energy in rotational motion are completely analogous to work and energy in translational motion. • The equation for the work-energy theorem for rotational motion is, net = 1 22 − 1 20 2. 10.5 Angular Momentum and Its Conservation • Every rotational phenomenon has a direct translational analog , likewise angular momentum can be defined as = . • This equation is an analog to the definition of linear momentum as = . The relationship between torque and angular momentum is net = Δ Δ . • Angular momentum, like energy and linear momentum, is conserved. This universally applicable law is another sign of underlying unity in physical laws. Angular momentum is conserved when net external torque is zero, just as linear momentum is conserved when the net external force is zero. 10.6 Collisions of Extended Bodies in Two Dimensions • Angular momentum is analogous to linear momentum and is given by = . • Angular momentum is changed by torque, following the relationship net = Δ Δ . • Angular momentum is conserved if the net torque is zero = constant (net = 0) or = ′ (net = 0) . This equation is known as the law of conservation of angular momentum, which may be conserved in collisions. 10.7 Gyroscopic Effects: Vector Aspects of Angular Momentum • Torque is perpendicular to the plane formed by and F and is the direction your right thumb would point if you curled the fingers of your right hand in the direction of F . The direction of the torque is thus the same as that of the angular momentum it produces. • The gyroscope precesses around a vertical axis, since the torque is always horizontal and perpendicular to L . If the gyroscope is not spinning, it acquires angular momentum in the direction of the torque ( L = ΔL ), and it rotates about a horizontal axis, falling over just as we would expect. • Earth itself acts like a gigantic gyroscope. Its angular momentum is along its axis and points at Polaris, the North Star. Conceptual Questions 10.1 Angular Acceleration 1. Analogies exist between rotational and translational physical quantities. Identify the rotational term analogous to each of the following: acceleration, force, mass, work, translational kinetic energy, linear momentum, impulse. 2. Explain why centripetal acceleration changes the direction of velocity in circular motion but not its magnitude. 3. In circular motion, a tangential acceleration can change the magnitude of the velocity but not its direction. Explain your answer. 4. Suppose a piece of food is on the edge of a rotating microwave oven plate. Does it experience nonzero tangential acceleration, centripetal acceleration, or both when: (a) The plate starts to spin? (b) The plate rotates at constant angular velocity? (c) The plate slows to a halt? 10.3 Dynamics of Rotational Motion: Rotational Inertia 5. The moment of inertia of a long rod spun around an axis through one end perpendicular to its length is 2 /3 . Why is this moment of inertia greater than it would be if you spun a point mass at the location of the center of mass of the rod (at / 2 )? (That would be 2 /4 .) 428 Chapter 10 \| Rotational Motion and Angular Momentum 6. Why is the moment of inertia of a hoop that has a mass and a radius greater than the moment of inertia of a disk that has the same mass and radius? Why is the moment of inertia of a spherical shell that has a mass and a radius greater than that of a solid sphere that has the same mass and radius? 7. Give an example in which a small force exerts a large torque. Give another example in which a large force exerts a small torque. 8. While reducing the mass of a racing bike, the greatest benefit is realized from reducing the mass of the tires and wheel rims. Why does this allow a racer to achieve greater accelerations than would an identical reduction in the mass of the bicycle's frame? Figure 10.32 The image shows a side view of a racing bicycle. Can you see evidence in the design of the wheels on this r
acing bicycle that their moment of inertia has been purposely reduced? (credit: Jesús Rodriguez) 9. A ball slides up a frictionless ramp. It is then rolled without slipping and with the same initial velocity up another frictionless ramp (with the same slope angle). In which case does it reach a greater height, and why? 10.4 Rotational Kinetic Energy: Work and Energy Revisited 10. Describe the energy transformations involved when a yo-yo is thrown downward and then climbs back up its string to be caught in the user's hand. 11. What energy transformations are involved when a dragster engine is revved, its clutch let out rapidly, its tires spun, and it starts to accelerate forward? Describe the source and transformation of energy at each step. 12. The Earth has more rotational kinetic energy now than did the cloud of gas and dust from which it formed. Where did this energy come from? Figure 10.33 An immense cloud of rotating gas and dust contracted under the influence of gravity to form the Earth and in the process rotational kinetic energy increased. (credit: NASA) 10.5 Angular Momentum and Its Conservation 13. When you start the engine of your car with the transmission in neutral, you notice that the car rocks in the opposite sense of the engine's rotation. Explain in terms of conservation of angular momentum. Is the angular momentum of the car conserved for long (for more than a few seconds)? 14. Suppose a child walks from the outer edge of a rotating merry-go round to the inside. Does the angular velocity of the merrygo-round increase, decrease, or remain the same? Explain your answer. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 10 \| Rotational Motion and Angular Momentum 429 Figure 10.34 A child may jump off a merry-go-round in a variety of directions. 15. Suppose a child gets off a rotating merry-go-round. Does the angular velocity of the merry-go-round increase, decrease, or remain the same if: (a) He jumps off radially? (b) He jumps backward to land motionless? (c) He jumps straight up and hangs onto an overhead tree branch? (d) He jumps off forward, tangential to the edge? Explain your answers. (Refer to Figure 10.34). 16. Helicopters have a small propeller on their tail to keep them from rotating in the opposite direction of their main lifting blades. Explain in terms of Newton's third law why the helicopter body rotates in the opposite direction to the blades. 17. Whenever a helicopter has two sets of lifting blades, they rotate in opposite directions (and there will be no tail propeller). Explain why it is best to have the blades rotate in opposite directions. 18. Describe how work is done by a skater pulling in her arms during a spin. In particular, identify the force she exerts on each arm to pull it in and the distance each moves, noting that a component of the force is in the direction moved. Why is angular momentum not increased by this action? 19. When there is a global heating trend on Earth, the atmosphere expands and the length of the day increases very slightly. Explain why the length of a day increases. 20. Nearly all conventional piston engines have flywheels on them to smooth out engine vibrations caused by the thrust of individual piston firings. Why does the flywheel have this effect? 21. Jet turbines spin rapidly. They are designed to fly apart if something makes them seize suddenly, rather than transfer angular momentum to the plane's wing, possibly tearing it off. Explain how flying apart conserves angular momentum without transferring it to the wing. 22. An astronaut tightens a bolt on a satellite in orbit. He rotates in a direction opposite to that of the bolt, and the satellite rotates in the same direction as the bolt. Explain why. If a handhold is available on the satellite, can this counter-rotation be prevented? Explain your answer. 23. Competitive divers pull their limbs in and curl up their bodies when they do flips. Just before entering the water, they fully extend their limbs to enter straight down. Explain the effect of both actions on their angular velocities. Also explain the effect on their angular momenta. Figure 10.35 The diver spins rapidly when curled up and slows when she extends her limbs before entering the water. 24. Draw a free body diagram to show how a diver gains angular momentum when leaving the diving board. 25. In terms of angular momentum, what is the advantage of giving a football or a rifle bullet a spin when throwing or releasing it? 430 Chapter 10 \| Rotational Motion and Angular Momentum Figure 10.36 The image shows a view down the barrel of a cannon, emphasizing its rifling. Rifling in the barrel of a canon causes the projectile to spin just as is the case for rifles (hence the name for the grooves in the barrel). (credit: Elsie esq., Flickr) 10.6 Collisions of Extended Bodies in Two Dimensions 26. Describe two different collisions—one in which angular momentum is conserved, and the other in which it is not. Which condition determines whether or not angular momentum is conserved in a collision? 27. Suppose an ice hockey puck strikes a hockey stick that lies flat on the ice and is free to move in any direction. Which quantities are likely to be conserved: angular momentum, linear momentum, or kinetic energy (assuming the puck and stick are very resilient)? 28. While driving his motorcycle at highway speed, a physics student notices that pulling back lightly on the right handlebar tips the cycle to the left and produces a left turn. Explain why this happens. 10.7 Gyroscopic Effects: Vector Aspects of Angular Momentum 29. While driving his motorcycle at highway speed, a physics student notices that pulling back lightly on the right handlebar tips the cycle to the left and produces a left turn. Explain why this happens. 30. Gyroscopes used in guidance systems to indicate directions in space must have an angular momentum that does not change in direction. Yet they are often subjected to large forces and accelerations. How can the direction of their angular momentum be constant when they are accelerated? This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 10 \| Rotational Motion and Angular Momentum 431 Problems & Exercises 10.1 Angular Acceleration 1. At its peak, a tornado is 60.0 m in diameter and carries 500 km/h winds. What is its angular velocity in revolutions per second? 2. Integrated Concepts An ultracentrifuge accelerates from rest to 100,000 rpm in 2.00 min. (a) What is its angular acceleration in rad/s2 ? (b) What is the tangential acceleration of a point 9.50 cm from the axis of rotation? (c) What is the radial acceleration in m/s2 and multiples of of this point at full rpm? 3. Integrated Concepts You have a grindstone (a disk) that is 90.0 kg, has a 0.340-m radius, and is turning at 90.0 rpm, and you press a steel axe against it with a radial force of 20.0 N. (a) Assuming the kinetic coefficient of friction between steel and stone is 0.20, calculate the angular acceleration of the grindstone. (b) How many turns will the stone make before coming to rest? 4. Unreasonable Results You are told that a basketball player spins the ball with an angular acceleration of 100 rad/s2 . (a) What is the ball's final angular velocity if the ball starts from rest and the acceleration lasts 2.00 s? (b) What is unreasonable about the result? (c) Which premises are unreasonable or inconsistent? 10.2 Kinematics of Rotational Motion 5. With the aid of a string, a gyroscope is accelerated from rest to 32 rad/s in 0.40 s. (a) What is its angular acceleration in rad/s2? (b) How many revolutions does it go through in the process? 6. Suppose a piece of dust finds itself on a CD. If the spin rate of the CD is 500 rpm, and the piece of dust is 4.3 cm from the center, what is the total distance traveled by the dust in 3 minutes? (Ignore accelerations due to getting the CD rotating.) 7. A gyroscope slows from an initial rate of 32.0 rad/s at a rate of 0.700 rad/s2 . (a) How long does it take to come to rest? (b) How many revolutions does it make before stopping? 8. During a very quick stop, a car decelerates at 7.00 m/s2 . (a) What is the angular acceleration of its 0.280-m-radius tires, assuming they do not slip on the pavement? (b) How many revolutions do the tires make before coming to rest, given their initial angular velocity is 95.0 rad/s ? (c) How long does the car take to stop completely? (d) What distance does the car travel in this time? (e) What was the car's initial velocity? (f) Do the values obtained seem reasonable, considering that this stop happens very quickly? Figure 10.37 Yo-yos are amusing toys that display significant physics and are engineered to enhance performance based on physical laws. (credit: Beyond Neon, Flickr) 9. Everyday application: Suppose a yo-yo has a center shaft that has a 0.250 cm radius and that its string is being pulled. (a) If the string is stationary and the yo-yo accelerates away from it at a rate of 1.50 m/s2 , what is the angular acceleration of the yo-yo? (b) What is the angular velocity after 0.750 s if it starts from rest? (c) The outside radius of the yo-yo is 3.50 cm. What is the tangential acceleration of a point on its edge? 10.3 Dynamics of Rotational Motion: Rotational Inertia 10. This problem considers additional aspects of example Calculating the Effect of Mass Distribution on a MerryGo-Round. (a) How long does it take the father to give the merry-go-round an angular velocity of 1.50 rad/s? (b) How many revolutions must he go through to generate this velocity? (c) If he exerts a slowing force of 300 N at a radius of 1.35 m, how long would it take him to stop them? 11. Calculate the moment of inertia of a skater given the following information. (a) The 60.0-kg skater is approximated as a cylinder that has a 0.110-m radius. (b) The skater with arms extended is approximately a cylinder that is 52.5 kg, has a 0.110-m radius, and has two 0.900-m-lon
g arms which are 3.75 kg each and extend straight out from the cylinder like rods rotated about their ends. 12. The triceps muscle in the back of the upper arm extends the forearm. This muscle in a professional boxer exerts a force of 2.00×103 N with an effective perpendicular lever arm of 3.00 cm, producing an angular acceleration of the forearm of 120 rad/s2 . What is the moment of inertia of the boxer's forearm? 13. A soccer player extends her lower leg in a kicking motion by exerting a force with the muscle above the knee in the front of her leg. She produces an angular acceleration of 30.00 rad/s2 and her lower leg has a moment of inertia of 0.750 kg ⋅ m2 . What is the force exerted by the muscle if its effective perpendicular lever arm is 1.90 cm? 14. Suppose you exert a force of 180 N tangential to a 0.280-m-radius 75.0-kg grindstone (a solid disk). (a)What torque is exerted? (b) What is the angular acceleration assuming negligible opposing friction? (c) What is the angular acceleration if there is an opposing frictional force of 20.0 N exerted 1.50 cm from the axis? 432 Chapter 10 \| Rotational Motion and Angular Momentum 15. Consider the 12.0 kg motorcycle wheel shown in Figure 10.38. Assume it to be approximately an annular ring with an inner radius of 0.280 m and an outer radius of 0.330 m. The motorcycle is on its center stand, so that the wheel can spin freely. (a) If the drive chain exerts a force of 2200 N at a radius of 5.00 cm, what is the angular acceleration of the wheel? (b) What is the tangential acceleration of a point on the outer edge of the tire? (c) How long, starting from rest, does it take to reach an angular velocity of 80.0 rad/s? Figure 10.38 A motorcycle wheel has a moment of inertia approximately that of an annular ring. 16. Zorch, an archenemy of Superman, decides to slow Earth's rotation to once per 28.0 h by exerting an opposing force at and parallel to the equator. Superman is not immediately concerned, because he knows Zorch can only exert a force of 4.00×107 N (a little greater than a Saturn V rocket's thrust). How long must Zorch push with this force to accomplish his goal? (This period gives Superman time to devote to other villains.) Explicitly show how you follow the steps found in Problem-Solving Strategy for Rotational Dynamics. 17. An automobile engine can produce 200 N · m of torque. Calculate the angular acceleration produced if 95.0% of this torque is applied to the drive shaft, axle, and rear wheels of a car, given the following information. The car is suspended so that the wheels can turn freely. Each wheel acts like a 15.0 kg disk that has a 0.180 m radius. The walls of each tire act like a 2.00-kg annular ring that has inside radius of 0.180 m and outside radius of 0.320 m. The tread of each tire acts like a 10.0-kg hoop of radius 0.330 m. The 14.0-kg axle acts like a rod that has a 2.00-cm radius. The 30.0-kg drive shaft acts like a rod that has a 3.20-cm radius. = 2 / 3 18. Starting with the formula for the moment of inertia of a rod rotated around an axis through one end perpendicular to its length , prove that the moment of inertia of a rod rotated about an axis through its center perpendicular to its length is = 2 / 12 . You will find the graphics in Figure 10.12 useful in visualizing these rotations. 19. Unreasonable Results A gymnast doing a forward flip lands on the mat and exerts a 500-N · m torque to slow and then reverse her angular velocity. Her initial angular velocity is 10.0 rad/s, and her moment of inertia is 0.050 kg ⋅ m2 . (a) What time is required for her to exactly reverse her spin? (b) What is This content is available for free at http://cnx.org/content/col11844/1.13 unreasonable about the result? (c) Which premises are unreasonable or inconsistent? 20. Unreasonable Results An advertisement claims that an 800-kg car is aided by its 20.0-kg flywheel, which can accelerate the car from rest to a speed of 30.0 m/s. The flywheel is a disk with a 0.150-m radius. (a) Calculate the angular velocity the flywheel must have if 95.0% of its rotational energy is used to get the car up to speed. (b) What is unreasonable about the result? (c) Which premise is unreasonable or which premises are inconsistent? 10.4 Rotational Kinetic Energy: Work and Energy Revisited 21. This problem considers energy and work aspects of Example 10.7—use data from that example as needed. (a) Calculate the rotational kinetic energy in the merry-go-round plus child when they have an angular velocity of 20.0 rpm. (b) Using energy considerations, find the number of revolutions the father will have to push to achieve this angular velocity starting from rest. (c) Again, using energy considerations, calculate the force the father must exert to stop the merry-goround in two revolutions 22. What is the final velocity of a hoop that rolls without slipping down a 5.00-m-high hill, starting from rest? 23. (a) Calculate the rotational kinetic energy of Earth on its axis. (b) What is the rotational kinetic energy of Earth in its orbit around the Sun? 24. Calculate the rotational kinetic energy in the motorcycle wheel (Figure 10.38) if its angular velocity is 120 rad/s. Assume M = 12.0 kg, R1 = 0.280 m, and R2 = 0.330 m. 25. A baseball pitcher throws the ball in a motion where there is rotation of the forearm about the elbow joint as well as other movements. If the linear velocity of the ball relative to the elbow joint is 20.0 m/s at a distance of 0.480 m from the joint and the moment of inertia of the forearm is 0.500 kg ⋅ m2 , what is the rotational kinetic energy of the forearm? 26. While punting a football, a kicker rotates his leg about the hip joint. The moment of inertia of the leg is 3.75 kg ⋅ m2 and its rotational kinetic energy is 175 J. (a) What is the angular velocity of the leg? (b) What is the velocity of tip of the punter's shoe if it is 1.05 m from the hip joint? (c) Explain how the football can be given a velocity greater than the tip of the shoe (necessary for a decent kick distance). 27. A bus contains a 1500 kg flywheel (a disk that has a 0.600 m radius) and has a total mass of 10,000 kg. (a) Calculate the angular velocity the flywheel must have to contain enough energy to take the bus from rest to a speed of 20.0 m/s, assuming 90.0% of the rotational kinetic energy can be transformed into translational energy. (b) How high a hill can the bus climb with this stored energy and still have a speed of 3.00 m/s at the top of the hill? Explicitly show how you follow the steps in the Problem-Solving Strategy for Rotational Energy. 28. A ball with an initial velocity of 8.00 m/s rolls up a hill without slipping. Treating the ball as a spherical shell, calculate the vertical height it reaches. (b) Repeat the calculation for the same ball if it slides up the hill without rolling. Chapter 10 \| Rotational Motion and Angular Momentum 433 (b) How does this angular momentum compare with the angular momentum of the Moon on its axis? Remember that the Moon keeps one side toward Earth at all times. (c) Discuss whether the values found in parts (a) and (b) seem consistent with the fact that tidal effects with Earth have caused the Moon to rotate with one side always facing Earth. 38. Suppose you start an antique car by exerting a force of 300 N on its crank for 0.250 s. What angular momentum is given to the engine if the handle of the crank is 0.300 m from the pivot and the force is exerted to create maximum torque the entire time? 39. A playground merry-go-round has a mass of 120 kg and a radius of 1.80 m and it is rotating with an angular velocity of 0.500 rev/s. What is its angular velocity after a 22.0-kg child gets onto it by grabbing its outer edge? The child is initially at rest. 40. Three children are riding on the edge of a merry-go-round that is 100 kg, has a 1.60-m radius, and is spinning at 20.0 rpm. The children have masses of 22.0, 28.0, and 33.0 kg. If the child who has a mass of 28.0 kg moves to the center of the merry-go-round, what is the new angular velocity in rpm? 41. (a) Calculate the angular momentum of an ice skater spinning at 6.00 rev/s given his moment of inertia is 0.400 kg ⋅ m2 . (b) He reduces his rate of spin (his angular velocity) by extending his arms and increasing his moment of inertia. Find the value of his moment of inertia if his angular velocity decreases to 1.25 rev/s. (c) Suppose instead he keeps his arms in and allows friction of the ice to slow him to 3.00 rev/s. What average torque was exerted if this takes 15.0 s? 42. Consider the Earth-Moon system. Construct a problem in which you calculate the total angular momentum of the system including the spins of the Earth and the Moon on their axes and the orbital angular momentum of the Earth-Moon system in its nearly monthly rotation. Calculate what happens to the Moon's orbital radius if the Earth's rotation decreases due to tidal drag. Among the things to be considered are the amount by which the Earth's rotation slows and the fact that the Moon will continue to have one side always facing the Earth. 10.6 Collisions of Extended Bodies in Two Dimensions 43. Repeat Example 10.15 in which the disk strikes and adheres to the stick 0.100 m from the nail. 44. Repeat Example 10.15 in which the disk originally spins clockwise at 1000 rpm and has a radius of 1.50 cm. 45. Twin skaters approach one another as shown in Figure 10.39 and lock hands. (a) Calculate their final angular velocity, given each had an initial speed of 2.50 m/s relative to the ice. Each has a mass of 70.0 kg, and each has a center of mass located 0.800 m from their locked hands. You may approximate their moments of inertia to be that of point masses at this radius. (b) Compare the initial kinetic energy and final kinetic energy. 29. While exercising in a fitness center, a man lies face down on a bench and lifts a weight with one lower leg by contacting the muscles in the back of the upper leg. (a) Find the angular acceleration produce
d given the mass lifted is 10.0 kg at a distance of 28.0 cm from the knee joint, the moment of inertia of the lower leg is 0.900 kg ⋅ m2 , the muscle force is 1500 N, and its effective perpendicular lever arm is 3.00 cm. (b) How much work is done if the leg rotates through an angle of 20.0º with a constant force exerted by the muscle? 30. To develop muscle tone, a woman lifts a 2.00-kg weight held in her hand. She uses her biceps muscle to flex the lower arm through an angle of 60.0º . (a) What is the angular acceleration if the weight is 24.0 cm from the elbow joint, her forearm has a moment of inertia of 0.250 kg ⋅ m2 , and the net force she exerts is 750 N at an effective perpendicular lever arm of 2.00 cm? (b) How much work does she do? 31. Consider two cylinders that start down identical inclines from rest except that one is frictionless. Thus one cylinder rolls without slipping, while the other slides frictionlessly without rolling. They both travel a short distance at the bottom and then start up another incline. (a) Show that they both reach the same height on the other incline, and that this height is equal to their original height. (b) Find the ratio of the time the rolling cylinder takes to reach the height on the second incline to the time the sliding cylinder takes to reach the height on the second incline. (c) Explain why the time for the rolling motion is greater than that for the sliding motion. 32. What is the moment of inertia of an object that rolls without slipping down a 2.00-m-high incline starting from rest, and has a final velocity of 6.00 m/s? Express the moment of inertia as a multiple of 2 , where is the mass of the object and is its radius. 33. Suppose a 200-kg motorcycle has two wheels like, the one described in Example 10.15 and is heading toward a hill at a speed of 30.0 m/s. (a) How high can it coast up the hill, if you neglect friction? (b) How much energy is lost to friction if the motorcycle only gains an altitude of 35.0 m before coming to rest? 34. In softball, the pitcher throws with the arm fully extended (straight at the elbow). In a fast pitch the ball leaves the hand with a speed of 139 km/h. (a) Find the rotational kinetic energy of the pitcher's arm given its moment of inertia is 0.720 kg ⋅ m2 and the ball leaves the hand at a distance of 0.600 m from the pivot at the shoulder. (b) What force did the muscles exert to cause the arm to rotate if their effective perpendicular lever arm is 4.00 cm and the ball is 0.156 kg? 35. Construct Your Own Problem Consider the work done by a spinning skater pulling her arms in to increase her rate of spin. Construct a problem in which you calculate the work done with a “force multiplied by distance” calculation and compare it to the skater's increase in kinetic energy. 10.5 Angular Momentum and Its Conservation 36. (a) Calculate the angular momentum of the Earth in its orbit around the Sun. (b) Compare this angular momentum with the angular momentum of Earth on its axis. 37. (a) What is the angular momentum of the Moon in its orbit around Earth? 434 Chapter 10 \| Rotational Motion and Angular Momentum Figure 10.39 Twin skaters approach each other with identical speeds. Then, the skaters lock hands and spin. 46. Suppose a 0.250-kg ball is thrown at 15.0 m/s to a motionless person standing on ice who catches it with an outstretched arm as shown in Figure 10.40. (a) Calculate the final linear velocity of the person, given his mass is 70.0 kg. (b) What is his angular velocity if each arm is 5.00 kg? You may treat the ball as a point mass and treat the person's arms as uniform rods (each has a length of 0.900 m) and the rest of his body as a uniform cylinder of radius 0.180 m. Neglect the effect of the ball on his center of mass so that his center of mass remains in his geometrical center. (c) Compare the initial and final total kinetic energies. Figure 10.40 The figure shows the overhead view of a person standing motionless on ice about to catch a ball. Both arms are outstretched. After catching the ball, the skater recoils and rotates. 47. Repeat Example 10.15 in which the stick is free to have translational motion as well as rotational motion. 10.7 Gyroscopic Effects: Vector Aspects of Angular Momentum 48. Integrated Concepts The axis of Earth makes a 23.5° angle with a direction perpendicular to the plane of Earth's orbit. As shown in Figure 10.41, this axis precesses, making one complete rotation in 25,780 y. (a) Calculate the change in angular momentum in half this time. (b) What is the average torque producing this change in angular momentum? (c) If this torque were created by a single force (it is not) acting at the most effective point on the equator, what would its magnitude be? Test Prep for AP® Courses This content is available for free at http://cnx.org/content/col11844/1.13 Figure 10.41 The Earth's axis slowly precesses, always making an angle of 23.5° with the direction perpendicular to the plane of Earth's orbit. The change in angular momentum for the two shown positions is quite large, although the magnitude L is unchanged. 10.3 Dynamics of Rotational Motion: Rotational Inertia Chapter 10 \| Rotational Motion and Angular Momentum 435 9. Which measure would not be useful to help you determine the change in angular velocity when the torque on a fishing reel is increased? a. b. c. d. the radius of the reel the amount of line that unspools the angular momentum of the fishing line the time it takes the line to unspool 10. What data could you collect to study the change in angular velocity when two people push a merry-go-round instead of one, providing twice as much torque? How would you use the data you collect? 10.5 Angular Momentum and Its Conservation 11. Which rotational system would be best to use as a model to measure how angular momentum changes when forces on the system are changed? a. a fishing reel b. a planet and its moon c. a figure skater spinning d. a person's lower leg 12. You are collecting data to study changes in the angular momentum of a bicycle wheel when a force is applied to it. Which of the following measurements would be least helpful to you? a. b. c. d. the time for which the force is applied the radius at which the force is applied the angular velocity of the wheel when the force is applied the direction of the force 13. Which torque applied to a disk with radius 7.0 cm for 3.5 s will produce an angular momentum of 25 N•m•s? a. 7.1 N•m b. 357.1 N•m c. 3.6 N•m d. 612.5 N•m 14. Which of the following would be the best way to produce measurable amounts of torque on a system to test the relationship between the angular momentum of the system, the average torque applied to the system, and the time for which the torque is applied? a. having different numbers of people push on a merry-go- round b. placing known masses on one end of a seesaw c. touching the outer edge of a bicycle wheel to a treadmill that is moving at different speeds d. hanging known masses from a string that is wound around a spool suspended horizontally on an axle 15. 1. A piece of wood can be carved by spinning it on a motorized lathe and holding a sharp chisel to the edge of the wood as it spins. How does the angular velocity of a piece of wood with a radius of 0.2 m spinning on a lathe change when a chisel is held to the wood's edge with a force of 50 N? a. b. c. d. It increases by 0.1 N•m multiplied by the moment of inertia of the wood. It decreases by 0.1 N•m divided by the moment of inertia of the wood-and-lathe system. It decreases by 0.1 N•m multiplied by the moment of inertia of the wood. It decreases by 0.1 m/s2. 2. A Ferris wheel is loaded with people in the chairs at the following positions: 4 o'clock, 1 o'clock, 9 o'clock, and 6 o'clock. As the wheel begins to turn, what forces are acting on the system? How will each force affect the angular velocity and angular momentum? 3. A lever is placed on a fulcrum. A rock is placed on the left end of the lever and a downward (clockwise) force is applied to the right end of the lever. What measurements would be most effective to help you determine the angular momentum of the system? (Assume the lever itself has negligible mass.) a. b. c. d. the angular velocity and mass of the rock the angular velocity and mass of the rock, and the radius of the lever the velocity of the force, the radius of the lever, and the mass of the rock the mass of the rock, the length of the lever on both sides of the fulcrum, and the force applied on the right side of the lever 4. You can use the following setup to determine angular acceleration and angular momentum: A lever is placed on a fulcrum. A rock is placed on the left end of the lever and a known downward (clockwise) force is applied to the right end of the lever. What calculations would you perform? How would you account for gravity in your calculations? 5. Consider two sizes of disk, both of mass M. One size of disk has radius R; the other has radius 2R. System A consists of two of the larger disks rigidly connected to each other with a common axis of rotation. System B consists of one of the larger disks and a number of the smaller disks rigidly connected with a common axis of rotation. If the moment of inertia for system A equals the moment of inertia for system B, how many of the smaller disks are in system B? a. 1 b. 2 c. 3 d. 4 6. How do you arrange these objects so that the resulting system has the maximum possible moment of inertia? What is that moment of inertia? 10.4 Rotational Kinetic Energy: Work and Energy Revisited 7. Gear A, which turns clockwise, meshes with gear B, which turns counterclockwise. When more force is applied through gear A, torque is created. How does the angular velocity of gear B change as a result? a. b. c. d. It increases in magnitude. It decreases in magnitude. It changes direction. It stays the same. 8. Which will cause a greater increase in the angular velocity of a disk: doubling the torque applied or halving th
e radius at which the torque is applied? Explain. 436 Chapter 10 \| Rotational Motion and Angular Momentum Figure 10.42 A curved arrow lies at the side of a gray disk. There is a point at the center of the disk, and around the point there is a dashed circle. There is a point labeled “Child” on the dashed circle. Below the disc is a label saying “Top View”. The diagram above shows a top view of a child of mass M on a circular platform of mass 2M that is rotating counterclockwise. Assume the platform rotates without friction. Which of the following describes an action by the child that will increase the angular speed of the platformchild system and why? a. The child moves toward the center of the platform, increasing the total angular momentum of the system. b. The child moves toward the center of the platform, decreasing the rotational inertia of the system. c. The child moves away from the center of the platform, increasing the total angular momentum of the system. d. The child moves away from the center of the platform, decreasing the rotational inertia of the system. 16. Figure 10.43 A point labeled “Moon” lies on a dashed ellipse. Two other points, labeled “A” and “B”, lie at opposite ends of the ellipse. A point labeled “Planet” lies inside the ellipse. A moon is in an elliptical orbit about a planet as shown above. At point A the moon has speed uA and is at distance RA from the planet. At point B the moon has speed uB. Has the moon's angular momentum changed? Explain your answer. 17. A hamster sits 0.10 m from the center of a lazy Susan of negligible mass. The wheel has an angular velocity of 1.0 rev/ s. How will the angular velocity of the lazy Susan change if the hamster walks to 0.30 m from the center of rotation? Assume zero friction and no external torque. a. b. c. d. It will speed up to 2.0 rev/s. It will speed up to 9.0 rev/s. It will slow to 0.01 rev/s. It will slow to 0.02 rev/s. 18. Earth has a mass of 6.0 × 1024 kg, a radius of 6.4 × 106 m, and an angular velocity of 1.2 × 10–5 rev/s. How would the planet's angular velocity change if a layer of Earth with mass 1.0 × 1023 kg broke off of the Earth, decreasing Earth's radius by 0.2 × 106 m? Assume no friction. 19. Consider system A, consisting of two disks of radius R, with both rotating clockwise. Now consider system B, consisting of one disk of radius R rotating counterclockwise and another disk of radius 2R rotating clockwise. All of the disks have the same mass, and all have the same magnitude of angular velocity. Which system has the greatest angular momentum? a. A b. B c. They're equal. d. Not enough information This content is available for free at http://cnx.org/content/col11844/1.13 20. Assume that a baseball bat being swung at 3π rad/s by a batting machine is equivalent to a 1.1 m thin rod with a mass of 1.0 kg. How fast would a 0.15 kg baseball that squarely hits the very tip of the bat have to be going for the net angular momentum of the bat-ball system to be zero? 10.6 Collisions of Extended Bodies in Two Dimensions 21. A box with a mass of 2.0 kg rests on one end of a seesaw. The seesaw is 6.0 m long, and we can assume it has negligible mass. Approximately what angular momentum will the box have if someone with a mass of 65 kg sits on the other end of the seesaw quickly, with a velocity of 1.2 m/s? a. 702 kg•m2/s b. 39 kg•m2/s c. 18 kg•m2/s d. 1.2 kg•m2/s 22. A spinner in a board game can be thought of as a thin rod that spins about an axis at its center. The spinner in a certain game is 12 cm long and has a mass of 10 g. How will its angular velocity change when it is flicked at one end with a force equivalent to 15 g travelling at 5.0 m/s if all the energy of the collision is transferred to the spinner? (You can use the table in Figure 10.12 to estimate the rotational inertia of the spinner.) 23. A cyclist pedals to exert a torque on the rear wheel of the bicycle. When the cyclist changes to a higher gear, the torque increases. Which of the following would be the most effective strategy to help you determine the change in angular momentum of the bicycle wheel? a. multiplying the ratio between the two torques by the mass of the bicycle and rider b. adding the two torques together, and multiplying by the time for which both torques are applied c. multiplying the difference in the two torques by the time for which the new torque is applied d. multiplying both torques by the mass of the bicycle and rider 24. An electric screwdriver has two speeds, each of which exerts a different torque on a screw. Describe what calculations you could use to help you compare the angular momentum of a screw at each speed. What measurements would you need to make in order to calculate this? 25. Why is it important to consider the shape of an object when determining the object's angular momentum? a. The shape determines the location of the center of mass. The location of the center of mass in turn determines the angular velocity of the object. b. The shape helps you determine the location of the object's outer edge, where rotational velocity will be greatest. c. The shape helps you determine the location of the center of rotation. d. The shape determines the location of the center of mass. The location of the center of mass contributes to the object's rotational inertia, which contributes to its angular momentum. 26. How could you collect and analyze data to test the difference between the torques provided by two speeds on a tabletop fan? 27. Describe a rotational system you could use to demonstrate the effect on the system's angular momentum of applying different amounts of external torque. Chapter 10 \| Rotational Motion and Angular Momentum 437 28. How could you use simple equipment such as balls and string to study the changes in angular momentum of a system when it interacts with another system? 10.7 Gyroscopic Effects: Vector Aspects of Angular Momentum 29. A globe (model of the Earth) is a hollow sphere with a radius of 16 cm. By wrapping a cord around the equator of a globe and pulling on it, a person exerts a torque on the globe of 120 N • m for 1.2 s. What angular momentum does the globe have after 1.2 s? 30. How could you use a fishing reel to test the relationship between the torque applied to a system, the time for which the torque was applied, and the resulting angular momentum of the system? How would you measure angular momentum? Chapter 16 \| Oscillatory Motion and Waves 673 16 OSCILLATORY MOTION AND WAVES Figure 16.1 There are at least four types of waves in this picture—only the water waves are evident. There are also sound waves, light waves, and waves on the guitar strings. (credit: John Norton) Chapter Outline 16.1. Hooke’s Law: Stress and Strain Revisited 16.2. Period and Frequency in Oscillations 16.3. Simple Harmonic Motion: A Special Periodic Motion 16.4. The Simple Pendulum 16.5. Energy and the Simple Harmonic Oscillator 16.6. Uniform Circular Motion and Simple Harmonic Motion 16.7. Damped Harmonic Motion 16.8. Forced Oscillations and Resonance 16.9. Waves 16.10. Superposition and Interference 16.11. Energy in Waves: Intensity Connection for AP® Courses In this chapter, students are introduced to oscillation, the regular variation in the position of a system about a central point accompanied by transfer of energy and momentum, and to waves. A child’s swing, a pendulum, a spring, and a vibrating string are all examples of oscillations. This chapter will address simple harmonic motion and periods of vibration, aspects of oscillation that produce waves, a common phenomenon in everyday life. Waves carry energy from one place to another.” This chapter will show how harmonic oscillations produce waves that transport energy across space and through time. The information and examples presented support Big Ideas 1, 2, and 3 of the AP® Physics Curriculum Framework. The chapter opens by discussing the forces that govern oscillations and waves. It goes on to discuss important concepts such as simple harmonic motion, uniform harmonic motion, and damped harmonic motion. You will also learn about energy in simple harmonic motion and how it changes from kinetic to potential, and how the total sum, which would be the mechanical energy of the oscillator, remains constant or conserved at all times. The chapter also discusses characteristics of waves, such as their frequency, period of oscillation, and the forms in which they can exist, i.e., transverse or longitudinal. The chapter ends by discussing what happens when two or more waves overlap and how the amplitude of the resultant wave changes, leading to the phenomena of superposition and interference. 674 Chapter 16 \| Oscillatory Motion and Waves The concepts in this chapter support: Big Idea 3 The interactions of an object with other objects can be described by forces. Enduring Understanding 3.B Classically, the acceleration of an object interacting with other objects can be predicted by using → → = ∑ . Essential Knowledge 3.B.3 Restoring forces can result in oscillatory motion. When a linear restoring force is exerted on an object displaced from an equilibrium position, the object will undergo a special type of motion called simple harmonic motion. Examples should include gravitational force exerted by the Earth on a simple pendulum and a mass-spring oscillator. Big Idea 4 Interactions between systems can result in changes in those systems. Enduring Understanding 4.C Interactions with other objects or systems can change the total energy of a system. Essential Knowledge 4.C.1 The energy of a system includes its kinetic energy, potential energy, and microscopic internal energy. Examples should include gravitational potential energy, elastic potential energy, and kinetic energy. Essential Knowledge 4.C.2 Mechanical energy (the sum of kinetic and potential energy) is transferred into or out of a system when an external force is exerted on a system such that a component of the force is parallel to i
ts displacement. The process through which the energy is transferred is called work. Big Idea 5 Changes that occur as a result of interactions are constrained by conservation laws. Enduring Understanding 5.B The energy of a system is conserved. Essential Knowledge 5.B.2 A system with internal structure can have internal energy, and changes in a system’s internal structure can result in changes in internal energy. [Physics 1: includes mass-spring oscillators and simple pendulums. Physics 2: includes charged object in electric ﬁelds and examining changes in internal energy with changes in conﬁguration.] Big Idea 6 Waves can transfer energy and momentum from one location to another without the permanent transfer of mass and serve as a mathematical model for the description of other phenomena. Enduring Understanding 6.A A wave is a traveling disturbance that transfers energy and momentum. Essential Knowledge 6.A.1 Waves can propagate via different oscillation modes such as transverse and longitudinal. Essential Knowledge 6.A.2 For propagation, mechanical waves require a medium, while electromagnetic waves do not require a physical medium. Examples should include light traveling through a vacuum and sound not traveling through a vacuum. Essential Knowledge 6.A.3 The amplitude is the maximum displacement of a wave from its equilibrium value. Essential Knowledge 6.A.4 Classically, the energy carried by a wave depends on and increases with amplitude. Examples should include sound waves. Enduring Understanding 6.B A periodic wave is one that repeats as a function of both time and position and can be described by its amplitude, frequency, wavelength, speed, and energy. Essential Knowledge 6.B.1 The period is the repeat time of the wave. The frequency is the number of repetitions over a period of time. Essential Knowledge 6.B.2 The wavelength is the repeat distance of the wave. Essential Knowledge 6.B.3 A simple wave can be described by an equation involving one sine or cosine function involving the wavelength, amplitude, and frequency of the wave. Essential Knowledge 6.B.4 The wavelength is the ratio of speed over frequency. Enduring Understanding 6.C Only waves exhibit interference and diffraction. Essential Knowledge 6.C.1 When two waves cross, they travel through each other; they do not bounce off each other. Where the waves overlap, the resulting displacement can be determined by adding the displacements of the two waves. This is called superposition. Enduring Understanding 6.D Interference and superposition lead to standing waves and beats. Essential Knowledge 6.D.1 Two or more wave pulses can interact in such a way as to produce amplitude variations in the resultant wave. When two pulses cross, they travel through each other; they do not bounce off each other. Where the pulses overlap, the resulting displacement can be determined by adding the displacements of the two pulses. This is called superposition. Essential Knowledge 6.D.2 Two or more traveling waves can interact in such a way as to produce amplitude variations in the resultant wave. Essential Knowledge 6.D.3 Standing waves are the result of the addition of incident and reﬂected waves that are conﬁned to a region and have nodes and antinodes. Examples should include waves on a ﬁxed length of string, and sound waves in both closed and open tubes. Essential Knowledge 6.D.4 The possible wavelengths of a standing wave are determined by the size of the region to which it is conﬁned. Essential Knowledge 6.D.5 Beats arise from the addition of waves of slightly different frequency. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 16 \| Oscillatory Motion and Waves 675 16.1 Hooke’s Law: Stress and Strain Revisited Learning Objectives By the end of this section, you will be able to: • Explain Newton’s third law of motion with respect to stress and deformation. • Describe the restoring force and displacement. • Use Hooke’s law of deformation, and calculate stored energy in a spring. The information presented in this section supports the following AP® learning objectives and science practices: • 3.B.3.3 The student can analyze data to identify qualitative or quantitative relationships between given values and variables (i.e., force, displacement, acceleration, velocity, period of motion, frequency, spring constant, string length, mass) associated with objects in oscillatory motion to use that data to determine the value of an unknown. (S.P. 2.2, 5.1) • 3.B.3.4 The student is able to construct a qualitative and/or a quantitative explanation of oscillatory behavior given evidence of a restoring force. (S.P. 2.2, 6.2) Figure 16.2 When displaced from its vertical equilibrium position, this plastic ruler oscillates back and forth because of the restoring force opposing displacement. When the ruler is on the left, there is a force to the right, and vice versa. Newton’s first law implies that an object oscillating back and forth is experiencing forces. Without force, the object would move in a straight line at a constant speed rather than oscillate. Consider, for example, plucking a plastic ruler to the left as shown in Figure 16.2. The deformation of the ruler creates a force in the opposite direction, known as a restoring force. Once released, the restoring force causes the ruler to move back toward its stable equilibrium position, where the net force on it is zero. However, by the time the ruler gets there, it gains momentum and continues to move to the right, producing the opposite deformation. It is then forced to the left, back through equilibrium, and the process is repeated until dissipative forces dampen the motion. These forces remove mechanical energy from the system, gradually reducing the motion until the ruler comes to rest. The simplest oscillations occur when the restoring force is directly proportional to displacement. When stress and strain were covered in Newton’s Third Law of Motion, the name was given to this relationship between force and displacement was Hooke’s law: = − (16.1) Here, is the restoring force, is the displacement from equilibrium or deformation, and is a constant related to the difficulty in deforming the system. The minus sign indicates the restoring force is in the direction opposite to the displacement. Figure 16.3 (a) The plastic ruler has been released, and the restoring force is returning the ruler to its equilibrium position. (b) The net force is zero at the equilibrium position, but the ruler has momentum and continues to move to the right. (c) The restoring force is in the opposite direction. It stops the ruler and moves it back toward equilibrium again. (d) Now the ruler has momentum to the left. (e) In the absence of damping (caused by frictional forces), the ruler reaches its original position. From there, the motion will repeat itself. 676 Chapter 16 \| Oscillatory Motion and Waves The force constant is related to the rigidity (or stiffness) of a system—the larger the force constant, the greater the restoring force, and the stiffer the system. The units of are newtons per meter (N/m). For example, is directly related to Young’s modulus when we stretch a string. Figure 16.4 shows a graph of the absolute value of the restoring force versus the displacement for a system that can be described by Hooke’s law—a simple spring in this case. The slope of the graph equals the force constant in newtons per meter. A common physics laboratory exercise is to measure restoring forces created by springs, determine if they follow Hooke’s law, and calculate their force constants if they do. Figure 16.4 (a) A graph of absolute value of the restoring force versus displacement is displayed. The fact that the graph is a straight line means that the system obeys Hooke’s law. The slope of the graph is the force constant . (b) The data in the graph were generated by measuring the displacement of a spring from equilibrium while supporting various weights. The restoring force equals the weight supported, if the mass is stationary. Example 16.1 How Stiff Are Car Springs? Figure 16.5 The mass of a car increases due to the introduction of a passenger. This affects the displacement of the car on its suspension system. (credit: exfordy on Flickr) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 16 \| Oscillatory Motion and Waves 677 What is the force constant for the suspension system of a car that settles 1.20 cm when an 80.0-kg person gets in? Strategy Consider the car to be in its equilibrium position = 0 before the person gets in. The car then settles down 1.20 cm, which means it is displaced to a position = −1.20×10−2 m . At that point, the springs supply a restoring force equal to the person’s weight = = = 784 N . We take this force to be in Hooke’s law. Knowing and 9.80 m/s2 80.0 kg , we can then solve the force constant . Solution 1. Solve Hooke’s law, = − , for : Substitute known values and solve : = − . = − 784 N −1.20×10−2 m = 6.53×104 N/m. (16.2) (16.3) Discussion Note that and have opposite signs because they are in opposite directions—the restoring force is up, and the displacement is down. Also, note that the car would oscillate up and down when the person got in if it were not for damping (due to frictional forces) provided by shock absorbers. Bouncing cars are a sure sign of bad shock absorbers. Energy in Hooke’s Law of Deformation In order to produce a deformation, work must be done. That is, a force must be exerted through a distance, whether you pluck a guitar string or compress a car spring. If the only result is deformation, and no work goes into thermal, sound, or kinetic energy, then all the work is initially stored in the deformed object as some form of potential energy. The potential energy stored in a spring is PEel = 1 described by Hooke’s law. Hence, 2 2 . Here, we generalize the idea to elastic potential energy for a deformation of any system that can be 2 2, whe
re PEel is the elastic potential energy stored in any deformed system that obeys Hooke’s law and has a displacement from equilibrium and a force constant . PEel = 1 (16.4) It is possible to find the work done in deforming a system in order to find the energy stored. This work is performed by an applied force app . The applied force is exactly opposite to the restoring force (action-reaction), and so app = . Figure 16.6 shows a graph of the applied force versus deformation for a system that can be described by Hooke’s law. Work done on the system is force multiplied by distance, which equals the area under the curve or (1 / 2) 2 (Method A in the figure). Another way to determine the work is to note that the force increases linearly from 0 to , so that the average force is (1 / 2) , the distance moved is , and thus = app = [(1 / 2)]() = (1 / 2) 2 (Method B in the figure). 678 Chapter 16 \| Oscillatory Motion and Waves Figure 16.6 A graph of applied force versus distance for the deformation of a system that can be described by Hooke’s law is displayed. The work done on the system equals the area under the graph or the area of the triangle, which is half its base multiplied by its height, or = (1 / 2) 2 . Example 16.2 Calculating Stored Energy: A Tranquilizer Gun Spring We can use a toy gun’s spring mechanism to ask and answer two simple questions: (a) How much energy is stored in the spring of a tranquilizer gun that has a force constant of 50.0 N/m and is compressed 0.150 m? (b) If you neglect friction and the mass of the spring, at what speed will a 2.00-g projectile be ejected from the gun? Figure 16.7 (a) In this image of the gun, the spring is uncompressed before being cocked. (b) The spring has been compressed a distance , and the projectile is in place. (c) When released, the spring converts elastic potential energy PEel into kinetic energy. Strategy for a (a): The energy stored in the spring can be found directly from elastic potential energy equation, because and are given. Solution for a Entering the given values for and yields PEel = 1 2 2 = 1 2 = 0.563 J Strategy for b (50.0 N/m)(0.150 m)2 = 0.563 N ⋅ m (16.5) Because there is no friction, the potential energy is converted entirely into kinetic energy. The expression for kinetic energy can be solved for the projectile’s speed. Solution for b This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 16 \| Oscillatory Motion and Waves 1. Identify known quantities: 2. Solve for : 3. Convert units: 23.7 m / s Discussion KEf = PEel or 1 / 22 = (1 / 2)2 = PEel = 0.563 J 1 / 2 = 2PEel = 2(0.563 J) 0.002 kg 1 / 2 = 23.7 J/kg 1 / 2 679 (16.6) (16.7) (a) and (b): This projectile speed is impressive for a tranquilizer gun (more than 80 km/h). The numbers in this problem seem reasonable. The force needed to compress the spring is small enough for an adult to manage, and the energy imparted to the dart is small enough to limit the damage it might do. Yet, the speed of the dart is great enough for it to travel an acceptable distance. Check your Understanding Envision holding the end of a ruler with one hand and deforming it with the other. When you let go, you can see the oscillations of the ruler. In what way could you modify this simple experiment to increase the rigidity of the system? Solution You could hold the ruler at its midpoint so that the part of the ruler that oscillates is half as long as in the original experiment. Check your Understanding If you apply a deforming force on an object and let it come to equilibrium, what happened to the work you did on the system? Solution It was stored in the object as potential energy. 16.2 Period and Frequency in Oscillations By the end of this section, you will be able to: Learning Objectives • Relate recurring mechanical vibrations to the frequency and period of harmonic motion, such as the motion of a guitar string. • Compute the frequency and period of an oscillation. The information presented in this section supports the following AP® learning objectives and science practices: • 3.B.3.3 The student can analyze data to identify qualitative or quantitative relationships between given values and variables (i.e., force, displacement, acceleration, velocity, period of motion, frequency, spring constant, string length, mass) associated with objects in oscillatory motion to use that data to determine the value of an unknown. (S.P. 2.2, 5.1) Figure 16.8 The strings on this guitar vibrate at regular time intervals. (credit: JAR) 680 Chapter 16 \| Oscillatory Motion and Waves When you pluck a guitar string, the resulting sound has a steady tone and lasts a long time. Each successive vibration of the string takes the same time as the previous one. We define periodic motion to be a motion that repeats itself at regular time intervals, such as exhibited by the guitar string or by an object on a spring moving up and down. The time to complete one oscillation remains constant and is called the period . Its units are usually seconds, but may be any convenient unit of time. The word period refers to the time for some event whether repetitive or not; but we shall be primarily interested in periodic motion, which is by definition repetitive. A concept closely related to period is the frequency of an event. For example, if you get a paycheck twice a month, the frequency of payment is two per month and the period between checks is half a month. Frequency is defined to be the number of events per unit time. For periodic motion, frequency is the number of oscillations per unit time. The relationship between frequency and period is The SI unit for frequency is the cycle per second, which is defined to be a hertz (Hz): 1 Hz = 1 cycle sec or 1 Hz = 1 s = 1 . (16.8) (16.9) A cycle is one complete oscillation. Note that a vibration can be a single or multiple event, whereas oscillations are usually repetitive for a significant number of cycles. Example 16.3 Determine the Frequency of Two Oscillations: Medical Ultrasound and the Period of Middle C We can use the formulas presented in this module to determine both the frequency based on known oscillations and the oscillation based on a known frequency. Let’s try one example of each. (a) A medical imaging device produces ultrasound by oscillating with a period of 0.400 µs. What is the frequency of this oscillation? (b) The frequency of middle C on a typical musical instrument is 264 Hz. What is the time for one complete oscillation? Strategy Both questions (a) and (b) can be answered using the relationship between period and frequency. In question (a), the period is given and we are asked to find frequency . In question (b), the frequency period . is given and we are asked to find the Solution a 1. Substitute 0.400 μs for in = 1 : Solve to find Discussion a = 1 = 1 0.400×10−6 s . = 2.50×106 Hz. (16.10) (16.11) The frequency of sound found in (a) is much higher than the highest frequency that humans can hear and, therefore, is called ultrasound. Appropriate oscillations at this frequency generate ultrasound used for noninvasive medical diagnoses, such as observations of a fetus in the womb. Solution b 1. Identify the known values: The time for one complete oscillation is the period : 2. Solve for : = 1 . = 1 . 3. Substitute the given value for the frequency into the resulting expression: = 1 = 1 264 Hz = 1 264 cycles/s = 3.79×10−3 s = 3.79 ms. Discussion (16.12) (16.13) (16.14) The period found in (b) is the time per cycle, but this value is often quoted as simply the time in convenient units (ms or milliseconds in this case). This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 16 \| Oscillatory Motion and Waves 681 Check your Understanding Identify an event in your life (such as receiving a paycheck) that occurs regularly. Identify both the period and frequency of this event. Solution I visit my parents for dinner every other Sunday. The frequency of my visits is 26 per calendar year. The period is two weeks. 16.3 Simple Harmonic Motion: A Special Periodic Motion By the end of this section, you will be able to: Learning Objectives • Describe a simple harmonic oscillator. • Relate physical characteristics of a vibrating system to aspects of simple harmonic motion and any resulting waves. The information presented in this section supports the following AP® learning objectives and science practices: • 3.B.3.1 The student is able to predict which properties determine the motion of a simple harmonic oscillator and what the dependence of the motion is on those properties. (S.P. 6.4, 7.2) • 3.B.3.4 The student is able to construct a qualitative and/or a quantitative explanation of oscillatory behavior given evidence of a restoring force. (S.P. 2.2, 6.2) • 6.A.3.1 The student is able to use graphical representation of a periodic mechanical wave to determine the amplitude of the wave. (S.P. 1.4) • 6.B.1.1 The student is able to use a graphical representation of a periodic mechanical wave (position versus time) to determine the period and frequency of the wave and describe how a change in the frequency would modify features of the representation. (S.P. 1.4, 2.2) The oscillations of a system in which the net force can be described by Hooke’s law are of special importance, because they are very common. They are also the simplest oscillatory systems. Simple Harmonic Motion (SHM) is the name given to oscillatory motion for a system where the net force can be described by Hooke’s law, and such a system is called a simple harmonic oscillator. If the net force can be described by Hooke’s law and there is no damping (by friction or other non-conservative forces), then a simple harmonic oscillator will oscillate with equal displacement on either side of the equilibrium position, as shown for an object on a spring in Figure 16.9. The maximum displacement from equilibrium is called the amplitude . The units for amplitude and displacement are the same, but depen
d on the type of oscillation. For the object on the spring, the units of amplitude and displacement are meters; whereas for sound oscillations, they have units of pressure (and other types of oscillations have yet other units). Because amplitude is the maximum displacement, it is related to the energy in the oscillation. Take-Home Experiment: SHM and the Marble Find a bowl or basin that is shaped like a hemisphere on the inside. Place a marble inside the bowl and tilt the bowl periodically so the marble rolls from the bottom of the bowl to equally high points on the sides of the bowl. Get a feel for the force required to maintain this periodic motion. What is the restoring force and what role does the force you apply play in the simple harmonic motion (SHM) of the marble? 682 Chapter 16 \| Oscillatory Motion and Waves Figure 16.9 An object attached to a spring sliding on a frictionless surface is an uncomplicated simple harmonic oscillator. When displaced from equilibrium, the object performs simple harmonic motion that has an amplitude and a period . The object’s maximum speed occurs as it passes through equilibrium. The stiffer the spring is, the smaller the period . The greater the mass of the object is, the greater the period . What is so significant about simple harmonic motion? One special thing is that the period and frequency of a simple harmonic oscillator are independent of amplitude. The string of a guitar, for example, will oscillate with the same frequency whether plucked gently or hard. Because the period is constant, a simple harmonic oscillator can be used as a clock. Two important factors do affect the period of a simple harmonic oscillator. The period is related to how stiff the system is. A very stiff object has a large force constant , which causes the system to have a smaller period. For example, you can adjust a diving board’s stiffness—the stiffer it is, the faster it vibrates, and the shorter its period. Period also depends on the mass of the oscillating system. The more massive the system is, the longer the period. For example, a heavy person on a diving board bounces up and down more slowly than a light one. In fact, the mass and the force constant are the only factors that affect the period and frequency of simple harmonic motion. Period of Simple Harmonic Oscillator The period of a simple harmonic oscillator is given by = 2π and, because = 1 / , the frequency of a simple harmonic oscillator is Note that neither nor has any dependence on amplitude. = 1 2π . Example 16.4 Mechanical Waves (16.15) (16.16) What do sound waves, water waves, and seismic waves have in common? They are all governed by Newton’s laws and they can exist only when traveling in a medium, such as air, water, or rocks. Waves that require a medium to travel are collectively known as “mechanical waves.” This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 16 \| Oscillatory Motion and Waves 683 Take-Home Experiment: Mass and Ruler Oscillations Find two identical wooden or plastic rulers. Tape one end of each ruler firmly to the edge of a table so that the length of each ruler that protrudes from the table is the same. On the free end of one ruler tape a heavy object such as a few large coins. Pluck the ends of the rulers at the same time and observe which one undergoes more cycles in a time period, and measure the period of oscillation of each of the rulers. Example 16.5 Calculate the Frequency and Period of Oscillations: Bad Shock Absorbers in a Car If the shock absorbers in a car go bad, then the car will oscillate at the least provocation, such as when going over bumps in the road and after stopping (See Figure 16.10). Calculate the frequency and period of these oscillations for such a car if the car’s mass (including its load) is 900 kg and the force constant ( ) of the suspension system is 6.53×104 N/m . Strategy The frequency of the car’s oscillations will be that of a simple harmonic oscillator as given in the equation = 1 2π . The mass and the force constant are both given. Solution 1. Enter the known values of k and m: = 1 2π = 1 2π 6.53×104 N/m 900 kg . 1 2π 72.6 / s–2 = 1.3656 / s–1 ≈ 1.36 / s–1 = 1.36 Hz. (16.17) (16.18) to calculate the period, but it is simpler to use the relationship = 1 / and substitute = 1 = 1 1.356 Hz = 0.738 s. (16.19) 2. Calculate the frequency: 3. You could use = 2π : the value just found for Discussion The values of and both seem about right for a bouncing car. You can observe these oscillations if you push down hard on the end of a car and let go. The Link between Simple Harmonic Motion and Waves If a time-exposure photograph of the bouncing car were taken as it drove by, the headlight would make a wavelike streak, as shown in Figure 16.10. Similarly, Figure 16.11 shows an object bouncing on a spring as it leaves a wavelike "trace of its position on a moving strip of paper. Both waves are sine functions. All simple harmonic motion is intimately related to sine and cosine waves. Figure 16.10 The bouncing car makes a wavelike motion. If the restoring force in the suspension system can be described only by Hooke’s law, then the wave is a sine function. (The wave is the trace produced by the headlight as the car moves to the right.) 684 Chapter 16 \| Oscillatory Motion and Waves Figure 16.11 The vertical position of an object bouncing on a spring is recorded on a strip of moving paper, leaving a sine wave. The displacement as a function of time t in any simple harmonic motion—that is, one in which the net restoring force can be described by Hooke’s law, is given by () = cos2 , (16.20) where is amplitude. At = 0 , the initial position is 0 = , and the displacement oscillates back and forth with a period . (When = , we get = again because cos 2π = 1 .). Furthermore, from this expression for , the velocity as a function of time is given by: () = −max sin 2π , (16.21) where max = 2π / = / . The object has zero velocity at maximum displacement—for example, = 0 when = 0 , and at that time = . The minus sign in the first equation for () gives the correct direction for the velocity. Just after the start of the motion, for instance, the velocity is negative because the system is moving back toward the equilibrium point. Finally, we can get an expression for acceleration using Newton’s second law. [Then we have () () , and () , the quantities needed for kinematics and a description of simple harmonic motion.] According to Newton’s second law, the acceleration is = / = / . So, () is also a cosine function: () = − cos2π . (16.22) Hence, () is directly proportional to and in the opposite direction to () . Figure 16.12 shows the simple harmonic motion of an object on a spring and presents graphs of ()(), and () versus time. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 16 \| Oscillatory Motion and Waves 685 Figure 16.12 Graphs of () () and () versus for the motion of an object on a spring. The net force on the object can be described by Hooke’s law, and so the object undergoes simple harmonic motion. Note that the initial position has the vertical displacement at its maximum value ; is initially zero and then negative as the object moves down; and the initial acceleration is negative, back toward the equilibrium position and becomes zero at that point. The most important point here is that these equations are mathematically straightforward and are valid for all simple harmonic motion. They are very useful in visualizing waves associated with simple harmonic motion, including visualizing how waves add with one another. Check Your Understanding Suppose you pluck a banjo string. You hear a single note that starts out loud and slowly quiets over time. Describe what happens to the sound waves in terms of period, frequency and amplitude as the sound decreases in volume. Solution Frequency and period remain essentially unchanged. Only amplitude decreases as volume decreases. Check Your Understanding A babysitter is pushing a child on a swing. At the point where the swing reaches , where would the corresponding point on a wave of this motion be located? Solution 686 Chapter 16 \| Oscillatory Motion and Waves is the maximum deformation, which corresponds to the amplitude of the wave. The point on the wave would either be at the very top or the very bottom of the curve. PhET Explorations: Masses and Springs A realistic mass and spring laboratory. Hang masses from springs and adjust the spring stiffness and damping. You can even slow time. Transport the lab to different planets. A chart shows the kinetic, potential, and thermal energy for each spring. Figure 16.13 Masses and Springs (http://cnx.org/content/m55273/1.2/mass-spring-lab_en.jar) 16.4 The Simple Pendulum By the end of this section, you will be able to: • Determine the period of oscillation of a hanging pendulum. Learning Objectives The information presented in this section supports the following AP® learning objectives and science practices: • 3.B.3.1 The student is able to predict which properties determine the motion of a simple harmonic oscillator and what the dependence of the motion is on those properties. (S.P. 6.4, 7.2) • 3.B.3.2 The student is able to design a plan and collect data in order to ascertain the characteristics of the motion of a system undergoing oscillatory motion caused by a restoring force. (S.P. 4.2) • 3.B.3.3 The student can analyze data to identify qualitative or quantitative relationships between given values and variables (i.e., force, displacement, acceleration, velocity, period of motion, frequency, spring constant, string length, mass) associated with objects in oscillatory motion to use that data to determine the value of an unknown. (S.P. 2.2, 5.1) • 3.B.3.4 The student is able to construct a qualitative and/or a quantitative explanation of oscillatory behavior given evidence of a restoring force. (S.P. 2.2, 6.2) Figure 16.14 A simple pendulum has a s
mall-diameter bob and a string that has a very small mass but is strong enough not to stretch appreciably. The linear displacement from equilibrium is , the length of the arc. Also shown are the forces on the bob, which result in a net force of − sin toward the equilibrium position—that is, a restoring force. Pendulums are in common usage. Some have crucial uses, such as in clocks; some are for fun, such as a child’s swing; and some are just there, such as the sinker on a fishing line. For small displacements, a pendulum is a simple harmonic oscillator. A simple pendulum is defined to have an object that has a small mass, also known as the pendulum bob, which is suspended from a light wire or string, such as shown in Figure 16.14. Exploring the simple pendulum a bit further, we can discover the conditions under which it performs simple harmonic motion, and we can derive an interesting expression for its period. We begin by defining the displacement to be the arc length . We see from Figure 16.14 that the net force on the bob is tangent to the arc and equals − sin . (The weight has components cos along the string and sin tangent to the arc.) Tension in the string exactly cancels the component cos parallel to the string. This leaves a net restoring force back toward the equilibrium position at = 0 . This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 16 \| Oscillatory Motion and Waves 687 Now, if we can show that the restoring force is directly proportional to the displacement, then we have a simple harmonic oscillator. In trying to determine if we have a simple harmonic oscillator, we should note that for small angles (less than about 15º ), sin ≈ ( sin and differ by about 1% or less at smaller angles). Thus, for angles less than about 15º , the restoring force is ≈ −θ. (16.23) The displacement is directly proportional to . When is expressed in radians, the arc length in a circle is related to its radius ( in this instance) by: so that = , = . For small angles, then, the expression for the restoring force is: This expression is of the form: ≈ − = −, (16.24) (16.25) (16.26) (16.27) where the force constant is given by = / and the displacement is given by = . For angles less than about 15º , the restoring force is directly proportional to the displacement, and the simple pendulum is a simple harmonic oscillator. Using this equation, we can find the period of a pendulum for amplitudes less than about 15º . For the simple pendulum: Thus, = 2π = 2π / = 2π (16.28) (16.29) for the period of a simple pendulum. This result is interesting because of its simplicity. The only things that affect the period of a simple pendulum are its length and the acceleration due to gravity. The period is completely independent of other factors, such as mass. As with simple harmonic oscillators, the period for a pendulum is nearly independent of amplitude, especially if is less than about 15º . Even simple pendulum clocks can be finely adjusted and accurate. Note the dependence of on . If the length of a pendulum is precisely known, it can actually be used to measure the acceleration due to gravity. Consider the following example. Example 16.6 Measuring Acceleration due to Gravity: The Period of a Pendulum What is the acceleration due to gravity in a region where a simple pendulum having a length 75.000 cm has a period of 1.7357 s? Strategy We are asked to find given the period and the length of a pendulum. We can solve = 2π for , assuming only that the angle of deflection is less than 15º . Solution 1. Square = 2π and solve for : 2. Substitute known values into the new equation: = 4π2 2 . 3. Calculate to find : = 4π2 0.75000 m (1.7357 s)2 . = 9.8281 m / s2. (16.30) (16.31) (16.32) 688 Discussion Chapter 16 \| Oscillatory Motion and Waves This method for determining can be very accurate. This is why length and period are given to five digits in this example. For the precision of the approximation sin θ ≈ to be better than the precision of the pendulum length and period, the maximum displacement angle should be kept below about 0.5º . Making Career Connections Knowing can be important in geological exploration; for example, a map of over large geographical regions aids the study of plate tectonics and helps in the search for oil fields and large mineral deposits. Take Home Experiment: Determining Use a simple pendulum to determine the acceleration due to gravity in your own locale. Cut a piece of a string or dental floss so that it is about 1 m long. Attach a small object of high density to the end of the string (for example, a metal nut or a car key). Starting at an angle of less than 10º , allow the pendulum to swing and measure the pendulum’s period for 10 oscillations using a stopwatch. Calculate . How accurate is this measurement? How might it be improved? Check Your Understanding An engineer builds two simple pendula. Both are suspended from small wires secured to the ceiling of a room. Each pendulum hovers 2 cm above the floor. Pendulum 1 has a bob with a mass of 10 kg . Pendulum 2 has a bob with a mass of 100 kg . Describe how the motion of the pendula will differ if the bobs are both displaced by 12º . Solution The movement of the pendula will not differ at all because the mass of the bob has no effect on the motion of a simple pendulum. The pendula are only affected by the period (which is related to the pendulum’s length) and by the acceleration due to gravity. PhET Explorations: Pendulum Lab Play with one or two pendulums and discover how the period of a simple pendulum depends on the length of the string, the mass of the pendulum bob, and the amplitude of the swing. It’s easy to measure the period using the photogate timer. You can vary friction and the strength of gravity. Use the pendulum to find the value of on planet X. Notice the anharmonic behavior at large amplitude. Figure 16.15 Pendulum Lab (http://cnx.org/content/m55274/1.2/pendulum-lab_en.jar) 16.5 Energy and the Simple Harmonic Oscillator By the end of this section, you will be able to: Learning Objectives • Describe the changes in energy that occur while a system undergoes simple harmonic motion. To study the energy of a simple harmonic oscillator, we first consider all the forms of energy it can have We know from Hooke’s Law: Stress and Strain Revisited that the energy stored in the deformation of a simple harmonic oscillator is a form of potential energy given by: PEel = 1 22. (16.33) Because a simple harmonic oscillator has no dissipative forces, the other important form of energy is kinetic energy KE . Conservation of energy for these two forms is: This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 16 \| Oscillatory Motion and Waves or KE + PEel = constant 22 + 1 1 22 = constant. 689 (16.34) (16.35) This statement of conservation of energy is valid for all simple harmonic oscillators, including ones where the gravitational force plays a role Namely, for a simple pendulum we replace the velocity with = , the spring constant with = / , and the displacement term with = . Thus 222 + 1 1 2 2 = constant. (16.36) In the case of undamped simple harmonic motion, the energy oscillates back and forth between kinetic and potential, going completely from one to the other as the system oscillates. So for the simple example of an object on a frictionless surface attached to a spring, as shown again in Figure 16.16, the motion starts with all of the energy stored in the spring. As the object starts to move, the elastic potential energy is converted to kinetic energy, becoming entirely kinetic energy at the equilibrium position. It is then converted back into elastic potential energy by the spring, the velocity becomes zero when the kinetic energy is completely converted, and so on. This concept provides extra insight here and in later applications of simple harmonic motion, such as alternating current circuits. Figure 16.16 The transformation of energy in simple harmonic motion is illustrated for an object attached to a spring on a frictionless surface. The conservation of energy principle can be used to derive an expression for velocity . If we start our simple harmonic motion with zero velocity and maximum displacement ( = ), then the total energy is 1 2 2. (16.37) This total energy is constant and is shifted back and forth between kinetic energy and potential energy, at most times being shared by each. The conservation of energy for this system in equation form is thus: 22 + 1 1 22 = 1 2 2. Solving this equation for yields: Manipulating this expression algebraically gives: = ± 2 − 2 . (16.38) (16.39) 690 and so where = ± 1 − 2 2 = ±max 1 − 2 2 , max = . Chapter 16 \| Oscillatory Motion and Waves (16.40) (16.41) (16.42) From this expression, we see that the velocity is a maximum ( max ) at = 0 , as stated earlier in () = − max sin 2π Notice that the maximum velocity depends on three factors. Maximum velocity is directly proportional to amplitude. As you might guess, the greater the maximum displacement the greater the maximum velocity. Maximum velocity is also greater for stiffer systems, because they exert greater force for the same displacement. This observation is seen in the expression for max; it is proportional to the square root of the force constant . Finally, the maximum velocity is smaller for objects that have larger masses, because the maximum velocity is inversely proportional to the square root of . For a given force, objects that have large masses accelerate more slowly. . A similar calculation for the simple pendulum produces a similar result, namely: max = max. (16.43) Making Connections: Mass Attached to a Spring Consider a mass m attached to a spring, with spring constant k, fixed to a wall. When the mass is displaced from its equilibrium position and released, the mass undergoes simple harmonic motion. The spring exerts a force = − on the mass. The potential energy of the system is stored in th
e spring. It will be zero when the spring is in the equilibrium position. All the internal energy exists in the form of kinetic energy, given by = 1 22 . As the system oscillates, which means that the spring compresses and expands, there is a change in the structure of the system and a corresponding change in its internal energy. Its kinetic energy is converted to potential energy and vice versa. This occurs at an equal rate, which means that a loss of kinetic energy yields a gain in potential energy, thus preserving the work-energy theorem and the law of conservation of energy. Example 16.7 Determine the Maximum Speed of an Oscillating System: A Bumpy Road Suppose that a car is 900 kg and has a suspension system that has a force constant = 6.53×104 N/m . The car hits a bump and bounces with an amplitude of 0.100 m. What is its maximum vertical velocity if you assume no damping occurs? Strategy We can use the expression for max given in max = and are given in the problem statement, and the maximum displacement is 0.100 m. to determine the maximum vertical velocity. The variables Solution 1. Identify known. 2. Substitute known values into max = : max = 6.53×104 N/m 900 kg (0.100 m). (16.44) 3. Calculate to find max= 0.852 m/s. Discussion This answer seems reasonable for a bouncing car. There are other ways to use conservation of energy to find max . We could use it directly, as was done in the example featured in Hooke’s Law: Stress and Strain Revisited. The small vertical displacement of an oscillating simple pendulum, starting from its equilibrium position, is given as This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 16 \| Oscillatory Motion and Waves where is the amplitude, is the angular velocity and is the time taken. Substituting = 2π , we have () = sin , = sin 2π . Thus, the displacement of pendulum is a function of time as shown above. Also the velocity of the pendulum is given by () = 2 cos 2π , so the motion of the pendulum is a function of time. Check Your Understanding 691 (16.45) (16.46) (16.47) Why does it hurt more if your hand is snapped with a ruler than with a loose spring, even if the displacement of each system is equal? Solution The ruler is a stiffer system, which carries greater force for the same amount of displacement. The ruler snaps your hand with greater force, which hurts more. Check Your Understanding You are observing a simple harmonic oscillator. Identify one way you could decrease the maximum velocity of the system. Solution You could increase the mass of the object that is oscillating. 16.6 Uniform Circular Motion and Simple Harmonic Motion Learning Objectives By the end of this section, you will be able to: • Compare simple harmonic motion with uniform circular motion. Figure 16.17 The horses on this merry-go-round exhibit uniform circular motion. (credit: Wonderlane, Flickr) There is an easy way to produce simple harmonic motion by using uniform circular motion. Figure 16.18 shows one way of using this method. A ball is attached to a uniformly rotating vertical turntable, and its shadow is projected on the floor as shown. The shadow undergoes simple harmonic motion. Hooke’s law usually describes uniform circular motions ( constant) rather than systems that have large visible displacements. So observing the projection of uniform circular motion, as in Figure 16.18, is often easier than observing a precise large-scale simple harmonic oscillator. If studied in sufficient depth, simple harmonic motion produced in this manner can give considerable insight into many aspects of oscillations and waves and is very useful mathematically. In our brief treatment, we shall indicate some of the major features of this relationship and how they might be useful. 692 Chapter 16 \| Oscillatory Motion and Waves Figure 16.18 The shadow of a ball rotating at constant angular velocity on a turntable goes back and forth in precise simple harmonic motion. Figure 16.19 shows the basic relationship between uniform circular motion and simple harmonic motion. The point P travels around the circle at constant angular velocity . The point P is analogous to an object on the merry-go-round. The projection of the position of P onto a fixed axis undergoes simple harmonic motion and is analogous to the shadow of the object. At the time shown in the figure, the projection has position and moves to the left with velocity . The velocity of the point P around the ¯ circle equals ¯ max .The projection of max on the -axis is the velocity of the simple harmonic motion along the -axis. Figure 16.19 A point P moving on a circular path with a constant angular velocity is undergoing uniform circular motion. Its projection on the x-axis ¯ undergoes simple harmonic motion. Also shown is the velocity of this point around the circle, velocities form a similar triangle to the displacement triangle. max , and its projection, which is . Note that these To see that the projection undergoes simple harmonic motion, note that its position is given by = cos , where = , is the constant angular velocity, and is the radius of the circular path. Thus, = cos . The angular velocity is in radians per unit time; in this case 2π radians is the time for one revolution . That is, = 2π / . Substituting this expression for , we see that the position is given by: () = cos 2π . (16.48) (16.49) (16.50) This expression is the same one we had for the position of a simple harmonic oscillator in Simple Harmonic Motion: A Special Periodic Motion. If we make a graph of position versus time as in Figure 16.20, we see again the wavelike character (typical of simple harmonic motion) of the projection of uniform circular motion onto the -axis. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 16 \| Oscillatory Motion and Waves 693 Figure 16.20 The position of the projection of uniform circular motion performs simple harmonic motion, as this wavelike graph of versus indicates. Now let us use Figure 16.19 to do some further analysis of uniform circular motion as it relates to simple harmonic motion. The triangle formed by the velocities in the figure and the triangle formed by the displacements ( and 2 − 2 ) are similar right triangles. Taking ratios of similar sides, we see that We can solve this equation for the speed or max = 2 − 2 = 1 − 2 2 . = max 1 − 2 2 . (16.51) (16.52) This expression for the speed of a simple harmonic oscillator is exactly the same as the equation obtained from conservation of energy considerations in Energy and the Simple Harmonic Oscillator. You can begin to see that it is possible to get all of the characteristics of simple harmonic motion from an analysis of the projection of uniform circular motion. Finally, let us consider the period of the motion of the projection. This period is the time it takes the point P to complete one revolution. That time is the circumference of the circle 2π divided by the velocity around the circle, max . Thus, the period is We know from conservation of energy considerations that = 2πX max . Solving this equation for / max gives max = . max = . Substituting this expression into the equation for yields = 2π . (16.53) (16.54) (16.55) (16.56) Thus, the period of the motion is the same as for a simple harmonic oscillator. We have determined the period for any simple harmonic oscillator using the relationship between uniform circular motion and simple harmonic motion. Some modules occasionally refer to the connection between uniform circular motion and simple harmonic motion. Moreover, if you carry your study of physics and its applications to greater depths, you will find this relationship useful. It can, for example, help to analyze how waves add when they are superimposed. 700 Chapter 16 \| Oscillatory Motion and Waves Check Your Understanding A famous magic trick involves a performer singing a note toward a crystal glass until the glass shatters. Explain why the trick works in terms of resonance and natural frequency. Solution The performer must be singing a note that corresponds to the natural frequency of the glass. As the sound wave is directed at the glass, the glass responds by resonating at the same frequency as the sound wave. With enough energy introduced into the system, the glass begins to vibrate and eventually shatters. 16.9 Waves Learning Objectives By the end of this section, you will be able to: • Describe various characteristics associated with a wave. • Differentiate between transverse and longitudinal waves. Figure 16.29 Waves in the ocean behave similarly to all other types of waves. (credit: Steve Jurveston, Flickr) What do we mean when we say something is a wave? The most intuitive and easiest wave to imagine is the familiar water wave. More precisely, a wave is a disturbance that propagates, or moves from the place it was created. For water waves, the disturbance is in the surface of the water, perhaps created by a rock thrown into a pond or by a swimmer splashing the surface repeatedly. For sound waves, the disturbance is a change in air pressure, perhaps created by the oscillating cone inside a speaker. For earthquakes, there are several types of disturbances, including disturbance of Earth’s surface and pressure disturbances under the surface. Even radio waves are most easily understood using an analogy with water waves. Visualizing water waves is useful because there is more to it than just a mental image. Water waves exhibit characteristics common to all waves, such as amplitude, period, frequency and energy. All wave characteristics can be described by a small set of underlying principles. A wave is a disturbance that propagates, or moves from the place it was created. The simplest waves repeat themselves for several cycles and are associated with simple harmonic motion. Let us start by considering the simplified water wave in Figure 16.30. The wave is an up and down disturbance of the water surface. It causes a
sea gull to move up and down in simple harmonic motion as the wave crests and troughs (peaks and valleys) pass under the bird. The time for one complete up and down motion is the wave’s period . The wave’s frequency is = 1 / , as usual. The wave itself moves to the right in the figure. This movement of the wave is actually the disturbance moving to the right, not the water itself (or the bird would move to the right). We define wave velocity w to be the speed at which the disturbance moves. Wave velocity is sometimes also called the propagation velocity or propagation speed, because the disturbance propagates from one location to another. Misconception Alert Many people think that water waves push water from one direction to another. In fact, the particles of water tend to stay in one location, save for moving up and down due to the energy in the wave. The energy moves forward through the water, but the water stays in one place. If you feel yourself pushed in an ocean, what you feel is the energy of the wave, not a rush of water. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 16 \| Oscillatory Motion and Waves 701 Figure 16.30 An idealized ocean wave passes under a sea gull that bobs up and down in simple harmonic motion. The wave has a wavelength , which is the distance between adjacent identical parts of the wave. The up and down disturbance of the surface propagates parallel to the surface at a speed w . The water wave in the figure also has a length associated with it, called its wavelength , the distance between adjacent identical parts of a wave. ( is the distance parallel to the direction of propagation.) The speed of propagation w is the distance the wave travels in a given time, which is one wavelength in the time of one period. In equation form, that is or w = w = . (16.66) (16.67) This fundamental relationship holds for all types of waves. For water waves, w is the speed of a surface wave; for sound, w is the speed of sound; and for visible light, w is the speed of light, for example. Applying the Science Practices: Different Types of Waves Consider a spring fixed to a wall with a mass connected to its end. This fixed point on the wall exerts a force on the complete spring-and-mass system, and this implies that the momentum of the complete system is not conserved. Now, consider energy. Since the system is fixed to a point on the wall, it does not do any work; hence, the total work done is conserved, which means that the energy is conserved. Consequently, we have an oscillator in which energy is conserved but momentum is not. Now, consider a system of two masses connected to each other by a spring. This type of system also forms an oscillator. Since there is no fixed point, momentum is conserved as the forces acting on the two masses are equal and opposite. Energy for such a system will be conserved, because there are no external forces acting on the spring-twomasses system. It is clear from above that, for momentum to be conserved, momentum needs to be carried by waves. This is a typical example of a mechanical oscillator producing mechanical waves that need a medium in which to propagate. Sound waves are also examples of mechanical waves. There are some waves that can travel in the absence of a medium of propagation. Such waves are called “electromagnetic waves.” Light waves are examples of electromagnetic waves. Electromagnetic waves are created by the vibration of electric charge. This vibration creates a wave with both electric and magnetic field components. Take-Home Experiment: Waves in a Bowl Fill a large bowl or basin with water and wait for the water to settle so there are no ripples. Gently drop a cork into the middle of the bowl. Estimate the wavelength and period of oscillation of the water wave that propagates away from the cork. Remove the cork from the bowl and wait for the water to settle again. Gently drop the cork at a height that is different from the first drop. Does the wavelength depend upon how high above the water the cork is dropped? Example 16.9 Calculate the Velocity of Wave Propagation: Gull in the Ocean Calculate the wave velocity of the ocean wave in Figure 16.30 if the distance between wave crests is 10.0 m and the time for a sea gull to bob up and down is 5.00 s. Strategy 702 Chapter 16 \| Oscillatory Motion and Waves We are asked to find w . The given information tells us that = 10.0 m and = 5.00 s . Therefore, we can use w = to find the wave velocity. Solution 1. Enter the known values into w = : 2. Solve for w to find w = 2.00 m/s. Discussion w = 10.0 m 5.00 s . (16.68) This slow speed seems reasonable for an ocean wave. Note that the wave moves to the right in the figure at this speed, not the varying speed at which the sea gull moves up and down. Transverse and Longitudinal Waves A simple wave consists of a periodic disturbance that propagates from one place to another. The wave in Figure 16.31 propagates in the horizontal direction while the surface is disturbed in the vertical direction. Such a wave is called a transverse wave or shear wave; in such a wave, the disturbance is perpendicular to the direction of propagation. In contrast, in a longitudinal wave or compressional wave, the disturbance is parallel to the direction of propagation. Figure 16.32 shows an example of a longitudinal wave. The size of the disturbance is its amplitude X and is completely independent of the speed of propagation w . Figure 16.31 In this example of a transverse wave, the wave propagates horizontally, and the disturbance in the cord is in the vertical direction. Figure 16.32 In this example of a longitudinal wave, the wave propagates horizontally, and the disturbance in the cord is also in the horizontal direction. Waves may be transverse, longitudinal, or a combination of the two. (Water waves are actually a combination of transverse and longitudinal. The simplified water wave illustrated in Figure 16.30 shows no longitudinal motion of the bird.) The waves on the strings of musical instruments are transverse—so are electromagnetic waves, such as visible light. Sound waves in air and water are longitudinal. Their disturbances are periodic variations in pressure that are transmitted in fluids. Fluids do not have appreciable shear strength, and thus the sound waves in them must be longitudinal or compressional. Sound in solids can be both longitudinal and transverse. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 16 \| Oscillatory Motion and Waves 703 Figure 16.33 The wave on a guitar string is transverse. The sound wave rattles a sheet of paper in a direction that shows the sound wave is longitudinal. Earthquake waves under Earth’s surface also have both longitudinal and transverse components (called compressional or Pwaves and shear or S-waves, respectively). These components have important individual characteristics—they propagate at different speeds, for example. Earthquakes also have surface waves that are similar to surface waves on water. Applying the Science Practices: Electricity in Your Home The source of electricity is of a sinusoidal nature. If we appropriately probe using an oscilloscope (an instrument used to display and analyze electronic signals), we can precisely determine the frequency and wavelength of the waveform. Inquire about the maximum voltage current that you get in your house and plot a sinusoidal waveform representing the frequency, wavelength, and period for it. Check Your Understanding Why is it important to differentiate between longitudinal and transverse waves? Solution In the different types of waves, energy can propagate in a different direction relative to the motion of the wave. This is important to understand how different types of waves affect the materials around them. PhET Explorations: Wave on a String Watch a string vibrate in slow motion. Wiggle the end of the string and make waves, or adjust the frequency and amplitude of an oscillator. Adjust the damping and tension. The end can be fixed, loose, or open. Figure 16.34 Wave on a String (http://cnx.org/content/m55281/1.2/wave-on-a-string_en.jar) 16.10 Superposition and Interference By the end of this section, you will be able to: Learning Objectives • Determine the resultant waveform when two waves act in superposition relative to each other. • Explain standing waves. • Describe the mathematical representation of overtones and beat frequency. 704 Chapter 16 \| Oscillatory Motion and Waves Figure 16.35 These waves result from the superposition of several waves from different sources, producing a complex pattern. (credit: waterborough, Wikimedia Commons) Most waves do not look very simple. They look more like the waves in Figure 16.35 than like the simple water wave considered in Waves. (Simple waves may be created by a simple harmonic oscillation, and thus have a sinusoidal shape). Complex waves are more interesting, even beautiful, but they look formidable. Most waves appear complex because they result from several simple waves adding together. Luckily, the rules for adding waves are quite simple. When two or more waves arrive at the same point, they superimpose themselves on one another. More specifically, the disturbances of waves are superimposed when they come together—a phenomenon called superposition. Each disturbance corresponds to a force, and forces add. If the disturbances are along the same line, then the resulting wave is a simple addition of the disturbances of the individual waves—that is, their amplitudes add. Figure 16.36 and Figure 16.37 illustrate superposition in two special cases, both of which produce simple results. Figure 16.36 shows two identical waves that arrive at the same point exactly in phase. The crests of the two waves are precisely aligned, as are the troughs. This superposition produces pure constructive interference. Because the disturbances add, pure constructive interference produces
a wave that has twice the amplitude of the individual waves, but has the same wavelength. Figure 16.37 shows two identical waves that arrive exactly out of phase—that is, precisely aligned crest to trough—producing pure destructive interference. Because the disturbances are in the opposite direction for this superposition, the resulting amplitude is zero for pure destructive interference—the waves completely cancel. Figure 16.36 Pure constructive interference of two identical waves produces one with twice the amplitude, but the same wavelength. Figure 16.37 Pure destructive interference of two identical waves produces zero amplitude, or complete cancellation. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 16 \| Oscillatory Motion and Waves 705 While pure constructive and pure destructive interference do occur, they require precisely aligned identical waves. The superposition of most waves produces a combination of constructive and destructive interference and can vary from place to place and time to time. Sound from a stereo, for example, can be loud in one spot and quiet in another. Varying loudness means the sound waves add partially constructively and partially destructively at different locations. A stereo has at least two speakers creating sound waves, and waves can reflect from walls. All these waves superimpose. An example of sounds that vary over time from constructive to destructive is found in the combined whine of airplane jets heard by a stationary passenger. The combined sound can fluctuate up and down in volume as the sound from the two engines varies in time from constructive to destructive. These examples are of waves that are similar. An example of the superposition of two dissimilar waves is shown in Figure 16.38. Here again, the disturbances add and subtract, producing a more complicated looking wave. Figure 16.38 Superposition of non-identical waves exhibits both constructive and destructive interference. Standing Waves Sometimes waves do not seem to move; rather, they just vibrate in place. Unmoving waves can be seen on the surface of a glass of milk in a refrigerator, for example. Vibrations from the refrigerator motor create waves on the milk that oscillate up and down but do not seem to move across the surface. These waves are formed by the superposition of two or more moving waves, such as illustrated in Figure 16.39 for two identical waves moving in opposite directions. The waves move through each other with their disturbances adding as they go by. If the two waves have the same amplitude and wavelength, then they alternate between constructive and destructive interference. The resultant looks like a wave standing in place and, thus, is called a standing wave. Waves on the glass of milk are one example of standing waves. There are other standing waves, such as on guitar strings and in organ pipes. With the glass of milk, the two waves that produce standing waves may come from reflections from the side of the glass. A closer look at earthquakes provides evidence for conditions appropriate for resonance, standing waves, and constructive and destructive interference. A building may be vibrated for several seconds with a driving frequency matching that of the natural frequency of vibration of the building—producing a resonance resulting in one building collapsing while neighboring buildings do not. Often buildings of a certain height are devastated while other taller buildings remain intact. The building height matches the condition for setting up a standing wave for that particular height. As the earthquake waves travel along the surface of Earth and reflect off denser rocks, constructive interference occurs at certain points. Often areas closer to the epicenter are not damaged while areas farther away are damaged. Figure 16.39 Standing wave created by the superposition of two identical waves moving in opposite directions. The oscillations are at fixed locations in space and result from alternately constructive and destructive interference. Standing waves are also found on the strings of musical instruments and are due to reflections of waves from the ends of the string. Figure 16.40 and Figure 16.41 show three standing waves that can be created on a string that is fixed at both ends. Nodes are the points where the string does not move; more generally, nodes are where the wave disturbance is zero in a 706 Chapter 16 \| Oscillatory Motion and Waves standing wave. The fixed ends of strings must be nodes, too, because the string cannot move there. The word antinode is used to denote the location of maximum amplitude in standing waves. Standing waves on strings have a frequency that is related to the propagation speed w of the disturbance on the string. The wavelength is determined by the distance between the points where the string is fixed in place. The lowest frequency, called the fundamental frequency, is thus for the longest wavelength, which is seen to be 1 = 2 . Therefore, the fundamental frequency is 1 = w / 1 = w / 2 . In this case, the overtones or harmonics are multiples of the fundamental frequency. As seen in Figure 16.41, the first harmonic can easily be calculated since 2 = . Thus . Similarly, 3 = 3 1 , and so on. All of these frequencies can be changed by adjusting the tension in the string. The greater the tension, the greater w is and the higher the frequencies. This observation is familiar to anyone who has ever observed a string instrument being tuned. We will see in later chapters that standing waves are crucial to many resonance phenomena, such as in sounding boxes on string instruments. Figure 16.40 The figure shows a string oscillating at its fundamental frequency. Figure 16.41 First and second harmonic frequencies are shown. Beats Striking two adjacent keys on a piano produces a warbling combination usually considered to be unpleasant. The superposition of two waves of similar but not identical frequencies is the culprit. Another example is often noticeable in jet aircraft, particularly the two-engine variety, while taxiing. The combined sound of the engines goes up and down in loudness. This varying loudness happens because the sound waves have similar but not identical frequencies. The discordant warbling of the piano and the fluctuating loudness of the jet engine noise are both due to alternately constructive and destructive interference as the two waves go in and out of phase. Figure 16.42 illustrates this graphically. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 16 \| Oscillatory Motion and Waves 707 Figure 16.42 Beats are produced by the superposition of two waves of slightly different frequencies but identical amplitudes. The waves alternate in time between constructive interference and destructive interference, giving the resulting wave a time-varying amplitude. The wave resulting from the superposition of two similar-frequency waves has a frequency that is the average of the two. This wave fluctuates in amplitude, or beats, with a frequency called the beat frequency. We can determine the beat frequency by adding two waves together mathematically. Note that a wave can be represented at one point in space as = cos 2π = cos , 2π where = 1 / is the frequency of the wave. Adding two waves that have different frequencies but identical amplitudes produces a resultant More specifically, = 1 + 2. = cos 2π 1 + cos . 2π 2 Using a trigonometric identity, it can be shown that = 2 cos B cos , 2π ave where B = ∣ 1 − 2 ∣ (16.69) (16.70) (16.71) (16.72) (16.73) is the beat frequency, and ave is the average of amplitude and the average frequency of the two superimposed waves, but it also fluctuates in overall amplitude at the beat frequency B . The first cosine term in the expression effectively causes the amplitude to go up and down. The second cosine term is the wave with frequency ave . This result is valid for all types of waves. However, if it is a sound wave, providing the two frequencies are similar, then what we hear is an average frequency that gets louder and softer (or warbles) at the beat frequency. 1 and 2 . These results mean that the resultant wave has twice the Real World Connections: Tuning Forks The MIT physics demo (http://openstaxcollege.org/l/31tuningforks/) entitled “Tuning Forks: Resonance and Beat Frequency” provides a qualitative picture of how wave interference produces beats. Description: Two identical forks and sounding boxes are placed next to each other. Striking one tuning fork will cause the other to resonate at the same frequency. When a weight is attached to one tuning fork, they are no longer identical. Thus, one will not cause the other to resonate. When two different forks are struck at the same time, the interference of their pitches produces beats. Real World Connections: Jump Rop This is a fun activity with which to learn about interference and superposition. Take a jump rope and hold it at the two ends with one of your friends. While each of you is holding the rope, snap your hands to produce a wave from each side. Record your observations and see if they match with the following: a. One wave starts from the right end and travels to the left end of the rope. b. Another wave starts at the left end and travels to the right end of the rope. c. The waves travel at the same speed. d. The shape of the waves depends on the way the person snaps his or her hands. e. There is a region of overlap. f. The shapes of the waves are identical to their original shapes after they overlap. Now, snap the rope up and down and ask your friend to snap his or her end of the rope sideways. The resultant that one sees here is the vector sum of two individual displacements. 708 Chapter 16 \| Oscillatory Motion and Waves This activity illustrates superposition and interference. When two or more waves interact with each other at a point, the disturbance at that point is given by the sum of the
disturbances each wave will produce in the absence of the other. This is the principle of superposition. Interference is a result of superposition of two or more waves to form a resultant wave of greater or lower amplitude. While beats may sometimes be annoying in audible sounds, we will find that beats have many applications. Observing beats is a very useful way to compare similar frequencies. There are applications of beats as apparently disparate as in ultrasonic imaging and radar speed traps. Check Your Understanding Imagine you are holding one end of a jump rope, and your friend holds the other. If your friend holds her end still, you can move your end up and down, creating a transverse wave. If your friend then begins to move her end up and down, generating a wave in the opposite direction, what resultant wave forms would you expect to see in the jump rope? Solution The rope would alternate between having waves with amplitudes two times the original amplitude and reaching equilibrium with no amplitude at all. The wavelengths will result in both constructive and destructive interference Check Your Understanding Define nodes and antinodes. Solution Nodes are areas of wave interference where there is no motion. Antinodes are areas of wave interference where the motion is at its maximum point. Check Your Understanding You hook up a stereo system. When you test the system, you notice that in one corner of the room, the sounds seem dull. In another area, the sounds seem excessively loud. Describe how the sound moving about the room could result in these effects. Solution With multiple speakers putting out sounds into the room, and these sounds bouncing off walls, there is bound to be some wave interference. In the dull areas, the interference is probably mostly destructive. In the louder areas, the interference is probably mostly constructive. PhET Explorations: Wave Interference Make waves with a dripping faucet, audio speaker, or laser! Add a second source or a pair of slits to create an interference pattern. Figure 16.43 Wave Interference (http://cnx.org/content/m55282/1.2/wave-interference_en.jar) 16.11 Energy in Waves: Intensity Learning Objectives By the end of this section, you will be able to: • Calculate the intensity and the power of rays and waves. This content is available for free at http://cnx.org/content/col11844/1.13 712 Chapter 16 \| Oscillatory Motion and Waves frequency: number of events per unit of time fundamental frequency: the lowest frequency of a periodic waveform intensity: power per unit area longitudinal wave: a wave in which the disturbance is parallel to the direction of propagation natural frequency: the frequency at which a system would oscillate if there were no driving and no damping forces nodes: the points where the string does not move; more generally, nodes are where the wave disturbance is zero in a standing wave oscillate: moving back and forth regularly between two points over damping: the condition in which damping of an oscillator causes it to return to equilibrium without oscillating; oscillator moves more slowly toward equilibrium than in the critically damped system overtones: multiples of the fundamental frequency of a sound period: time it takes to complete one oscillation periodic motion: motion that repeats itself at regular time intervals resonance: the phenomenon of driving a system with a frequency equal to the system's natural frequency resonate: a system being driven at its natural frequency restoring force: force acting in opposition to the force caused by a deformation simple harmonic motion: the oscillatory motion in a system where the net force can be described by Hooke’s law simple harmonic oscillator: a device that implements Hooke’s law, such as a mass that is attached to a spring, with the other end of the spring being connected to a rigid support such as a wall simple pendulum: an object with a small mass suspended from a light wire or string superposition: the phenomenon that occurs when two or more waves arrive at the same point transverse wave: a wave in which the disturbance is perpendicular to the direction of propagation under damping: the condition in which damping of an oscillator causes it to return to equilibrium with the amplitude gradually decreasing to zero; system returns to equilibrium faster but overshoots and crosses the equilibrium position one or more times wave: a disturbance that moves from its source and carries energy wave velocity: the speed at which the disturbance moves. Also called the propagation velocity or propagation speed wavelength: the distance between adjacent identical parts of a wave Section Summary 16.1 Hooke’s Law: Stress and Strain Revisited • An oscillation is a back and forth motion of an object between two points of deformation. • An oscillation may create a wave, which is a disturbance that propagates from where it was created. • The simplest type of oscillations and waves are related to systems that can be described by Hooke’s law: = −, where is the restoring force, is the displacement from equilibrium or deformation, and is the force constant of the system. • Elastic potential energy PEel stored in the deformation of a system that can be described by Hooke’s law is given by PEel = (1 / 2)2. 16.2 Period and Frequency in Oscillations • Periodic motion is a repetitious oscillation. • The time for one oscillation is the period . • The number of oscillations per unit time is the frequency . • These quantities are related by This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 16 \| Oscillatory Motion and Waves 713 = 1 . 16.3 Simple Harmonic Motion: A Special Periodic Motion • Simple harmonic motion is oscillatory motion for a system that can be described only by Hooke’s law. Such a system is also called a simple harmonic oscillator. • Maximum displacement is the amplitude . The period and frequency of a simple harmonic oscillator are given by = 2π and = 1 2π , where is the mass of the system. • Displacement in simple harmonic motion as a function of time is given by () = cos 2π . • The velocity is given by () = − max sin2πt cos 2π • The acceleration is found to be () = − . , where max = / . 16.4 The Simple Pendulum • A mass suspended by a wire of length is a simple pendulum and undergoes simple harmonic motion for amplitudes less than about 15º. The period of a simple pendulum is where is the length of the string and is the acceleration due to gravity. = 2π , 16.5 Energy and the Simple Harmonic Oscillator • Energy in the simple harmonic oscillator is shared between elastic potential energy and kinetic energy, with the total being constant: • Maximum velocity depends on three factors: it is directly proportional to amplitude, it is greater for stiffer systems, and it is 22 + 1 1 22 = constant. smaller for objects that have larger masses: max = . 16.6 Uniform Circular Motion and Simple Harmonic Motion A projection of uniform circular motion undergoes simple harmonic oscillation. 16.7 Damped Harmonic Motion • Damped harmonic oscillators have non-conservative forces that dissipate their energy. • Critical damping returns the system to equilibrium as fast as possible without overshooting. • An underdamped system will oscillate through the equilibrium position. • An overdamped system moves more slowly toward equilibrium than one that is critically damped. 16.8 Forced Oscillations and Resonance • A system’s natural frequency is the frequency at which the system will oscillate if not affected by driving or damping forces. • A periodic force driving a harmonic oscillator at its natural frequency produces resonance. The system is said to resonate. • The less damping a system has, the higher the amplitude of the forced oscillations near resonance. The more damping a system has, the broader response it has to varying driving frequencies. 16.9 Waves • A wave is a disturbance that moves from the point of creation with a wave velocity w . • A wave has a wavelength , which is the distance between adjacent identical parts of the wave. • Wave velocity and wavelength are related to the wave’s frequency and period by w = or w = . • A transverse wave has a disturbance perpendicular to its direction of propagation, whereas a longitudinal wave has a disturbance parallel to its direction of propagation. 16.10 Superposition and Interference • Superposition is the combination of two waves at the same location. 714 Chapter 16 \| Oscillatory Motion and Waves • Constructive interference occurs when two identical waves are superimposed in phase. • Destructive interference occurs when two identical waves are superimposed exactly out of phase. • A standing wave is one in which two waves superimpose to produce a wave that varies in amplitude but does not propagate. • Nodes are points of no motion in standing waves. • An antinode is the location of maximum amplitude of a standing wave. • Waves on a string are resonant standing waves with a fundamental frequency and can occur at higher multiples of the fundamental, called overtones or harmonics. • Beats occur when waves of similar frequencies 1 and 2 are superimposed. The resulting amplitude oscillates with a beat frequency given by B = ∣ 1 − 2 ∣ . 16.11 Energy in Waves: Intensity Intensity is defined to be the power per unit area: = and has units of W/m2 . Conceptual Questions 16.1 Hooke’s Law: Stress and Strain Revisited 1. Describe a system in which elastic potential energy is stored. 16.3 Simple Harmonic Motion: A Special Periodic Motion 2. What conditions must be met to produce simple harmonic motion? 3. (a) If frequency is not constant for some oscillation, can the oscillation be simple harmonic motion? (b) Can you think of any examples of harmonic motion where the frequency may depend on the amplitude? 4. Give an example of a simple harmonic oscillator, specifically noting how its frequency is independent of amplitude. 5. Explain why you expect an object ma
de of a stiff material to vibrate at a higher frequency than a similar object made of a spongy material. 6. As you pass a freight truck with a trailer on a highway, you notice that its trailer is bouncing up and down slowly. Is it more likely that the trailer is heavily loaded or nearly empty? Explain your answer. 7. Some people modify cars to be much closer to the ground than when manufactured. Should they install stiffer springs? Explain your answer. 16.4 The Simple Pendulum 8. Pendulum clocks are made to run at the correct rate by adjusting the pendulum’s length. Suppose you move from one city to another where the acceleration due to gravity is slightly greater, taking your pendulum clock with you, will you have to lengthen or shorten the pendulum to keep the correct time, other factors remaining constant? Explain your answer. 16.5 Energy and the Simple Harmonic Oscillator 9. Explain in terms of energy how dissipative forces such as friction reduce the amplitude of a harmonic oscillator. Also explain how a driving mechanism can compensate. (A pendulum clock is such a system.) 16.7 Damped Harmonic Motion 10. Give an example of a damped harmonic oscillator. (They are more common than undamped or simple harmonic oscillators.) 11. How would a car bounce after a bump under each of these conditions? • overdamping • underdamping • critical damping 12. Most harmonic oscillators are damped and, if undriven, eventually come to a stop. How is this observation related to the second law of thermodynamics? 16.8 Forced Oscillations and Resonance 13. Why are soldiers in general ordered to “route step” (walk out of step) across a bridge? 16.9 Waves 14. Give one example of a transverse wave and another of a longitudinal wave, being careful to note the relative directions of the disturbance and wave propagation in each. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 16 \| Oscillatory Motion and Waves 715 15. What is the difference between propagation speed and the frequency of a wave? Does one or both affect wavelength? If so, how? 16.10 Superposition and Interference 16. Speakers in stereo systems have two color-coded terminals to indicate how to hook up the wires. If the wires are reversed, the speaker moves in a direction opposite that of a properly connected speaker. Explain why it is important to have both speakers connected the same way. 16.11 Energy in Waves: Intensity 17. Two identical waves undergo pure constructive interference. Is the resultant intensity twice that of the individual waves? Explain your answer. 18. Circular water waves decrease in amplitude as they move away from where a rock is dropped. Explain why. 716 Chapter 16 \| Oscillatory Motion and Waves Problems & Exercises 16.1 Hooke’s Law: Stress and Strain Revisited 1. Fish are hung on a spring scale to determine their mass (most fishermen feel no obligation to truthfully report the mass). (a) What is the force constant of the spring in such a scale if it the spring stretches 8.00 cm for a 10.0 kg load? (b) What is the mass of a fish that stretches the spring 5.50 cm? (c) How far apart are the half-kilogram marks on the scale? 2. It is weigh-in time for the local under-85-kg rugby team. The bathroom scale used to assess eligibility can be described by Hooke’s law and is depressed 0.75 cm by its maximum load of 120 kg. (a) What is the spring’s effective spring constant? (b) A player stands on the scales and depresses it by 0.48 cm. Is he eligible to play on this under-85 kg team? 3. One type of BB gun uses a spring-driven plunger to blow the BB from its barrel. (a) Calculate the force constant of its plunger’s spring if you must compress it 0.150 m to drive the 0.0500-kg plunger to a top speed of 20.0 m/s. (b) What force must be exerted to compress the spring? 4. (a) The springs of a pickup truck act like a single spring with a force constant of 1.30×105 N/m . By how much will the truck be depressed by its maximum load of 1000 kg? (b) If the pickup truck has four identical springs, what is the force constant of each? 5. When an 80.0-kg man stands on a pogo stick, the spring is compressed 0.120 m. (a) What is the force constant of the spring? (b) Will the spring be compressed more when he hops down the road? 6. A spring has a length of 0.200 m when a 0.300-kg mass hangs from it, and a length of 0.750 m when a 1.95-kg mass hangs from it. (a) What is the force constant of the spring? (b) What is the unloaded length of the spring? 16.2 Period and Frequency in Oscillations 7. What is the period of 60.0 Hz electrical power? 8. If your heart rate is 150 beats per minute during strenuous exercise, what is the time per beat in units of seconds? 9. Find the frequency of a tuning fork that takes 2.50×10−3 s to complete one oscillation. 10. A stroboscope is set to flash every 8.00×10−5 s . What is the frequency of the flashes? 11. A tire has a tread pattern with a crevice every 2.00 cm. Each crevice makes a single vibration as the tire moves. What is the frequency of these vibrations if the car moves at 30.0 m/s? 12. Engineering Application Each piston of an engine makes a sharp sound every other revolution of the engine. (a) How fast is a race car going if its eight-cylinder engine emits a sound of frequency 750 Hz, given that the engine makes 2000 revolutions per kilometer? (b) At how many revolutions per minute is the engine rotating? This content is available for free at http://cnx.org/content/col11844/1.13 16.3 Simple Harmonic Motion: A Special Periodic Motion 13. A type of cuckoo clock keeps time by having a mass bouncing on a spring, usually something cute like a cherub in a chair. What force constant is needed to produce a period of 0.500 s for a 0.0150-kg mass? 14. If the spring constant of a simple harmonic oscillator is doubled, by what factor will the mass of the system need to change in order for the frequency of the motion to remain the same? 15. A 0.500-kg mass suspended from a spring oscillates with a period of 1.50 s. How much mass must be added to the object to change the period to 2.00 s? 16. By how much leeway (both percentage and mass) would you have in the selection of the mass of the object in the previous problem if you did not wish the new period to be greater than 2.01 s or less than 1.99 s? 17. Suppose you attach the object with mass to a vertical spring originally at rest, and let it bounce up and down. You release the object from rest at the spring’s original rest length. (a) Show that the spring exerts an upward force of 2.00 on the object at its lowest point. (b) If the spring has a force constant of 10.0 N/m and a 0.25-kg-mass object is set in motion as described, find the amplitude of the oscillations. (c) Find the maximum velocity. 18. A diver on a diving board is undergoing simple harmonic motion. Her mass is 55.0 kg and the period of her motion is 0.800 s. The next diver is a male whose period of simple harmonic oscillation is 1.05 s. What is his mass if the mass of the board is negligible? 19. Suppose a diving board with no one on it bounces up and down in a simple harmonic motion with a frequency of 4.00 Hz. The board has an effective mass of 10.0 kg. What is the frequency of the simple harmonic motion of a 75.0-kg diver on the board? 20. Figure 16.46 This child’s toy relies on springs to keep infants entertained. (credit: By Humboldthead, Flickr) The device pictured in Figure 16.46 entertains infants while keeping them from wandering. The child bounces in a harness suspended from a door frame by a spring constant. Chapter 16 \| Oscillatory Motion and Waves 717 (a) If the spring stretches 0.250 m while supporting an 8.0-kg child, what is its spring constant? (b) What is the effect on the period of a pendulum if you decrease its length by 5.00%? (b) What is the time for one complete bounce of this child? (c) What is the child’s maximum velocity if the amplitude of her bounce is 0.200 m? 21. A 90.0-kg skydiver hanging from a parachute bounces up and down with a period of 1.50 s. What is the new period of oscillation when a second skydiver, whose mass is 60.0 kg, hangs from the legs of the first, as seen in Figure 16.47. Figure 16.47 The oscillations of one skydiver are about to be affected by a second skydiver. (credit: U.S. Army, www.army.mil) 16.4 The Simple Pendulum As usual, the acceleration due to gravity in these problems is taken to be = 9.80 m / s2 , unless otherwise specified. 22. What is the length of a pendulum that has a period of 0.500 s? 23. Some people think a pendulum with a period of 1.00 s can be driven with “mental energy” or psycho kinetically, because its period is the same as an average heartbeat. True or not, what is the length of such a pendulum? 24. What is the period of a 1.00-m-long pendulum? 25. How long does it take a child on a swing to complete one swing if her center of gravity is 4.00 m below the pivot? 26. The pendulum on a cuckoo clock is 5.00 cm long. What is its frequency? 27. Two parakeets sit on a swing with their combined center of mass 10.0 cm below the pivot. At what frequency do they swing? 28. (a) A pendulum that has a period of 3.00000 s and that is located where the acceleration due to gravity is 9.79 m/s2 is moved to a location where it the acceleration due to gravity is 9.82 m/s2 . What is its new period? (b) Explain why so many digits are needed in the value for the period, based on the relation between the period and the acceleration due to gravity. 29. A pendulum with a period of 2.00000 s in one location = 9.80 m/s2 is moved to a new location where the period is now 1.99796 s. What is the acceleration due to gravity at its new location? 30. (a) What is the effect on the period of a pendulum if you double its length? 31. Find the ratio of the new/old periods of a pendulum if the pendulum were transported from Earth to the Moon, where the acceleration due to gravity is 1.63 m/s2 . 32. At what rate will a pendulum clock run on the Moon, where the ac
celeration due to gravity is 1.63 m/s2 , if it keeps time accurately on Earth? That is, find the time (in hours) it takes the clock’s hour hand to make one revolution on the Moon. 33. Suppose the length of a clock’s pendulum is changed by 1.000%, exactly at noon one day. What time will it read 24.00 hours later, assuming it the pendulum has kept perfect time before the change? Note that there are two answers, and perform the calculation to four-digit precision. 34. If a pendulum-driven clock gains 5.00 s/day, what fractional change in pendulum length must be made for it to keep perfect time? 16.5 Energy and the Simple Harmonic Oscillator 35. The length of nylon rope from which a mountain climber is suspended has a force constant of 1.40×104 N/m . (a) What is the frequency at which he bounces, given his mass plus and the mass of his equipment are 90.0 kg? (b) How much would this rope stretch to break the climber’s fall if he free-falls 2.00 m before the rope runs out of slack? Hint: Use conservation of energy. (c) Repeat both parts of this problem in the situation where twice this length of nylon rope is used. 36. Engineering Application Near the top of the Citigroup Center building in New York City, there is an object with mass of 4.00×105 kg on springs that have adjustable force constants. Its function is to dampen wind-driven oscillations of the building by oscillating at the same frequency as the building is being driven—the driving force is transferred to the object, which oscillates instead of the entire building. (a) What effective force constant should the springs have to make the object oscillate with a period of 2.00 s? (b) What energy is stored in the springs for a 2.00-m displacement from equilibrium? 16.6 Uniform Circular Motion and Simple Harmonic Motion 37. (a)What is the maximum velocity of an 85.0-kg person bouncing on a bathroom scale having a force constant of 1.50×106 N/m , if the amplitude of the bounce is 0.200 cm? (b)What is the maximum energy stored in the spring? 38. A novelty clock has a 0.0100-kg mass object bouncing on a spring that has a force constant of 1.25 N/m. What is the maximum velocity of the object if the object bounces 3.00 cm above and below its equilibrium position? (b) How many joules of kinetic energy does the object have at its maximum velocity? 39. At what positions is the speed of a simple harmonic oscillator half its maximum? That is, what values of / give = ±max / 2 , where is the amplitude of the motion? 718 Chapter 16 \| Oscillatory Motion and Waves 40. A ladybug sits 12.0 cm from the center of a Beatles music album spinning at 33.33 rpm. What is the maximum velocity of its shadow on the wall behind the turntable, if illuminated parallel to the record by the parallel rays of the setting Sun? 51. Scouts at a camp shake the rope bridge they have just crossed and observe the wave crests to be 8.00 m apart. If they shake it the bridge twice per second, what is the propagation speed of the waves? 16.7 Damped Harmonic Motion 41. The amplitude of a lightly damped oscillator decreases by 3.0% during each cycle. What percentage of the mechanical energy of the oscillator is lost in each cycle? 16.8 Forced Oscillations and Resonance 42. How much energy must the shock absorbers of a 1200-kg car dissipate in order to damp a bounce that initially has a velocity of 0.800 m/s at the equilibrium position? Assume the car returns to its original vertical position. 43. If a car has a suspension system with a force constant of 5.00×104 N/m , how much energy must the car’s shocks remove to dampen an oscillation starting with a maximum displacement of 0.0750 m? 44. (a) How much will a spring that has a force constant of 40.0 N/m be stretched by an object with a mass of 0.500 kg when hung motionless from the spring? (b) Calculate the decrease in gravitational potential energy of the 0.500-kg object when it descends this distance. (c) Part of this gravitational energy goes into the spring. Calculate the energy stored in the spring by this stretch, and compare it with the gravitational potential energy. Explain where the rest of the energy might go. 45. Suppose you have a 0.750-kg object on a horizontal surface connected to a spring that has a force constant of 150 N/m. There is simple friction between the object and surface with a static coefficient of friction s = 0.100 . (a) How far can the spring be stretched without moving the mass? (b) If the object is set into oscillation with an amplitude twice the distance found in part (a), and the kinetic coefficient of friction is k = 0.0850 , what total distance does it travel before stopping? Assume it starts at the maximum amplitude. 46. Engineering Application: A suspension bridge oscillates with an effective force constant of 1.00×108 N/m . (a) How much energy is needed to make it oscillate with an amplitude of 0.100 m? (b) If soldiers march across the bridge with a cadence equal to the bridge’s natural frequency and impart 1.00×104 J of energy each second, how long does it take for the bridge’s oscillations to go from 0.100 m to 0.500 m amplitude? 16.9 Waves 47. Storms in the South Pacific can create waves that travel all the way to the California coast, which are 12,000 km away. How long does it take them if they travel at 15.0 m/s? 48. Waves on a swimming pool propagate at 0.750 m/s. You splash the water at one end of the pool and observe the wave go to the opposite end, reflect, and return in 30.0 s. How far away is the other end of the pool? 49. Wind gusts create ripples on the ocean that have a wavelength of 5.00 cm and propagate at 2.00 m/s. What is their frequency? 50. How many times a minute does a boat bob up and down on ocean waves that have a wavelength of 40.0 m and a propagation speed of 5.00 m/s? This content is available for free at http://cnx.org/content/col11844/1.13 52. What is the wavelength of the waves you create in a swimming pool if you splash your hand at a rate of 2.00 Hz and the waves propagate at 0.800 m/s? 53. What is the wavelength of an earthquake that shakes you with a frequency of 10.0 Hz and gets to another city 84.0 km away in 12.0 s? 54. Radio waves transmitted through space at 3.00×108 m/s by the Voyager spacecraft have a wavelength of 0.120 m. What is their frequency? 55. Your ear is capable of differentiating sounds that arrive at the ear just 1.00 ms apart. What is the minimum distance between two speakers that produce sounds that arrive at noticeably different times on a day when the speed of sound is 340 m/s? 56. (a) Seismographs measure the arrival times of earthquakes with a precision of 0.100 s. To get the distance to the epicenter of the quake, they compare the arrival times of S- and P-waves, which travel at different speeds. Figure 16.48) If S- and P-waves travel at 4.00 and 7.20 km/s, respectively, in the region considered, how precisely can the distance to the source of the earthquake be determined? (b) Seismic waves from underground detonations of nuclear bombs can be used to locate the test site and detect violations of test bans. Discuss whether your answer to (a) implies a serious limit to such detection. (Note also that the uncertainty is greater if there is an uncertainty in the propagation speeds of the S- and P-waves.) Figure 16.48 A seismograph as described in above problem.(credit: Oleg Alexandrov) 16.10 Superposition and Interference 57. A car has two horns, one emitting a frequency of 199 Hz and the other emitting a frequency of 203 Hz. What beat frequency do they produce? 58. The middle-C hammer of a piano hits two strings, producing beats of 1.50 Hz. One of the strings is tuned to 260.00 Hz. What frequencies could the other string have? 59. Two tuning forks having frequencies of 460 and 464 Hz are struck simultaneously. What average frequency will you hear, and what will the beat frequency be? Chapter 16 \| Oscillatory Motion and Waves 719 (a) What is the intensity in W/m2 of a laser beam used to burn away cancerous tissue that, when 90.0% absorbed, puts 500 J of energy into a circular spot 2.00 mm in diameter in 4.00 s? (b) Discuss how this intensity compares to the average intensity of sunlight (about 700 W/m2 ) and the implications that would have if the laser beam entered your eye. Note how your answer depends on the time duration of the exposure. 60. Twin jet engines on an airplane are producing an average sound frequency of 4100 Hz with a beat frequency of 0.500 Hz. What are their individual frequencies? 61. A wave traveling on a Slinky® that is stretched to 4 m takes 2.4 s to travel the length of the Slinky and back again. (a) What is the speed of the wave? (b) Using the same Slinky stretched to the same length, a standing wave is created which consists of three antinodes and four nodes. At what frequency must the Slinky be oscillating? 62. Three adjacent keys on a piano (F, F-sharp, and G) are struck simultaneously, producing frequencies of 349, 370, and 392 Hz. What beat frequencies are produced by this discordant combination? 16.11 Energy in Waves: Intensity 63. Medical Application Ultrasound of intensity 1.50×102 W/m2 is produced by the rectangular head of a medical imaging device measuring 3.00 by 5.00 cm. What is its power output? 64. The low-frequency speaker of a stereo set has a surface area of 0.05 m2 and produces 1W of acoustical power. What is the intensity at the speaker? If the speaker projects sound uniformly in all directions, at what distance from the speaker is the intensity 0.1 W/m2 ? 65. To increase intensity of a wave by a factor of 50, by what factor should the amplitude be increased? 66. Engineering Application A device called an insolation meter is used to measure the intensity of sunlight has an area of 100 cm2 and registers 6.50 W. What is the intensity in W/m2 ? 67. Astronomy Application Energy from the Sun arrives at the top of the Earth’s atmosphere with an intensity of 1.30 kW/m2. How long does it take for 1.8×109 J to arrive on an area of
1.00 m2 ? 68. Suppose you have a device that extracts energy from ocean breakers in direct proportion to their intensity. If the device produces 10.0 kW of power on a day when the breakers are 1.20 m high, how much will it produce when they are 0.600 m high? 69. Engineering Application (a) A photovoltaic array of (solar cells) is 10.0% efficient in gathering solar energy and converting it to electricity. If the average intensity of sunlight on one day is 700 W/m2, what area should your array have to gather energy at the rate of 100 W? (b) What is the maximum cost of the array if it must pay for itself in two years of operation averaging 10.0 hours per day? Assume that it earns money at the rate of 9.00 ¢ per kilowatt-hour. 70. A microphone receiving a pure sound tone feeds an oscilloscope, producing a wave on its screen. If the sound intensity is originally 2.00×10–5 W/m2, but is turned up until the amplitude increases by 30.0%, what is the new intensity? 71. Medical Application 720 Chapter 16 \| Oscillatory Motion and Waves Test Prep for AP® Courses 16.1 Hooke’s Law: Stress and Strain Revisited 1. Which of the following represents the distance (how much ground the particle covers) moved by a particle in a simple harmonic motion in one time period? (Here, A represents the amplitude of the oscillation.) a. 0 cm b. A cm c. 2A cm d. 4A cm 2. A spring has a spring constant of 80 N·m−1. What is the force required to (a) compress the spring by 5 cm and (b) expand the spring by 15 cm? 3. In the formula = − , what does the minus sign indicate? a. b. c. It indicates that the restoring force is in the direction of the displacement. It indicates that the restoring force is in the direction opposite the displacement. It indicates that mechanical energy in the system decreases when a system undergoes oscillation. d. None of the above 4. The splashing of a liquid resembles an oscillation. The restoring force in this scenario will be due to which of the following? a. Potential energy b. Kinetic energy c. Gravity d. Mechanical energy 16.2 Period and Frequency in Oscillations 5. A mass attached to a spring oscillates and completes 50 full cycles in 30 s. What is the time period and frequency of this system? 16.3 Simple Harmonic Motion: A Special Periodic Motion 6. Use these figures to answer the following questions. Figure 16.49 a. Which of the two pendulums oscillates with larger amplitude? b. Which of the two pendulums oscillates at a higher frequency? 7. A particle of mass 100 g undergoes a simple harmonic motion. The restoring force is provided by a spring with a spring constant of 40 N·m−1. What is the period of oscillation? a. 10 s b. 0.5 s c. 0.1 s d. 1 8. The graph shows the simple harmonic motion of a mass m attached to a spring with spring constant k. Figure 16.50 What is the displacement at time 8π? a. 1 m b. 0 m c. Not defined d. −1 m 9. A pendulum of mass 200 g undergoes simple harmonic motion when acted upon by a force of 15 N. The pendulum crosses the point of equilibrium at a speed of 5 m·s−1. What is the energy of the pendulum at the center of the oscillation? 16.4 The Simple Pendulum 10. A ball is attached to a string of length 4 m to make a pendulum. The pendulum is placed at a location that is away from the Earth’s surface by twice the radius of the Earth. What is the acceleration due to gravity at that height and what is the period of the oscillations? 11. Which of the following gives the correct relation between the acceleration due to gravity and period of a pendulum? a. b. = 2 2 = 4 2 2 c. = 2 = 2 2 d. 12. Tom has two pendulums with him. Pendulum 1 has a ball of mass 0.1 kg attached to it and has a length of 5 m. Pendulum 2 has a ball of mass 0.5 kg attached to a string of length 1 m. How does mass of the ball affect the frequency of the pendulum? Which pendulum will have a higher frequency and why? 16.5 Energy and the Simple Harmonic Oscillator This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 16 \| Oscillatory Motion and Waves 721 13. A mass of 1 kg undergoes simple harmonic motion with amplitude of 1 m. If the period of the oscillation is 1 s, calculate the internal energy of the system. 16.6 Uniform Circular Motion and Simple Harmonic Motion 14. In the equation = sin t, what values can the position take? a. −1 to +1 b. –A to +A c. 0 d. –t to t 16.7 Damped Harmonic Motion 15. The non-conservative damping force removes energy from a system in which form? a. Mechanical energy b. Electrical energy c. Thermal energy d. None of the above 16. The time rate of change of mechanical energy for a damped oscillator is always: a. 0 b. Negative c. Positive d. Undefined 17. A 0.5-kg object is connected to a spring that undergoes oscillatory motion. There is friction between the object and the surface it is kept on given by coefficient of friction = 0.06 . If the object is released 0.2 m from equilibrium, what is the distance that the object travels? Given that the force constant of the spring is 50 N m-1 and the frictional force between the objects is 0.294 N. 16.8 Forced Oscillations and Resonance 18. How is constant amplitude sustained in forced oscillations? 16.9 Waves 19. What is the difference between the waves coming from a tuning fork and electromagnetic waves? 20. Represent longitudinal and transverse waves in a graphical form. 21. Why is the sound produced by a tambourine different from that produced by drums? 22. A transverse wave is traveling left to right. Which of the following is correct about the motion of particles in the wave? a. The particles move up and down when the wave travels in a vacuum. b. The particles move left and right when the wave travels in a medium. c. The particles move up and down when the wave travels in a medium. d. The particles move right and left when the wave travels in a vacuum. 23. Figure 16.51 The graph shows propagation of a mechanical wave. What is the wavelength of this wave? 16.10 Superposition and Interference 24. A guitar string has a number of frequencies at which it vibrates naturally. Which of the following is true in this context? a. The resonant frequencies of the string are integer multiples of fundamental frequencies. b. The resonant frequencies of the string are not integer multiples of fundamental frequencies. c. They have harmonic overtones. d. None of the above 25. Explain the principle of superposition with figures that show the changes in the wave amplitude. 26. In this figure which points represent the points of constructive interference? Figure 16.52 a. A, B, F b. A, B, C, D, E, F c. A, C, D, E d. A, B, D 27. A string is fixed on both sides. It is snapped from both ends at the same time by applying an equal force. What happens to the shape of the waves generated in the string? Also, will you observe an overlap of waves? 28. In the preceding question, what would happen to the amplitude of the waves generated in this way? Also, consider another scenario where the string is snapped up from one end and down from the other end. What will happen in this situation? 29. Two sine waves travel in the same direction in a medium. The amplitude of each wave is A, and the phase difference between the two is 180°. What is the resultant amplitude? a. 2A b. 3A c. 0 d. 9A 30. Standing wave patterns consist of nodes and antinodes formed by repeated interference between two waves of the same frequency traveling in opposite directions. What are nodes and antinodes and how are they produced? 722 Chapter 16 \| Oscillatory Motion and Waves This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 17 \| Physics of Hearing 723 17 PHYSICS OF HEARING Figure 17.1 This tree fell some time ago. When it fell, atoms in the air were disturbed. Physicists would call this disturbance sound whether someone was around to hear it or not. (credit: B.A. Bowen Photography) Chapter Outline 17.1. Sound 17.2. Speed of Sound, Frequency, and Wavelength 17.3. Sound Intensity and Sound Level 17.4. Doppler Effect and Sonic Booms 17.5. Sound Interference and Resonance: Standing Waves in Air Columns 17.6. Hearing 17.7. Ultrasound Connection for AP® Courses In this chapter, the concept of waves is specifically applied to the phenomena of sound. As such, Big Idea 6 continues to be supported, as sound waves carry energy and momentum from one location to another without the permanent transfer of mass. This energy is carried through vibrations caused by disturbances in air pressure (Enduring Understanding 6.A). As air pressure increases, amplitudes of vibration and energy transfer do as well. This idea (Enduring Understanding 6.A.4) explains why a very loud sound can break glass. The chapter continues the fundamental analysis of waves addressed in Chapter 16. Sound waves are periodic, and can therefore be expressed as a function of position and time. Furthermore, sound waves are described by amplitude, frequency, wavelength, and speed (Enduring Understanding 6.B). The relationship between speed and frequency is analyzed further in Section 17.4, as the frequency of sound depends upon the relative motion between the source and observer. This concept, known as the Doppler effect, supports Essential Knowledge 6.B.5. Like all other waves, sound waves can overlap. When they do so, their interaction will produce an amplitude variation within the resultant wave. This amplitude can be determined by adding the displacement of the two pulses, through a process called superposition. This process, covered in Section 17.5, reinforces the content in Enduring Understanding 6.D.1. In situations where the interfering waves are confined, such as on a fixed length of string or in a tube, standing waves can result. These waves are the result of interference between the incident and reflecting wave. Standing waves are described using nodes and antinodes, and their wavelengths are determined by the size of the region to which they are confined. This chapter’s 724 Chapter 17
\| Physics of Hearing description of both standing waves and the concept of beats strongly support Enduring Understanding 6.D, as well as Essential Knowledge 6.D.1, 6.D.3, and 6.D.4. The concepts in this chapter support: Big Idea 6 Waves can transfer energy and momentum from one location to another without the permanent transfer of mass and serve as a mathematical model for the description of other phenomena. Enduring Understanding 6.B A periodic wave is one that repeats as a function of both time and position and can be described by its amplitude, frequency, wavelength, speed, and energy. Essential Knowledge 6.B.5 The observed frequency of a wave depends on the relative motion of the source and the observer. This is a qualitative measurement only. Enduring Understanding 6.D Interference and superposition lead to standing waves and beats. Essential Knowledge 6.D.1 Two or more wave pulses can interact in such a way as to produce amplitude variations in the resultant wave. When two pulses cross, they travel through each other; they do not bounce off each other. Where the pulses overlap, the resulting displacement can be determined by adding the displacements of the two pulses. This is called superposition. Essential Knowledge 6.D.3 Standing waves are the result of the addition of incident and reflected waves that are conﬁned to a region and have nodes and antinodes. Examples should include waves on a fixed length of string, and sound waves in both closed and open tubes. Essential Knowledge 6.D.4 The possible wavelengths of a standing wave are determined by the size of the region in which it is conﬁned. 17.1 Sound Learning Objectives By the end of this section, you will be able to: • Define sound and hearing. • Describe sound as a longitudinal wave. Figure 17.2 This glass has been shattered by a high-intensity sound wave of the same frequency as the resonant frequency of the glass. While the sound is not visible, the effects of the sound prove its existence. (credit: \|\|read\|\|, Flickr) Sound can be used as a familiar illustration of waves. Because hearing is one of our most important senses, it is interesting to see how the physical properties of sound correspond to our perceptions of it. Hearing is the perception of sound, just as vision is the perception of visible light. But sound has important applications beyond hearing. Ultrasound, for example, is not heard but can be employed to form medical images and is also used in treatment. The physical phenomenon of sound is defined to be a disturbance of matter that is transmitted from its source outward. Sound is a wave. On the atomic scale, it is a disturbance of atoms that is far more ordered than their thermal motions. In many instances, sound is a periodic wave, and the atoms undergo simple harmonic motion. In this text, we shall explore such periodic sound waves. A vibrating string produces a sound wave as illustrated in Figure 17.3, Figure 17.4, and Figure 17.5. As the string oscillates back and forth, it transfers energy to the air, mostly as thermal energy created by turbulence. But a small part of the string’s energy goes into compressing and expanding the surrounding air, creating slightly higher and lower local pressures. These compressions (high pressure regions) and rarefactions (low pressure regions) move out as longitudinal pressure waves having the same frequency as the string—they are the disturbance that is a sound wave. (Sound waves in air and most fluids are longitudinal, because fluids have almost no shear strength. In solids, sound waves can be both transverse and longitudinal.) Figure 17.5 shows a graph of gauge pressure versus distance from the vibrating string. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 17 \| Physics of Hearing 725 Figure 17.3 A vibrating string moving to the right compresses the air in front of it and expands the air behind it. Figure 17.4 As the string moves to the left, it creates another compression and rarefaction as the ones on the right move away from the string. Figure 17.5 After many vibrations, there are a series of compressions and rarefactions moving out from the string as a sound wave. The graph shows gauge pressure versus distance from the source. Pressures vary only slightly from atmospheric for ordinary sounds. The amplitude of a sound wave decreases with distance from its source, because the energy of the wave is spread over a larger and larger area. But it is also absorbed by objects, such as the eardrum in Figure 17.6, and converted to thermal energy by the viscosity of air. In addition, during each compression a little heat transfers to the air and during each rarefaction even less heat transfers from the air, so that the heat transfer reduces the organized disturbance into random thermal motions. (These processes can be viewed as a manifestation of the second law of thermodynamics presented in Introduction to the Second Law of Thermodynamics: Heat Engines and Their Efficiency.) Whether the heat transfer from compression to rarefaction is significant depends on how far apart they are—that is, it depends on wavelength. Wavelength, frequency, amplitude, and speed of propagation are important for sound, as they are for all waves. Figure 17.6 Sound wave compressions and rarefactions travel up the ear canal and force the eardrum to vibrate. There is a net force on the eardrum, since the sound wave pressures differ from the atmospheric pressure found behind the eardrum. A complicated mechanism converts the vibrations to nerve impulses, which are perceived by the person. 726 Chapter 17 \| Physics of Hearing PhET Explorations: Wave Interference Make waves with a dripping faucet, audio speaker, or laser! Add a second source or a pair of slits to create an interference pattern. Figure 17.7 Wave Interference (http://cnx.org/content/m55288/1.2/wave-interference_en.jar) 17.2 Speed of Sound, Frequency, and Wavelength Learning Objectives By the end of this section, you will be able to: • Define pitch. • Describe the relationship between the speed of sound, its frequency, and its wavelength. • Describe the effects on the speed of sound as it travels through various media. • Describe the effects of temperature on the speed of sound. The information presented in this section supports the following AP® learning objectives and science practices: • 6.B.4.1 The student is able to design an experiment to determine the relationship between periodic wave speed, wavelength, and frequency, and relate these concepts to everyday examples. (S.P. 4.2, 5.1, 7.2) Figure 17.8 When a firework explodes, the light energy is perceived before the sound energy. Sound travels more slowly than light does. (credit: Dominic Alves, Flickr) Sound, like all waves, travels at a certain speed and has the properties of frequency and wavelength. You can observe direct evidence of the speed of sound while watching a fireworks display. The flash of an explosion is seen well before its sound is heard, implying both that sound travels at a finite speed and that it is much slower than light. You can also directly sense the frequency of a sound. Perception of frequency is called pitch. The wavelength of sound is not directly sensed, but indirect evidence is found in the correlation of the size of musical instruments with their pitch. Small instruments, such as a piccolo, typically make high-pitch sounds, while large instruments, such as a tuba, typically make low-pitch sounds. High pitch means small wavelength, and the size of a musical instrument is directly related to the wavelengths of sound it produces. So a small instrument creates short-wavelength sounds. Similar arguments hold that a large instrument creates long-wavelength sounds. The relationship of the speed of sound, its frequency, and wavelength is the same as for all waves: w = (17.1) where w is the speed of sound, between adjacent identical parts of a wave—for example, between adjacent compressions as illustrated in Figure 17.9. The frequency is the same as that of the source and is the number of waves that pass a point per unit time. is its frequency, and is its wavelength. The wavelength of a sound is the distance This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 17 \| Physics of Hearing 727 Figure 17.9 A sound wave emanates from a source vibrating at a frequency , propagates at w , and has a wavelength . Table 17.4 makes it apparent that the speed of sound varies greatly in different media. The speed of sound in a medium is determined by a combination of the medium’s rigidity (or compressibility in gases) and its density. The more rigid (or less compressible) the medium, the faster the speed of sound. This observation is analogous to the fact that the frequency of a simple harmonic motion is directly proportional to the stiffness of the oscillating object. The greater the density of a medium, the slower the speed of sound. This observation is analogous to the fact that the frequency of a simple harmonic motion is inversely proportional to the mass of the oscillating object. The speed of sound in air is low, because air is compressible. Because liquids and solids are relatively rigid and very difficult to compress, the speed of sound in such media is generally greater than in gases. Applying the Science Practices: Bottle Music When liquid is poured into a small-necked container like a soda bottle, it can make for a fun musical experience! Find a small-necked bottle and pour water into it. When you blow across the surface of the bottle, a musical pitch should be created. This pitch, which corresponds to the resonant frequency of the air remaining in the bottle, can be determined using Equation 17.1. Your task is to design an experiment and collect data to confirm this relationship between the frequency created by blowing into the bottle and the depth of air remaining. 1. Use the explanation above to design an experiment that w
ill yield data on depth of air column and frequency of pitch. Use the data table below to record your data. Table 17.1 Depth of air column (λ) Frequency of pitch generated (f) 2. Construct a graph using the information collected above. The graph should include all five data points and should display frequency on the dependent axis. 3. What type of relationship is displayed on your graph? (direct, inverse, quadratic, etc.) 4. Does your graph align with equation 17.1, given earlier in this section? Explain. Note: For an explanation of why a frequency is created when you blow across a small-necked container, explore Section 17.5 later in this chapter. Answer 1. As the depth of the air column increases, the frequency values must decrease. A sample set of data is displayed below. 728 Chapter 17 \| Physics of Hearing Table 17.2 Depth of air column (λ) Frequency of pitch generated (f) 24 cm 22 cm 20 cm 18 cm 16 cm 689.6 Hz 752.3 Hz 827.5 Hz 919.4 Hz 1034.4 Hz 2. The graph drawn should have frequency on the vertical axis, contain five data points, and trend downward and to the right. A graph using the sample data from above is displayed below. Figure 17.10 A graph of the depth of air column versus the frequency of pitch generated. 3. Inverse relationship. Table 17.3 Depth of air column (λ) Frequency of pitch generated (f) Product of wavelength and frequency 24 cm 22 cm 20 cm 18 cm 16 cm 689.6 Hz 752.3 Hz 827.5 Hz 919.4 Hz 1034.4 Hz 165.5 165.5 165.5 165.5 165.5 4. The graph does align with the equation v = f λ. As the wavelength decreases, the frequency of the pitch generated increases. This relationship is validated by both the sample data table and the sample graph. Additionally, as Table 17.1 demonstrates, the product of λ and f is constant across all five data points. In addition to these explanations, the student may use the formula as given in the problem statement to show that the product f × air column height is consistently 165.5. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 17 \| Physics of Hearing 729 Table 17.4 Speed of Sound in Various Media Medium vw(m/s) Gases at 0ºC Air Carbon dioxide Oxygen Helium 331 259 316 965 Hydrogen 1290 Liquids at 20ºC Ethanol Mercury Water, fresh Sea water Human tissue 1160 1450 1480 1540 1540 Solids (longitudinal or bulk) Vulcanized rubber 54 Polyethylene Marble Glass, Pyrex Lead Aluminum Steel 920 3810 5640 1960 5120 5960 Earthquakes, essentially sound waves in Earth’s crust, are an interesting example of how the speed of sound depends on the rigidity of the medium. Earthquakes have both longitudinal and transverse components, and these travel at different speeds. The bulk modulus of granite is greater than its shear modulus. For that reason, the speed of longitudinal or pressure waves (Pwaves) in earthquakes in granite is significantly higher than the speed of transverse or shear waves (S-waves). Both components of earthquakes travel slower in less rigid material, such as sediments. P-waves have speeds of 4 to 7 km/s, and S-waves correspondingly range in speed from 2 to 5 km/s, both being faster in more rigid material. The P-wave gets progressively farther ahead of the S-wave as they travel through Earth’s crust. The time between the P- and S-waves is routinely used to determine the distance to their source, the epicenter of the earthquake. The speed of sound is affected by temperature in a given medium. For air at sea level, the speed of sound is given by where the temperature (denoted as ) is in units of kelvin. The speed of sound in gases is related to the average speed of particles in the gas, rms , and that w = (331 m/s) 273 K , rms = 3 , (17.2) (17.3) where is the Boltzmann constant ( 1.38×10−23 J/K ) and is the mass of each (identical) particle in the gas. So, it is reasonable that the speed of sound in air and other gases should depend on the square root of temperature. While not negligible, this is not a strong dependence. At 0ºC , the speed of sound is 331 m/s, whereas at 20.0ºC it is 343 m/s, less than a 4% increase. Figure 17.11 shows a use of the speed of sound by a bat to sense distances. Echoes are also used in medical imaging. 730 Chapter 17 \| Physics of Hearing Figure 17.11 A bat uses sound echoes to find its way about and to catch prey. The time for the echo to return is directly proportional to the distance. One of the more important properties of sound is that its speed is nearly independent of frequency. This independence is certainly true in open air for sounds in the audible range of 20 to 20,000 Hz. If this independence were not true, you would certainly notice it for music played by a marching band in a football stadium, for example. Suppose that high-frequency sounds traveled faster—then the farther you were from the band, the more the sound from the low-pitch instruments would lag that from the high-pitch ones. But the music from all instruments arrives in cadence independent of distance, and so all frequencies must travel at nearly the same speed. Recall that w = (17.4) In a given medium under fixed conditions, w is constant, so that there is a relationship between and ; the higher the frequency, the smaller the wavelength. See Figure 17.12 and consider the following example. Figure 17.12 Because they travel at the same speed in a given medium, low-frequency sounds must have a greater wavelength than high-frequency sounds. Here, the lower-frequency sounds are emitted by the large speaker, called a woofer, while the higher-frequency sounds are emitted by the small speaker, called a tweeter. Example 17.1 Calculating Wavelengths: What Are the Wavelengths of Audible Sounds? Calculate the wavelengths of sounds at the extremes of the audible range, 20 and 20,000 Hz, in 30.0ºC air. (Assume that the frequency values are accurate to two significant figures.) Strategy To find wavelength from frequency, we can use w = . Solution 1. Identify knowns. The value for w , is given by w = (331 m/s) 273 K . 2. Convert the temperature into kelvin and then enter the temperature into the equation w = (331 m/s) 303 K 273 K 3. Solve the relationship between speed and wavelength for : = w . = 348.7 m/s. 4. Enter the speed and the minimum frequency to give the maximum wavelength: max = 348.7 m/s 20 Hz = 17 m. This content is available for free at http://cnx.org/content/col11844/1.13 (17.5) (17.6) (17.7) (17.8) Chapter 17 \| Physics of Hearing 5. Enter the speed and the maximum frequency to give the minimum wavelength: min = 348.7 m/s 20,000 Hz = 0.017 m = 1.7 cm. Discussion Because the product of multiplied by equals a constant, the smaller is, the larger must be, and vice versa. 731 (17.9) The speed of sound can change when sound travels from one medium to another. However, the frequency usually remains the same because it is like a driven oscillation and has the frequency of the original source. If w changes and same, then the wavelength must change. That is, because w = , the higher the speed of a sound, the greater its wavelength for a given frequency. remains the Making Connections: Take-Home Investigation—Voice as a Sound Wave Suspend a sheet of paper so that the top edge of the paper is fixed and the bottom edge is free to move. You could tape the top edge of the paper to the edge of a table. Gently blow near the edge of the bottom of the sheet and note how the sheet moves. Speak softly and then louder such that the sounds hit the edge of the bottom of the paper, and note how the sheet moves. Explain the effects. Check Your Understanding Imagine you observe two fireworks explode. You hear the explosion of one as soon as you see it. However, you see the other firework for several milliseconds before you hear the explosion. Explain why this is so. Solution Sound and light both travel at definite speeds. The speed of sound is slower than the speed of light. The first firework is probably very close by, so the speed difference is not noticeable. The second firework is farther away, so the light arrives at your eyes noticeably sooner than the sound wave arrives at your ears. Check Your Understanding You observe two musical instruments that you cannot identify. One plays high-pitch sounds and the other plays low-pitch sounds. How could you determine which is which without hearing either of them play? Solution Compare their sizes. High-pitch instruments are generally smaller than low-pitch instruments because they generate a smaller wavelength. 17.3 Sound Intensity and Sound Level Learning Objectives By the end of this section, you will be able to: • Define intensity, sound intensity, and sound pressure level. • Calculate sound intensity levels in decibels (dB). The information presented in this section supports the following AP® learning objectives and science practices: • 6.A.4.1 The student is able to explain and/or predict qualitatively how the energy carried by a sound wave relates to the amplitude of the wave, and/or apply this concept to a real-world example. (S.P. 6.4) 732 Chapter 17 \| Physics of Hearing Figure 17.13 Noise on crowded roadways like this one in Delhi makes it hard to hear others unless they shout. (credit: Lingaraj G J, Flickr) In a quiet forest, you can sometimes hear a single leaf fall to the ground. After settling into bed, you may hear your blood pulsing through your ears. But when a passing motorist has his stereo turned up, you cannot even hear what the person next to you in your car is saying. We are all very familiar with the loudness of sounds and aware that they are related to how energetically the source is vibrating. In cartoons depicting a screaming person (or an animal making a loud noise), the cartoonist often shows an open mouth with a vibrating uvula, the hanging tissue at the back of the mouth, to suggest a loud sound coming from the throat Figure 17.14. High noise exposure is hazardous to hearing, and it is common for musicians to have hearing losses that are sufficiently severe that they interf
ere with the musicians’ abilities to perform. The relevant physical quantity is sound intensity, a concept that is valid for all sounds whether or not they are in the audible range. Intensity is defined to be the power per unit area carried by a wave. Power is the rate at which energy is transferred by the wave. In equation form, intensity is = where is the power through an area . The SI unit for is W/m2 . The intensity of a sound wave is related to its amplitude squared by the following relationship: , (17.10) = 2 Δ 2w . (17.11) Here Δ is the pressure variation or pressure amplitude (half the difference between the maximum and minimum pressure in the sound wave) in units of pascals (Pa) or N/m2 . (We are using a lower case for pressure to distinguish it from power, denoted by above.) The energy (as kinetic energy 2 2 proportional to its amplitude squared. In this equation, is the density of the material in which the sound wave travels, in units of kg/m3 , and w is the speed of sound in the medium, in units of m/s. The pressure variation is proportional to the amplitude of the oscillation, and so varies as (Δ)2 (Figure 17.14). This relationship is consistent with the fact that the sound wave is produced by some vibration; the greater its pressure amplitude, the more the air is compressed in the sound it creates. ) of an oscillating element of air due to a traveling sound wave is Figure 17.14 Graphs of the gauge pressures in two sound waves of different intensities. The more intense sound is produced by a source that has larger-amplitude oscillations and has greater pressure maxima and minima. Because pressures are higher in the greater-intensity sound, it can exert larger forces on the objects it encounters. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 17 \| Physics of Hearing 733 Sound intensity levels are quoted in decibels (dB) much more often than sound intensities in watts per meter squared. Decibels are the unit of choice in the scientific literature as well as in the popular media. The reasons for this choice of units are related to how we perceive sounds. How our ears perceive sound can be more accurately described by the logarithm of the intensity rather than directly to the intensity. The sound intensity level in decibels of a sound having an intensity in watts per meter squared is defined to be (dB) = 10 log10 0 , (17.12) where 0 = 10–12 W/m2 is a reference intensity. In particular, 0 is the lowest or threshold intensity of sound a person with normal hearing can perceive at a frequency of 1000 Hz. Sound intensity level is not the same as intensity. Because is defined in terms of a ratio, it is a unitless quantity telling you the level of the sound relative to a fixed standard ( 10–12 W/m2 , in this case). The units of decibels (dB) are used to indicate this ratio is multiplied by 10 in its definition. The bel, upon which the decibel is based, is named for Alexander Graham Bell, the inventor of the telephone. Table 17.5 Sound Intensity Levels and Intensities Sound intensity level β (dB) Intensity I(W/m2) Example/effect 0 10 20 30 40 50 60 70 80 90 100 110 120 140 160 1×10–12 1×10–11 1×10–10 1×10–9 1×10–8 1×10–7 1×10–6 1×10–5 1×10–4 1×10–3 1×10–2 1×10–1 1 1×102 1×104 Threshold of hearing at 1000 Hz Rustle of leaves Whisper at 1 m distance Quiet home Average home Average office, soft music Normal conversation Noisy office, busy traffic Loud radio, classroom lecture Inside a heavy truck; damage from prolonged exposure[1] Noisy factory, siren at 30 m; damage from 8 h per day exposure Damage from 30 min per day exposure Loud rock concert, pneumatic chipper at 2 m; threshold of pain Jet airplane at 30 m; severe pain, damage in seconds Bursting of eardrums The decibel level of a sound having the threshold intensity of 10 – 12 W/m2 is = 0 dB , because log10 1 = 0 . That is, the threshold of hearing is 0 decibels. Table 17.5 gives levels in decibels and intensities in watts per meter squared for some familiar sounds. One of the more striking things about the intensities in Table 17.5 is that the intensity in watts per meter squared is quite small for most sounds. The ear is sensitive to as little as a trillionth of a watt per meter squared—even more impressive when you realize that the area of the eardrum is only about 1 cm2 , so that only 10 – 16 W falls on it at the threshold of hearing! Air molecules in a sound wave of this intensity vibrate over a distance of less than one molecular diameter, and the gauge pressures involved are less than 10 – 9 atm. 1. Several government agencies and health-related professional associations recommend that 85 dB not be exceeded for 8-hour daily exposures in the absence of hearing protection. 734 Chapter 17 \| Physics of Hearing Another impressive feature of the sounds in Table 17.5 is their numerical range. Sound intensity varies by a factor of 1012 from threshold to a sound that causes damage in seconds. You are unaware of this tremendous range in sound intensity because how your ears respond can be described approximately as the logarithm of intensity. Thus, sound intensity levels in decibels fit your experience better than intensities in watts per meter squared. The decibel scale is also easier to relate to because most people are more accustomed to dealing with numbers such as 0, 53, or 120 than numbers such as 1.00×10 – 11 . One more observation readily verified by examining Table 17.5 or using = 2 is that each factor of 10 in intensity Δ 2w corresponds to 10 dB. For example, a 90 dB sound compared with a 60 dB sound is 30 dB greater, or three factors of 10 (that is, 103 times) as intense. Another example is that if one sound is 107 as intense as another, it is 70 dB higher. See Table 17.6. Table 17.6 Ratios of Intensities and Corresponding Differences in Sound Intensity Levels β2 – β1 I2 / I1 2.0 5.0 10.0 3.0 dB 7.0 dB 10.0 dB Example 17.2 Calculating Sound Intensity Levels: Sound Waves Calculate the sound intensity level in decibels for a sound wave traveling in air at 0ºC and having a pressure amplitude of 0.656 Pa. Strategy We are given Δ , so we can calculate using the equation = from its definition in (dB) = 10 log10 / 0 . Δ Solution (1) Identify knowns: Sound travels at 331 m/s in air at 0ºC . Air has a density of 1.29 kg/m3 at atmospheric pressure and 0ºC . 2 / 2w 2 . Using , we can calculate straight (2) Enter these values and the pressure amplitude into = Δ 2 / 2w : = 2 Δ 2w = (0.656 Pa)2 1.29 kg/m3 2 (331 m/s) = 5.04×10−4 W/m2. (17.13) (3) Enter the value for and the known value for 0 into (dB) = 10 log10 level in decibels: / 0 . Calculate to find the sound intensity Discussion 10 log10 5.04×108 = 10 8.70 dB = 87 dB. (17.14) This 87 dB sound has an intensity five times as great as an 80 dB sound. So a factor of five in intensity corresponds to a difference of 7 dB in sound intensity level. This value is true for any intensities differing by a factor of five. Example 17.3 Change Intensity Levels of a Sound: What Happens to the Decibel Level? Show that if one sound is twice as intense as another, it has a sound level about 3 dB higher. Strategy You are given that the ratio of two intensities is 2 to 1, and are then asked to find the difference in their sound levels in decibels. You can solve this problem using of the properties of logarithms. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 17 \| Physics of Hearing 735 Solution (1) Identify knowns: The ratio of the two intensities is 2 to 1, or: 2 1 = 2.00. We wish to show that the difference in sound levels is about 3 dB. That is, we want to show: Note that: (2) Use the definition of to get: 2 − 1 = 3 dB. log10 − log10 = log10 . 2 − 1 = 10 log10 2 1 = 10 log10 2.00 = 10 (0.301) dB. 2 − 1 = 3.01 dB. Thus, Discussion (17.15) (17.16) (17.17) (17.18) (17.19) This means that the two sound intensity levels differ by 3.01 dB, or about 3 dB, as advertised. Note that because only the ratio 2 / 1 is given (and not the actual intensities), this result is true for any intensities that differ by a factor of two. For example, a 56.0 dB sound is twice as intense as a 53.0 dB sound, a 97.0 dB sound is half as intense as a 100 dB sound, and so on. It should be noted at this point that there is another decibel scale in use, called the sound pressure level, based on the ratio of the pressure amplitude to a reference pressure. This scale is used particularly in applications where sound travels in water. It is beyond the scope of most introductory texts to treat this scale because it is not commonly used for sounds in air, but it is important to note that very different decibel levels may be encountered when sound pressure levels are quoted. For example, ocean noise pollution produced by ships may be as great as 200 dB expressed in the sound pressure level, where the more familiar sound intensity level we use here would be something under 140 dB for the same sound. Take-Home Investigation: Feeling Sound Find a CD player and a CD that has rock music. Place the player on a light table, insert the CD into the player, and start playing the CD. Place your hand gently on the table next to the speakers. Increase the volume and note the level when the table just begins to vibrate as the rock music plays. Increase the reading on the volume control until it doubles. What has happened to the vibrations? Check Your Understanding Describe how amplitude is related to the loudness of a sound. Solution Amplitude is directly proportional to the experience of loudness. As amplitude increases, loudness increases. Check Your Understanding Identify common sounds at the levels of 10 dB, 50 dB, and 100 dB. Solution 10 dB: Running fingers through your hair. 50 dB: Inside a quiet home with no television or radio. 100 dB: Take-off of a jet plane. 736 Chapter 17 \| Physics of Hearing 17.4 Doppler Effect and Sonic Booms Learning Objectives By the end of this section, you will be able to:
• Define Doppler effect, Doppler shift, and sonic boom. • Calculate the frequency of a sound heard by someone observing Doppler shift. • Describe the sounds produced by objects moving faster than the speed of sound. The information presented in this section supports the following AP® learning objectives and science practices: • 6.B.5.1 The student is able to create or use a wave front diagram to demonstrate or interpret qualitatively the observed frequency of a wave, dependent upon relative motions of source and observer. (S.P. 1.4) The characteristic sound of a motorcycle buzzing by is an example of the Doppler effect. The high-pitch scream shifts dramatically to a lower-pitch roar as the motorcycle passes by a stationary observer. The closer the motorcycle brushes by, the more abrupt the shift. The faster the motorcycle moves, the greater the shift. We also hear this characteristic shift in frequency for passing race cars, airplanes, and trains. It is so familiar that it is used to imply motion and children often mimic it in play. The Doppler effect is an alteration in the observed frequency of a sound due to motion of either the source or the observer. Although less familiar, this effect is easily noticed for a stationary source and moving observer. For example, if you ride a train past a stationary warning bell, you will hear the bell’s frequency shift from high to low as you pass by. The actual change in frequency due to relative motion of source and observer is called a Doppler shift. The Doppler effect and Doppler shift are named for the Austrian physicist and mathematician Christian Johann Doppler (1803–1853), who did experiments with both moving sources and moving observers. Doppler, for example, had musicians play on a moving open train car and also play standing next to the train tracks as a train passed by. Their music was observed both on and off the train, and changes in frequency were measured. What causes the Doppler shift? Figure 17.15, Figure 17.16, and Figure 17.17 compare sound waves emitted by stationary and moving sources in a stationary air mass. Each disturbance spreads out spherically from the point where the sound was emitted. If the source is stationary, then all of the spheres representing the air compressions in the sound wave centered on the same point, and the stationary observers on either side see the same wavelength and frequency as emitted by the source, as in Figure 17.15. If the source is moving, as in Figure 17.16, then the situation is different. Each compression of the air moves out in a sphere from the point where it was emitted, but the point of emission moves. This moving emission point causes the air compressions to be closer together on one side and farther apart on the other. Thus, the wavelength is shorter in the direction the source is moving (on the right in Figure 17.16), and longer in the opposite direction (on the left in Figure 17.16). Finally, if the observers move, as in Figure 17.17, the frequency at which they receive the compressions changes. The observer moving toward the source receives them at a higher frequency, and the person moving away from the source receives them at a lower frequency. Figure 17.15 Sounds emitted by a source spread out in spherical waves. Because the source, observers, and air are stationary, the wavelength and frequency are the same in all directions and to all observers. Figure 17.16 Sounds emitted by a source moving to the right spread out from the points at which they were emitted. The wavelength is reduced and, consequently, the frequency is increased in the direction of motion, so that the observer on the right hears a higher-pitch sound. The opposite is true for the observer on the left, where the wavelength is increased and the frequency is reduced. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 17 \| Physics of Hearing 737 Figure 17.17 The same effect is produced when the observers move relative to the source. Motion toward the source increases frequency as the observer on the right passes through more wave crests than she would if stationary. Motion away from the source decreases frequency as the observer on the left passes through fewer wave crests than he would if stationary. We know that wavelength and frequency are related by w = , where w is the fixed speed of sound. The sound moves in a medium and has the same speed w in that medium whether the source is moving or not. Thus multiplied by is a constant. Because the observer on the right in Figure 17.16 receives a shorter wavelength, the frequency she receives must be higher. Similarly, the observer on the left receives a longer wavelength, and hence he hears a lower frequency. The same thing happens in Figure 17.17. A higher frequency is received by the observer moving toward the source, and a lower frequency is received by an observer moving away from the source. In general, then, relative motion of source and observer toward one another increases the received frequency. Relative motion apart decreases frequency. The greater the relative speed is, the greater the effect. The Doppler Effect The Doppler effect occurs not only for sound but for any wave when there is relative motion between the observer and the source. There are Doppler shifts in the frequency of sound, light, and water waves, for example. Doppler shifts can be used to determine velocity, such as when ultrasound is reflected from blood in a medical diagnostic. The recession of galaxies is determined by the shift in the frequencies of light received from them and has implied much about the origins of the universe. Modern physics has been profoundly affected by observations of Doppler shifts. For a stationary observer and a moving source, the frequency fobs received by the observer can be shown to be obs = s w w ± s , (17.20) where s is the frequency of the source, s is the speed of the source along a line joining the source and observer, and w is the speed of sound. The minus sign is used for motion toward the observer and the plus sign for motion away from the observer, producing the appropriate shifts up and down in frequency. Note that the greater the speed of the source, the greater the effect. Similarly, for a stationary source and moving observer, the frequency received by the observer obs is given by obs = s w ± obs w , (17.21) where obs is the speed of the observer along a line joining the source and observer. Here the plus sign is for motion toward the source, and the minus is for motion away from the source. Example 17.4 Calculate Doppler Shift: A Train Horn Suppose a train that has a 150-Hz horn is moving at 35.0 m/s in still air on a day when the speed of sound is 340 m/s. (a) What frequencies are observed by a stationary person at the side of the tracks as the train approaches and after it passes? (b) What frequency is observed by the train’s engineer traveling on the train? Strategy To find the observed frequency in (a), obs = s w w ± s , must be used because the source is moving. The minus sign is used for the approaching train, and the plus sign for the receding train. In (b), there are two Doppler shifts—one for a moving source and the other for a moving observer. Solution for (a) 738 Chapter 17 \| Physics of Hearing (1) Enter known values into obs = s w w – s . obs = s w w − s = (150 Hz) 340 m/s 340 m/s – 35.0 m/s (2) Calculate the frequency observed by a stationary person as the train approaches. obs = (150 Hz)(1.11) = 167 Hz (3) Use the same equation with the plus sign to find the frequency heard by a stationary person as the train recedes. obs = s w w + s = (150 Hz) 340 m/s 340 m/s + 35.0 m/s (4) Calculate the second frequency. Discussion on (a) obs = (150 Hz)(0.907) = 136 Hz (17.22) (17.23) (17.24) (17.25) The numbers calculated are valid when the train is far enough away that the motion is nearly along the line joining train and observer. In both cases, the shift is significant and easily noticed. Note that the shift is 17.0 Hz for motion toward and 14.0 Hz for motion away. The shifts are not symmetric. Solution for (b) (1) Identify knowns: • It seems reasonable that the engineer would receive the same frequency as emitted by the horn, because the relative velocity between them is zero. • Relative to the medium (air), the speeds are s = obs = 35.0 m/s. • The first Doppler shift is for the moving observer; the second is for the moving source. (2) Use the following equation: obs = s w ± obs w w w ± s . (17.26) The quantity in the square brackets is the Doppler-shifted frequency due to a moving observer. The factor on the right is the effect of the moving source. (3) Because the train engineer is moving in the direction toward the horn, we must use the plus sign for obs; however, because the horn is also moving in the direction away from the engineer, we also use the plus sign for s . But the train is carrying both the engineer and the horn at the same velocity, so s = obs . As a result, everything but s cancels, yielding obs = s. (17.27) Discussion for (b) We may expect that there is no change in frequency when source and observer move together because it fits your experience. For example, there is no Doppler shift in the frequency of conversations between driver and passenger on a motorcycle. People talking when a wind moves the air between them also observe no Doppler shift in their conversation. The crucial point is that source and observer are not moving relative to each other. Sonic Booms to Bow Wakes What happens to the sound produced by a moving source, such as a jet airplane, that approaches or even exceeds the speed of sound? The answer to this question applies not only to sound but to all other waves as well. Suppose a jet airplane is coming nearly straight at you, emitting a sound of frequency s . The greater the plane’s speed s , the greater the Doppler shift and the greater the value observed for approaches infinity, because the de
nominator in obs = s obs . Now, as s approaches the speed of sound, w w ± s approaches zero. At the speed of sound, this result obs means that in front of the source, each successive wave is superimposed on the previous one because the source moves forward at the speed of sound. The observer gets them all at the same instant, and so the frequency is infinite. (Before airplanes exceeded the speed of sound, some people argued it would be impossible because such constructive superposition would produce pressures great enough to destroy the airplane.) If the source exceeds the speed of sound, no sound is received by the observer until the source has passed, so that the sounds from the approaching source are mixed with those from it when receding. This mixing appears messy, but something interesting happens—a sonic boom is created. (See Figure 17.18.) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 17 \| Physics of Hearing 739 Figure 17.18 Sound waves from a source that moves faster than the speed of sound spread spherically from the point where they are emitted, but the source moves ahead of each. Constructive interference along the lines shown (actually a cone in three dimensions) creates a shock wave called a sonic boom. The faster the speed of the source, the smaller the angle . There is constructive interference along the lines shown (a cone in three dimensions) from similar sound waves arriving there simultaneously. This superposition forms a disturbance called a sonic boom, a constructive interference of sound created by an object moving faster than sound. Inside the cone, the interference is mostly destructive, and so the sound intensity there is much less than on the shock wave. An aircraft creates two sonic booms, one from its nose and one from its tail. (See Figure 17.19.) During television coverage of space shuttle landings, two distinct booms could often be heard. These were separated by exactly the time it would take the shuttle to pass by a point. Observers on the ground often do not see the aircraft creating the sonic boom, because it has passed by before the shock wave reaches them, as seen in Figure 17.19. If the aircraft flies close by at low altitude, pressures in the sonic boom can be destructive and break windows as well as rattle nerves. Because of how destructive sonic booms can be, supersonic flights are banned over populated areas of the United States. Figure 17.19 Two sonic booms, created by the nose and tail of an aircraft, are observed on the ground after the plane has passed by. Sonic booms are one example of a broader phenomenon called bow wakes. A bow wake, such as the one in Figure 17.20, is created when the wave source moves faster than the wave propagation speed. Water waves spread out in circles from the point where created, and the bow wake is the familiar V-shaped wake trailing the source. A more exotic bow wake is created when a subatomic particle travels through a medium faster than the speed of light travels in that medium. (In a vacuum, the maximum speed of light will be = 3.00×108 m/s ; in the medium of water, the speed of light is closer to 0.75 . If the particle creates light in its passage, that light spreads on a cone with an angle indicative of the speed of the particle, as illustrated in Figure 17.21. Such a bow wake is called Cerenkov radiation and is commonly observed in particle physics. Figure 17.20 Bow wake created by a duck. Constructive interference produces the rather structured wake, while there is relatively little wave action inside the wake, where interference is mostly destructive. (credit: Horia Varlan, Flickr) 740 Chapter 17 \| Physics of Hearing Figure 17.21 The blue glow in this research reactor pool is Cerenkov radiation caused by subatomic particles traveling faster than the speed of light in water. (credit: U.S. Nuclear Regulatory Commission) Doppler shifts and sonic booms are interesting sound phenomena that occur in all types of waves. They can be of considerable use. For example, the Doppler shift in ultrasound can be used to measure blood velocity, while police use the Doppler shift in radar (a microwave) to measure car velocities. In meteorology, the Doppler shift is used to track the motion of storm clouds; such “Doppler Radar” can give velocity and direction and rain or snow potential of imposing weather fronts. In astronomy, we can examine the light emitted from distant galaxies and determine their speed relative to ours. As galaxies move away from us, their light is shifted to a lower frequency, and so to a longer wavelength—the so-called red shift. Such information from galaxies far, far away has allowed us to estimate the age of the universe (from the Big Bang) as about 14 billion years. Check Your Understanding Why did scientist Christian Doppler observe musicians both on a moving train and also from a stationary point not on the train? Solution Doppler needed to compare the perception of sound when the observer is stationary and the sound source moves, as well as when the sound source and the observer are both in motion. Check Your Understanding Describe a situation in your life when you might rely on the Doppler shift to help you either while driving a car or walking near traffic. Solution If I am driving and I hear Doppler shift in an ambulance siren, I would be able to tell when it was getting closer and also if it has passed by. This would help me to know whether I needed to pull over and let the ambulance through. 17.5 Sound Interference and Resonance: Standing Waves in Air Columns By the end of this section, you will be able to: Learning Objectives • Define antinode, node, fundamental, overtones, and harmonics. • Identify instances of sound interference in everyday situations. • Describe how sound interference occurring inside open and closed tubes changes the characteristics of the sound, and how this applies to sounds produced by musical instruments. • Calculate the length of a tube using sound wave measurements. The information presented in this section supports the following AP® learning objectives and science practices: • 6.D.1.1 The student is able to use representations of individual pulses and construct representations to model the interaction of two wave pulses to analyze the superposition of two pulses. (S.P. 1.1, 1.4) • 6.D.1.2 The student is able to design a suitable experiment and analyze data illustrating the superposition of mechanical waves (only for wave pulses or standing waves). (S.P. 4.2, 5.1) • 6.D.1.3 The student is able to design a plan for collecting data to quantify the amplitude variations when two or more traveling waves or wave pulses interact in a given medium. (S.P. 4.2) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 17 \| Physics of Hearing 741 • 6.D.3.1 The student is able to refine a scientific question related to standing waves and design a detailed plan for the experiment that can be conducted to examine the phenomenon qualitatively or quantitatively. (S.P. 2.1, 2.2, 4.2) • 6.D.3.2 The student is able to predict properties of standing waves that result from the addition of incident and reflected waves that are confined to a region and have nodes and antinodes. (S.P. 6.4) • 6.D.3.3 The student is able to plan data collection strategies, predict the outcome based on the relationship under test, perform data analysis, evaluate evidence compared to the prediction, explain any discrepancy and, if necessary, revise the relationship among variables responsible for establishing standing waves on a string or in a column of air. (S.P. 3.2, 4.1, 5.1, 5.2, 5.3) • 6.D.3.4 The student is able to describe representations and models of situations in which standing waves result from the addition of incident and reflected waves confined to a region. (S.P. 1.2) • 6.D.4.2 The student is able to calculate wavelengths and frequencies (if given wave speed) of standing waves based on boundary conditions and length of region within which the wave is confined, and calculate numerical values of wavelengths and frequencies. Examples should include musical instruments. (S.P. 2.2) Figure 17.22 Some types of headphones use the phenomena of constructive and destructive interference to cancel out outside noises. (credit: JVC America, Flickr) Interference is the hallmark of waves, all of which exhibit constructive and destructive interference exactly analogous to that seen for water waves. In fact, one way to prove something “is a wave” is to observe interference effects. So, sound being a wave, we expect it to exhibit interference; we have already mentioned a few such effects, such as the beats from two similar notes played simultaneously. Figure 17.23 shows a clever use of sound interference to cancel noise. Larger-scale applications of active noise reduction by destructive interference are contemplated for entire passenger compartments in commercial aircraft. To obtain destructive interference, a fast electronic analysis is performed, and a second sound is introduced with its maxima and minima exactly reversed from the incoming noise. Sound waves in fluids are pressure waves and consistent with Pascal’s principle; pressures from two different sources add and subtract like simple numbers; that is, positive and negative gauge pressures add to a much smaller pressure, producing a lower-intensity sound. Although completely destructive interference is possible only under the simplest conditions, it is possible to reduce noise levels by 30 dB or more using this technique. Figure 17.23 Headphones designed to cancel noise with destructive interference create a sound wave exactly opposite to the incoming sound. These headphones can be more effective than the simple passive attenuation used in most ear protection. Such headphones were used on the recordsetting, around the world nonstop flight of the Voyager aircraft to protect the pilots’ hearing from engine noise. W
here else can we observe sound interference? All sound resonances, such as in musical instruments, are due to constructive and destructive interference. Only the resonant frequencies interfere constructively to form standing waves, while others interfere destructively and are absent. From the toot made by blowing over a bottle, to the characteristic flavor of a violin’s sounding box, to the recognizability of a great singer’s voice, resonance and standing waves play a vital role. 742 Interference Chapter 17 \| Physics of Hearing Interference is such a fundamental aspect of waves that observing interference is proof that something is a wave. The wave nature of light was established by experiments showing interference. Similarly, when electrons scattered from crystals exhibited interference, their wave nature was confirmed to be exactly as predicted by symmetry with certain wave characteristics of light. Applying the Science Practices: Standing Wave Figure 17.24 The standing wave pattern of a rubber tube attached to a doorknob. Tie one end of a strip of long rubber tubing to a stable object (doorknob, fence post, etc.) and shake the other end up and down until a standing wave pattern is achieved. Devise a method to determine the frequency and wavelength generated by your arm shaking. Do your results align with the equation? Do you find that the velocity of the wave generated is consistent for each trial? If not, explain why this is the case. Answer This task will likely require two people. The frequency of the wave pattern can be found by timing how long it takes the student shaking the rubber tubing to move his or her hand up and down one full time. (It may be beneficial to time how long it takes the student to do this ten times, and then divide by ten to reduce error.) The wavelength of the standing wave can be measured with a meter stick by measuring the distance between two nodes and multiplying by two. This information should be gathered for standing wave patterns of multiple different wavelengths. As students collect their data, they can use the equation to determine if the wave velocity is consistent. There will likely be some error in the experiment yielding velocities of slightly different value. This error is probably due to an inaccuracy in the wavelength and/or frequency measurements. Suppose we hold a tuning fork near the end of a tube that is closed at the other end, as shown in Figure 17.25, Figure 17.26, Figure 17.27, and Figure 17.28. If the tuning fork has just the right frequency, the air column in the tube resonates loudly, but at most frequencies it vibrates very little. This observation just means that the air column has only certain natural frequencies. The figures show how a resonance at the lowest of these natural frequencies is formed. A disturbance travels down the tube at the speed of sound and bounces off the closed end. If the tube is just the right length, the reflected sound arrives back at the tuning fork exactly half a cycle later, and it interferes constructively with the continuing sound produced by the tuning fork. The incoming and reflected sounds form a standing wave in the tube as shown. Figure 17.25 Resonance of air in a tube closed at one end, caused by a tuning fork. A disturbance moves down the tube. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 17 \| Physics of Hearing 743 Figure 17.26 Resonance of air in a tube closed at one end, caused by a tuning fork. The disturbance reflects from the closed end of the tube. Figure 17.27 Resonance of air in a tube closed at one end, caused by a tuning fork. If the length of the tube is just right, the disturbance gets back to the tuning fork half a cycle later and interferes constructively with the continuing sound from the tuning fork. This interference forms a standing wave, and the air column resonates. Figure 17.28 Resonance of air in a tube closed at one end, caused by a tuning fork. A graph of air displacement along the length of the tube shows none at the closed end, where the motion is constrained, and a maximum at the open end. This standing wave has one-fourth of its wavelength in the tube, so that = 4 . The standing wave formed in the tube has its maximum air displacement (an antinode) at the open end, where motion is unconstrained, and no displacement (a node) at the closed end, where air movement is halted. The distance from a node to an antinode is one-fourth of a wavelength, and this equals the length of the tube; thus, = 4 . This same resonance can be produced by a vibration introduced at or near the closed end of the tube, as shown in Figure 17.29. It is best to consider this a natural vibration of the air column independently of how it is induced. 744 Chapter 17 \| Physics of Hearing Figure 17.29 The same standing wave is created in the tube by a vibration introduced near its closed end. Given that maximum air displacements are possible at the open end and none at the closed end, there are other, shorter wavelengths that can resonate in the tube, such as the one shown in Figure 17.30. Here the standing wave has three-fourths of its wavelength in the tube, or = (3 / 4)′ , so that ′ = 4 / 3 . Continuing this process reveals a whole series of shorterwavelength and higher-frequency sounds that resonate in the tube. We use specific terms for the resonances in any system. The lowest resonant frequency is called the fundamental, while all higher resonant frequencies are called overtones. All resonant frequencies are integral multiples of the fundamental, and they are collectively called harmonics. The fundamental is the first harmonic, the first overtone is the second harmonic, and so on. Figure 17.31 shows the fundamental and the first three overtones (the first four harmonics) in a tube closed at one end. Figure 17.30 Another resonance for a tube closed at one end. This has maximum air displacements at the open end, and none at the closed end. The wavelength is shorter, with three-fourths ′ equaling the length of the tube, so that ′ = 4 / 3 . This higher-frequency vibration is the first overtone. Figure 17.31 The fundamental and three lowest overtones for a tube closed at one end. All have maximum air displacements at the open end and none at the closed end. The fundamental and overtones can be present simultaneously in a variety of combinations. For example, middle C on a trumpet has a sound distinctively different from middle C on a clarinet, both instruments being modified versions of a tube closed at one end. The fundamental frequency is the same (and usually the most intense), but the overtones and their mix of intensities are different and subject to shading by the musician. This mix is what gives various musical instruments (and human voices) their distinctive characteristics, whether they have air columns, strings, sounding boxes, or drumheads. In fact, much of our speech is determined by shaping the cavity formed by the throat and mouth and positioning the tongue to adjust the fundamental and combination of overtones. Simple resonant cavities can be made to resonate with the sound of the vowels, for example. (See Figure 17.32.) In boys, at puberty, the larynx grows and the shape of the resonant cavity changes giving rise to the difference in predominant frequencies in speech between men and women. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 17 \| Physics of Hearing 745 Figure 17.32 The throat and mouth form an air column closed at one end that resonates in response to vibrations in the voice box. The spectrum of overtones and their intensities vary with mouth shaping and tongue position to form different sounds. The voice box can be replaced with a mechanical vibrator, and understandable speech is still possible. Variations in basic shapes make different voices recognizable. Now let us look for a pattern in the resonant frequencies for a simple tube that is closed at one end. The fundamental has = 4 , and frequency is related to wavelength and the speed of sound as given by: Solving for in this equation gives w = = w = w 4 , where w is the speed of sound in air. Similarly, the first overtone has ′ = 4 / 3 (see Figure 17.31), so that ′ = 3w 4 = 3 . Because ′ = 3 , we call the first overtone the third harmonic. Continuing this process, we see a pattern that can be generalized in a single expression. The resonant frequencies of a tube closed at one end are = w 4 , = 1,3,5, (17.28) (17.29) (17.30) (17.31) where 1 is the fundamental, speed of sound and, hence, on temperature. This dependence poses a noticeable problem for organs in old unheated cathedrals, and it is also the reason why musicians commonly bring their wind instruments to room temperature before playing them. 3 is the first overtone, and so on. It is interesting that the resonant frequencies depend on the Example 17.5 Find the Length of a Tube with a 128 Hz Fundamental (a) What length should a tube closed at one end have on a day when the air temperature, is 22.0ºC , if its fundamental frequency is to be 128 Hz (C below middle C)? (b) What is the frequency of its fourth overtone? Strategy The length can be found from the relationship in = w 4 Solution for (a) (1) Identify knowns: • • the fundamental frequency is 128 Hz the air temperature is 22.0ºC , but we will first need to find the speed of sound w . (2) Use = w 4 to find the fundamental frequency ( = 1 ). (3) Solve this equation for length17.32) (17.33) 746 Chapter 17 \| Physics of Hearing (4) Find the speed of sound using w = (331 m/s) 273 K . w = (331 m/s) 295 K 273 K = 344 m/s (5) Enter the values of the speed of sound and frequency into the expression for . = w 4 1 = 344 m/s 4(128 Hz) = 0.672 m Discussion on (a) (17.34) (17.35) Many wind instruments are modified tubes that have finger holes, valves, and other devices for changing the length of the resonating air column and hence, the frequency of the note played. Horns pr
oducing very low frequencies, such as tubas, require tubes so long that they are coiled into loops. Solution for (b) (1) Identify knowns: • • • • the first overtone has = 3 the second overtone has = 5 the third overtone has = 7 the fourth overtone has = 9 (2) Enter the value for the fourth overtone into = w 4 . 9 = 9w 4 = 9 1 = 1.15 kHz (17.36) Discussion on (b) Whether this overtone occurs in a simple tube or a musical instrument depends on how it is stimulated to vibrate and the details of its shape. The trombone, for example, does not produce its fundamental frequency and only makes overtones. Another type of tube is one that is open at both ends. Examples are some organ pipes, flutes, and oboes. The resonances of tubes open at both ends can be analyzed in a very similar fashion to those for tubes closed at one end. The air columns in tubes open at both ends have maximum air displacements at both ends, as illustrated in Figure 17.33. Standing waves form as shown. Figure 17.33 The resonant frequencies of a tube open at both ends are shown, including the fundamental and the first three overtones. In all cases the maximum air displacements occur at both ends of the tube, giving it different natural frequencies than a tube closed at one end. Based on the fact that a tube open at both ends has maximum air displacements at both ends, and using Figure 17.33 as a guide, we can see that the resonant frequencies of a tube open at both ends are: = w 2 , = 1, 2, 3..., (17.37) 2 is the first overtone, where 1 is the fundamental, 3 is the second overtone, and so on. Note that a tube open at both ends has a fundamental frequency twice what it would have if closed at one end. It also has a different spectrum of overtones than a tube closed at one end. So if you had two tubes with the same fundamental frequency but one was open at both ends and the other was closed at one end, they would sound different when played because they have different overtones. Middle C, for example, would sound richer played on an open tube, because it has even multiples of the fundamental as well as odd. A closed tube has only odd multiples. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 17 \| Physics of Hearing 747 Applying the Science Practices: Closed- and Open-Ended Tubes Strike an open-ended length of plastic pipe while holding it in the air. Now place one end of the pipe on a hard surface, sealing one opening, and strike it again. How does the sound change? Further investigate the sound created by the pipe by striking pipes of different lengths and composition. Answer When the pipe is placed on the ground, the standing wave within the pipe changes from being open on both ends to being closed on one end. As a result, the fundamental frequency will change from = 2 to = 4 . This decrease in frequency results in a decrease in observed pitch. Real-World Applications: Resonance in Everyday Systems Resonance occurs in many different systems, including strings, air columns, and atoms. Resonance is the driven or forced oscillation of a system at its natural frequency. At resonance, energy is transferred rapidly to the oscillating system, and the amplitude of its oscillations grows until the system can no longer be described by Hooke’s law. An example of this is the distorted sound intentionally produced in certain types of rock music. Wind instruments use resonance in air columns to amplify tones made by lips or vibrating reeds. Other instruments also use air resonance in clever ways to amplify sound. Figure 17.34 shows a violin and a guitar, both of which have sounding boxes but with different shapes, resulting in different overtone structures. The vibrating string creates a sound that resonates in the sounding box, greatly amplifying the sound and creating overtones that give the instrument its characteristic flavor. The more complex the shape of the sounding box, the greater its ability to resonate over a wide range of frequencies. The marimba, like the one shown in Figure 17.35 uses pots or gourds below the wooden slats to amplify their tones. The resonance of the pot can be adjusted by adding water. Figure 17.34 String instruments such as violins and guitars use resonance in their sounding boxes to amplify and enrich the sound created by their vibrating strings. The bridge and supports couple the string vibrations to the sounding boxes and air within. (credits: guitar, Feliciano Guimares, Fotopedia; violin, Steve Snodgrass, Flickr) 748 Chapter 17 \| Physics of Hearing Figure 17.35 Resonance has been used in musical instruments since prehistoric times. This marimba uses gourds as resonance chambers to amplify its sound. (credit: APC Events, Flickr) We have emphasized sound applications in our discussions of resonance and standing waves, but these ideas apply to any system that has wave characteristics. Vibrating strings, for example, are actually resonating and have fundamentals and overtones similar to those for air columns. More subtle are the resonances in atoms due to the wave character of their electrons. Their orbitals can be viewed as standing waves, which have a fundamental (ground state) and overtones (excited states). It is fascinating that wave characteristics apply to such a wide range of physical systems. Check Your Understanding Describe how noise-canceling headphones differ from standard headphones used to block outside sounds. Solution Regular headphones only block sound waves with a physical barrier. Noise-canceling headphones use destructive interference to reduce the loudness of outside sounds. Check Your Understanding How is it possible to use a standing wave's node and antinode to determine the length of a closed-end tube? Solution When the tube resonates at its natural frequency, the wave's node is located at the closed end of the tube, and the antinode is located at the open end. The length of the tube is equal to one-fourth of the wavelength of this wave. Thus, if we know the wavelength of the wave, we can determine the length of the tube. PhET Explorations: Sound This simulation lets you see sound waves. Adjust the frequency or volume and you can see and hear how the wave changes. Move the listener around and hear what she hears. Figure 17.36 Sound (http://cnx.org/content/m55293/1.2/sound_en.jar) Applying the Science Practices: Variables Affecting Superposition In the PhET Interactive Simulation above, select the tab titled ‘Two Source Interference.’ Within this tab, manipulate the variables present (frequency, amplitude, and speaker separation) to investigate the relationship the variables have with the superposition pattern constructed on the screen. Record all observations. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 17 \| Physics of Hearing 761 (3) Calculate to find the frequency returning to the source: 2,500,649 Hz. Solution for (c) (1) Identify knowns: • The beat frequency is simply the absolute value of the difference between s and obs , as stated in: (2) Substitute known values: B = ∣ obs − s ∣ . ∣ 2,500649 Hz − 2,500000 Hz ∣ (3) Calculate to find the beat frequency: 649 Hz. Discussion (17.47) (17.48) The Doppler shifts are quite small compared with the original frequency of 2.50 MHz. It is far easier to measure the beat frequency than it is to measure the echo frequency with an accuracy great enough to see shifts of a few hundred hertz out of a couple of megahertz. Furthermore, variations in the source frequency do not greatly affect the beat frequency, because both s and obs would increase or decrease. Those changes subtract out in B = ∣ obs − s ∣ . Industrial and Other Applications of Ultrasound Industrial, retail, and research applications of ultrasound are common. A few are discussed here. Ultrasonic cleaners have many uses. Jewelry, machined parts, and other objects that have odd shapes and crevices are immersed in a cleaning fluid that is agitated with ultrasound typically about 40 kHz in frequency. The intensity is great enough to cause cavitation, which is responsible for most of the cleansing action. Because cavitation-produced shock pressures are large and well transmitted in a fluid, they reach into small crevices where even a low-surface-tension cleaning fluid might not penetrate. Sonar is a familiar application of ultrasound. Sonar typically employs ultrasonic frequencies in the range from 30.0 to 100 kHz. Bats, dolphins, submarines, and even some birds use ultrasonic sonar. Echoes are analyzed to give distance and size information both for guidance and finding prey. In most sonar applications, the sound reflects quite well because the objects of interest have significantly different density than the medium in which they travel. When the Doppler shift is observed, velocity information can also be obtained. Submarine sonar can be used to obtain such information, and there is evidence that some bats also sense velocity from their echoes. Similarly, there are a range of relatively inexpensive devices that measure distance by timing ultrasonic echoes. Many cameras, for example, use such information to focus automatically. Some doors open when their ultrasonic ranging devices detect a nearby object, and certain home security lights turn on when their ultrasonic rangers observe motion. Ultrasonic “measuring tapes” also exist to measure such things as room dimensions. Sinks in public restrooms are sometimes automated with ultrasound devices to turn faucets on and off when people wash their hands. These devices reduce the spread of germs and can conserve water. Ultrasound is used for nondestructive testing in industry and by the military. Because ultrasound reflects well from any large change in density, it can reveal cracks and voids in solids, such as aircraft wings, that are too small to be seen with x-rays. For similar reasons, ultrasound is also good for measuring the thickness of coatings, particularly where there are several la
yers involved. Basic research in solid state physics employs ultrasound. Its attenuation is related to a number of physical characteristics, making it a useful probe. Among these characteristics are structural changes such as those found in liquid crystals, the transition of a material to a superconducting phase, as well as density and other properties. These examples of the uses of ultrasound are meant to whet the appetites of the curious, as well as to illustrate the underlying physics of ultrasound. There are many more applications, as you can easily discover for yourself. Check Your Understanding Why is it possible to use ultrasound both to observe a fetus in the womb and also to destroy cancerous tumors in the body? Solution Ultrasound can be used medically at different intensities. Lower intensities do not cause damage and are used for medical imaging. Higher intensities can pulverize and destroy targeted substances in the body, such as tumors. Glossary acoustic impedance: property of medium that makes the propagation of sound waves more difficult antinode: point of maximum displacement bow wake: V-shaped disturbance created when the wave source moves faster than the wave propagation speed 762 Chapter 17 \| Physics of Hearing Doppler effect: an alteration in the observed frequency of a sound due to motion of either the source or the observer Doppler shift: the actual change in frequency due to relative motion of source and observer Doppler-shifted ultrasound: a medical technique to detect motion and determine velocity through the Doppler shift of an echo fundamental: the lowest-frequency resonance harmonics: the term used to refer collectively to the fundamental and its overtones hearing: the perception of sound infrasound: sounds below 20 Hz intensity: the power per unit area carried by a wave intensity reflection coefficient: a measure of the ratio of the intensity of the wave reflected off a boundary between two media relative to the intensity of the incident wave loudness: the perception of sound intensity node: point of zero displacement note: basic unit of music with specific names, combined to generate tunes overtones: all resonant frequencies higher than the fundamental phon: the numerical unit of loudness pitch: the perception of the frequency of a sound sonic boom: a constructive interference of sound created by an object moving faster than sound sound: a disturbance of matter that is transmitted from its source outward sound intensity level: a unitless quantity telling you the level of the sound relative to a fixed standard sound pressure level: the ratio of the pressure amplitude to a reference pressure timbre: number and relative intensity of multiple sound frequencies tone: number and relative intensity of multiple sound frequencies ultrasound: sounds above 20,000 Hz Section Summary 17.1 Sound • Sound is a disturbance of matter that is transmitted from its source outward. • Sound is one type of wave. • Hearing is the perception of sound. 17.2 Speed of Sound, Frequency, and Wavelength The relationship of the speed of sound w , its frequency , and its wavelength is given by which is the same relationship given for all waves. In air, the speed of sound is related to air temperature by w = , w = (331 m/s) 273 K . w is the same for all frequencies and wavelengths. 17.3 Sound Intensity and Sound Level • Intensity is the same for a sound wave as was defined for all waves; it is This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 17 \| Physics of Hearing 763 where is the power crossing area . The SI unit for is watts per meter squared. The intensity of a sound wave is also related to the pressure amplitude Δ = , = 2 Δ 2w , where is the density of the medium in which the sound wave travels and w is the speed of sound in the medium. • Sound intensity level in units of decibels (dB) is where 0 = 10–12 W/m2 is the threshold intensity of hearing. (dB) = 10 log10 0 , 17.4 Doppler Effect and Sonic Booms • The Doppler effect is an alteration in the observed frequency of a sound due to motion of either the source or the observer. • The actual change in frequency is called the Doppler shift. • A sonic boom is constructive interference of sound created by an object moving faster than sound. • A sonic boom is a type of bow wake created when any wave source moves faster than the wave propagation speed. • For a stationary observer and a moving source, the observed frequency obs is: obs = s w w ± s , where s is the frequency of the source, s is the speed of the source, and w is the speed of sound. The minus sign is used for motion toward the observer and the plus sign for motion away. • For a stationary source and moving observer, the observed frequency is: w ± obs w obs = s , where obs is the speed of the observer. 17.5 Sound Interference and Resonance: Standing Waves in Air Columns • Sound interference and resonance have the same properties as defined for all waves. • In air columns, the lowest-frequency resonance is called the fundamental, whereas all higher resonant frequencies are called overtones. Collectively, they are called harmonics. • The resonant frequencies of a tube closed at one end are: w 4 , = 1, 3, 5..., = 1 is the fundamental and is the length of the tube. • The resonant frequencies of a tube open at both ends are: w 2 = , = 1, 2, 3... 17.6 Hearing • The range of audible frequencies is 20 to 20,000 Hz. • Those sounds above 20,000 Hz are ultrasound, whereas those below 20 Hz are infrasound. • The perception of frequency is pitch. • The perception of intensity is loudness. • Loudness has units of phons. 17.7 Ultrasound • The acoustic impedance is defined as: = , is the density of a medium through which the sound travels and is the speed of sound through that medium. • The intensity reflection coefficient , a measure of the ratio of the intensity of the wave reflected off a boundary between two media relative to the intensity of the incident wave, is given by 764 Chapter 17 \| Physics of Hearing • The intensity reflection coefficient is a unitless quantity. Conceptual Questions = 2 − 1 1 + 2 2 2 . 17.2 Speed of Sound, Frequency, and Wavelength 1. How do sound vibrations of atoms differ from thermal motion? 2. When sound passes from one medium to another where its propagation speed is different, does its frequency or wavelength change? Explain your answer briefly. 17.3 Sound Intensity and Sound Level 3. Six members of a synchronized swim team wear earplugs to protect themselves against water pressure at depths, but they can still hear the music and perform the combinations in the water perfectly. One day, they were asked to leave the pool so the dive team could practice a few dives, and they tried to practice on a mat, but seemed to have a lot more difficulty. Why might this be? 4. A community is concerned about a plan to bring train service to their downtown from the town’s outskirts. The current sound intensity level, even though the rail yard is blocks away, is 70 dB downtown. The mayor assures the public that there will be a difference of only 30 dB in sound in the downtown area. Should the townspeople be concerned? Why? 17.4 Doppler Effect and Sonic Booms 5. Is the Doppler shift real or just a sensory illusion? 6. Due to efficiency considerations related to its bow wake, the supersonic transport aircraft must maintain a cruising speed that is a constant ratio to the speed of sound (a constant Mach number). If the aircraft flies from warm air into colder air, should it increase or decrease its speed? Explain your answer. 7. When you hear a sonic boom, you often cannot see the plane that made it. Why is that? 17.5 Sound Interference and Resonance: Standing Waves in Air Columns 8. How does an unamplified guitar produce sounds so much more intense than those of a plucked string held taut by a simple stick? 9. You are given two wind instruments of identical length. One is open at both ends, whereas the other is closed at one end. Which is able to produce the lowest frequency? 10. What is the difference between an overtone and a harmonic? Are all harmonics overtones? Are all overtones harmonics? 17.6 Hearing 11. Why can a hearing test show that your threshold of hearing is 0 dB at 250 Hz, when Figure 17.39 implies that no one can hear such a frequency at less than 20 dB? 17.7 Ultrasound 12. If audible sound follows a rule of thumb similar to that for ultrasound, in terms of its absorption, would you expect the high or low frequencies from your neighbor’s stereo to penetrate into your house? How does this expectation compare with your experience? 13. Elephants and whales are known to use infrasound to communicate over very large distances. What are the advantages of infrasound for long distance communication? 14. It is more difficult to obtain a high-resolution ultrasound image in the abdominal region of someone who is overweight than for someone who has a slight build. Explain why this statement is accurate. 15. Suppose you read that 210-dB ultrasound is being used to pulverize cancerous tumors. You calculate the intensity in watts per centimeter squared and find it is unreasonably high ( 105 W/cm2 ). What is a possible explanation? This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 17 \| Physics of Hearing 765 Problems & Exercises 17.2 Speed of Sound, Frequency, and Wavelength 15. What intensity level does the sound in the preceding problem correspond to? 16. What sound intensity level in dB is produced by earphones that create an intensity of 4.00×10−2 W/m2 ? 1. When poked by a spear, an operatic soprano lets out a 1200-Hz shriek. What is its wavelength if the speed of sound is 345 m/s? 17. Show that an intensity of 10–12 W/m2 is the same as 10–16 W/cm2 . 2. What frequency sound has a 0.10-m wavelength when the speed of sound is 340 m/s? 3. Calculate the speed of sound on a day when a 1500 Hz frequency has a wavelength of 0.221 m. 4. (a)
What is the speed of sound in a medium where a 100-kHz frequency produces a 5.96-cm wavelength? (b) Which substance in Table 17.4 is this likely to be? 5. Show that the speed of sound in 20.0ºC air is 343 m/s, as claimed in the text. 6. Air temperature in the Sahara Desert can reach 56.0ºC (about 134ºF ). What is the speed of sound in air at that temperature? 7. Dolphins make sounds in air and water. What is the ratio of the wavelength of a sound in air to its wavelength in seawater? Assume air temperature is 20.0ºC . 8. A sonar echo returns to a submarine 1.20 s after being emitted. What is the distance to the object creating the echo? (Assume that the submarine is in the ocean, not in fresh water.) 9. (a) If a submarine’s sonar can measure echo times with a precision of 0.0100 s, what is the smallest difference in distances it can detect? (Assume that the submarine is in the ocean, not in fresh water.) (b) Discuss the limits this time resolution imposes on the ability of the sonar system to detect the size and shape of the object creating the echo. 10. A physicist at a fireworks display times the lag between seeing an explosion and hearing its sound, and finds it to be 0.400 s. (a) How far away is the explosion if air temperature is 24.0ºC and if you neglect the time taken for light to reach the physicist? (b) Calculate the distance to the explosion taking the speed of light into account. Note that this distance is negligibly greater. 11. Suppose a bat uses sound echoes to locate its insect prey, 3.00 m away. (See Figure 17.11.) (a) Calculate the echo times for temperatures of 5.00ºC and 35.0ºC . (b) What percent uncertainty does this cause for the bat in locating the insect? (c) Discuss the significance of this uncertainty and whether it could cause difficulties for the bat. (In practice, the bat continues to use sound as it closes in, eliminating most of any difficulties imposed by this and other effects, such as motion of the prey.) 17.3 Sound Intensity and Sound Level 12. What is the intensity in watts per meter squared of 85.0-dB sound? 13. The warning tag on a lawn mower states that it produces noise at a level of 91.0 dB. What is this in watts per meter squared? 14. A sound wave traveling in 20ºC air has a pressure amplitude of 0.5 Pa. What is the intensity of the wave? 18. (a) What is the decibel level of a sound that is twice as intense as a 90.0-dB sound? (b) What is the decibel level of a sound that is one-fifth as intense as a 90.0-dB sound? 19. (a) What is the intensity of a sound that has a level 7.00 dB lower than a 4.00×10–9 W/m2 sound? (b) What is the intensity of a sound that is 3.00 dB higher than a 4.00×10–9 W/m2 sound? 20. (a) How much more intense is a sound that has a level 17.0 dB higher than another? (b) If one sound has a level 23.0 dB less than another, what is the ratio of their intensities? 21. People with good hearing can perceive sounds as low in level as –8.00 dB at a frequency of 3000 Hz. What is the intensity of this sound in watts per meter squared? 22. If a large housefly 3.0 m away from you makes a noise of 40.0 dB, what is the noise level of 1000 flies at that distance, assuming interference has a negligible effect? 23. Ten cars in a circle at a boom box competition produce a 120-dB sound intensity level at the center of the circle. What is the average sound intensity level produced there by each stereo, assuming interference effects can be neglected? 24. The amplitude of a sound wave is measured in terms of its maximum gauge pressure. By what factor does the amplitude of a sound wave increase if the sound intensity level goes up by 40.0 dB? 25. If a sound intensity level of 0 dB at 1000 Hz corresponds to a maximum gauge pressure (sound amplitude) of 10–9 atm , what is the maximum gauge pressure in a 60-dB sound? What is the maximum gauge pressure in a 120-dB sound? 26. An 8-hour exposure to a sound intensity level of 90.0 dB may cause hearing damage. What energy in joules falls on a 0.800-cm-diameter eardrum so exposed? 27. (a) Ear trumpets were never very common, but they did aid people with hearing losses by gathering sound over a large area and concentrating it on the smaller area of the eardrum. What decibel increase does an ear trumpet produce if its sound gathering area is 900 cm2 and the area of the eardrum is 0.500 cm2 , but the trumpet only has an efficiency of 5.00% in transmitting the sound to the eardrum? (b) Comment on the usefulness of the decibel increase found in part (a). 28. Sound is more effectively transmitted into a stethoscope by direct contact than through the air, and it is further intensified by being concentrated on the smaller area of the eardrum. It is reasonable to assume that sound is transmitted into a stethoscope 100 times as effectively compared with transmission though the air. What, then, is the gain in decibels produced by a stethoscope that has a sound gathering area of 15.0 cm2 , and concentrates the sound 766 Chapter 17 \| Physics of Hearing onto two eardrums with a total area of 0.900 cm2 with an efficiency of 40.0%? 29. Loudspeakers can produce intense sounds with surprisingly small energy input in spite of their low efficiencies. Calculate the power input needed to produce a 90.0-dB sound intensity level for a 12.0-cm-diameter speaker that has an efficiency of 1.00%. (This value is the sound intensity level right at the speaker.) 17.4 Doppler Effect and Sonic Booms 30. (a) What frequency is received by a person watching an oncoming ambulance moving at 110 km/h and emitting a steady 800-Hz sound from its siren? The speed of sound on this day is 345 m/s. (b) What frequency does she receive after the ambulance has passed? 31. (a) At an air show a jet flies directly toward the stands at a speed of 1200 km/h, emitting a frequency of 3500 Hz, on a day when the speed of sound is 342 m/s. What frequency is received by the observers? (b) What frequency do they receive as the plane flies directly away from them? 32. What frequency is received by a mouse just before being dispatched by a hawk flying at it at 25.0 m/s and emitting a screech of frequency 3500 Hz? Take the speed of sound to be 331 m/s. 33. A spectator at a parade receives an 888-Hz tone from an oncoming trumpeter who is playing an 880-Hz note. At what speed is the musician approaching if the speed of sound is 338 m/s? 34. A commuter train blows its 200-Hz horn as it approaches a crossing. The speed of sound is 335 m/s. (a) An observer waiting at the crossing receives a frequency of 208 Hz. What is the speed of the train? (b) What frequency does the observer receive as the train moves away? 35. Can you perceive the shift in frequency produced when you pull a tuning fork toward you at 10.0 m/s on a day when the speed of sound is 344 m/s? To answer this question, calculate the factor by which the frequency shifts and see if it is greater than 0.300%. 36. Two eagles fly directly toward one another, the first at 15.0 m/s and the second at 20.0 m/s. Both screech, the first one emitting a frequency of 3200 Hz and the second one emitting a frequency of 3800 Hz. What frequencies do they receive if the speed of sound is 330 m/s? 37. What is the minimum speed at which a source must travel toward you for you to be able to hear that its frequency is Doppler shifted? That is, what speed produces a shift of 0.300% on a day when the speed of sound is 331 m/s? 17.5 Sound Interference and Resonance: Standing Waves in Air Columns 38. A “showy” custom-built car has two brass horns that are supposed to produce the same frequency but actually emit 263.8 and 264.5 Hz. What beat frequency is produced? 39. What beat frequencies will be present: (a) If the musical notes A and C are played together (frequencies of 220 and 264 Hz)? (b) If D and F are played together (frequencies of 297 and 352 Hz)? (c) If all four are played together? 40. What beat frequencies result if a piano hammer hits three strings that emit frequencies of 127.8, 128.1, and 128.3 Hz? 41. A piano tuner hears a beat every 2.00 s when listening to a 264.0-Hz tuning fork and a single piano string. What are the two possible frequencies of the string? This content is available for free at http://cnx.org/content/col11844/1.13 42. (a) What is the fundamental frequency of a 0.672-m-long tube, open at both ends, on a day when the speed of sound is 344 m/s? (b) What is the frequency of its second harmonic? 43. If a wind instrument, such as a tuba, has a fundamental frequency of 32.0 Hz, what are its first three overtones? It is closed at one end. (The overtones of a real tuba are more complex than this example, because it is a tapered tube.) 44. What are the first three overtones of a bassoon that has a fundamental frequency of 90.0 Hz? It is open at both ends. (The overtones of a real bassoon are more complex than this example, because its double reed makes it act more like a tube closed at one end.) 45. How long must a flute be in order to have a fundamental frequency of 262 Hz (this frequency corresponds to middle C on the evenly tempered chromatic scale) on a day when air temperature is 20.0ºC ? It is open at both ends. 46. What length should an oboe have to produce a fundamental frequency of 110 Hz on a day when the speed of sound is 343 m/s? It is open at both ends. 47. What is the length of a tube that has a fundamental frequency of 176 Hz and a first overtone of 352 Hz if the speed of sound is 343 m/s? 48. (a) Find the length of an organ pipe closed at one end that produces a fundamental frequency of 256 Hz when air temperature is 18.0ºC . (b) What is its fundamental frequency at 25.0ºC ? 49. By what fraction will the frequencies produced by a wind instrument change when air temperature goes from 10.0ºC to 30.0ºC ? That is, find the ratio of the frequencies at those temperatures. 50. The ear canal resonates like a tube closed at one end. (See Figure 17.41.) If ear canals range in length from 1.80 to 2.60 cm in an average population, w
hat is the range of fundamental resonant frequencies? Take air temperature to be 37.0ºC , which is the same as body temperature. How does this result correlate with the intensity versus frequency graph (Figure 17.39 of the human ear? 51. Calculate the first overtone in an ear canal, which resonates like a 2.40-cm-long tube closed at one end, by taking air temperature to be 37.0ºC . Is the ear particularly sensitive to such a frequency? (The resonances of the ear canal are complicated by its nonuniform shape, which we shall ignore.) 52. A crude approximation of voice production is to consider the breathing passages and mouth to be a resonating tube closed at one end. (See Figure 17.32.) (a) What is the fundamental frequency if the tube is 0.240-m long, by taking air temperature to be 37.0ºC ? (b) What would this frequency become if the person replaced the air with helium? Assume the same temperature dependence for helium as for air. 53. (a) Students in a physics lab are asked to find the length of an air column in a tube closed at one end that has a fundamental frequency of 256 Hz. They hold the tube vertically and fill it with water to the top, then lower the water while a 256-Hz tuning fork is rung and listen for the first resonance. What is the air temperature if the resonance occurs for a length of 0.336 m? (b) At what length will they observe the second resonance (first overtone)? Chapter 17 \| Physics of Hearing 767 54. What frequencies will a 1.80-m-long tube produce in the audible range at 20.0ºC if: (a) The tube is closed at one end? (b) It is open at both ends? 17.6 Hearing 55. The factor of 10−12 in the range of intensities to which the ear can respond, from threshold to that causing damage after brief exposure, is truly remarkable. If you could measure distances over the same range with a single instrument and the smallest distance you could measure was 1 mm, what would the largest be? 56. The frequencies to which the ear responds vary by a factor of 103 . Suppose the speedometer on your car measured speeds differing by the same factor of 103 , and the greatest speed it reads is 90.0 mi/h. What would be the slowest nonzero speed it could read? 57. What are the closest frequencies to 500 Hz that an average person can clearly distinguish as being different in frequency from 500 Hz? The sounds are not present simultaneously. 58. Can the average person tell that a 2002-Hz sound has a different frequency than a 1999-Hz sound without playing them simultaneously? 59. If your radio is producing an average sound intensity level of 85 dB, what is the next lowest sound intensity level that is clearly less intense? 60. Can you tell that your roommate turned up the sound on the TV if its average sound intensity level goes from 70 to 73 dB? 61. Based on the graph in Figure 17.38, what is the threshold of hearing in decibels for frequencies of 60, 400, 1000, 4000, and 15,000 Hz? Note that many AC electrical appliances produce 60 Hz, music is commonly 400 Hz, a reference frequency is 1000 Hz, your maximum sensitivity is near 4000 Hz, and many older TVs produce a 15,750 Hz whine. 62. What sound intensity levels must sounds of frequencies 60, 3000, and 8000 Hz have in order to have the same loudness as a 40-dB sound of frequency 1000 Hz (that is, to have a loudness of 40 phons)? 63. What is the approximate sound intensity level in decibels of a 600-Hz tone if it has a loudness of 20 phons? If it has a loudness of 70 phons? 64. (a) What are the loudnesses in phons of sounds having frequencies of 200, 1000, 5000, and 10,000 Hz, if they are all at the same 60.0-dB sound intensity level? (b) If they are all at 110 dB? (c) If they are all at 20.0 dB? 65. Suppose a person has a 50-dB hearing loss at all frequencies. By how many factors of 10 will low-intensity sounds need to be amplified to seem normal to this person? Note that smaller amplification is appropriate for more intense sounds to avoid further hearing damage. 66. If a woman needs an amplification of 5.0×1012 times the threshold intensity to enable her to hear at all frequencies, what is her overall hearing loss in dB? Note that smaller amplification is appropriate for more intense sounds to avoid further damage to her hearing from levels above 90 dB. 67. (a) What is the intensity in watts per meter squared of a just barely audible 200-Hz sound? (b) What is the intensity in watts per meter squared of a barely audible 4000-Hz sound? 68. (a) Find the intensity in watts per meter squared of a 60.0-Hz sound having a loudness of 60 phons. (b) Find the intensity in watts per meter squared of a 10,000-Hz sound having a loudness of 60 phons. 69. A person has a hearing threshold 10 dB above normal at 100 Hz and 50 dB above normal at 4000 Hz. How much more intense must a 100-Hz tone be than a 4000-Hz tone if they are both barely audible to this person? 70. A child has a hearing loss of 60 dB near 5000 Hz, due to noise exposure, and normal hearing elsewhere. How much more intense is a 5000-Hz tone than a 400-Hz tone if they are both barely audible to the child? 71. What is the ratio of intensities of two sounds of identical frequency if the first is just barely discernible as louder to a person than the second? 17.7 Ultrasound Unless otherwise indicated, for problems in this section, assume that the speed of sound through human tissues is 1540 m/s. 72. What is the sound intensity level in decibels of ultrasound of intensity 105 W/m2 , used to pulverize tissue during surgery? 73. Is 155-dB ultrasound in the range of intensities used for deep heating? Calculate the intensity of this ultrasound and compare this intensity with values quoted in the text. 74. Find the sound intensity level in decibels of 2.00×10–2 W/m2 ultrasound used in medical diagnostics. 75. The time delay between transmission and the arrival of the reflected wave of a signal using ultrasound traveling through a piece of fat tissue was 0.13 ms. At what depth did this reflection occur? 76. In the clinical use of ultrasound, transducers are always coupled to the skin by a thin layer of gel or oil, replacing the air that would otherwise exist between the transducer and the skin. (a) Using the values of acoustic impedance given in Table 17.8 calculate the intensity reflection coefficient between transducer material and air. (b) Calculate the intensity reflection coefficient between transducer material and gel (assuming for this problem that its acoustic impedance is identical to that of water). (c) Based on the results of your calculations, explain why the gel is used. 77. (a) Calculate the minimum frequency of ultrasound that will allow you to see details as small as 0.250 mm in human tissue. (b) What is the effective depth to which this sound is effective as a diagnostic probe? 78. (a) Find the size of the smallest detail observable in human tissue with 20.0-MHz ultrasound. (b) Is its effective penetration depth great enough to examine the entire eye (about 3.00 cm is needed)? (c) What is the wavelength of such ultrasound in 0ºC air? 79. (a) Echo times are measured by diagnostic ultrasound scanners to determine distances to reflecting surfaces in a patient. What is the difference in echo times for tissues that are 3.50 and 3.60 cm beneath the surface? (This difference is the minimum resolving time for the scanner to see details as small as 0.100 cm, or 1.00 mm. Discrimination of smaller time differences is needed to see smaller details.) (b) Discuss whether the period of this ultrasound must be smaller than the minimum time resolution. If so, what is the minimum 768 Chapter 17 \| Physics of Hearing frequency of the ultrasound and is that out of the normal range for diagnostic ultrasound? 80. (a) How far apart are two layers of tissue that produce echoes having round-trip times (used to measure distances) that differ by 0.750 μs ? (b) What minimum frequency must the ultrasound have to see detail this small? 81. (a) A bat uses ultrasound to find its way among trees. If this bat can detect echoes 1.00 ms apart, what minimum distance between objects can it detect? (b) Could this distance explain the difficulty that bats have finding an open door when they accidentally get into a house? 82. A dolphin is able to tell in the dark that the ultrasound echoes received from two sharks come from two different objects only if the sharks are separated by 3.50 m, one being that much farther away than the other. (a) If the ultrasound has a frequency of 100 kHz, show this ability is not limited by its wavelength. (b) If this ability is due to the dolphin’s ability to detect the arrival times of echoes, what is the minimum time difference the dolphin can perceive? 83. A diagnostic ultrasound echo is reflected from moving blood and returns with a frequency 500 Hz higher than its original 2.00 MHz. What is the velocity of the blood? (Assume that the frequency of 2.00 MHz is accurate to seven significant figures and 500 Hz is accurate to three significant figures.) 84. Ultrasound reflected from an oncoming bloodstream that is moving at 30.0 cm/s is mixed with the original frequency of 2.50 MHz to produce beats. What is the beat frequency? (Assume that the frequency of 2.50 MHz is accurate to seven significant figures.) Test Prep for AP® Courses 17.2 Speed of Sound, Frequency, and Wavelength 1. A teacher wants to demonstrate that the speed of sound is not a constant value. Considering her regular classroom voice as the control, which of the following will increase the speed of sound leaving her mouth? I. Submerge her mouth underwater and speak at the same volume. Increase the temperature of the room and speak at the same volume. Increase the pitch of her voice and speak at the same volume. I only I and II only I, II and III II and III III only II. III. a. b. c. d. e. 2. All members of an orchestra begin tuning their instruments at the same time. While some woodwind instruments play high frequency notes, other stringed instruments pla
y notes of lower frequency. Yet an audience member will hear all notes simultaneously, in apparent contrast to the equation. Explain how a student could demonstrate the flaw in the above logic, using a slinky, stopwatch, and meter stick. Make sure to explain what relationship is truly demonstrated in the above equation, in addition to what would be necessary to get the speed of the slinky to actually change. You may include diagrams and equations as part of your explanation. 17.3 Sound Intensity and Sound Level This content is available for free at http://cnx.org/content/col11844/1.13 3. In order to waken a sleeping child, the volume on an alarm clock is tripled. Under this new scenario, how much more energy will be striking the child’s ear drums each second? twice as much three times as much a. b. c. approximately 4.8 times as much d. six times as much e. nine times as much 4. A musician strikes the strings of a guitar such that they vibrate with twice the amplitude. a. Explain why this requires an energy input greater than twice the original value. b. Explain why the sound leaving the string will not result in a decibel level that is twice as great. 17.4 Doppler Effect and Sonic Booms 5. A baggage handler stands on the edge of a runway as a landing plane approaches. Compared to the pitch of the plane as heard by the plane’s pilot, which of the following correctly describes the sensation experienced by the handler? a. The frequency of the plane will be lower pitched according to the baggage handler and will become even lower pitched as the plane slows to a stop. b. The frequency of the plane will be lower pitched according to the baggage handler but will increase in pitch as the plane slows to a stop. c. The frequency of the plane will be higher pitched according to the baggage handler but will decrease in pitch as the plane slows to a stop. d. The frequency of the plane will be higher pitched according to the baggage handler and will further increase in pitch as the plane slows to a stop. 6. The following graph represents the perceived frequency of a car as it passes a student. Chapter 17 \| Physics of Hearing 769 c. Figure 17.55 d. Figure 17.56 9. A student sends a transverse wave pulse of amplitude A along a rope attached at one end. As the pulse returns to the student, a second pulse of amplitude 3A is sent along the opposite side of the rope. What is the resulting amplitude when the two pulses interact? a. 4A b. A c. 2A, on the side of the original wave pulse d. 2A, on the side of the second wave pulse 10. A student would like to demonstrate destructive interference using two sound sources. Explain how the student could set up this demonstration and what restrictions they would need to place upon their sources. Be sure to consider both the layout of space and the sounds created in your explanation. 11. A student is shaking a flexible string attached to a wooden board in a rhythmic manner. Which of the following choices will decrease the wavelength within the rope? I. The student could shake her hand back and forth with greater frequency. II. The student could shake her hand back in forth with a greater amplitude. III. The student could increase the tension within the rope by stepping backwards from the board. I only I and II I and III II and III I, II, and III a. b. c. d. e. 12. A ripple tank has two locations (L1 and L2) that vibrate in tandem as shown below. Both L1 and L2 vibrate in a plane perpendicular to the page, creating a two-dimensional interference pattern. Figure 17.51 Plot of time versus perceived frequency to illustrate the Doppler effect. a. If the true frequency of the car’s horn is 200 Hz, how fast was the car traveling? b. On the graph above, draw a line demonstrating the perceived frequency for a car traveling twice as fast. Label all intercepts, maximums, and minimums on the graph. 17.5 Sound Interference and Resonance: Standing Waves in Air Columns 7. A common misconception is that two wave pulses traveling in opposite directions will reflect off each other. Outline a procedure that you would use to convince someone that the two wave pulses do not reflect off each other, but instead travel through each other. You may use sketches to represent your understanding. Be sure to provide evidence to not only refute the original claim, but to support yours as well. 8. Two wave pulses are traveling toward each other on a string, as shown below. Which of the following representations correctly shows the string as the two pulses overlap? Figure 17.52 a. b. Figure 17.53 Figure 17.54 770 Chapter 17 \| Physics of Hearing c. Using information from the graph, determine the speed of sound within the student’s classroom, and explain what characteristic of the graph provides this evidence. d. Determine the temperature of the classroom. 15. A tube is open at one end. If the fundamental frequency f is created by a wavelength λ, then which of the following describes the frequency and wavelength associated with the tube’s fourth overtone? f λ (a) 4f λ/4 (b) 4f λ (c) 9f λ/9 (d) 9f λ 16. A group of students were tasked with collecting information about standing waves. Table 17.10 a series of their data, showing the length of an air column and a resonant frequency present when the column is struck. Table 17.10 Length (m) Resonant Frequency (Hz) 1 2 3 4 85.75 43 29 21.5 a. From their data, determine whether the air column was open or closed on each end. b. Predict the resonant frequency of the column at a length of 2.5 meters. 17. When a student blows across a glass half-full of water, a resonant frequency is created within the air column remaining in the glass. Which of the following can the student do to increase this resonant frequency? I. Add more water to the glass. II. Replace the water with a more dense fluid. Increase the temperature of the room. III. I only a. I and III b. c. II and III d. all of the above 18. A wooden ruler rests on a desk with half of its length protruding off the desk edge. A student holds one end in place and strikes the protruding end with his other hand, creating a musical sound. a. Explain, without using a sound meter, how the student could experimentally determine the speed of sound that travels within the ruler. b. A sound meter is then used to measure the true frequency of the ruler. It is found that the experimental result is lower than the true value. Explain a factor that may have caused this difference. Also explain what affect this result has on the calculated speed of sound. 19. A musician stands outside in a field and plucks a string on an acoustic guitar. Standing waves will most likely occur in which of the following media? Select two answers. a. The guitar string b. The air inside the guitar c. The air surrounding the guitar Figure 17.57 Describe an experimental procedure to determine the speed of the waves created within the water, including all additional equipment that you would need. You may use the diagram below to help your description, or you may create one of your own. Include enough detail so that another student could carry out your experiment. 13. A string is vibrating between two posts as shown above. Students are to determine the speed of the wave within this string. They have already measured the amount of time necessary for the wave to oscillate up and down. The students must also take what other measurements to determine the speed of the wave? a. The distance between the two posts. b. The amplitude of the wave c. The tension in the string d. The amplitude of the wave and the tension in the string e. The distance between the two posts, the amplitude of the wave, and the tension in the string 14. The accepted speed of sound in room temperature air is 346 m/s. Knowing that their school is colder than usual, a group of students is asked to determine the speed of sound in their room. They are permitted to use any materials necessary; however, their lab procedure must utilize standing wave patterns. The students collect the information Table 17.9. Table 17.9 Trial Number Wavelength (m) Frequency (Hz) 1 2 3 4 5 3.45 2.32 1.70 1.45 1.08 95 135 190 240 305 a. Describe an experimental procedure the group of students could have used to obtain this data. Include diagrams of the experimental setup and any equipment used in the process. b. Select a set of data points from the table and plot those points on a graph to determine the speed of sound within the classroom. Fill in the blank column in the table for any quantities you graph other than the given data. Label the axes and indicate the scale for each. Draw a best-fit line or curve through your data points. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 17 \| Physics of Hearing 771 d. The ground beneath the musician a. Based on the information above, what is the speed of 20. the wave within the string? b. The guitarist then slides her finger along the neck of the guitar, changing the string length as a result. Calculate the fundamental frequency of the string and wave speed present if the string length is reduced to 2/3 L. Figure 17.58 This figure shows two tubes that are identical except for their slightly different lengths. Both tubes have one open end and one closed end. A speaker connected to a variable frequency generator is placed in front of the tubes, as shown. The speaker is set to produce a note of very low frequency when turned on. The frequency is then slowly increased to produce resonances in the tubes. Students observe that at first only one of the tubes resonates at a time. Later, as the frequency gets very high, there are times when both tubes resonate. In a clear, coherent, paragraph-length answer, explain why there are some high frequencies, but no low frequencies, at which both tubes resonate. You may include diagrams and/or equations as part of your explanation. Figure 17.59 21. A student connects one end of a string with negligible mass to an oscillator.
The other end of the string is passed over a pulley and attached to a suspended weight, as shown above. The student finds that a standing wave with one antinode is formed on the string when the frequency of the oscillator is f0. The student then moves the oscillator to shorten the horizontal segment of string to half its original length. At what frequency will a standing wave with one antinode now be formed on the string? f0/2 f0 a. b. c. 2f0 d. There is no frequency at which a standing wave will be formed. 22. A guitar string of length L is bound at both ends. Table 17.11 shows the string’s harmonic frequencies when struck. Table 17.11 Harmonic Number Frequency 1 2 3 4 225/L 450/L 675/L 900/L Chapter 18 \| Electric Charge and Electric Field 773 18 ELECTRIC CHARGE AND ELECTRIC FIELD Figure 18.1 Static electricity from this plastic slide causes the child's hair to stand on end. The sliding motion stripped electrons away from the child's body, leaving an excess of positive charges, which repel each other along each strand of hair. (credit: Ken Bosma/Wikimedia Commons) Chapter Outline 18.1. Static Electricity and Charge: Conservation of Charge 18.2. Conductors and Insulators 18.3. Conductors and Electric Fields in Static Equilibrium 18.4. Coulomb’s Law 18.5. Electric Field: Concept of a Field Revisited 18.6. Electric Field Lines: Multiple Charges 18.7. Electric Forces in Biology 18.8. Applications of Electrostatics Connection for AP® Courses The image of American politician and scientist Benjamin Franklin (1706–1790) flying a kite in a thunderstorm (shown in Figure 18.2) is familiar to every schoolchild. In this experiment, Franklin demonstrated a connection between lightning and static electricity. Sparks were drawn from a key hung on a kite string during an electrical storm. These sparks were like those produced by static electricity, such as the spark that jumps from your finger to a metal doorknob after you walk across a wool carpet. Much has been written about Franklin. His experiments were only part of the life of a man who was a scientist, inventor, revolutionary, statesman, and writer. Franklin's experiments were not performed in isolation, nor were they the only ones to reveal connections. 774 Chapter 18 \| Electric Charge and Electric Field Figure 18.2 Benjamin Franklin, his kite, and electricity. When Benjamin Franklin demonstrated that lightning was related to static electricity, he made a connection that is now part of the evidence that all directly experienced forces (except gravitational force) are manifestations of the electromagnetic force. For example, the Italian scientist Luigi Galvani (1737-1798) performed a series of experiments in which static electricity was used to stimulate contractions of leg muscles of dead frogs, an effect already known in humans subjected to static discharges. But Galvani also found that if he joined one end of two metal wires (say copper and zinc) and touched the other ends of the wires to muscles; he produced the same effect in frogs as static discharge. Alessandro Volta (1745-1827), partly inspired by Galvani's work, experimented with various combinations of metals and developed the battery. During the same era, other scientists made progress in discovering fundamental connections. The periodic table was developed as systematic properties of the elements were discovered. This influenced the development and refinement of the concept of atoms as the basis of matter. Such submicroscopic descriptions of matter also help explain a great deal more. Atomic and molecular interactions, such as the forces of friction, cohesion, and adhesion, are now known to be manifestations of the electromagnetic force. Static electricity is just one aspect of the electromagnetic force, which also includes moving electricity and magnetism. All the macroscopic forces that we experience directly, such as the sensations of touch and the tension in a rope, are due to the electromagnetic force, one of the four fundamental forces in nature. The gravitational force, another fundamental force, is actually sensed through the electromagnetic interaction of molecules, such as between those in our feet and those on the top of a bathroom scale. (The other two fundamental forces, the strong nuclear force and the weak nuclear force, cannot be sensed on the human scale.) This chapter begins the study of electromagnetic phenomena at a fundamental level. The next several chapters will cover static electricity, moving electricity, and magnetism – collectively known as electromagnetism. In this chapter, we begin with the study of electric phenomena due to charges that are at least temporarily stationary, called electrostatics, or static electricity. The chapter introduces several very important concepts of charge, electric force, and electric field, as well as defining the relationships between these concepts. The charge is defined as a property of a system (Big Idea 1) that can affect its interaction with other charged systems (Enduring Understanding 1.B). The law of conservation of electric charge is also discussed (Essential Knowledge 1.B.1). The two kinds of electric charge are defined as positive and negative, providing an explanation for having positively charged, negatively charged, or neutral objects (containing equal quantities of positive and negative charges) (Essential Knowledge 1.B.2). The discrete nature of the electric charge is introduced in this chapter by defining the elementary charge as the smallest observed unit of charge that can be isolated, which is the electron charge (Essential Knowledge 1.B.3). The concepts of a system (having internal structure) and of an object (having no internal structure) are implicitly introduced to explain charges carried by the electron and proton (Enduring Understanding 1.A, Essential Knowledge 1.A.1). An electric field is caused by the presence of charged objects (Enduring Understanding 2.C) and can be used to explain interactions between electrically charged objects (Big Idea 2). The electric force represents the effect of an electric field on a charge placed in the field. The magnitude and direction of the electric force are defined by the magnitude and direction of the electric field and magnitude and sign of the charge (Essential Knowledge 2.C.1). The magnitude of the electric field is proportional to the net charge of the objects that created that field (Essential Knowledge 2.C.2). For the special case of a spherically symmetric charged object, the electric field outside the object is radial, and its magnitude varies as the inverse square of the radial distance from the center of that object (Essential Knowledge 2.C.3). The chapter provides examples of vector field maps for various charged systems, including point charges, spherically symmetric charge distributions, and uniformly charged parallel plates (Essential Knowledge 2.C.1, Essential Knowledge 2.C.2). For multiple point charges, the chapter explains how to This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 18 \| Electric Charge and Electric Field 775 find the vector field map by adding the electric field vectors of each individual object, including the special case of two equal charges having opposite signs, known as an electric dipole (Essential Knowledge 2.C.4). The special case of two oppositely charged parallel plates with uniformly distributed electric charge when the electric field is perpendicular to the plates and is constant in both magnitude and direction is described in detail, providing many opportunities for problem solving and applications (Essential Knowledge 2.C.5). The idea that interactions can be described by forces is also reinforced in this chapter (Big Idea 3). Like all other forces that you have learned about so far, electric force is a vector that affects the motion according to Newton's laws (Enduring Understanding 3.A). It is clearly stated in the chapter that electric force appears as a result of interactions between two charged objects (Essential Knowledge 3.A.3, Essential Knowledge 3.C.2). At the macroscopic level the electric force is a long-range force (Enduring Understanding 3.C); however, at the microscopic level many contact forces, such as friction, can be explained by interatomic electric forces (Essential Knowledge 3.C.4). This understanding of friction is helpful when considering properties of conductors and insulators and the transfer of charge by conduction. Interactions between systems can result in changes in those systems (Big Idea 4). In the case of charged systems, such interactions can lead to changes of electric properties (Enduring Understanding 4.E), such as charge distribution (Essential Knowledge 4.E.3). Any changes are governed by conservation laws (Big Idea 5). Depending on whether the system is closed or open, certain quantities of the system remain the same or changes in those quantities are equal to the amount of transfer of this quantity from or to the system (Enduring Understanding 5.A). The electric charge is one of these quantities (Essential Knowledge 5.A.2). Therefore, the electric charge of a system is conserved (Enduring Understanding 5.C) and the exchange of electric charge between objects in a system does not change the total electric charge of the system (Essential Knowledge 5.C.2). Big Idea 1 Objects and systems have properties such as mass and charge. Systems may have internal structure. Enduring Understanding 1.A The internal structure of a system determines many properties of the system. Essential Knowledge 1.A.1 A system is an object or a collection of objects. Objects are treated as having no internal structure. Enduring Understanding 1.B Electric charge is a property of an object or system that affects its interactions with other objects or systems containing charge. Essential Knowledge 1.B.1 Electric charge is conserved. The net charge of a system is equal to the sum of the charges of all t
he objects in the system. Essential Knowledge 1.B.2 There are only two kinds of electric charge. Neutral objects or systems contain equal quantities of positive and negative charge, with the exception of some fundamental particles that have no electric charge. Essential Knowledge 1.B.3 The smallest observed unit of charge that can be isolated is the electron charge, also known as the elementary charge. Big Idea 2 Fields existing in space can be used to explain interactions. Enduring Understanding 2.C An electric field is caused by an object with electric charge. Essential Knowledge 2.C.1 The magnitude of the electric force F exerted on an object with electric charge q by an electric field ( → . The direction of the force is determined by the direction of the field and the sign of the charge, with → = → is positively charged objects accelerating in the direction of the field and negatively charged objects accelerating in the direction opposite the field. This should include a vector field map for positive point charges, negative point charges, spherically symmetric charge distribution, and uniformly charged parallel plates. Essential Knowledge 2.C.2 The magnitude of the electric field vector is proportional to the net electric charge of the object(s) creating that field. This includes positive point charges, negative point charges, spherically symmetric charge distributions, and uniformly charged parallel plates. Essential Knowledge 2.C.3 The electric field outside a spherically symmetric charged object is radial, and its magnitude varies as the inverse square of the radial distance from the center of that object. Electric field lines are not in the curriculum. Students will be expected to rely only on the rough intuitive sense underlying field lines, wherein the field is viewed as analogous to something emanating uniformly from a source. Essential Knowledge 2.C.4 The electric field around dipoles and other systems of electrically charged objects (that can be modeled as point objects) is found by vector addition of the field of each individual object. Electric dipoles are treated qualitatively in this course as a teaching analogy to facilitate student understanding of magnetic dipoles. Essential Knowledge 2.C.5 Between two oppositely charged parallel plates with uniformly distributed electric charge, at points far from the edges of the plates, the electric field is perpendicular to the plates and is constant in both magnitude and direction. Big Idea 3 The interactions of an object with other objects can be described by forces. Enduring Understanding 3.A All forces share certain common characteristics when considered by observers in inertial reference frames. Essential Knowledge 3.A.3 A force exerted on an object is always due to the interaction of that object with another object. Enduring Understanding 3.C At the macroscopic level, forces can be categorized as either long-range (action-at-a-distance) forces or contact forces. Essential Knowledge 3.C.2 Electric force results from the interaction of one object that has an electric charge with another object that has an electric charge. 776 Chapter 18 \| Electric Charge and Electric Field Essential Knowledge 3.C.4 Contact forces result from the interaction of one object touching another object, and they arise from interatomic electric forces. These forces include tension, friction, normal, spring (Physics 1), and buoyant (Physics 2). Big Idea 4 Interactions between systems can result in changes in those systems. Enduring Understanding 4.E The electric and magnetic properties of a system can change in response to the presence of, or changes in, other objects or systems. Essential Knowledge 4.E.3 The charge distribution in a system can be altered by the effects of electric forces produced by a charged object. Big Idea 5 Changes that occur as a result of interactions are constrained by conservation laws. Enduring Understanding 5.A Certain quantities are conserved, in the sense that the changes of those quantities in a given system are always equal to the transfer of that quantity to or from the system by all possible interactions with other systems. Essential Knowledge 5.A.2 For all systems under all circumstances, energy, charge, linear momentum, and angular momentum are conserved. Enduring Understanding 5.C The electric charge of a system is conserved. Essential Knowledge 5.C.2 The exchange of electric charges among a set of objects in a system conserves electric charge. 18.1 Static Electricity and Charge: Conservation of Charge Learning Objectives By the end of this section, you will be able to: • Define electric charge, and describe how the two types of charge interact. • Describe three common situations that generate static electricity. • State the law of conservation of charge. The information presented in this section supports the following AP® learning objectives and science practices: • 1.B.1.1 The student is able to make claims about natural phenomena based on conservation of electric charge. (S.P. 6.4) • 1.B.1.2 The student is able to make predictions, using the conservation of electric charge, about the sign and relative quantity of net charge of objects or systems after various charging processes, including conservation of charge in simple circuits. (S.P. 6.4, 7.2) • 1.B.2.1 The student is able to construct an explanation of the two-charge model of electric charge based on evidence produced through scientific practices. (S.P. 6.4) • 1.B.3.1 The student is able to challenge the claim that an electric charge smaller than the elementary charge has been isolated. (S.P. 1.5, 6.1, 7.2) • 5.A.2.1 The student is able to define open and closed systems for everyday situations and apply conservation concepts for energy, charge, and linear momentum to those situations. (S.P. 6.4, 7.2) • 5.C.2.1 The student is able to predict electric charges on objects within a system by application of the principle of charge conservation within a system. (S.P. 6.4) • 5.C.2.2 The student is able to design a plan to collect data on the electrical charging of objects and electric charge induction on neutral objects and qualitatively analyze that data. (S.P. 4.2, 5.1) • 5.C.2.3 The student is able to justify the selection of data relevant to an investigation of the electrical charging of objects and electric charge induction on neutral objects. (S.P. 4.1) Figure 18.3 Borneo amber was mined in Sabah, Malaysia, from shale-sandstone-mudstone veins. When a piece of amber is rubbed with a piece of silk, the amber gains more electrons, giving it a net negative charge. At the same time, the silk, having lost electrons, becomes positively charged. (credit: Sebakoamber, Wikimedia Commons) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 18 \| Electric Charge and Electric Field 777 What makes plastic wrap cling? Static electricity. Not only are applications of static electricity common these days, its existence has been known since ancient times. The first record of its effects dates to ancient Greeks who noted more than 500 years B.C. that polishing amber temporarily enabled it to attract bits of straw (see Figure 18.3). The very word electric derives from the Greek word for amber (electron). Many of the characteristics of static electricity can be explored by rubbing things together. Rubbing creates the spark you get from walking across a wool carpet, for example. Static cling generated in a clothes dryer and the attraction of straw to recently polished amber also result from rubbing. Similarly, lightning results from air movements under certain weather conditions. You can also rub a balloon on your hair, and the static electricity created can then make the balloon cling to a wall. We also have to be cautious of static electricity, especially in dry climates. When we pump gasoline, we are warned to discharge ourselves (after sliding across the seat) on a metal surface before grabbing the gas nozzle. Attendants in hospital operating rooms must wear booties with aluminum foil on the bottoms to avoid creating sparks which may ignite the oxygen being used. Some of the most basic characteristics of static electricity include: • The effects of static electricity are explained by a physical quantity not previously introduced, called electric charge. • There are only two types of charge, one called positive and the other called negative. • Like charges repel, whereas unlike charges attract. • The force between charges decreases with distance. How do we know there are two types of electric charge? When various materials are rubbed together in controlled ways, certain combinations of materials always produce one type of charge on one material and the opposite type on the other. By convention, we call one type of charge “positive”, and the other type “negative.” For example, when glass is rubbed with silk, the glass becomes positively charged and the silk negatively charged. Since the glass and silk have opposite charges, they attract one another like clothes that have rubbed together in a dryer. Two glass rods rubbed with silk in this manner will repel one another, since each rod has positive charge on it. Similarly, two silk cloths so rubbed will repel, since both cloths have negative charge. Figure 18.4 shows how these simple materials can be used to explore the nature of the force between charges. Figure 18.4 A glass rod becomes positively charged when rubbed with silk, while the silk becomes negatively charged. (a) The glass rod is attracted to the silk because their charges are opposite. (b) Two similarly charged glass rods repel. (c) Two similarly charged silk cloths repel. More sophisticated questions arise. Where do these charges come from? Can you create or destroy charge? Is there a smallest unit of charge? Exactly how does the force depend on the amount of charge and the distance between charges? Such questions obviously occurred to Benjamin Franklin and other
early researchers, and they interest us even today. Charge Carried by Electrons and Protons Franklin wrote in his letters and books that he could see the effects of electric charge but did not understand what caused the phenomenon. Today we have the advantage of knowing that normal matter is made of atoms, and that atoms contain positive and negative charges, usually in equal amounts. Figure 18.5 shows a simple model of an atom with negative electrons orbiting its positive nucleus. The nucleus is positive due to the presence of positively charged protons. Nearly all charge in nature is due to electrons and protons, which are two of the three building blocks of most matter. (The third is the neutron, which is neutral, carrying no charge.) Other charge-carrying particles are observed in cosmic rays and nuclear decay, and are created in particle accelerators. All but the electron and proton survive only a short time and are quite rare by comparison. 778 Chapter 18 \| Electric Charge and Electric Field Figure 18.5 This simplified (and not to scale) view of an atom is called the planetary model of the atom. Negative electrons orbit a much heavier positive nucleus, as the planets orbit the much heavier sun. There the similarity ends, because forces in the atom are electromagnetic, whereas those in the planetary system are gravitational. Normal macroscopic amounts of matter contain immense numbers of atoms and molecules and, hence, even greater numbers of individual negative and positive charges. The charges of electrons and protons are identical in magnitude but opposite in sign. Furthermore, all charged objects in nature are integral multiples of this basic quantity of charge, meaning that all charges are made of combinations of a basic unit of charge. Usually, charges are formed by combinations of electrons and protons. The magnitude of this basic charge is The symbol is commonly used for charge and the subscript indicates the charge of a single electron (or proton). The SI unit of charge is the coulomb (C). The number of protons needed to make a charge of 1.00 C is ∣ ∣ = 1.60×10−19 C. 1.00 C× 1 proton 1.60×10−19 C = 6.25×1018 protons. (18.1) (18.2) Similarly, 6.25×1018 atom), there is a smallest bit of charge. There is no directly observed charge smaller than ∣ ∣ Small: The Submicroscopic Origin of Charge), and all observed charges are integral multiples of electrons have a combined charge of −1.00 coulomb. Just as there is a smallest bit of an element (an (see Things Great and ∣ ∣ . Things Great and Small: The Submicroscopic Origin of Charge With the exception of exotic, short-lived particles, all charge in nature is carried by electrons and protons. Electrons carry the charge we have named negative. Protons carry an equal-magnitude charge that we call positive. (See Figure 18.6.) Electron and proton charges are considered fundamental building blocks, since all other charges are integral multiples of those carried by electrons and protons. Electrons and protons are also two of the three fundamental building blocks of ordinary matter. The neutron is the third and has zero total charge. Figure 18.6 shows a person touching a Van de Graaff generator and receiving excess positive charge. The expanded view of a hair shows the existence of both types of charges but an excess of positive. The repulsion of these positive like charges causes the strands of hair to repel other strands of hair and to stand up. The further blowup shows an artist's conception of an electron and a proton perhaps found in an atom in a strand of hair. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 18 \| Electric Charge and Electric Field 779 Figure 18.6 When this person touches a Van de Graaff generator, she receives an excess of positive charge, causing her hair to stand on end. The charges in one hair are shown. An artist's conception of an electron and a proton illustrate the particles carrying the negative and positive charges. We cannot really see these particles with visible light because they are so small (the electron seems to be an infinitesimal point), but we know a great deal about their measurable properties, such as the charges they carry. The electron seems to have no substructure; in contrast, when the substructure of protons is explored by scattering extremely energetic electrons from them, it appears that there are point-like particles inside the proton. These sub-particles, named quarks, have never been directly observed, but they are believed to carry fractional charges as seen in Figure 18.7. Charges on electrons and protons and all other directly observable particles are unitary, but these quark substructures carry charges of either − 1 3 . There are continuing attempts to observe fractional charge directly and to learn of the properties of quarks, which or + 2 3 are perhaps the ultimate substructure of matter. Figure 18.7 Artist's conception of fractional quark charges inside a proton. A group of three quark charges add up to the single positive charge on the proton1 . Separation of Charge in Atoms Charges in atoms and molecules can be separated—for example, by rubbing materials together. Some atoms and molecules have a greater affinity for electrons than others and will become negatively charged by close contact in rubbing, leaving the other material positively charged. (See Figure 18.8.) Positive charge can similarly be induced by rubbing. Methods other than rubbing can also separate charges. Batteries, for example, use combinations of substances that interact in such a way as to separate charges. Chemical interactions may transfer negative charge from one substance to the other, making one battery terminal negative and leaving the first one positive. 780 Chapter 18 \| Electric Charge and Electric Field Figure 18.8 When materials are rubbed together, charges can be separated, particularly if one material has a greater affinity for electrons than another. (a) Both the amber and cloth are originally neutral, with equal positive and negative charges. Only a tiny fraction of the charges are involved, and only a few of them are shown here. (b) When rubbed together, some negative charge is transferred to the amber, leaving the cloth with a net positive charge. (c) When separated, the amber and cloth now have net charges, but the absolute value of the net positive and negative charges will be equal. No charge is actually created or destroyed when charges are separated as we have been discussing. Rather, existing charges are moved about. In fact, in all situations the total amount of charge is always constant. This universally obeyed law of nature is called the law of conservation of charge. Law of Conservation of Charge Total charge is constant in any process. Making Connections: Net Charge Hence if a closed system is neutral, it will remain neutral. Similarly, if a closed system has a charge, say, −10e, it will always have that charge. The only way to change the charge of a system is to transfer charge outside, either by bringing in charge or removing charge. If it is possible to transfer charge outside, the system is no longer closed/isolated and is known as an open system. However, charge is always conserved, for both open and closed systems. Consequently, the charge transferred to/from an open system is equal to the change in the system's charge. For example, each of the two materials (amber and cloth) discussed in Figure 18.8 have no net charge initially. The only way to change their charge is to transfer charge from outside each object. When they are rubbed together, negative charge is transferred to the amber and the final charge of the amber is the sum of the initial charge and the charge transferred to it. On the other hand, the final charge on the cloth is equal to its initial charge minus the charge transferred out. Similarly when glass is rubbed with silk, the net charge on the silk is its initial charge plus the incoming charge and the charge on the glass is the initial charge minus the outgoing charge. Also the charge gained by the silk will be equal to the charge lost by the glass, which means that if the silk gains –5e charge, the glass would have lost −5e charge. In more exotic situations, such as in particle accelerators, mass, Δ , can be created from energy in the amount Δ = 2 . Sometimes, the created mass is charged, such as when an electron is created. Whenever a charged particle is created, another having an opposite charge is always created along with it, so that the total charge created is zero. Usually, the two particles are “matter-antimatter” counterparts. For example, an antielectron would usually be created at the same time as an electron. The antielectron has a positive charge (it is called a positron), and so the total charge created is zero. (See Figure 18.9.) All particles have antimatter counterparts with opposite signs. When matter and antimatter counterparts are brought together, they completely annihilate one another. By annihilate, we mean that the mass of the two particles is converted to energy E, again obeying the relationship Δ = 2 annihilation; thus, total charge is conserved. . Since the two particles have equal and opposite charge, the total charge is zero before and after the Making Connections: Conservation Laws Only a limited number of physical quantities are universally conserved. Charge is one—energy, momentum, and angular momentum are others. Because they are conserved, these physical quantities are used to explain more phenomena and form more connections than other, less basic quantities. We find that conserved quantities give us great insight into the rules followed by nature and hints to the organization of nature. Discoveries of conservation laws have led to further discoveries, such as the weak nuclear force and the quark substructure of protons and other particles. This content is available for free at http://cnx.org/content/col11844/1.13
Chapter 18 \| Electric Charge and Electric Field 781 Figure 18.9 (a) When enough energy is present, it can be converted into matter. Here the matter created is an electron–antielectron pair. ( is the electron's mass.) The total charge before and after this event is zero. (b) When matter and antimatter collide, they annihilate each other; the total charge is conserved at zero before and after the annihilation. The law of conservation of charge is absolute—it has never been observed to be violated. Charge, then, is a special physical quantity, joining a very short list of other quantities in nature that are always conserved. Other conserved quantities include energy, momentum, and angular momentum. PhET Explorations: Balloons and Static Electricity Why does a balloon stick to your sweater? Rub a balloon on a sweater, then let go of the balloon and it flies over and sticks to the sweater. View the charges in the sweater, balloons, and the wall. Figure 18.10 Balloons and Static Electricity (http://cnx.org/content/m55300/1.2/balloons_en.jar) Applying the Science Practices: Electrical Charging Design an experiment to demonstrate the electrical charging of objects, by using a glass rod, a balloon, small bits of paper, and different pieces of cloth (like silk, wool, or nylon). Also show that like charges repel each other whereas unlike charges attract each other. 18.2 Conductors and Insulators Learning Objectives By the end of this section, you will be able to: • Define conductor and insulator, explain the difference, and give examples of each. • Describe three methods for charging an object. • Explain what happens to an electric force as you move farther from the source. • Define polarization. The information presented in this section supports the following AP® learning objectives and science practices: 782 Chapter 18 \| Electric Charge and Electric Field • 1.B.2.2 The student is able to make a qualitative prediction about the distribution of positive and negative electric charges within neutral systems as they undergo various processes. (S.P. 6.4, 7.2) • 1.B.2.3 The student is able to challenge claims that polarization of electric charge or separation of charge must result in a net charge on the object. (S.P. 6.1) • 4.E.3.1 The student is able to make predictions about the redistribution of charge during charging by friction, conduction, and induction. (S.P. 6.4) • 4.E.3.2 The student is able to make predictions about the redistribution of charge caused by the electric field due to other systems, resulting in charged or polarized objects. (S.P. 6.4, 7.2) • 4.E.3.3 The student is able to construct a representation of the distribution of fixed and mobile charge in insulators and conductors. (S.P. 1.1, 1.4, 6.4) • 4.E.3.4 The student is able to construct a representation of the distribution of fixed and mobile charge in insulators and conductors that predicts charge distribution in processes involving induction or conduction. (S.P. 1.1, 1.4, 6.4) • 4.E.3.5 The student is able to plan and/or analyze the results of experiments in which electric charge rearrangement occurs by electrostatic induction, or is able to refine a scientific question relating to such an experiment by identifying anomalies in a data set or procedure. (S.P. 3.2, 4.1, 4.2, 5.1, 5.3) Figure 18.11 This power adapter uses metal wires and connectors to conduct electricity from the wall socket to a laptop computer. The conducting wires allow electrons to move freely through the cables, which are shielded by rubber and plastic. These materials act as insulators that don't allow electric charge to escape outward. (credit: Evan-Amos, Wikimedia Commons) Some substances, such as metals and salty water, allow charges to move through them with relative ease. Some of the electrons in metals and similar conductors are not bound to individual atoms or sites in the material. These free electrons can move through the material much as air moves through loose sand. Any substance that has free electrons and allows charge to move relatively freely through it is called a conductor. The moving electrons may collide with fixed atoms and molecules, losing some energy, but they can move in a conductor. Superconductors allow the movement of charge without any loss of energy. Salty water and other similar conducting materials contain free ions that can move through them. An ion is an atom or molecule having a positive or negative (nonzero) total charge. In other words, the total number of electrons is not equal to the total number of protons. Other substances, such as glass, do not allow charges to move through them. These are called insulators. Electrons and ions in insulators are bound in the structure and cannot move easily—as much as 1023 water and dry table salt are insulators, for example, whereas molten salt and salty water are conductors. times more slowly than in conductors. Pure Figure 18.12 An electroscope is a favorite instrument in physics demonstrations and student laboratories. It is typically made with gold foil leaves hung from a (conducting) metal stem and is insulated from the room air in a glass-walled container. (a) A positively charged glass rod is brought near the tip of the electroscope, attracting electrons to the top and leaving a net positive charge on the leaves. Like charges in the light flexible gold leaves repel, separating them. (b) When the rod is touched against the ball, electrons are attracted and transferred, reducing the net charge on the glass rod but leaving the electroscope positively charged. (c) The excess charges are evenly distributed in the stem and leaves of the electroscope once the glass rod is removed. Charging by Contact Figure 18.12 shows an electroscope being charged by touching it with a positively charged glass rod. Because the glass rod is an insulator, it must actually touch the electroscope to transfer charge to or from it. (Note that the extra positive charges reside on the surface of the glass rod as a result of rubbing it with silk before starting the experiment.) Since only electrons move in This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 18 \| Electric Charge and Electric Field 783 metals, we see that they are attracted to the top of the electroscope. There, some are transferred to the positive rod by touch, leaving the electroscope with a net positive charge. Electrostatic repulsion in the leaves of the charged electroscope separates them. The electrostatic force has a horizontal component that results in the leaves moving apart as well as a vertical component that is balanced by the gravitational force. Similarly, the electroscope can be negatively charged by contact with a negatively charged object. Charging by Induction It is not necessary to transfer excess charge directly to an object in order to charge it. Figure 18.13 shows a method of induction wherein a charge is created in a nearby object, without direct contact. Here we see two neutral metal spheres in contact with one another but insulated from the rest of the world. A positively charged rod is brought near one of them, attracting negative charge to that side, leaving the other sphere positively charged. This is an example of induced polarization of neutral objects. Polarization is the separation of charges in an object that remains neutral. If the spheres are now separated (before the rod is pulled away), each sphere will have a net charge. Note that the object closest to the charged rod receives an opposite charge when charged by induction. Note also that no charge is removed from the charged rod, so that this process can be repeated without depleting the supply of excess charge. Another method of charging by induction is shown in Figure 18.14. The neutral metal sphere is polarized when a charged rod is brought near it. The sphere is then grounded, meaning that a conducting wire is run from the sphere to the ground. Since the earth is large and most ground is a good conductor, it can supply or accept excess charge easily. In this case, electrons are attracted to the sphere through a wire called the ground wire, because it supplies a conducting path to the ground. The ground connection is broken before the charged rod is removed, leaving the sphere with an excess charge opposite to that of the rod. Again, an opposite charge is achieved when charging by induction and the charged rod loses none of its excess charge. Figure 18.13 Charging by induction. (a) Two uncharged or neutral metal spheres are in contact with each other but insulated from the rest of the world. (b) A positively charged glass rod is brought near the sphere on the left, attracting negative charge and leaving the other sphere positively charged. (c) The spheres are separated before the rod is removed, thus separating negative and positive charge. (d) The spheres retain net charges after the inducing rod is removed—without ever having been touched by a charged object. 784 Chapter 18 \| Electric Charge and Electric Field Figure 18.14 Charging by induction, using a ground connection. (a) A positively charged rod is brought near a neutral metal sphere, polarizing it. (b) The sphere is grounded, allowing electrons to be attracted from the earth's ample supply. (c) The ground connection is broken. (d) The positive rod is removed, leaving the sphere with an induced negative charge. Figure 18.15 Both positive and negative objects attract a neutral object by polarizing its molecules. (a) A positive object brought near a neutral insulator polarizes its molecules. There is a slight shift in the distribution of the electrons orbiting the molecule, with unlike charges being brought nearer and like charges moved away. Since the electrostatic force decreases with distance, there is a net attraction. (b) A negative object produces the opposite polarization, but again attracts the neutral object. (c) The same effect occurs for a conductor; since the unlike charges are c
loser, there is a net attraction. Neutral objects can be attracted to any charged object. The pieces of straw attracted to polished amber are neutral, for example. If you run a plastic comb through your hair, the charged comb can pick up neutral pieces of paper. Figure 18.15 shows how the polarization of atoms and molecules in neutral objects results in their attraction to a charged object. When a charged rod is brought near a neutral substance, an insulator in this case, the distribution of charge in atoms and molecules is shifted slightly. Opposite charge is attracted nearer the external charged rod, while like charge is repelled. Since the electrostatic force decreases with distance, the repulsion of like charges is weaker than the attraction of unlike charges, and so there is a net attraction. Thus a positively charged glass rod attracts neutral pieces of paper, as will a negatively charged rubber rod. Some molecules, like water, are polar molecules. Polar molecules have a natural or inherent separation of charge, although they are neutral overall. Polar molecules are particularly affected by other charged objects and show greater polarization effects than molecules with naturally uniform charge distributions. Check Your Understanding Can you explain the attraction of water to the charged rod in the figure below? This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 18 \| Electric Charge and Electric Field 785 Figure 18.16 Solution Water molecules are polarized, giving them slightly positive and slightly negative sides. This makes water even more susceptible to a charged rod's attraction. As the water flows downward, due to the force of gravity, the charged conductor exerts a net attraction to the opposite charges in the stream of water, pulling it closer. Applying the Science Practices: Electrostatic Induction Plan an experiment to demonstrate electrostatic induction using household items, like balloons, woolen cloth, aluminum drink cans, or foam cups. Explain the process of induction in your experiment by discussing details of (and making diagrams relating to) the movement and alignment of charges. PhET Explorations: John Travoltage Make sparks fly with John Travoltage. Wiggle Johnnie's foot and he picks up charges from the carpet. Bring his hand close to the door knob and get rid of the excess charge. Figure 18.17 John Travoltage (http://cnx.org/content/m55301/1.2/travoltage_en.jar) 18.3 Conductors and Electric Fields in Static Equilibrium Learning Objectives By the end of this section, you will be able to: • List the three properties of a conductor in electrostatic equilibrium. • Explain the effect of an electric field on free charges in a conductor. • Explain why no electric field may exist inside a conductor. • Describe the electric field surrounding Earth. • Explain what happens to an electric field applied to an irregular conductor. • Describe how a lightning rod works. • Explain how a metal car may protect passengers inside from the dangerous electric fields caused by a downed line touching the car. The information presented in this section supports the following AP learning objectives: • 2.C.3.1 The student is able to explain the inverse square dependence of the electric field surrounding a spherically symmetric electrically charged object. 786 Chapter 18 \| Electric Charge and Electric Field • 2.C.5.1 The student is able to create representations of the magnitude and direction of the electric field at various distances (small compared to plate size) from two electrically charged plates of equal magnitude and opposite signs and is able to recognize that the assumption of uniform field is not appropriate near edges of plates. Conductors contain free charges that move easily. When excess charge is placed on a conductor or the conductor is put into a static electric field, charges in the conductor quickly respond to reach a steady state called electrostatic equilibrium. Figure 18.18 shows the effect of an electric field on free charges in a conductor. The free charges move until the field is perpendicular to the conductor's surface. There can be no component of the field parallel to the surface in electrostatic equilibrium, since, if there were, it would produce further movement of charge. A positive free charge is shown, but free charges can be either positive or negative and are, in fact, negative in metals. The motion of a positive charge is equivalent to the motion of a negative charge in the opposite direction. Figure 18.18 When an electric field E is applied to a conductor, free charges inside the conductor move until the field is perpendicular to the surface. (a) The electric field is a vector quantity, with both parallel and perpendicular components. The parallel component ( E∥ ) exerts a force ( F∥ ) on the free charge , which moves the charge until F∥ = 0 . (b) The resulting field is perpendicular to the surface. The free charge has been brought to the conductor's surface, leaving electrostatic forces in equilibrium. A conductor placed in an electric field will be polarized. Figure 18.19 shows the result of placing a neutral conductor in an originally uniform electric field. The field becomes stronger near the conductor but entirely disappears inside it. Figure 18.19 This illustration shows a spherical conductor in static equilibrium with an originally uniform electric field. Free charges move within the conductor, polarizing it, until the electric field lines are perpendicular to the surface. The field lines end on excess negative charge on one section of the surface and begin again on excess positive charge on the opposite side. No electric field exists inside the conductor, since free charges in the conductor would continue moving in response to any field until it was neutralized. Misconception Alert: Electric Field inside a Conductor Excess charges placed on a spherical conductor repel and move until they are evenly distributed, as shown in Figure 18.20. Excess charge is forced to the surface until the field inside the conductor is zero. Outside the conductor, the field is exactly the same as if the conductor were replaced by a point charge at its center equal to the excess charge. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 18 \| Electric Charge and Electric Field 787 Figure 18.20 The mutual repulsion of excess positive charges on a spherical conductor distributes them uniformly on its surface. The resulting electric field is perpendicular to the surface and zero inside. Outside the conductor, the field is identical to that of a point charge at the center equal to the excess charge. Properties of a Conductor in Electrostatic Equilibrium 1. The electric field is zero inside a conductor. 2. Just outside a conductor, the electric field lines are perpendicular to its surface, ending or beginning on charges on the surface. 3. Any excess charge resides entirely on the surface or surfaces of a conductor. The properties of a conductor are consistent with the situations already discussed and can be used to analyze any conductor in electrostatic equilibrium. This can lead to some interesting new insights, such as described below. How can a very uniform electric field be created? Consider a system of two metal plates with opposite charges on them, as shown in Figure 18.21. The properties of conductors in electrostatic equilibrium indicate that the electric field between the plates will be uniform in strength and direction. Except near the edges, the excess charges distribute themselves uniformly, producing field lines that are uniformly spaced (hence uniform in strength) and perpendicular to the surfaces (hence uniform in direction, since the plates are flat). The edge effects are less important when the plates are close together. Figure 18.21 Two metal plates with equal, but opposite, excess charges. The field between them is uniform in strength and direction except near the edges. One use of such a field is to produce uniform acceleration of charges between the plates, such as in the electron gun of a TV tube. Earth's Electric Field A near uniform electric field of approximately 150 N/C, directed downward, surrounds Earth, with the magnitude increasing slightly as we get closer to the surface. What causes the electric field? At around 100 km above the surface of Earth we have a layer of charged particles, called the ionosphere. The ionosphere is responsible for a range of phenomena including the electric field surrounding Earth. In fair weather the ionosphere is positive and the Earth largely negative, maintaining the electric field (Figure 18.22(a)). In storm conditions clouds form and localized electric fields can be larger and reversed in direction (Figure 18.22(b)). The exact charge distributions depend on the local conditions, and variations of Figure 18.22(b) are possible. If the electric field is sufficiently large, the insulating properties of the surrounding material break down and it becomes conducting. For air this occurs at around 3×106 form of lightning sparks and corona discharge. N/C. Air ionizes ions and electrons recombine, and we get discharge in the 788 Chapter 18 \| Electric Charge and Electric Field Figure 18.22 Earth's electric field. (a) Fair weather field. Earth and the ionosphere (a layer of charged particles) are both conductors. They produce a uniform electric field of about 150 N/C. (credit: D. H. Parks) (b) Storm fields. In the presence of storm clouds, the local electric fields can be larger. At very high fields, the insulating properties of the air break down and lightning can occur. (credit: Jan-Joost Verhoef) Electric Fields on Uneven Surfaces So far we have considered excess charges on a smooth, symmetrical conductor surface. What happens if a conductor has sharp corners or is pointed? Excess charges on a nonuniform conductor become concentrated at the sharpest points. Additionally, excess charge
may move on or off the conductor at the sharpest points. To see how and why this happens, consider the charged conductor in Figure 18.23. The electrostatic repulsion of like charges is most effective in moving them apart on the flattest surface, and so they become least concentrated there. This is because the forces between identical pairs of charges at either end of the conductor are identical, but the components of the forces parallel to the surfaces are different. The component parallel to the surface is greatest on the flattest surface and, hence, more effective in moving the charge. The same effect is produced on a conductor by an externally applied electric field, as seen in Figure 18.23 (c). Since the field lines must be perpendicular to the surface, more of them are concentrated on the most curved parts. Figure 18.23 Excess charge on a nonuniform conductor becomes most concentrated at the location of greatest curvature. (a) The forces between identical pairs of charges at either end of the conductor are identical, but the components of the forces parallel to the surface are different. It is F∥ that moves the charges apart once they have reached the surface. (b) F∥ producing the electric field shown. (c) An uncharged conductor in an originally uniform electric field is polarized, with the most concentrated charge at its most pointed end. is smallest at the more pointed end, the charges are left closer together, Applications of Conductors On a very sharply curved surface, such as shown in Figure 18.24, the charges are so concentrated at the point that the resulting electric field can be great enough to remove them from the surface. This can be useful. Lightning rods work best when they are most pointed. The large charges created in storm clouds induce an opposite charge on a building that can result in a lightning bolt hitting the building. The induced charge is bled away continually by a lightning rod, preventing the more dramatic lightning strike. Of course, we sometimes wish to prevent the transfer of charge rather than to facilitate it. In that case, the conductor should be very smooth and have as large a radius of curvature as possible. (See Figure 18.25.) Smooth surfaces are used on high-voltage transmission lines, for example, to avoid leakage of charge into the air. Another device that makes use of some of these principles is a Faraday cage. This is a metal shield that encloses a volume. All electrical charges will reside on the outside surface of this shield, and there will be no electrical field inside. A Faraday cage is used to prohibit stray electrical fields in the environment from interfering with sensitive measurements, such as the electrical signals inside a nerve cell. During electrical storms if you are driving a car, it is best to stay inside the car as its metal body acts as a Faraday cage with zero electrical field inside. If in the vicinity of a lightning strike, its effect is felt on the outside of the car and the inside is unaffected, provided you remain totally inside. This is also true if an active (“hot”) electrical wire was broken (in a storm or an accident) and fell on your car. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 18 \| Electric Charge and Electric Field 789 Figure 18.24 A very pointed conductor has a large charge concentration at the point. The electric field is very strong at the point and can exert a force large enough to transfer charge on or off the conductor. Lightning rods are used to prevent the buildup of large excess charges on structures and, thus, are pointed. Figure 18.25 (a) A lightning rod is pointed to facilitate the transfer of charge. (credit: Romaine, Wikimedia Commons) (b) This Van de Graaff generator has a smooth surface with a large radius of curvature to prevent the transfer of charge and allow a large voltage to be generated. The mutual repulsion of like charges is evident in the person's hair while touching the metal sphere. (credit: Jon ‘ShakataGaNai' Davis/Wikimedia Commons). 18.4 Coulomb’s Law By the end of this section, you will be able to: Learning Objectives • State Coulomb's law in terms of how the electrostatic force changes with the distance between two objects. • Calculate the electrostatic force between two point charges, such as electrons or protons. • Compare the electrostatic force to the gravitational attraction for a proton and an electron; for a human and the Earth. The information presented in this section supports the following AP® learning objectives and science practices: • 3.A.3.3 The student is able to describe a force as an interaction between two objects and identify both objects for any force. (S.P. 1.4) • 3.A.3.4 The student is able to make claims about the force on an object due to the presence of other objects with the same property: mass, electric charge. (S.P. 6.1, 6.4) • 3.C.2.1 The student is able to use Coulomb's law qualitatively and quantitatively to make predictions about the interaction between two electric point charges (interactions between collections of electric point charges are not covered in Physics 1 and instead are restricted to Physics 2). (S.P. 2.2, 6.4) • 3.C.2.2 The student is able to connect the concepts of gravitational force and electric force to compare similarities and differences between the forces. (S.P. 7.2) 790 Chapter 18 \| Electric Charge and Electric Field Figure 18.26 This NASA image of Arp 87 shows the result of a strong gravitational attraction between two galaxies. In contrast, at the subatomic level, the electrostatic attraction between two objects, such as an electron and a proton, is far greater than their mutual attraction due to gravity. (credit: NASA/HST) Through the work of scientists in the late 18th century, the main features of the electrostatic force—the existence of two types of charge, the observation that like charges repel, unlike charges attract, and the decrease of force with distance—were eventually refined, and expressed as a mathematical formula. The mathematical formula for the electrostatic force is called Coulomb's law after the French physicist Charles Coulomb (1736–1806), who performed experiments and first proposed a formula to calculate it. Coulomb's Law = \|1 2\| 2 Coulomb's law calculates the magnitude of the force between two point charges, 1 and 2 , separated by a distance . In SI units, the constant is equal to . (18.3) = 8.988×109N ⋅ m2 C2 ≈ 8.99×109N ⋅ m2 C2 . (18.4) The electrostatic force is a vector quantity and is expressed in units of newtons. The force is understood to be along the line joining the two charges. (See Figure 18.27.) Although the formula for Coulomb's law is simple, it was no mean task to prove it. The experiments Coulomb did, with the primitive equipment then available, were difficult. Modern experiments have verified Coulomb's law to great precision. For ∝ 1 / 2 example, it has been shown that the force is inversely proportional to distance between two objects squared to an accuracy of 1 part in 1016 . No exceptions have ever been found, even at the small distances within the atom. Figure 18.27 The magnitude of the electrostatic force between point charges 1 and 2 separated by a distance is given by Coulomb's law. Note that Newton's third law (every force exerted creates an equal and opposite force) applies as usual—the force on 1 is equal in magnitude and opposite in direction to the force it exerts on 2 . (a) Like charges. (b) Unlike charges. Making Connections: Comparing Gravitational and Electrostatic Forces Recall that the gravitational force (Newton's law of gravitation) quantifies force as = 2 . The comparison between the two forces—gravitational and electrostatic—shows some similarities and differences. Gravitational force is proportional to the masses of interacting objects, and the electrostatic force is proportional to the magnitudes of the charges of interacting objects. Hence both forces are proportional to a property that represents the strength of interaction for a given field. In addition, both forces are inversely proportional to the square of the distances between them. It may seem that the two forces are related but that is not the case. In fact, there are huge variations in the magnitudes of the two forces as they depend on different parameters and different mechanisms. For electrons (or protons), electrostatic force is dominant and is much greater than the gravitational force. On the other hand, gravitational force is This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 18 \| Electric Charge and Electric Field 791 generally dominant for objects with large masses. Another major difference between the two forces is that gravitational force can only be attractive, whereas electrostatic could be attractive or repulsive (depending on the sign of charges; unlike charges attract and like charges repel). Example 18.1 How Strong is the Coulomb Force Relative to the Gravitational Force? Compare the electrostatic force between an electron and proton separated by 0.530×10−10 m with the gravitational force between them. This distance is their average separation in a hydrogen atom. Strategy To compare the two forces, we first compute the electrostatic force using Coulomb's law, = \|1 2\| 2 . We then calculate the gravitational force using Newton's universal law of gravitation. Finally, we take a ratio to see how the forces compare in magnitude. Solution Entering the given and known information about the charges and separation of the electron and proton into the expression of Coulomb's law yields = \|1 2\| 2 = 8.99×109 N ⋅ m2 / C2 × (1.60×10–19 C)(1.60×10–19 C) (0.530×10–10 m)2 (18.5) (18.6) Thus the Coulomb force is = 8.19×10–8 N. The charges are opposite in sign, so this is an attractive force. This is a very large force for an electron—it would cause an acceleration of 8.99×1022 m / s2 (verification is left as an end-of-section problem).The gravitationa
l force is given by Newton's law of gravitation as: (18.7) = 2 where = 6.67×10−11 N ⋅ m2 / kg2 . Here and represent the electron and proton masses, which can be found in the appendices. Entering values for the knowns yields , (18.8) = (6.67×10 – 11 N ⋅ m2 / kg2)× (9.11×10–31 kg)(1.67×10–27 kg) (0.530×10–10 m)2 = 3.61×10–47 N (18.9) This is also an attractive force, although it is traditionally shown as positive since gravitational force is always attractive. The ratio of the magnitude of the electrostatic force to gravitational force in this case is, thus, = 2.27×1039. (18.10) Discussion This is a remarkably large ratio! Note that this will be the ratio of electrostatic force to gravitational force for an electron and a proton at any distance (taking the ratio before entering numerical values shows that the distance cancels). This ratio gives some indication of just how much larger the Coulomb force is than the gravitational force between two of the most common particles in nature. As the example implies, gravitational force is completely negligible on a small scale, where the interactions of individual charged particles are important. On a large scale, such as between the Earth and a person, the reverse is true. Most objects are nearly electrically neutral, and so attractive and repulsive Coulomb forces nearly cancel. Gravitational force on a large scale dominates interactions between large objects because it is always attractive, while Coulomb forces tend to cancel. Chapter 18 \| Electric Charge and Electric Field 801 The Van de Graaff Generator Van de Graaff generators (or Van de Graaffs) are not only spectacular devices used to demonstrate high voltage due to static electricity—they are also used for serious research. The first was built by Robert Van de Graaff in 1931 (based on original suggestions by Lord Kelvin) for use in nuclear physics research. Figure 18.38 shows a schematic of a large research version. Van de Graaffs utilize both smooth and pointed surfaces, and conductors and insulators to generate large static charges and, hence, large voltages. A very large excess charge can be deposited on the sphere, because it moves quickly to the outer surface. Practical limits arise because the large electric fields polarize and eventually ionize surrounding materials, creating free charges that neutralize excess charge or allow it to escape. Nevertheless, voltages of 15 million volts are well within practical limits. Figure 18.38 Schematic of Van de Graaff generator. A battery (A) supplies excess positive charge to a pointed conductor, the points of which spray the charge onto a moving insulating belt near the bottom. The pointed conductor (B) on top in the large sphere picks up the charge. (The induced electric field at the points is so large that it removes the charge from the belt.) This can be done because the charge does not remain inside the conducting sphere but moves to its outside surface. An ion source inside the sphere produces positive ions, which are accelerated away from the positive sphere to high velocities. Take-Home Experiment: Electrostatics and Humidity Rub a comb through your hair and use it to lift pieces of paper. It may help to tear the pieces of paper rather than cut them neatly. Repeat the exercise in your bathroom after you have had a long shower and the air in the bathroom is moist. Is it easier to get electrostatic effects in dry or moist air? Why would torn paper be more attractive to the comb than cut paper? Explain your observations. Xerography Most copy machines use an electrostatic process called xerography—a word coined from the Greek words xeros for dry and graphos for writing. The heart of the process is shown in simplified form in Figure 18.39. A selenium-coated aluminum drum is sprayed with positive charge from points on a device called a corotron. Selenium is a substance with an interesting property—it is a photoconductor. That is, selenium is an insulator when in the dark and a conductor when exposed to light. In the first stage of the xerography process, the conducting aluminum drum is grounded so that a negative charge is induced under the thin layer of uniformly positively charged selenium. In the second stage, the surface of the drum is exposed to the image of whatever is to be copied. Where the image is light, the selenium becomes conducting, and the positive charge is neutralized. In dark areas, the positive charge remains, and so the image has been transferred to the drum. The third stage takes a dry black powder, called toner, and sprays it with a negative charge so that it will be attracted to the positive regions of the drum. Next, a blank piece of paper is given a greater positive charge than on the drum so that it will pull Chapter 18 \| Electric Charge and Electric Field 805 electric field is given to be upward, the electric force is upward. We thus have a one-dimensional (vertical direction) problem, and we can state Newton's second law as where net = − . Entering this and the known values into the expression for Newton's second law yields = net . = − = 9.60×10−14 N − 3.92×10−14 N 4.00×10−15 kg = 14.2 m/s2. (18.24) (18.25) Discussion for (c) This is an upward acceleration great enough to carry the drop to places where you might not wish to have gasoline. This worked example illustrates how to apply problem-solving strategies to situations that include topics in different chapters. The first step is to identify the physical principles involved in the problem. The second step is to solve for the unknown using familiar problem-solving strategies. These are found throughout the text, and many worked examples show how to use them for single topics. In this integrated concepts example, you can see how to apply them across several topics. You will find these techniques useful in applications of physics outside a physics course, such as in your profession, in other science disciplines, and in everyday life. The following problems will build your skills in the broad application of physical principles. Unreasonable Results The Unreasonable Results exercises for this module have results that are unreasonable because some premise is unreasonable or because certain of the premises are inconsistent with one another. Physical principles applied correctly then produce unreasonable results. The purpose of these problems is to give practice in assessing whether nature is being accurately described, and if it is not to trace the source of difficulty. Problem-Solving Strategy To determine if an answer is reasonable, and to determine the cause if it is not, do the following. 1. Solve the problem using strategies as outlined above. Use the format followed in the worked examples in the text to solve the problem as usual. 2. Check to see if the answer is reasonable. Is it too large or too small, or does it have the wrong sign, improper units, and so on? 3. If the answer is unreasonable, look for what specifically could cause the identified difficulty. Usually, the manner in which the answer is unreasonable is an indication of the difficulty. For example, an extremely large Coulomb force could be due to the assumption of an excessively large separated charge. Glossary conductor: a material that allows electrons to move separately from their atomic orbits conductor: an object with properties that allow charges to move about freely within it Coulomb force: another term for the electrostatic force Coulomb interaction: the interaction between two charged particles generated by the Coulomb forces they exert on one another Coulomb's law: the mathematical equation calculating the electrostatic force vector between two charged particles dipole: a molecule's lack of symmetrical charge distribution, causing one side to be more positive and another to be more negative electric charge: a physical property of an object that causes it to be attracted toward or repelled from another charged object; each charged object generates and is influenced by a force called an electromagnetic force electric field: a three-dimensional map of the electric force extended out into space from a point charge electric field lines: a series of lines drawn from a point charge representing the magnitude and direction of force exerted by that charge 806 Chapter 18 \| Electric Charge and Electric Field electromagnetic force: one of the four fundamental forces of nature; the electromagnetic force consists of static electricity, moving electricity and magnetism electron: a particle orbiting the nucleus of an atom and carrying the smallest unit of negative charge electrostatic equilibrium: an electrostatically balanced state in which all free electrical charges have stopped moving about electrostatic force: the amount and direction of attraction or repulsion between two charged bodies electrostatic precipitators: filters that apply charges to particles in the air, then attract those charges to a filter, removing them from the airstream electrostatic repulsion: the phenomenon of two objects with like charges repelling each other electrostatics: the study of electric forces that are static or slow-moving Faraday cage: a metal shield which prevents electric charge from penetrating its surface field: a map of the amount and direction of a force acting on other objects, extending out into space free charge: an electrical charge (either positive or negative) which can move about separately from its base molecule free electron: an electron that is free to move away from its atomic orbit grounded: when a conductor is connected to the Earth, allowing charge to freely flow to and from Earth's unlimited reservoir grounded: connected to the ground with a conductor, so that charge flows freely to and from the Earth to the grounded object induction: the process by which an electrically charged object brought near a neutral object creates a charge in that object ink-jet printer: small ink droplets sprayed with an electric charge are
controlled by electrostatic plates to create images on paper insulator: a material that holds electrons securely within their atomic orbits ionosphere: a layer of charged particles located around 100 km above the surface of Earth, which is responsible for a range of phenomena including the electric field surrounding Earth laser printer: uses a laser to create a photoconductive image on a drum, which attracts dry ink particles that are then rolled onto a sheet of paper to print a high-quality copy of the image law of conservation of charge: is created simultaneously states that whenever a charge is created, an equal amount of charge with the opposite sign photoconductor: a substance that is an insulator until it is exposed to light, when it becomes a conductor point charge: A charged particle, designated , generating an electric field polar molecule: a molecule with an asymmetrical distribution of positive and negative charge polarization: slight shifting of positive and negative charges to opposite sides of an atom or molecule polarized: a state in which the positive and negative charges within an object have collected in separate locations proton: a particle in the nucleus of an atom and carrying a positive charge equal in magnitude and opposite in sign to the amount of negative charge carried by an electron screening: the dilution or blocking of an electrostatic force on a charged object by the presence of other charges nearby static electricity: a buildup of electric charge on the surface of an object test charge: A particle (designated ) with either a positive or negative charge set down within an electric field generated by a point charge Van de Graaff generator: a machine that produces a large amount of excess charge, used for experiments with high voltage vector: a quantity with both magnitude and direction vector addition: mathematical combination of two or more vectors, including their magnitudes, directions, and positions xerography: a dry copying process based on electrostatics This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 18 \| Electric Charge and Electric Field 807 Section Summary 18.1 Static Electricity and Charge: Conservation of Charge • There are only two types of charge, which we call positive and negative. • Like charges repel, unlike charges attract, and the force between charges decreases with the square of the distance. • The vast majority of positive charge in nature is carried by protons, while the vast majority of negative charge is carried by electrons. • The electric charge of one electron is equal in magnitude and opposite in sign to the charge of one proton. • An ion is an atom or molecule that has nonzero total charge due to having unequal numbers of electrons and protons. • The SI unit for charge is the coulomb (C), with protons and electrons having charges of opposite sign but equal magnitude; the magnitude of this basic charge ∣ ∣ is • Whenever charge is created or destroyed, equal amounts of positive and negative are involved. • Most often, existing charges are separated from neutral objects to obtain some net charge. • Both positive and negative charges exist in neutral objects and can be separated by rubbing one object with another. For macroscopic objects, negatively charged means an excess of electrons and positively charged means a depletion of electrons. ∣ ∣ = 1.60×10−19 C. • The law of conservation of charge ensures that whenever a charge is created, an equal charge of the opposite sign is created at the same time. 18.2 Conductors and Insulators • Polarization is the separation of positive and negative charges in a neutral object. • A conductor is a substance that allows charge to flow freely through its atomic structure. • An insulator holds charge within its atomic structure. • Objects with like charges repel each other, while those with unlike charges attract each other. • A conducting object is said to be grounded if it is connected to the Earth through a conductor. Grounding allows transfer of charge to and from the earth's large reservoir. • Objects can be charged by contact with another charged object and obtain the same sign charge. • • Polarized objects have their positive and negative charges concentrated in different areas, giving them a non-symmetrical If an object is temporarily grounded, it can be charged by induction, and obtains the opposite sign charge. charge. • Polar molecules have an inherent separation of charge. 18.3 Conductors and Electric Fields in Static Equilibrium • A conductor allows free charges to move about within it. • The electrical forces around a conductor will cause free charges to move around inside the conductor until static equilibrium is reached. • Any excess charge will collect along the surface of a conductor. • Conductors with sharp corners or points will collect more charge at those points. • A lightning rod is a conductor with sharply pointed ends that collect excess charge on the building caused by an electrical storm and allow it to dissipate back into the air. • Electrical storms result when the electrical field of Earth's surface in certain locations becomes more strongly charged, due to changes in the insulating effect of the air. • A Faraday cage acts like a shield around an object, preventing electric charge from penetrating inside. 18.4 Coulomb’s Law • Frenchman Charles Coulomb was the first to publish the mathematical equation that describes the electrostatic force between two objects. • Coulomb's law gives the magnitude of the force between point charges. It is = \|1 2\| 2 , where 1 and 2 are two point charges separated by a distance , and ≈ 8.99×109 N · m2/ C2 • This Coulomb force is extremely basic, since most charges are due to point-like particles. It is responsible for all electrostatic effects and underlies most macroscopic forces. • The Coulomb force is extraordinarily strong compared with the gravitational force, another basic force—but unlike gravitational force it can cancel, since it can be either attractive or repulsive. • The electrostatic force between two subatomic particles is far greater than the gravitational force between the same two particles. 18.5 Electric Field: Concept of a Field Revisited • The electrostatic force field surrounding a charged object extends out into space in all directions. 808 Chapter 18 \| Electric Charge and Electric Field • The electrostatic force exerted by a point charge on a test charge at a distance depends on the charge of both charges, as well as the distance between the two. • The electric field E is defined to be E = F where F is the Coulomb or electrostatic force exerted on a small positive test charge . E has units of N/C. • The magnitude of the electric field E created by a point charge is E = \|\| 2 . where is the distance from . The electric field E is a vector and fields due to multiple charges add like vectors. 18.6 Electric Field Lines: Multiple Charges • Drawings of electric field lines are useful visual tools. The properties of electric field lines for any charge distribution are that: • Field lines must begin on positive charges and terminate on negative charges, or at infinity in the hypothetical case of isolated charges. • The number of field lines leaving a positive charge or entering a negative charge is proportional to the magnitude of the charge. • The strength of the field is proportional to the closeness of the field lines—more precisely, it is proportional to the number of lines per unit area perpendicular to the lines. • The direction of the electric field is tangent to the field line at any point in space. • Field lines can never cross. 18.7 Electric Forces in Biology • Many molecules in living organisms, such as DNA, carry a charge. • An uneven distribution of the positive and negative charges within a polar molecule produces a dipole. • The effect of a Coulomb field generated by a charged object may be reduced or blocked by other nearby charged objects. • Biological systems contain water, and because water molecules are polar, they have a strong effect on other molecules in living systems. 18.8 Applications of Electrostatics • Electrostatics is the study of electric fields in static equilibrium. • In addition to research using equipment such as a Van de Graaff generator, many practical applications of electrostatics exist, including photocopiers, laser printers, ink-jet printers and electrostatic air filters. Conceptual Questions 18.1 Static Electricity and Charge: Conservation of Charge 1. There are very large numbers of charged particles in most objects. Why, then, don't most objects exhibit static electricity? 2. Why do most objects tend to contain nearly equal numbers of positive and negative charges? 18.2 Conductors and Insulators 3. An eccentric inventor attempts to levitate by first placing a large negative charge on himself and then putting a large positive charge on the ceiling of his workshop. Instead, while attempting to place a large negative charge on himself, his clothes fly off. Explain. 4. If you have charged an electroscope by contact with a positively charged object, describe how you could use it to determine the charge of other objects. Specifically, what would the leaves of the electroscope do if other charged objects were brought near its knob? 5. When a glass rod is rubbed with silk, it becomes positive and the silk becomes negative—yet both attract dust. Does the dust have a third type of charge that is attracted to both positive and negative? Explain. 6. Why does a car always attract dust right after it is polished? (Note that car wax and car tires are insulators.) 7. Describe how a positively charged object can be used to give another object a negative charge. What is the name of this process? 8. What is grounding? What effect does it have on a charged conductor? On a charged insulator? 18.3 Conductors and Electric Fields in Static Equilibrium 9. Is the object in a conductor or an ins
ulator? Justify your answer. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 18 \| Electric Charge and Electric Field 809 Figure 18.43 10. If the electric field lines in the figure above were perpendicular to the object, would it necessarily be a conductor? Explain. 11. The discussion of the electric field between two parallel conducting plates, in this module states that edge effects are less important if the plates are close together. What does close mean? That is, is the actual plate separation crucial, or is the ratio of plate separation to plate area crucial? 12. Would the self-created electric field at the end of a pointed conductor, such as a lightning rod, remove positive or negative charge from the conductor? Would the same sign charge be removed from a neutral pointed conductor by the application of a similar externally created electric field? (The answers to both questions have implications for charge transfer utilizing points.) 13. Why is a golfer with a metal club over her shoulder vulnerable to lightning in an open fairway? Would she be any safer under a tree? 14. Can the belt of a Van de Graaff accelerator be a conductor? Explain. 15. Are you relatively safe from lightning inside an automobile? Give two reasons. 16. Discuss pros and cons of a lightning rod being grounded versus simply being attached to a building. 17. Using the symmetry of the arrangement, show that the net Coulomb force on the charge at the center of the square below (Figure 18.44) is zero if the charges on the four corners are exactly equal. Figure 18.44 Four point charges , , , and lie on the corners of a square and is located at its center. 18. (a) Using the symmetry of the arrangement, show that the electric field at the center of the square in Figure 18.44 is zero if the charges on the four corners are exactly equal. (b) Show that this is also true for any combination of charges in which = and = 19. (a) What is the direction of the total Coulomb force on in Figure 18.44 if is negative, = and both are negative, and = and both are positive? (b) What is the direction of the electric field at the center of the square in this situation? 20. Considering Figure 18.44, suppose that = and = . First show that is in static equilibrium. (You may neglect the gravitational force.) Then discuss whether the equilibrium is stable or unstable, noting that this may depend on the signs of the charges and the direction of displacement of from the center of the square. 21. If = 0 in Figure 18.44, under what conditions will there be no net Coulomb force on ? 22. In regions of low humidity, one develops a special “grip” when opening car doors, or touching metal door knobs. This involves placing as much of the hand on the device as possible, not just the ends of one's fingers. Discuss the induced charge and explain why this is done. 23. Tollbooth stations on roadways and bridges usually have a piece of wire stuck in the pavement before them that will touch a car as it approaches. Why is this done? 810 Chapter 18 \| Electric Charge and Electric Field 24. Suppose a woman carries an excess charge. To maintain her charged status can she be standing on ground wearing just any pair of shoes? How would you discharge her? What are the consequences if she simply walks away? 18.4 Coulomb’s Law 25. Figure 18.45 shows the charge distribution in a water molecule, which is called a polar molecule because it has an inherent separation of charge. Given water's polar character, explain what effect humidity has on removing excess charge from objects. Figure 18.45 Schematic representation of the outer electron cloud of a neutral water molecule. The electrons spend more time near the oxygen than the hydrogens, giving a permanent charge separation as shown. Water is thus a polar molecule. It is more easily affected by electrostatic forces than molecules with uniform charge distributions. 26. Using Figure 18.45, explain, in terms of Coulomb's law, why a polar molecule (such as in Figure 18.45) is attracted by both positive and negative charges. 27. Given the polar character of water molecules, explain how ions in the air form nucleation centers for rain droplets. 18.5 Electric Field: Concept of a Field Revisited 28. Why must the test charge in the definition of the electric field be vanishingly small? 29. Are the direction and magnitude of the Coulomb force unique at a given point in space? What about the electric field? 18.6 Electric Field Lines: Multiple Charges 30. Compare and contrast the Coulomb force field and the electric field. To do this, make a list of five properties for the Coulomb force field analogous to the five properties listed for electric field lines. Compare each item in your list of Coulomb force field properties with those of the electric field—are they the same or different? (For example, electric field lines cannot cross. Is the same true for Coulomb field lines?) 31. Figure 18.46 shows an electric field extending over three regions, labeled I, II, and III. Answer the following questions. (a) Are there any isolated charges? If so, in what region and what are their signs? (b) Where is the field strongest? (c) Where is it weakest? (d) Where is the field the most uniform? Figure 18.46 18.7 Electric Forces in Biology This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 18 \| Electric Charge and Electric Field 811 32. A cell membrane is a thin layer enveloping a cell. The thickness of the membrane is much less than the size of the cell. In a static situation the membrane has a charge distribution of −2.5×10−6 C/m2 on its outer surface. Draw a diagram of the cell and the surrounding cell membrane. Include on this diagram the charge distribution and the corresponding electric field. Is there any electric field inside the cell? Is there any electric field outside the cell? C/m 2 on its inner surface and +2.5×10−6 812 Chapter 18 \| Electric Charge and Electric Field Problems & Exercises 18.1 Static Electricity and Charge: Conservation of Charge 1. Common static electricity involves charges ranging from nanocoulombs to microcoulombs. (a) How many electrons are needed to form a charge of –2.00 nC (b) How many electrons must be removed from a neutral object to leave a net charge of 0.500 C ? electrons move through a pocket calculator 2. If 1.80×1020 during a full day's operation, how many coulombs of charge moved through it? 3. To start a car engine, the car battery moves 3.75×1021 electrons through the starter motor. How many coulombs of charge were moved? 4. A certain lightning bolt moves 40.0 C of charge. How many fundamental units of charge ∣ ∣ is this? 18.2 Conductors and Insulators 5. Suppose a speck of dust in an electrostatic precipitator has 1.0000×1012 protons in it and has a net charge of –5.00 nC (a very large charge for a small speck). How many electrons does it have? 6. An amoeba has 1.00×1016 0.300 pC. (a) How many fewer electrons are there than protons? (b) If you paired them up, what fraction of the protons would have no electrons? 7. A 50.0 g ball of copper has a net charge of 2.00 C . What fraction of the copper's electrons has been removed? (Each copper atom has 29 protons, and copper has an atomic mass of 63.5.) protons and a net charge of 8. What net charge would you place on a 100 g piece of sulfur if you put an extra electron on 1 in 1012 of its atoms? (Sulfur has an atomic mass of 32.1.) 9. How many coulombs of positive charge are there in 4.00 kg of plutonium, given its atomic mass is 244 and that each plutonium atom has 94 protons? 18.3 Conductors and Electric Fields in Static Equilibrium 10. Sketch the electric field lines in the vicinity of the conductor in Figure 18.47 given the field was originally uniform and parallel to the object's long axis. Is the resulting field small near the long side of the object? Figure 18.48 12. Sketch the electric field between the two conducting plates shown in Figure 18.49, given the top plate is positive and an equal amount of negative charge is on the bottom plate. Be certain to indicate the distribution of charge on the plates. Figure 18.49 13. Sketch the electric field lines in the vicinity of the charged insulator in Figure 18.50 noting its nonuniform charge distribution. Figure 18.50 A charged insulating rod such as might be used in a classroom demonstration. 14. What is the force on the charge located at = 8.00 cm in Figure 18.51(a) given that = 1.00 μC ? Figure 18.47 11. Sketch the electric field lines in the vicinity of the conductor in Figure 18.48 given the field was originally uniform and parallel to the object's long axis. Is the resulting field small near the long side of the object? Figure 18.51 (a) Point charges located at 3.00, 8.00, and 11.0 cm along the x-axis. (b) Point charges located at 1.00, 5.00, 8.00, and 14.0 cm along the x-axis. 15. (a) Find the total electric field at = 1.00 cm in Figure 18.51(b) given that = 5.00 nC . (b) Find the total electric field at = 11.00 cm in Figure 18.51(b). (c) If the charges are allowed to move and eventually be brought to rest by friction, what will the final charge configuration be? (That is, This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 18 \| Electric Charge and Electric Field 813 will there be a single charge, double charge, etc., and what will its value(s) be?) 16. (a) Find the electric field at = 5.00 cm in Figure 18.51(a), given that = 1.00 μC . (b) At what position between 3.00 and 8.00 cm is the total electric field the same as that for –2 alone? (c) Can the electric field be zero anywhere between 0.00 and 8.00 cm? (d) At very large positive or negative values of x, the electric field approaches zero in both (a) and (b). In which does it most rapidly approach zero and why? (e) At what position to the right of 11.0 cm is the total electric field zero, other than at infinity? (Hint: A graphing calculator can yield considerable insight
in this problem.) 17. (a) Find the total Coulomb force on a charge of 2.00 nC located at = 4.00 cm in Figure 18.51 (b), given that = 1.00 μC . (b) Find the x-position at which the electric field is zero in Figure 18.51 (b). 18. Using the symmetry of the arrangement, determine the direction of the force on in the figure below, given that = =+7.50 μC and = = −7.50 μC . (b) Calculate the magnitude of the force on the charge , given that the square is 10.0 cm on a side and = 2.00 μC . Figure 18.52 19. (a) Using the symmetry of the arrangement, determine the direction of the electric field at the center of the square in Figure 18.52, given that = = −1.00 μC and = =+1.00 μC . (b) Calculate the magnitude of the electric field at the location of , given that the square is 5.00 cm on a side. 20. Find the electric field at the location of in Figure 18.52 given that = = =+2.00 nC , = −1.00 nC , and the square is 20.0 cm on a side. 21. Find the total Coulomb force on the charge in Figure 18.52, given that = 1.00 μC , = 2.00 μC , = −3.00 μC , = −4.00 μC , and =+1.00 μC . The square is 50.0 cm on a side. 22. (a) Find the electric field at the location of in Figure 18.53, given that b = +10.00 C and c = –5.00 C . (b) What is the force on , given that a = +1.50 nC ? Figure 18.53 Point charges located at the corners of an equilateral triangle 25.0 cm on a side. 23. (a) Find the electric field at the center of the triangular configuration of charges in Figure 18.53, given that =+2.50 nC , = −8.00 nC , and =+1.50 nC . (b) Is there any combination of charges, other than = = , that will produce a zero strength electric field at the center of the triangular configuration? 18.4 Coulomb’s Law 24. What is the repulsive force between two pith balls that are 8.00 cm apart and have equal charges of – 30.0 nC? 25. (a) How strong is the attractive force between a glass rod with a 0.700 C charge and a silk cloth with a –0.600 C charge, which are 12.0 cm apart, using the approximation that they act like point charges? (b) Discuss how the answer to this problem might be affected if the charges are distributed over some area and do not act like point charges. 26. Two point charges exert a 5.00 N force on each other. What will the force become if the distance between them is increased by a factor of three? 27. Two point charges are brought closer together, increasing the force between them by a factor of 25. By what factor was their separation decreased? 28. How far apart must two point charges of 75.0 nC (typical of static electricity) be to have a force of 1.00 N between them? 29. If two equal charges each of 1 C each are separated in air by a distance of 1 km, what is the magnitude of the force acting between them? You will see that even at a distance as large as 1 km, the repulsive force is substantial because 1 C is a very significant amount of charge. 30. A test charge of +2 C is placed halfway between a charge of +6 C and another of +4 C separated by 10 cm. (a) What is the magnitude of the force on the test charge? (b) What is the direction of this force (away from or toward the +6 C charge)? 31. Bare free charges do not remain stationary when close together. To illustrate this, calculate the acceleration of two isolated protons separated by 2.00 nm (a typical distance between gas atoms). Explicitly show how you follow the steps in the Problem-Solving Strategy for electrostatics. 32. (a) By what factor must you change the distance between two point charges to change the force between them by a factor of 10? (b) Explain how the distance can either increase or decrease by this factor and still cause a factor of 10 change in the force. 814 Chapter 18 \| Electric Charge and Electric Field (b) What magnitude and direction force does this field exert on a proton? 18.6 Electric Field Lines: Multiple Charges 47. (a) Sketch the electric field lines near a point charge + . (b) Do the same for a point charge –3.00 . 48. Sketch the electric field lines a long distance from the charge distributions shown in Figure 18.34 (a) and (b) 49. Figure 18.54 shows the electric field lines near two charges 1 and 2 . What is the ratio of their magnitudes? (b) Sketch the electric field lines a long distance from the charges shown in the figure. Figure 18.54 The electric field near two charges. 50. Sketch the electric field lines in the vicinity of two opposite charges, where the negative charge is three times greater in magnitude than the positive. (See Figure 18.54 for a similar situation). 18.8 Applications of Electrostatics 51. (a) What is the electric field 5.00 m from the center of the terminal of a Van de Graaff with a 3.00 mC charge, noting that the field is equivalent to that of a point charge at the center of the terminal? (b) At this distance, what force does the field exert on a 2.00 C charge on the Van de Graaff's belt? 52. (a) What is the direction and magnitude of an electric field that supports the weight of a free electron near the surface of Earth? (b) Discuss what the small value for this field implies regarding the relative strength of the gravitational and electrostatic forces. 53. A simple and common technique for accelerating electrons is shown in Figure 18.55, where there is a uniform electric field between two plates. Electrons are released, usually from a hot filament, near the negative plate, and there is a small hole in the positive plate that allows the electrons to continue moving. (a) Calculate the acceleration of the electron if the field strength is 2.50×104 N/C . (b) Explain why the electron will not be pulled back to the positive plate once it moves through the hole. 33. Suppose you have a total charge tot that you can split in any manner. Once split, the separation distance is fixed. How do you split the charge to achieve the greatest force? 34. (a) Common transparent tape becomes charged when pulled from a dispenser. If one piece is placed above another, the repulsive force can be great enough to support the top piece's weight. Assuming equal point charges (only an approximation), calculate the magnitude of the charge if electrostatic force is great enough to support the weight of a 10.0 mg piece of tape held 1.00 cm above another. (b) Discuss whether the magnitude of this charge is consistent with what is typical of static electricity. 35. (a) Find the ratio of the electrostatic to gravitational force between two electrons. (b) What is this ratio for two protons? (c) Why is the ratio different for electrons and protons? 36. At what distance is the electrostatic force between two protons equal to the weight of one proton? 37. A certain five cent coin contains 5.00 g of nickel. What fraction of the nickel atoms' electrons, removed and placed 1.00 m above it, would support the weight of this coin? The atomic mass of nickel is 58.7, and each nickel atom contains 28 electrons and 28 protons. 38. (a) Two point charges totaling 8.00 C exert a repulsive force of 0.150 N on one another when separated by 0.500 m. What is the charge on each? (b) What is the charge on each if the force is attractive? 39. Point charges of 5.00 C and –3.00 C are placed 0.250 m apart. (a) Where can a third charge be placed so that the net force on it is zero? (b) What if both charges are positive? 40. Two point charges 1 and 2 are 3.00 m apart, and their total charge is 20 C . (a) If the force of repulsion between them is 0.075N, what are magnitudes of the two charges? (b) If one charge attracts the other with a force of 0.525N, what are the magnitudes of the two charges? Note that you may need to solve a quadratic equation to reach your answer. 18.5 Electric Field: Concept of a Field Revisited 41. What is the magnitude and direction of an electric field that exerts a 2.00×10-5 N upward force on a –1.75 C charge? 42. What is the magnitude and direction of the force exerted on a 3.50 C charge by a 250 N/C electric field that points due east? 43. Calculate the magnitude of the electric field 2.00 m from a point charge of 5.00 mC (such as found on the terminal of a Van de Graaff). 44. (a) What magnitude point charge creates a 10,000 N/C electric field at a distance of 0.250 m? (b) How large is the field at 10.0 m? 45. Calculate the initial (from rest) acceleration of a proton in a 5.00×106 N/C electric field (such as created by a research Van de Graaff). Explicitly show how you follow the steps in the Problem-Solving Strategy for electrostatics. 46. (a) Find the direction and magnitude of an electric field that exerts a 4.80×10−17 N westward force on an electron. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 18 \| Electric Charge and Electric Field 815 because air begins to ionize and charges flow, reducing the field. (a) Calculate the distance a free proton must travel in this field to reach 3.00% of the speed of light, starting from rest. (b) Is this practical in air, or must it occur in a vacuum? 60. Integrated Concepts A 5.00 g charged insulating ball hangs on a 30.0 cm long string in a uniform horizontal electric field as shown in Figure 18.56. Given the charge on the ball is 1.00 C , find the strength of the field. Figure 18.55 Parallel conducting plates with opposite charges on them create a relatively uniform electric field used to accelerate electrons to the right. Those that go through the hole can be used to make a TV or computer screen glow or to produce X-rays. 54. Earth has a net charge that produces an electric field of approximately 150 N/C downward at its surface. (a) What is the magnitude and sign of the excess charge, noting the electric field of a conducting sphere is equivalent to a point charge at its center? (b) What acceleration will the field produce on a free electron near Earth's surface? (c) What mass object with a single extra electron will have its weight supported by this field? 55. Point charges of 25.0 C and 45.0 C are placed 0.500 m apart. (a) At what point along the line between them is the
electric field zero? (b) What is the electric field halfway between them? 56. What can you say about two charges 1 and 2 , if the electric field one-fourth of the way from 1 to 2 is zero? 57. Integrated Concepts Calculate the angular velocity ω of an electron orbiting a proton in the hydrogen atom, given the radius of the orbit is 0.530×10–10 m . You may assume that the proton is stationary and the centripetal force is supplied by Coulomb attraction. 58. Integrated Concepts An electron has an initial velocity of 5.00×106 m/s in a uniform 2.00×105 N/C strength electric field. The field accelerates the electron in the direction opposite to its initial velocity. (a) What is the direction of the electric field? (b) How far does the electron travel before coming to rest? (c) How long does it take the electron to come to rest? (d) What is the electron's velocity when it returns to its starting point? 59. Integrated Concepts The practical limit to an electric field in air is about 3.00×106 N/C . Above this strength, sparking takes place Figure 18.56 A horizontal electric field causes the charged ball to hang at an angle of 8.00º . 61. Integrated Concepts Figure 18.57 shows an electron passing between two charged metal plates that create an 100 N/C vertical electric field perpendicular to the electron's original horizontal velocity. (These can be used to change the electron's direction, such as in an oscilloscope.) The initial speed of the electron is 3.00×106 m/s , and the horizontal distance it travels in the uniform field is 4.00 cm. (a) What is its vertical deflection? (b) What is the vertical component of its final velocity? (c) At what angle does it exit? Neglect any edge effects. Figure 18.57 62. Integrated Concepts The classic Millikan oil drop experiment was the first to obtain an accurate measurement of the charge on an electron. In it, oil drops were suspended against the gravitational force by a vertical electric field. (See Figure 18.58.) Given the oil drop to be 1.00 m in radius and have a density of 920 kg/m3 (a) Find the weight of the drop. (b) If the drop has a single excess electron, find the electric field strength needed to balance its weight. : 816 Chapter 18 \| Electric Charge and Electric Field 67. Construct Your Own Problem Consider two insulating balls with evenly distributed equal and opposite charges on their surfaces, held with a certain distance between the centers of the balls. Construct a problem in which you calculate the electric field (magnitude and direction) due to the balls at various points along a line running through the centers of the balls and extending to infinity on either side. Choose interesting points and comment on the meaning of the field at those points. For example, at what points might the field be just that due to one ball and where does the field become negligibly small? Among the things to be considered are the magnitudes of the charges and the distance between the centers of the balls. Your instructor may wish for you to consider the electric field off axis or for a more complex array of charges, such as those in a water molecule. 68. Construct Your Own Problem Consider identical spherical conducting space ships in deep space where gravitational fields from other bodies are negligible compared to the gravitational attraction between the ships. Construct a problem in which you place identical excess charges on the space ships to exactly counter their gravitational attraction. Calculate the amount of excess charge needed. Examine whether that charge depends on the distance between the centers of the ships, the masses of the ships, or any other factors. Discuss whether this would be an easy, difficult, or even impossible thing to do in practice. Figure 18.58 In the Millikan oil drop experiment, small drops can be suspended in an electric field by the force exerted on a single excess electron. Classically, this experiment was used to determine the electron charge e by measuring the electric field and mass of the drop. 63. Integrated Concepts (a) In Figure 18.59, four equal charges lie on the corners of a square. A fifth charge is on a mass directly above the center of the square, at a height equal to the length of one side of the square. Determine the magnitude of in terms of , , and , if the Coulomb force is to equal the weight of . (b) Is this equilibrium stable or unstable? Discuss. Figure 18.59 Four equal charges on the corners of a horizontal square support the weight of a fifth charge located directly above the center of the square. 64. Unreasonable Results (a) Calculate the electric field strength near a 10.0 cm diameter conducting sphere that has 1.00 C of excess charge on it. (b) What is unreasonable about this result? (c) Which assumptions are responsible? 65. Unreasonable Results (a) Two 0.500 g raindrops in a thunderhead are 1.00 cm apart when they each acquire 1.00 mC charges. Find their acceleration. (b) What is unreasonable about this result? (c) Which premise or assumption is responsible? 66. Unreasonable Results A wrecking yard inventor wants to pick up cars by charging a 0.400 m diameter ball and inducing an equal and opposite charge on the car. If a car has a 1000 kg mass and the ball is to be able to lift it from a distance of 1.00 m: (a) What minimum charge must be used? (b) What is the electric field near the surface of the ball? (c) Why are these results unreasonable? (d) Which premise or assumption is responsible? This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 18 \| Electric Charge and Electric Field 817 Test Prep for AP® Courses Z is attracted to balloon Y. Which of the following can be the charge on Z? Select two answers. 18.1 Static Electricity and Charge: Conservation of Charge 1. When a glass rod is rubbed against silk, which of the following statements is true? a. Electrons are removed from the silk. b. Electrons are removed from the rod. c. Protons are removed from the silk. d. Protons are removed from the rod. 2. In an experiment, three microscopic latex spheres are sprayed into a chamber and become charged with +3e, +5e, and −3e, respectively. Later, all three spheres collide simultaneously and then separate. Which of the following are possible values for the final charges on the spheres? Select two answers. X Y Z (a) +4e −4e +5e (b) −4e +4.5e +5.5e (c) +5e −8e (d) +6e +6e +7e −7e 3. If objects X and Y attract each other, which of the following will be false? a. X has positive charge and Y has negative charge. b. X has negative charge and Y has positive charge. c. X and Y both have positive charge. d. X is neutral and Y has a charge. 4. Suppose a positively charged object A is brought in contact with an uncharged object B in a closed system. What type of charge will be left on object B? a. negative b. positive c. neutral d. cannot be determined 5. What will be the net charge on an object which attracts neutral pieces of paper but repels a negatively charged balloon? a. negative b. positive c. neutral d. cannot be determined 6. When two neutral objects are rubbed against each other, the first one gains a net charge of 3e. Which of the following statements is true? a. The second object gains 3e and is negatively charged. b. The second object loses 3e and is negatively charged. c. The second object gains 3e and is positively charged. d. The second object loses 3e and is positively charged. 7. In an experiment, a student runs a comb through his hair several times and brings it close to small pieces of paper. Which of the following will he observe? a. Pieces of paper repel the comb. b. Pieces of paper are attracted to the comb. c. Some pieces of paper are attracted and some repel the comb. d. There is no attraction or repulsion between the pieces of paper and the comb. 8. In an experiment a negatively charged balloon (balloon X) is repelled by another charged balloon Y. However, an object a. negative b. positive c. neutral d. cannot be determined 9. Suppose an object has a charge of 1 C and gains 6.88×1018 electrons. a. What will be the net charge of the object? b. If the object has gained electrons from a neutral object, what will be the charge on the neutral object? c. Find and explain the relationship between the total charges of the two objects before and after the transfer. d. When a third object is brought in contact with the first object (after it gains the electrons), the resulting charge on the third object is 0.4 C. What was its initial charge? 10. The charges on two identical metal spheres (placed in a closed system) are -2.4×10−17 C and -4.8×10−17 C. a. How many electrons will be equivalent to the charge on b. each sphere? If the two spheres are brought in contact and then separated, find the charge on each sphere. c. Calculate the number of electrons that would be equivalent to the resulting charge on each sphere. 11. In an experiment the following observations are made by a student for four charged objects W, X, Y, and Z: • A glass rod rubbed with silk attracts W. • W attracts Z but repels X. • X attracts Z but repels Y. • Y attracts W and Z. Estimate whether the charges on each of the four objects are positive, negative, or neutral. 18.2 Conductors and Insulators 12. Some students experimenting with an uncharged metal sphere want to give the sphere a net charge using a charged aluminum pie plate. Which of the following steps would give the sphere a net charge of the same sign as the pie plate? a. bringing the pie plate close to, but not touching, the metal sphere, then moving the pie plate away. b. bringing the pie plate close to, but not touching, the metal sphere, then momentarily touching a grounding wire to the metal sphere. c. bringing the pie plate close to, but not touching, the metal sphere, then momentarily touching a grounding wire to the pie plate. touching the pie plate to the metal sphere. d. 13. Figure 18.60 Balloon and sphere. When the balloon is brought closer to the sphere,
there will be a redistribution of charges. What is this phenomenon called? a. electrostatic repulsion 818 Chapter 18 \| Electric Charge and Electric Field b. conduction c. polarization d. none of the above 14. What will be the charge at Y (i.e., the part of the sphere furthest from the balloon)? a. positive b. negative c. zero d. It can be positive or negative depending on the material. 15. What will be the net charge on the sphere? a. positive b. negative c. zero d. It can be positive or negative depending on the material. second experiment the rod is only brought close to the electroscope but not in contact. However, while the rod is close, the electroscope is momentarily grounded and then the rod is removed. In both experiments the needles of the electroscopes deflect, which indicates the presence of charges. a. What is the charging method in each of the two experiments? b. What is the net charge on the electroscope in the first c. experiment? Explain how the electroscope obtains that charge. Is the net charge on the electroscope in the second experiment different from that of the first experiment? Explain why. 16. If Y is grounded while the balloon is still close to X, which of the following will be true? 18.3 Conductors and Electric Fields in Static Equilibrium a. Electrons will flow from the sphere to the ground. b. Electrons will flow from the ground to the sphere. c. Protons will flow from the sphere to the ground. d. Protons will flow from the ground to the sphere. 21. 17. If the balloon is moved away after grounding, what will be the net charge on the sphere? a. positive b. negative c. zero d. It can be positive or negative depending on the material. 18. A positively charged rod is used to charge a sphere by induction. Which of the following is true? a. The sphere must be a conductor. b. The sphere must be an insulator. c. The sphere can be a conductor or insulator but must be connected to ground. d. The sphere can be a conductor or insulator but must be already charged. 19. Figure 18.62 A sphere conductor. An electric field due to a positively charged spherical conductor is shown above. Where will the electric field be weakest? a. Point A b. Point B c. Point C d. Same at all points 22. Figure 18.63 Electric field between two parallel metal plates. The electric field created by two parallel metal plates is shown above. Where will the electric field be strongest? a. Point A b. Point B c. Point C d. Same at all points 23. Suppose that the electric field experienced due to a positively charged small spherical conductor at a certain distance is E. What will be the percentage change in electric field experienced at thrice the distance if the charge on the conductor is doubled? 24. Figure 18.61 Rod and metal balls. As shown in the figure above, two metal balls are suspended and a negatively charged rod is brought close to them. a. If the two balls are in contact with each other what will be the charges on each ball? b. Explain how the balls get these charges. c. What will happen to the charge on the second ball (i.e., the ball further away from the rod) if it is momentarily grounded while the rod is still there? If (instead of grounding) the second ball is moved away and then the rod is removed from the first ball, will the two balls have induced charges? If yes, what will be the charges? If no, why not? d. 20. Two experiments are performed using positively charged glass rods and neutral electroscopes. In the first experiment the rod is brought in contact with the electroscope. In the This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 18 \| Electric Charge and Electric Field 819 b. Will this ratio change if the two electrons are replaced by protons? If yes, find the new ratio. 18.5 Electric Field: Concept of a Field Revisited 31. Two particles with charges +2q and +q are separated by a distance r. The +2q particle has an electric field E at distance r and exerts a force F on the +q particle. Use this information to answer questions 31–32. What is the electric field of the +q particle at the same distance and what force does it exert on the +2q particle? a. E/2, F/2 b. E, F/2 c. E/2, F d. E, F 32. When the +q particle is replaced by a +3q particle, what will be the electric field and force from the +2q particle experienced by the +3q particle? a. E/3, 3F b. E, 3F c. E/3, F d. E, F 33. The direction of the electric field of a negative charge is inward for both positive and negative charges. a. b. outward for both positive and negative charges. c. inward for other positive charges and outward for other negative charges. d. outward for other positive charges and inward for other negative charges. 34. The force responsible for holding an atom together is frictional a. b. electric c. gravitational d. magnetic 35. When a positively charged particle exerts an inward force on another particle P, what will be the charge of P? a. positive b. negative c. neutral d. cannot be determined 36. Find the force exerted due to a particle having a charge of 3.2×10−19 C on another identical particle 5 cm away. 37. Suppose that the force exerted on an electron is 5.6×10−17 N, directed to the east. a. Find the magnitude of the electric field that exerts the force. b. What will be the direction of the electric field? c. If the electron is replaced by a proton, what will be the magnitude of force exerted? d. What will be the direction of force on the proton? 18.6 Electric Field Lines: Multiple Charges 38. Figure 18.65 An electric dipole (with +2q and –2q as the two charges) is shown in the figure above. A third charge, −q is Figure 18.64 Millikan oil drop experiment. The classic Millikan oil drop experiment setup is shown above. In this experiment oil drops are suspended in a vertical electric field against the gravitational force to measure their charge. If the mass of a negatively charged drop suspended in an electric field of 1.18×10−4 N/C strength is 3.85×10−21 g, find the number of excess electrons in the drop. 18.4 Coulomb’s Law 25. For questions 25–27, suppose that the electrostatics force between two charges is F. What will be the force if the distance between them is halved? a. 4F b. 2F c. F/4 d. F/2 26. Which of the following is false? a. b. If the charge of one of the particles is doubled and that of the second is unchanged, the force will become 2F. If the charge of one of the particles is doubled and that of the second is halved, the force will remain F. If the charge of both the particles is doubled, the force will become 4F. d. None of the above. c. 27. Which of the following is true about the gravitational force between the particles? a. b. c. d. It will be 3.25×10−38 F. It will be 3.25×1038 F. It will be equal to F. It is not possible to determine the gravitational force as the masses of the particles are not given. 28. Two massive, positively charged particles are initially held a fixed distance apart. When they are moved farther apart, the magnitude of their mutual gravitational force changes by a factor of n. Which of the following indicates the factor by which the magnitude of their mutual electrostatic force changes? a. 1/n2 b. 1/n c. n d. n2 29. a. What is the electrostatic force between two charges of 1 C each, separated by a distance of 0.5 m? b. How will this force change if the distance is increased to 1 m? 30. a. Find the ratio of the electrostatic force to the gravitational force between two electrons. 820 Chapter 18 \| Electric Charge and Electric Field placed equidistant from the dipole charges. What will be the direction of the net force on the third charge? negative and that the sign of the charge of object S is positive. a. → b. ← c. ↓ d. ↑ 39. ii) Briefly describe the characteristics of the field diagram that indicate that the magnitudes of the charges of objects R and T are equal and that the magnitude of the charge of object S is about twice that of objects R and T. For the following parts, an electric field directed to the right is defined to be positive. (b) On the axes below, sketch a graph of the electric field E along the x-axis as a function of position x. Figure 18.68 An Electric field (E) axis and Position (x) axis. (c) Write an expression for the electric field E along the x-axis as a function of position x in the region between objects S and T in terms of q, d, and fundamental constants, as appropriate. (d) Your classmate tells you there is a point between S and T where the electric field is zero. Determine whether this statement is true, and explain your reasoning using two of the representations from parts (a), (b), or (c). Figure 18.66 Four objects, each with charge +q, are held fixed on a square with sides of length d, as shown in the figure. Objects X and Z are at the midpoints of the sides of the square. The electrostatic force exerted by object W on object X is F. Use this information to answer questions 39–40. What is the magnitude of force exerted by object W on Z? a. F/7 b. F/5 c. F/3 d. F/2 40. What is the magnitude of the net force exerted on object X by objects W, Y, and Z? a. F/4 b. F/2 c. 9F/4 d. 3F 41. Figure 18.67 Electric field with three charged objects. The figure above represents the electric field in the vicinity of three small charged objects, R, S, and T. The objects have charges −q, +2q, and −q, respectively, and are located on the x-axis at −d, 0, and d. Field vectors of very large magnitude are omitted for clarity. (a) i) Briefly describe the characteristics of the field diagram that indicate that the sign of the charges of objects R and T is This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 20 \| Electric Current, Resistance, and Ohm's Law 867 20 ELECTRIC CURRENT, RESISTANCE, AND OHM'S LAW Figure 20.1 Electric energy in massive quantities is transmitted from this hydroelectric facility, the Srisailam power station located along the Krishna River in India (http://en.wikipedia.org/wiki/Srisailam_Da
m) , by the movement of charge—that is, by electric current. (credit: Chintohere, Wikimedia Commons) Chapter Outline 20.1. Current 20.2. Ohm’s Law: Resistance and Simple Circuits 20.3. Resistance and Resistivity 20.4. Electric Power and Energy 20.5. Alternating Current versus Direct Current 20.6. Electric Hazards and the Human Body 20.7. Nerve Conduction–Electrocardiograms Connection for AP® Courses In our daily lives, we see and experience many examples of electricity which involve electric current, the movement of charge. These include the flicker of numbers on a handheld calculator, nerve impulses carrying signals of vision to the brain, an ultrasound device sending a signal to a computer screen, the brain sending a message for a baby to twitch its toes, an electric train pulling its load over a mountain pass, and a hydroelectric plant sending energy to metropolitan and rural users. Humankind has indeed harnessed electricity, the basis of technology, to improve the quality of life. While the previous two chapters concentrated on static electricity and the fundamental force underlying its behavior, the next few chapters will be 868 Chapter 20 \| Electric Current, Resistance, and Ohm's Law devoted to electric and magnetic phenomena involving electric current. In addition to exploring applications of electricity, we shall gain new insights into its nature – in particular, the fact that all magnetism results from electric current. This chapter supports learning objectives covered under Big Ideas 1, 4, and 5 of the AP Physics Curriculum Framework. Electric charge is a property of a system (Big Idea 1) that affects its interaction with other charged systems (Enduring Understanding 1.B), whereas electric current is fundamentally the movement of charge through a conductor and is based on the fact that electric charge is conserved within a system (Essential Knowledge 1.B.1). The conservation of charge also leads to the concept of an electric circuit as a closed loop of electrical current. In addition, this chapter discusses examples showing that the current in a circuit is resisted by the elements of the circuit and the strength of the resistance depends on the material of the elements. The macroscopic properties of materials, including resistivity, depend on their molecular and atomic structure (Enduring Understanding 1.E). In addition, resistivity depends on the temperature of the material (Essential Knowledge 1.E.2). The chapter also describes how the interaction of systems of objects can result in changes in those systems (Big Idea 4). For example, electric properties of a system of charged objects can change in response to the presence of, or changes in, other charged objects or systems (Enduring Understanding 4.E). A simple circuit with a resistor and an energy source is an example of such a system. The current through the resistor in the circuit is equal to the difference of potentials across the resistor divided by its resistance (Essential Knowledge 4.E.4). The unifying theme of the physics curriculum is that any changes in the systems due to interactions are governed by laws of conservation (Big Idea 5). This chapter applies the idea of energy conservation (Enduring Understanding 5.B) to electric circuits and connects concepts of electric energy and electric power as rates of energy use (Essential Knowledge 5.B.5). While the laws of conservation of energy in electric circuits are fully described by Kirchoff's rules, which are introduced in the next chapter (Essential Knowledge 5.B.9), the specific definition of power (based on Essential Knowledge 5.B.9) is that it is the rate at which energy is transferred from a resistor as the product of the electric potential difference across the resistor and the current through the resistor. Big Idea 1 Objects and systems have properties such as mass and charge. Systems may have internal structure. Enduring Understanding 1.B Electric charge is a property of an object or system that affects its interactions with other objects or systems containing charge. Essential Knowledge 1.B.1 Electric charge is conserved. The net charge of a system is equal to the sum of the charges of all the objects in the system. Enduring Understanding 1.E Materials have many macroscopic properties that result from the arrangement and interactions of the atoms and molecules that make up the material. Essential Knowledge 1.E.2 Matter has a property called resistivity. Big Idea 4 Interactions between systems can result in changes in those systems. Enduring Understanding 4.E The electric and magnetic properties of a system can change in response to the presence of, or changes in, other objects or systems. Essential Knowledge 4.E.4 The resistance of a resistor, and the capacitance of a capacitor, can be understood from the basic properties of electric fields and forces, as well as the properties of materials and their geometry. Big Idea 5: Changes that occur as a result of interactions are constrained by conservation laws. Enduring Understanding 5.B The energy of a system is conserved. Essential Knowledge 5.B.5 Energy can be transferred by an external force exerted on an object or system that moves the object or system through a distance; this energy transfer is called work. Energy transfer in mechanical or electrical systems may occur at different rates. Power is deﬁned as the rate of energy transfer into, out of, or within a system. [A piston ﬁlled with gas getting compressed or expanded is treated in Physics 2 as a part of thermodynamics.] Essential Knowledge 5.B.9 Kirchhoff's loop rule describes conservation of energy in electrical circuits. [The application of Kirchhoff's laws to circuits is introduced in Physics 1 and further developed in Physics 2 in the context of more complex circuits, including those with capacitors.] 20.1 Current Learning Objectives By the end of this section, you will be able to: • Define electric current, ampere, and drift velocity. • Describe the direction of charge flow in conventional current. • Use drift velocity to calculate current and vice versa. The information presented in this section supports the following AP® learning objectives and science practices: • 1.B.1.1 The student is able to make claims about natural phenomena based on conservation of electric charge. (S.P. 6.4) • 1.B.1.2 The student is able to make predictions, using the conservation of electric charge, about the sign and relative quantity of net charge of objects or systems after various charging processes, including conservation of charge in simple circuits. (S.P. 6.4, 7.2) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 20 \| Electric Current, Resistance, and Ohm's Law 869 Electric Current Electric current is defined to be the rate at which charge flows. A large current, such as that used to start a truck engine, moves a large amount of charge in a small time, whereas a small current, such as that used to operate a hand-held calculator, moves a small amount of charge over a long period of time. In equation form, electric current is defined to be = Δ Δ , (20.1) where Δ is the amount of charge passing through a given area in time Δ . (As in previous chapters, initial time is often taken to be zero, in which case Δ = .) (See Figure 20.2.) The SI unit for current is the ampere (A), named for the French physicist André-Marie Ampère (1775–1836). Since = Δ / Δ , we see that an ampere is one coulomb per second: 1 A = 1 C/s (20.2) Not only are fuses and circuit breakers rated in amperes (or amps), so are many electrical appliances. Figure 20.2 The rate of flow of charge is current. An ampere is the flow of one coulomb through an area in one second. Example 20.1 Calculating Currents: Current in a Truck Battery and a Handheld Calculator (a) What is the current involved when a truck battery sets in motion 720 C of charge in 4.00 s while starting an engine? (b) How long does it take 1.00 C of charge to flow through a handheld calculator if a 0.300-mA current is flowing? Strategy We can use the definition of current in the equation = Δ / Δ to find the current in part (a), since charge and time are given. In part (b), we rearrange the definition of current and use the given values of charge and current to find the time required. Solution for (a) Entering the given values for charge and time into the definition of current gives = Δ Δ = 180 A. = 720 C 4.00 s = 180 C/s (20.3) Discussion for (a) This large value for current illustrates the fact that a large charge is moved in a small amount of time. The currents in these “starter motors” are fairly large because large frictional forces need to be overcome when setting something in motion. Solution for (b) Solving the relationship = Δ / Δ for time Δ , and entering the known values for charge and current gives Δ = Δ = 1.00 C 0.30010-3 C/s = 3.33103 s. (20.4) Discussion for (b) This time is slightly less than an hour. The small current used by the hand-held calculator takes a much longer time to move a smaller charge than the large current of the truck starter. So why can we operate our calculators only seconds after turning them on? It's because calculators require very little energy. Such small current and energy demands allow handheld calculators to operate from solar cells or to get many hours of use out of small batteries. Remember, calculators do not have moving parts in the same way that a truck engine has with cylinders and pistons, so the technology requires smaller currents. 870 Chapter 20 \| Electric Current, Resistance, and Ohm's Law Figure 20.3 shows a simple circuit and the standard schematic representation of a battery, conducting path, and load (a resistor). Schematics are very useful in visualizing the main features of a circuit. A single schematic can represent a wide variety of situations. The schematic in Figure 20.3 (b), for example, can represent anything from a truck battery connected to a headlight lighting th
e street in front of the truck to a small battery connected to a penlight lighting a keyhole in a door. Such schematics are useful because the analysis is the same for a wide variety of situations. We need to understand a few schematics to apply the concepts and analysis to many more situations. Figure 20.3 (a) A simple electric circuit. A closed path for current to flow through is supplied by conducting wires connecting a load to the terminals of a battery. (b) In this schematic, the battery is represented by the two parallel red lines, conducting wires are shown as straight lines, and the zigzag represents the load. The schematic represents a wide variety of similar circuits. Note that the direction of current in Figure 20.3 is from positive to negative. The direction of conventional current is the direction that positive charge would flow. In a single loop circuit (as shown in Figure 20.3), the value for current at all points of the circuit should be the same if there are no losses. This is because current is the flow of charge and charge is conserved, i.e., the charge flowing out from the battery will be the same as the charge flowing into the battery. Depending on the situation, positive charges, negative charges, or both may move. In metal wires, for example, current is carried by electrons—that is, negative charges move. In ionic solutions, such as salt water, both positive and negative charges move. This is also true in nerve cells. A Van de Graaff generator used for nuclear research can produce a current of pure positive charges, such as protons. Figure 20.4 illustrates the movement of charged particles that compose a current. The fact that conventional current is taken to be in the direction that positive charge would flow can be traced back to American politician and scientist Benjamin Franklin in the 1700s. He named the type of charge associated with electrons negative, long before they were known to carry current in so many situations. Franklin, in fact, was totally unaware of the small-scale structure of electricity. It is important to realize that there is an electric field in conductors responsible for producing the current, as illustrated in Figure 20.4. Unlike static electricity, where a conductor in equilibrium cannot have an electric field in it, conductors carrying a current have an electric field and are not in static equilibrium. An electric field is needed to supply energy to move the charges. Making Connections: Take-Home Investigation—Electric Current Illustration Find a straw and little peas that can move freely in the straw. Place the straw flat on a table and fill the straw with peas. When you pop one pea in at one end, a different pea should pop out the other end. This demonstration is an analogy for an electric current. Identify what compares to the electrons and what compares to the supply of energy. What other analogies can you find for an electric current? Note that the flow of peas is based on the peas physically bumping into each other; electrons flow due to mutually repulsive electrostatic forces. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 20 \| Electric Current, Resistance, and Ohm's Law 871 Figure 20.4 Current is the rate at which charge moves through an area , such as the cross-section of a wire. Conventional current is defined to move in the direction of the electric field. (a) Positive charges move in the direction of the electric field and the same direction as conventional current. (b) Negative charges move in the direction opposite to the electric field. Conventional current is in the direction opposite to the movement of negative charge. The flow of electrons is sometimes referred to as electronic flow. Example 20.2 Calculating the Number of Electrons that Move through a Calculator If the 0.300-mA current through the calculator mentioned in the Example 20.1 example is carried by electrons, how many electrons per second pass through it? Strategy The current calculated in the previous example was defined for the flow of positive charge. For electrons, the magnitude is the same, but the sign is opposite, electrons = −0.300×10−3 C/s .Since each electron (−) has a charge of –1.60×10−19 C , we can convert the current in coulombs per second to electrons per second. Solution Starting with the definition of current, we have electrons = Δelectrons Δ = –0.300×10−3 C s . We divide this by the charge per electron, so that s = –0.30010 – 3 C – = 1.881015 – s . s 1 – –1.6010−19 C (20.5) (20.6) Discussion There are so many charged particles moving, even in small currents, that individual charges are not noticed, just as individual water molecules are not noticed in water flow. Even more amazing is that they do not always keep moving forward like soldiers in a parade. Rather they are like a crowd of people with movement in different directions but a general trend to move forward. There are lots of collisions with atoms in the metal wire and, of course, with other electrons. Drift Velocity Electrical signals are known to move very rapidly. Telephone conversations carried by currents in wires cover large distances without noticeable delays. Lights come on as soon as a switch is flicked. Most electrical signals carried by currents travel at speeds on the order of 108 m/s , a significant fraction of the speed of light. Interestingly, the individual charges that make up the current move much more slowly on average, typically drifting at speeds on the order of 10−4 m/s . How do we reconcile these two speeds, and what does it tell us about standard conductors? The high speed of electrical signals results from the fact that the force between charges acts rapidly at a distance. Thus, when a free charge is forced into a wire, as in Figure 20.5, the incoming charge pushes other charges ahead of it, which in turn push on charges farther down the line. The density of charge in a system cannot easily be increased, and so the signal is passed on 872 Chapter 20 \| Electric Current, Resistance, and Ohm's Law rapidly. The resulting electrical shock wave moves through the system at nearly the speed of light. To be precise, this rapidly moving signal or shock wave is a rapidly propagating change in electric field. Figure 20.5 When charged particles are forced into this volume of a conductor, an equal number are quickly forced to leave. The repulsion between like charges makes it difficult to increase the number of charges in a volume. Thus, as one charge enters, another leaves almost immediately, carrying the signal rapidly forward. Good conductors have large numbers of free charges in them. In metals, the free charges are free electrons. Figure 20.6 shows how free electrons move through an ordinary conductor. The distance that an individual electron can move between collisions with atoms or other electrons is quite small. The electron paths thus appear nearly random, like the motion of atoms in a gas. But there is an electric field in the conductor that causes the electrons to drift in the direction shown (opposite to the field, since they are negative). The drift velocity d is the average velocity of the free charges. Drift velocity is quite small, since there are so many free charges. If we have an estimate of the density of free electrons in a conductor, we can calculate the drift velocity for a given current. The larger the density, the lower the velocity required for a given current. Figure 20.6 Free electrons moving in a conductor make many collisions with other electrons and atoms. The path of one electron is shown. The average velocity of the free charges is called the drift velocity, d , and it is in the direction opposite to the electric field for electrons. The collisions normally transfer energy to the conductor, requiring a constant supply of energy to maintain a steady current. Conduction of Electricity and Heat Good electrical conductors are often good heat conductors, too. This is because large numbers of free electrons can carry electrical current and can transport thermal energy. The free-electron collisions transfer energy to the atoms of the conductor. The electric field does work in moving the electrons through a distance, but that work does not increase the kinetic energy (nor speed, therefore) of the electrons. The work is transferred to the conductor's atoms, possibly increasing temperature. Thus a continuous power input is required to maintain current. An exception, of course, is found in superconductors, for reasons we shall explore in a later chapter. Superconductors can have a steady current without a continual supply of energy—a great energy savings. In contrast, the supply of energy can be useful, such as in a lightbulb filament. The supply of energy is necessary to increase the temperature of the tungsten filament, so that the filament glows. Making Connections: Take-Home Investigation—Filament Observations Find a lightbulb with a filament. Look carefully at the filament and describe its structure. To what points is the filament connected? We can obtain an expression for the relationship between current and drift velocity by considering the number of free charges in a segment of wire, as illustrated in Figure 20.7. The number of free charges per unit volume is given the symbol and depends on the material. The shaded segment has a volume , so that the number of free charges in it is . The charge Δ in this segment is thus , where is the amount of charge on each carrier. (Recall that for electrons, is This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 20 \| Electric Current, Resistance, and Ohm's Law 873 −1.60×10−19 C .) Current is charge moved per unit time; thus, if all the original charges move out of this segment in time Δ , the current is = Δ Δ = Δ . Note that / Δ is the magnitude of the drift velocity, d , since the charges move an average distance in a time Δ . Rearranging terms gives = d, (20.7)
(20.8) where is the current through a wire of cross-sectional area made of a material with a free charge density . The carriers of the current each have charge and move with a drift velocity of magnitude d . Figure 20.7 All the charges in the shaded volume of this wire move out in a time , having a drift velocity of magnitude d = / . See text for further discussion. Note that simple drift velocity is not the entire story. The speed of an electron is much greater than its drift velocity. In addition, not all of the electrons in a conductor can move freely, and those that do might move somewhat faster or slower than the drift velocity. So what do we mean by free electrons? Atoms in a metallic conductor are packed in the form of a lattice structure. Some electrons are far enough away from the atomic nuclei that they do not experience the attraction of the nuclei as much as the inner electrons do. These are the free electrons. They are not bound to a single atom but can instead move freely among the atoms in a “sea” of electrons. These free electrons respond by accelerating when an electric field is applied. Of course as they move they collide with the atoms in the lattice and other electrons, generating thermal energy, and the conductor gets warmer. In an insulator, the organization of the atoms and the structure do not allow for such free electrons. Example 20.3 Calculating Drift Velocity in a Common Wire Calculate the drift velocity of electrons in a 12-gauge copper wire (which has a diameter of 2.053 mm) carrying a 20.0-A current, given that there is one free electron per copper atom. (Household wiring often contains 12-gauge copper wire, and the maximum current allowed in such wire is usually 20 A.) The density of copper is 8.80×103 kg/m3 . Strategy We can calculate the drift velocity using the equation = d . The current = 20.0 A is given, and = – 1.60×10 – 19 C is the charge of an electron. We can calculate the area of a cross-section of the wire using the formula = 2, where is one-half the given diameter, 2.053 mm. We are given the density of copper, 8.80×103 kg/m3, and the periodic table shows that the atomic mass of copper is 63.54 g/mol. We can use these two quantities along with Avogadro's number, 6.02×1023 atoms/mol, cubic meter. the number of free electrons per to determine , Solution First, calculate the density of free electrons in copper. There is one free electron per copper atom. Therefore, is the same as the number of copper atoms per m3 . We can now find as follows: = 1 − atom× 6.02×1023 atoms mol × 1 mol 63.54 g × 1000 g kg × 8.80×103 kg 1 m3 (20.9) = 8.342×1028 − /m3 . The cross-sectional area of the wire is 874 Chapter 20 \| Electric Current, Resistance, and Ohm's Law = 2 = 2.053×10−3 m 2 2 = 3.310×10–6 m2 . Rearranging = d to isolate drift velocity gives = d = 20.0 A (8.342×1028/m3)(–1.60×10–19 C)(3.310×10–6 m2) = –4.53×10–4 m/s. (20.10) (20.11) Discussion The minus sign indicates that the negative charges are moving in the direction opposite to conventional current. The small value for drift velocity (on the order of 10−4 m/s ) confirms that the signal moves on the order of 1012 times faster (about 108 m/s ) than the charges that carry it. 20.2 Ohm’s Law: Resistance and Simple Circuits Learning Objectives By the end of this section, you will be able to: • Explain the origin of Ohm's law. • Calculate voltages, currents, and resistances with Ohm's law. • Explain the difference between ohmic and non-ohmic materials. • Describe a simple circuit. The information presented in this section supports the following AP® learning objectives and science practices: • 4.E.4.1 The student is able to make predictions about the properties of resistors and/or capacitors when placed in a simple circuit based on the geometry of the circuit element and supported by scientific theories and mathematical relationships. (S.P. 2.2, 6.4) What drives current? We can think of various devices—such as batteries, generators, wall outlets, and so on—which are necessary to maintain a current. All such devices create a potential difference and are loosely referred to as voltage sources. When a voltage source is connected to a conductor, it applies a potential difference that creates an electric field. The electric field in turn exerts force on charges, causing current. Ohm's Law The current that flows through most substances is directly proportional to the voltage applied to it. The German physicist Georg Simon Ohm (1787–1854) was the first to demonstrate experimentally that the current in a metal wire is directly proportional to the voltage applied: ∝ . (20.12) This important relationship is known as Ohm's law. It can be viewed as a cause-and-effect relationship, with voltage the cause and current the effect. This is an empirical law like that for friction—an experimentally observed phenomenon. Such a linear relationship doesn't always occur. Resistance and Simple Circuits If voltage drives current, what impedes it? The electric property that impedes current (crudely similar to friction and air resistance) is called resistance . Collisions of moving charges with atoms and molecules in a substance transfer energy to the substance and limit current. Resistance is defined as inversely proportional to current, or ∝ 1 . (20.13) Thus, for example, current is cut in half if resistance doubles. Combining the relationships of current to voltage and current to resistance gives This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 20 \| Electric Current, Resistance, and Ohm's Law = . 875 (20.14) This relationship is also called Ohm's law. Ohm's law in this form really defines resistance for certain materials. Ohm's law (like Hooke's law) is not universally valid. The many substances for which Ohm's law holds are called ohmic. These include good conductors like copper and aluminum, and some poor conductors under certain circumstances. Ohmic materials have a resistance that is independent of voltage and current . An object that has simple resistance is called a resistor, even if its resistance is small. The unit for resistance is an ohm and is given the symbol Ω (upper case Greek omega). Rearranging = gives = , and so the units of resistance are 1 ohm = 1 volt per ampere: Figure 20.8 shows the schematic for a simple circuit. A simple circuit has a single voltage source and a single resistor. The wires connecting the voltage source to the resistor can be assumed to have negligible resistance, or their resistance can be included in . 1 Ω = 1 . (20.15) Figure 20.8 A simple electric circuit in which a closed path for current to flow is supplied by conductors (usually metal wires) connecting a load to the terminals of a battery, represented by the red parallel lines. The zigzag symbol represents the single resistor and includes any resistance in the connections to the voltage source. Making Connections: Real World Connections Ohm's law ( = ) is a fundamental relationship that could be presented by a linear function with the slope of the line being the resistance. The resistance represents the voltage that needs to be applied to the resistor to create a current of 1 A through the circuit. The graph (in the figure below) shows this representation for two simple circuits with resistors that have different resistances and thus different slopes. Figure 20.9 The figure illustrates the relationship between current and voltage for two different resistors. The slope of the graph represents the resistance value, which is 2Ω and 4Ω for the two lines shown. 876 Chapter 20 \| Electric Current, Resistance, and Ohm's Law Making Connections: Real World Connections The materials which follow Ohm's law by having a linear relationship between voltage and current are known as ohmic materials. On the other hand, some materials exhibit a nonlinear voltage-current relationship and hence are known as nonohmic materials. The figure below shows current voltage relationships for the two types of materials. Figure 20.10 The relationship between voltage and current for ohmic and non-ohmic materials are shown. (a) (b) Clearly the resistance of an ohmic material (shown in (a)) remains constant and can be calculated by finding the slope of the graph but that is not true for a non-ohmic material (shown in (b)). Example 20.4 Calculating Resistance: An Automobile Headlight What is the resistance of an automobile headlight through which 2.50 A flows when 12.0 V is applied to it? Strategy We can rearrange Ohm's law as stated by = and use it to find the resistance. Solution Rearranging = and substituting known values gives = = 12.0 V 2.50 A = 4.80 Ω. (20.16) Discussion This is a relatively small resistance, but it is larger than the cold resistance of the headlight. As we shall see in Resistance and Resistivity, resistance usually increases with temperature, and so the bulb has a lower resistance when it is first switched on and will draw considerably more current during its brief warm-up period. Resistances range over many orders of magnitude. Some ceramic insulators, such as those used to support power lines, have resistances of 1012 Ω or more. A dry person may have a hand-to-foot resistance of 105 Ω , whereas the resistance of the human heart is about 103 Ω . A meter-long piece of large-diameter copper wire may have a resistance of 10−5 Ω , and superconductors have no resistance at all (they are non-ohmic). Resistance is related to the shape of an object and the material of which it is composed, as will be seen in Resistance and Resistivity. Additional insight is gained by solving = for , yielding = (20.17) This expression for can be interpreted as the voltage drop across a resistor produced by the current . The phrase drop is often used for this voltage. For instance, the headlight in Example 20.4 has an drop of 12.0 V. If voltage is measured at various points in a circuit, it will be seen to increase at the voltage source and decrease at the resis
tor. Voltage is similar to fluid pressure. The voltage source is like a pump, creating a pressure difference, causing current—the flow of charge. The resistor is like a pipe that reduces pressure and limits flow because of its resistance. Conservation of energy has important consequences here. The voltage source supplies energy (causing an electric field and a current), and the resistor converts it to another form (such as thermal energy). In a simple circuit (one with a single simple resistor), the voltage supplied by the source equals the voltage drop across the resistor, since PE = Δ , and the same flows through each. Thus the energy supplied by the voltage source and the energy converted by the resistor are equal. (See Figure 20.11.) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 20 \| Electric Current, Resistance, and Ohm's Law 877 Figure 20.11 The voltage drop across a resistor in a simple circuit equals the voltage output of the battery. Making Connections: Conservation of Energy In a simple electrical circuit, the sole resistor converts energy supplied by the source into another form. Conservation of energy is evidenced here by the fact that all of the energy supplied by the source is converted to another form by the resistor alone. We will find that conservation of energy has other important applications in circuits and is a powerful tool in circuit analysis. PhET Explorations: Ohm's Law See how the equation form of Ohm's law relates to a simple circuit. Adjust the voltage and resistance, and see the current change according to Ohm's law. The sizes of the symbols in the equation change to match the circuit diagram. Figure 20.12 Ohm's Law (http://cnx.org/content/m55356/1.2/ohms-law_en.jar) 20.3 Resistance and Resistivity Learning Objectives By the end of this section, you will be able to: • Explain the concept of resistivity. • Use resistivity to calculate the resistance of specified configurations of material. • Use the thermal coefficient of resistivity to calculate the change of resistance with temperature. The information presented in this section supports the following AP® learning objectives and science practices: • 1.E.2.1 The student is able to choose and justify the selection of data needed to determine resistivity for a given material. (S.P. 4.1) • 4.E.4.2 The student is able to design a plan for the collection of data to determine the effect of changing the geometry and/or materials on the resistance or capacitance of a circuit element and relate results to the basic properties of resistors and capacitors. (S.P. 4.1, 4.2) • 4.E.4.3 The student is able to analyze data to determine the effect of changing the geometry and/or materials on the resistance or capacitance of a circuit element and relate results to the basic properties of resistors and capacitors. (S.P. 5.1) Material and Shape Dependence of Resistance The resistance of an object depends on its shape and the material of which it is composed. The cylindrical resistor in Figure 20.13 is easy to analyze, and, by so doing, we can gain insight into the resistance of more complicated shapes. As you might expect, the cylinder's electric resistance is directly proportional to its length , similar to the resistance of a pipe to fluid flow. The longer the cylinder, the more collisions charges will make with its atoms. The greater the diameter of the cylinder, the more current it can carry (again similar to the flow of fluid through a pipe). In fact, is inversely proportional to the cylinder's crosssectional area . 878 Chapter 20 \| Electric Current, Resistance, and Ohm's Law Figure 20.13 A uniform cylinder of length and cross-sectional area . Its resistance to the flow of current is similar to the resistance posed by a pipe to fluid flow. The longer the cylinder, the greater its resistance. The larger its cross-sectional area , the smaller its resistance. For a given shape, the resistance depends on the material of which the object is composed. Different materials offer different resistance to the flow of charge. We define the resistivity of a substance so that the resistance of an object is directly proportional to . Resistivity is an intrinsic property of a material, independent of its shape or size. The resistance of a uniform cylinder of length , of cross-sectional area , and made of a material with resistivity , is = . (20.18) Table 20.1 gives representative values of . The materials listed in the table are separated into categories of conductors, semiconductors, and insulators, based on broad groupings of resistivities. Conductors have the smallest resistivities, and insulators have the largest; semiconductors have intermediate resistivities. Conductors have varying but large free charge densities, whereas most charges in insulators are bound to atoms and are not free to move. Semiconductors are intermediate, having far fewer free charges than conductors, but having properties that make the number of free charges depend strongly on the type and amount of impurities in the semiconductor. These unique properties of semiconductors are put to use in modern electronics, as will be explored in later chapters. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 20 \| Electric Current, Resistance, and Ohm's Law 879 Table 20.1 Resistivities of Various materials at 20ºC Material Conductors Silver Copper Gold Aluminum Tungsten Iron Platinum Steel Lead Resistivity ρ ( Ω ⋅ m ) 1.59×10−8 1.72×10−8 2.44×10−8 2.65×10−8 5.6×10−8 9.71×10−8 10.6×10−8 20×10−8 22×10−8 Manganin (Cu, Mn, Ni alloy) 44×10−8 Constantan (Cu, Ni alloy) Mercury Nichrome (Ni, Fe, Cr alloy) 49×10−8 96×10−8 100×10−8 Semiconductors[1] Carbon (pure) Carbon Germanium (pure) Germanium Silicon (pure) Silicon Insulators Amber Glass Lucite Mica Quartz (fused) Rubber (hard) Sulfur Teflon Wood 3.5×105 (3.5 − 60)×105 600×10−3 (1 − 600)×10−3 2300 0.1–2300 5×1014 109 − 1014 >1013 1011 − 1015 75×1016 1013 − 1016 1015 >1013 108 − 1014 880 Chapter 20 \| Electric Current, Resistance, and Ohm's Law Example 20.5 Calculating Resistor Diameter: A Headlight Filament A car headlight filament is made of tungsten and has a cold resistance of 0.350 Ω . If the filament is a cylinder 4.00 cm long (it may be coiled to save space), what is its diameter? Strategy We can rearrange the equation = to find the cross-sectional area of the filament from the given information. Then its diameter can be found by assuming it has a circular cross-section. Solution The cross-sectional area, found by rearranging the expression for the resistance of a cylinder given in = , is Substituting the given values, and taking from Table 20.1, yields = . = (5.610–8 Ω ⋅ m)(4.0010–2 m) 0.350 Ω = 6.4010–9 m2 . The area of a circle is related to its diameter by = 2 4 . Solving for the diameter , and substituting the value found for , gives 1 2 = 2 = 2 = 9.010–5 m. 6.4010–9 m2 3.14 1 2 (20.19) (20.20) (20.21) (20.22) Discussion The diameter is just under a tenth of a millimeter. It is quoted to only two digits, because is known to only two digits. Temperature Variation of Resistance The resistivity of all materials depends on temperature. Some even become superconductors (zero resistivity) at very low temperatures. (See Figure 20.14.) Conversely, the resistivity of conductors increases with increasing temperature. Since the atoms vibrate more rapidly and over larger distances at higher temperatures, the electrons moving through a metal make more collisions, effectively making the resistivity higher. Over relatively small temperature changes (about 100ºC or less), resistivity varies with temperature change Δ as expressed in the following equation = 0(1 + Δ), (20.23) where 0 is the original resistivity and is the temperature coefficient of resistivity. (See the values of in Table 20.2 below.) For larger temperature changes, may vary or a nonlinear equation may be needed to find . Note that is positive for metals, meaning their resistivity increases with temperature. Some alloys have been developed specifically to have a small temperature dependence. Manganin (which is made of copper, manganese and nickel), for example, has close to zero (to three digits on the scale in Table 20.2), and so its resistivity varies only slightly with temperature. This is useful for making a temperature-independent resistance standard, for example. 1. Values depend strongly on amounts and types of impurities This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 20 \| Electric Current, Resistance, and Ohm's Law 881 Figure 20.14 The resistance of a sample of mercury is zero at very low temperatures—it is a superconductor up to about 4.2 K. Above that critical temperature, its resistance makes a sudden jump and then increases nearly linearly with temperature. Table 20.2 Temperature Coefficients of Resistivity Material Conductors Silver Copper Gold Aluminum Tungsten Iron Platinum Lead Coefficient α (1/°C)[2] 3.8×10−3 3.9×10−3 3.4×10−3 3.9×10−3 4.5×10−3 5.0×10−3 3.93×10−3 4.3×10−3 Manganin (Cu, Mn, Ni alloy) 0.000×10−3 Constantan (Cu, Ni alloy) Mercury Nichrome (Ni, Fe, Cr alloy) Semiconductors Carbon (pure) Germanium (pure) Silicon (pure) 0.002×10−3 0.89×10−3 0.4×10−3 −0.5×10−3 −50×10−3 −70×10−3 Note also that is negative for the semiconductors listed in Table 20.2, meaning that their resistivity decreases with increasing temperature. They become better conductors at higher temperature, because increased thermal agitation increases the number of free charges available to carry current. This property of decreasing with temperature is also related to the type and amount of impurities present in the semiconductors. 2. Values at 20°C. 882 Chapter 20 \| Electric Current, Resistance, and Ohm's Law The resistance of an object also depends on temperature, since 0 is directly proportional to . For a cylinder we know = / , and so, if and do not change greatly with temperature, will have the s
ame temperature dependence as . (Examination of the coefficients of linear expansion shows them to be about two orders of magnitude less than typical temperature coefficients of resistivity, and so the effect of temperature on and is about two orders of magnitude less than on .) Thus, = 0(1 + Δ) (20.24) is the temperature dependence of the resistance of an object, where 0 is the original resistance and is the resistance after a temperature change Δ . Numerous thermometers are based on the effect of temperature on resistance. (See Figure 20.15.) One of the most common is the thermistor, a semiconductor crystal with a strong temperature dependence, the resistance of which is measured to obtain its temperature. The device is small, so that it quickly comes into thermal equilibrium with the part of a person it touches. Figure 20.15 These familiar thermometers are based on the automated measurement of a thermistor's temperature-dependent resistance. (credit: Biol, Wikimedia Commons) Example 20.6 Calculating Resistance: Hot-Filament Resistance Although caution must be used in applying = 0(1 + Δ) and = 0(1 + Δ) for temperature changes greater than 100ºC , for tungsten the equations work reasonably well for very large temperature changes. What, then, is the resistance of the tungsten filament in the previous example if its temperature is increased from room temperature ( 20ºC ) to a typical operating temperature of 2850ºC ? Strategy This is a straightforward application of = 0(1 + Δ) , since the original resistance of the filament was given to be 0 = 0.350 Ω , and the temperature change is Δ = 2830ºC . Solution The hot resistance is obtained by entering known values into the above equation: = 0(1 + Δ) = (0.350 Ω)[1 + (4.510–3 / ºC)(2830ºC)] = 4.8 Ω. (20.25) Discussion This value is consistent with the headlight resistance example in Ohm's Law: Resistance and Simple Circuits. PhET Explorations: Resistance in a Wire Learn about the physics of resistance in a wire. Change its resistivity, length, and area to see how they affect the wire's resistance. The sizes of the symbols in the equation change along with the diagram of a wire. Figure 20.16 Resistance in a Wire (http://cnx.org/content/m55357/1.2/resistance-in-a-wire_en.jar) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 20 \| Electric Current, Resistance, and Ohm's Law 883 Applying the Science Practices: Examining Resistance Using the PhET Simulation “Resistance in a Wire”, design an experiment to determine how different variables – resistivity, length, and area – affect the resistance of a resistor. For each variable, you should record your results in a table and then create a graph to determine the relationship. 20.4 Electric Power and Energy Learning Objectives By the end of this section, you will be able to: • Calculate the power dissipated by a resistor and the power supplied by a power supply. • Calculate the cost of electricity under various circumstances. The information presented in this section supports the following AP® learning objectives and science practices: • 5.B.9.8 The student is able to translate between graphical and symbolic representations of experimental data describing relationships among power, current, and potential difference across a resistor. (S.P. 1.5) Power in Electric Circuits Power is associated by many people with electricity. Knowing that power is the rate of energy use or energy conversion, what is the expression for electric power? Power transmission lines might come to mind. We also think of lightbulbs in terms of their power ratings in watts. Let us compare a 25-W bulb with a 60-W bulb. (See Figure 20.17(a).) Since both operate on the same voltage, the 60-W bulb must draw more current to have a greater power rating. Thus the 60-W bulb's resistance must be lower than that of a 25-W bulb. If we increase voltage, we also increase power. For example, when a 25-W bulb that is designed to operate on 120 V is connected to 240 V, it briefly glows very brightly and then burns out. Precisely how are voltage, current, and resistance related to electric power? Figure 20.17 (a) Which of these lightbulbs, the 25-W bulb (upper left) or the 60-W bulb (upper right), has the higher resistance? Which draws more current? Which uses the most energy? Can you tell from the color that the 25-W filament is cooler? Is the brighter bulb a different color and if so why? (credits: Dickbauch, Wikimedia Commons; Greg Westfall, Flickr) (b) This compact fluorescent light (CFL) puts out the same intensity of light as the 60-W bulb, but at 1/4 to 1/10 the input power. (credit: dbgg1979, Flickr) Electric energy depends on both the voltage involved and the charge moved. This is expressed most simply as PE = , where is the charge moved and is the voltage (or more precisely, the potential difference the charge moves through). Power is the rate at which energy is moved, and so electric power is = = . (20.26) Recognizing that current is = / (note that Δ = here), the expression for power becomes = (20.27) 884 Chapter 20 \| Electric Current, Resistance, and Ohm's Law Electric power ( ) is simply the product of current times voltage. Power has familiar units of watts. Since the SI unit for potential energy (PE) is the joule, power has units of joules per second, or watts. Thus, 1 A ⋅ V = 1 W . For example, cars often have one or more auxiliary power outlets with which you can charge a cell phone or other electronic devices. These outlets may be rated at 20 A, so that the circuit can deliver a maximum power = = (20 A)(12 V) = 240 W . In some applications, electric power may be expressed as volt-amperes or even kilovolt-amperes ( 1 kA ⋅ V = 1 kW ). To see the relationship of power to resistance, we combine Ohm's law with = . Substituting = gives = ( / ) = 2 / . Similarly, substituting = gives = () = 2 . Three expressions for electric power are listed together here for convenience: = = 2 = 2. Note that the first equation is always valid, whereas the other two can be used only for resistors. In a simple circuit, with one voltage source and a single resistor, the power supplied by the voltage source and that dissipated by the resistor are identical. (In more complicated circuits, can be the power dissipated by a single device and not the total power in the circuit.) (20.28) (20.29) (20.30) Making Connections: Using Graphs to Calculate Resistance As ∝ 2 and ∝ 2 , the graph for power versus current or voltage is quadratic. An example is shown in the figure below. Figure 20.18 The figure shows (a) power versus current and (b) power versus voltage relationships for simple resistor circuits. (a) (b) Using equations (20.29) and (20.30), we can calculate the resistance in each case. In graph (a), the power is 50 W when current is 5 A; hence, the resistance can be calculated as = / 2 = 50/ 52 = 2 Ω . Similarly, the resistance value can be calculated in graph (b) as = 2/ = 102/ 50 = 2 Ω Different insights can be gained from the three different expressions for electric power. For example, = 2 / implies that the lower the resistance connected to a given voltage source, the greater the power delivered. Furthermore, since voltage is squared in = 2 / , the effect of applying a higher voltage is perhaps greater than expected. Thus, when the voltage is doubled to a 25-W bulb, its power nearly quadruples to about 100 W, burning it out. If the bulb's resistance remained constant, its power would be exactly 100 W, but at the higher temperature its resistance is higher, too. Example 20.7 Calculating Power Dissipation and Current: Hot and Cold Power (a) Consider the examples given in Ohm's Law: Resistance and Simple Circuits and Resistance and Resistivity. Then find the power dissipated by the car headlight in these examples, both when it is hot and when it is cold. (b) What current does it draw when cold? Strategy for (a) For the hot headlight, we know voltage and current, so we can use = to find the power. For the cold headlight, we know the voltage and resistance, so we can use = 2 / to find the power. Solution for (a) Entering the known values of current and voltage for the hot headlight, we obtain This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 20 \| Electric Current, Resistance, and Ohm's Law The cold resistance was 0.350 Ω , and so the power it uses when first switched on is = = (2.50 A)(12.0 V) = 30.0 W. = 2 = (12.0 V)2 0.350 Ω = 411 W. 885 (20.31) (20.32) Discussion for (a) The 30 W dissipated by the hot headlight is typical. But the 411 W when cold is surprisingly higher. The initial power quickly decreases as the bulb's temperature increases and its resistance increases. Strategy and Solution for (b) The current when the bulb is cold can be found several different ways. We rearrange one of the power equations, = 2 , and enter known values, obtaining = = 411 W 0.350 Ω = 34.3 A. (20.33) Discussion for (b) The cold current is remarkably higher than the steady-state value of 2.50 A, but the current will quickly decline to that value as the bulb's temperature increases. Most fuses and circuit breakers (used to limit the current in a circuit) are designed to tolerate very high currents briefly as a device comes on. In some cases, such as with electric motors, the current remains high for several seconds, necessitating special “slow blow” fuses. The Cost of Electricity The more electric appliances you use and the longer they are left on, the higher your electric bill. This familiar fact is based on the relationship between energy and power. You pay for the energy used. Since = / , we see that = (20.34) is the energy used by a device using power for a time interval . For example, the more lightbulbs burning, the greater used; the longer they are on, the greater is. The energy unit on electric bills is the kilowatt-hour ( kW ⋅ h ), consistent with the relationship = . It is easy to estimate the cost of operating electric applian
ces if you have some idea of their power consumption rate in watts or kilowatts, the time they are on in hours, and the cost per kilowatt-hour for your electric utility. Kilowatt-hours, like all other specialized energy units such as food calories, can be converted to joules. You can prove to yourself that 1 kW ⋅ h = 3.6106 J . The electrical energy ( ) used can be reduced either by reducing the time of use or by reducing the power consumption of that appliance or fixture. This will not only reduce the cost, but it will also result in a reduced impact on the environment. Improvements to lighting are some of the fastest ways to reduce the electrical energy used in a home or business. About 20% of a home's use of energy goes to lighting, while the number for commercial establishments is closer to 40%. Fluorescent lights are about four times more efficient than incandescent lights—this is true for both the long tubes and the compact fluorescent lights (CFL). (See Figure 20.17(b).) Thus, a 60-W incandescent bulb can be replaced by a 15-W CFL, which has the same brightness and color. CFLs have a bent tube inside a globe or a spiral-shaped tube, all connected to a standard screw-in base that fits standard incandescent light sockets. (Original problems with color, flicker, shape, and high initial investment for CFLs have been addressed in recent years.) The heat transfer from these CFLs is less, and they last up to 10 times longer. The significance of an investment in such bulbs is addressed in the next example. New white LED lights (which are clusters of small LED bulbs) are even more efficient (twice that of CFLs) and last 5 times longer than CFLs. However, their cost is still high. Making Connections: Energy, Power, and Time The relationship = is one that you will find useful in many different contexts. The energy your body uses in exercise is related to the power level and duration of your activity, for example. The amount of heating by a power source is related to the power level and time it is applied. Even the radiation dose of an X-ray image is related to the power and time of exposure. Example 20.8 Calculating the Cost Effectiveness of Compact Fluorescent Lights (CFL) If the cost of electricity in your area is 12 cents per kWh, what is the total cost (capital plus operation) of using a 60-W incandescent bulb for 1000 hours (the lifetime of that bulb) if the bulb cost 25 cents? (b) If we replace this bulb with a compact fluorescent light that provides the same light output, but at one-quarter the wattage, and which costs $1.50 but lasts 10 times longer (10,000 hours), what will that total cost be? Strategy 886 Chapter 20 \| Electric Current, Resistance, and Ohm's Law To find the operating cost, we first find the energy used in kilowatt-hours and then multiply by the cost per kilowatt-hour. Solution for (a) The energy used in kilowatt-hours is found by entering the power and time into the expression for energy: = = (60 W)(1000 h) = 60,000 W ⋅ h. In kilowatt-hours, this is Now the electricity cost is = 60.0 kW ⋅ h. cost = (60.0 kW ⋅ h)($0.12/kW ⋅ h) = $7.20. The total cost will be $7.20 for 1000 hours (about one-half year at 5 hours per day). Solution for (b) (20.35) (20.36) (20.37) Since the CFL uses only 15 W and not 60 W, the electricity cost will be $7.20/4 = $1.80. The CFL will last 10 times longer than the incandescent, so that the investment cost will be 1/10 of the bulb cost for that time period of use, or 0.1($1.50) = $0.15. Therefore, the total cost will be $1.95 for 1000 hours. Discussion Therefore, it is much cheaper to use the CFLs, even though the initial investment is higher. The increased cost of labor that a business must include for replacing the incandescent bulbs more often has not been figured in here. Making Connections: Take-Home Experiment—Electrical Energy Use Inventory 1) Make a list of the power ratings on a range of appliances in your home or room. Explain why something like a toaster has a higher rating than a digital clock. Estimate the energy consumed by these appliances in an average day (by estimating their time of use). Some appliances might only state the operating current. If the household voltage is 120 V, then use = . 2) Check out the total wattage used in the rest rooms of your school's floor or building. (You might need to assume the long fluorescent lights in use are rated at 32 W.) Suppose that the building was closed all weekend and that these lights were left on from 6 p.m. Friday until 8 a.m. Monday. What would this oversight cost? How about for an entire year of weekends? 20.5 Alternating Current versus Direct Current Learning Objectives By the end of this section, you will be able to: • Explain the differences and similarities between AC and DC current. • Calculate rms voltage, current, and average power. • Explain why AC current is used for power transmission. Alternating Current Most of the examples dealt with so far, and particularly those utilizing batteries, have constant voltage sources. Once the current is established, it is thus also a constant. Direct current (DC) is the flow of electric charge in only one direction. It is the steady state of a constant-voltage circuit. Most well-known applications, however, use a time-varying voltage source. Alternating current (AC) is the flow of electric charge that periodically reverses direction. If the source varies periodically, particularly sinusoidally, the circuit is known as an alternating current circuit. Examples include the commercial and residential power that serves so many of our needs. Figure 20.19 shows graphs of voltage and current versus time for typical DC and AC power. The AC voltages and frequencies commonly used in homes and businesses vary around the world. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 20 \| Electric Current, Resistance, and Ohm's Law 901 Figure 20.38 This NASA scientist and NEEMO 5 aquanaut's heart rate and other vital signs are being recorded by a portable device while living in an underwater habitat. (credit: NASA, Life Sciences Data Archive at Johnson Space Center, Houston, Texas) PhET Explorations: Neuron Figure 20.39 Neuron (http://cnx.org/content/m55361/1.2/neuron_en.jar) Stimulate a neuron and monitor what happens. Pause, rewind, and move forward in time in order to observe the ions as they move across the neuron membrane. Glossary AC current: current that fluctuates sinusoidally with time, expressed as I = I0 sin 2πft, where I is the current at time t, I0 is the peak current, and f is the frequency in hertz AC voltage: voltage that fluctuates sinusoidally with time, expressed as V = V0 sin 2πft, where V is the voltage at time t, V0 is the peak voltage, and f is the frequency in hertz alternating current: (AC) the flow of electric charge that periodically reverses direction ampere: (amp) the SI unit for current; 1 A = 1 C/s bioelectricity: electrical effects in and created by biological systems direct current: (DC) the flow of electric charge in only one direction drift velocity: the average velocity at which free charges flow in response to an electric field electric current: the rate at which charge flows, I = ΔQ/Δt electric power: the rate at which electrical energy is supplied by a source or dissipated by a device; it is the product of current times voltage electrocardiogram (ECG): especially in the heart usually abbreviated ECG, a record of voltages created by depolarization and repolarization, microshock sensitive: a condition in which a person's skin resistance is bypassed, possibly by a medical procedure, rendering the person vulnerable to electrical shock at currents about 1/1000 the normally required level nerve conduction: the transport of electrical signals by nerve cells ohm: the unit of resistance, given by 1Ω = 1 V/A Ohm's law: an empirical relation stating that the current I is proportional to the potential difference V, ∝ V; it is often written as I = V/R, where R is the resistance ohmic: a type of a material for which Ohm's law is valid resistance: the electric property that impedes current; for ohmic materials, it is the ratio of voltage to current, R = V/I 902 resistivity: by ρ an intrinsic property of a material, independent of its shape or size, directly proportional to the resistance, denoted Chapter 20 \| Electric Current, Resistance, and Ohm's Law rms current: the root mean square of the current, rms = 0 / 2 , where I0 is the peak current, in an AC system rms voltage: the root mean square of the voltage, rms = 0 / 2 , where V0 is the peak voltage, in an AC system semipermeable: property of a membrane that allows only certain types of ions to cross it shock hazard: when electric current passes through a person short circuit: also known as a “short,” a low-resistance path between terminals of a voltage source simple circuit: a circuit with a single voltage source and a single resistor temperature coefficient of resistivity: an empirical quantity, denoted by α, which describes the change in resistance or resistivity of a material with temperature thermal hazard: a hazard in which electric current causes undesired thermal effects Section Summary 20.1 Current • Electric current is the rate at which charge flows, given by = Δ Δ , where Δ is the amount of charge passing through an area in time Δ . • The direction of conventional current is taken as the direction in which positive charge moves. • The SI unit for current is the ampere (A), where 1 A = 1 C/s. • Current is the flow of free charges, such as electrons and ions. • Drift velocity d is the average speed at which these charges move. • Current is proportional to drift velocity d , as expressed in the relationship = d . Here, is the current through a wire of cross-sectional area . The wire's material has a free-charge density , and each carrier has charge and a drift velocity d . • Electrical signals travel at speeds about 1012 times greater than the drift velocity
of free electrons. 20.2 Ohm’s Law: Resistance and Simple Circuits • A simple circuit is one in which there is a single voltage source and a single resistance. • One statement of Ohm's law gives the relationship between current , voltage , and resistance in a simple circuit to be = . • Resistance has units of ohms ( Ω ), related to volts and amperes by 1 Ω = 1 V/A . • There is a voltage or drop across a resistor, caused by the current flowing through it, given by = . 20.3 Resistance and Resistivity • The resistance of a cylinder of length and cross-sectional area is = , where is the resistivity of the material. • Values of in Table 20.1 show that materials fall into three groups—conductors, semiconductors, and insulators. • Temperature affects resistivity; for relatively small temperature changes Δ , resistivity is = 0(1 + Δ) , where 0 is the original resistivity and α is the temperature coefficient of resistivity. • Table 20.2 gives values for , the temperature coefficient of resistivity. • The resistance of an object also varies with temperature: = 0(1 + Δ) , where 0 is the original resistance, and is the resistance after the temperature change. 20.4 Electric Power and Energy • Electric power is the rate (in watts) that energy is supplied by a source or dissipated by a device. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 20 \| Electric Current, Resistance, and Ohm's Law 903 • Three expressions for electrical power are = = 2 , and • The energy used by a device with a power over a time is = . = 2. 20.5 Alternating Current versus Direct Current • Direct current (DC) is the flow of electric current in only one direction. It refers to systems where the source voltage is constant. • The voltage source of an alternating current (AC) system puts out = 0 sin 2 , where is the voltage at time , • 0 is the peak voltage, and In a simple circuit, = and AC current is = 0 sin 2 , where is the current at time , and 0 = 0 is the peak current. is the frequency in hertz. • The average AC power is ave = 1 20 0 . • Average (rms) current rms and average (rms) voltage rms are rms = 0 2 and rms = 0 2 , where rms stands for root mean square. • Thus, ave = rmsrms . • Ohm's law for AC is rms = rms . • Expressions for the average power of an AC circuit are ave = rmsrms , ave = rms 2 , and ave = rms 2 , analogous to the expressions for DC circuits. 20.6 Electric Hazards and the Human Body • The two types of electric hazards are thermal (excessive power) and shock (current through a person). • Shock severity is determined by current, path, duration, and AC frequency. • Table 20.3 lists shock hazards as a function of current. • Figure 20.28 graphs the threshold current for two hazards as a function of frequency. 20.7 Nerve Conduction–Electrocardiograms • Electric potentials in neurons and other cells are created by ionic concentration differences across semipermeable membranes. • Stimuli change the permeability and create action potentials that propagate along neurons. • Myelin sheaths speed this process and reduce the needed energy input. • This process in the heart can be measured with an electrocardiogram (ECG). Conceptual Questions 20.1 Current 1. Can a wire carry a current and still be neutral—that is, have a total charge of zero? Explain. 2. Car batteries are rated in ampere-hours ( A ⋅ h ). To what physical quantity do ampere-hours correspond (voltage, charge, . . .), and what relationship do ampere-hours have to energy content? 3. If two different wires having identical cross-sectional areas carry the same current, will the drift velocity be higher or lower in the better conductor? Explain in terms of the equation d = , by considering how the density of charge carriers relates to whether or not a material is a good conductor. 4. Why are two conducting paths from a voltage source to an electrical device needed to operate the device? 5. In cars, one battery terminal is connected to the metal body. How does this allow a single wire to supply current to electrical devices rather than two wires? 6. Why isn't a bird sitting on a high-voltage power line electrocuted? Contrast this with the situation in which a large bird hits two wires simultaneously with its wings. 20.2 Ohm’s Law: Resistance and Simple Circuits 904 Chapter 20 \| Electric Current, Resistance, and Ohm's Law 7. The drop across a resistor means that there is a change in potential or voltage across the resistor. Is there any change in current as it passes through a resistor? Explain. 8. How is the drop in a resistor similar to the pressure drop in a fluid flowing through a pipe? 20.3 Resistance and Resistivity 9. In which of the three semiconducting materials listed in Table 20.1 do impurities supply free charges? (Hint: Examine the range of resistivity for each and determine whether the pure semiconductor has the higher or lower conductivity.) 10. Does the resistance of an object depend on the path current takes through it? Consider, for example, a rectangular bar—is its resistance the same along its length as across its width? (See Figure 20.40.) Figure 20.40 Does current taking two different paths through the same object encounter different resistance? 11. If aluminum and copper wires of the same length have the same resistance, which has the larger diameter? Why? 12. Explain why = 0(1 + Δ) for the temperature variation of the resistance of an object is not as accurate as = 0(1 + Δ) , which gives the temperature variation of resistivity . 20.4 Electric Power and Energy 13. Why do incandescent lightbulbs grow dim late in their lives, particularly just before their filaments break? 14. The power dissipated in a resistor is given by = 2 / , which means power decreases if resistance increases. Yet this power is also given by = 2 , which means power increases if resistance increases. Explain why there is no contradiction here. 20.5 Alternating Current versus Direct Current 15. Give an example of a use of AC power other than in the household. Similarly, give an example of a use of DC power other than that supplied by batteries. 16. Why do voltage, current, and power go through zero 120 times per second for 60-Hz AC electricity? 17. You are riding in a train, gazing into the distance through its window. As close objects streak by, you notice that the nearby fluorescent lights make dashed streaks. Explain. 20.6 Electric Hazards and the Human Body 18. Using an ohmmeter, a student measures the resistance between various points on his body. He finds that the resistance between two points on the same finger is about the same as the resistance between two points on opposite hands—both are several hundred thousand ohms. Furthermore, the resistance decreases when more skin is brought into contact with the probes of the ohmmeter. Finally, there is a dramatic drop in resistance (to a few thousand ohms) when the skin is wet. Explain these observations and their implications regarding skin and internal resistance of the human body. 19. What are the two major hazards of electricity? 20. Why isn't a short circuit a shock hazard? 21. What determines the severity of a shock? Can you say that a certain voltage is hazardous without further information? 22. An electrified needle is used to burn off warts, with the circuit being completed by having the patient sit on a large butt plate. Why is this plate large? 23. Some surgery is performed with high-voltage electricity passing from a metal scalpel through the tissue being cut. Considering the nature of electric fields at the surface of conductors, why would you expect most of the current to flow from the sharp edge of the scalpel? Do you think high- or low-frequency AC is used? 24. Some devices often used in bathrooms, such as hairdryers, often have safety messages saying “Do not use when the bathtub or basin is full of water.” Why is this so? 25. We are often advised to not flick electric switches with wet hands, dry your hand first. We are also advised to never throw water on an electric fire. Why is this so? 26. Before working on a power transmission line, linemen will touch the line with the back of the hand as a final check that the voltage is zero. Why the back of the hand? This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 20 \| Electric Current, Resistance, and Ohm's Law 905 27. Why is the resistance of wet skin so much smaller than dry, and why do blood and other bodily fluids have low resistances? 28. Could a person on intravenous infusion (an IV) be microshock sensitive? 29. In view of the small currents that cause shock hazards and the larger currents that circuit breakers and fuses interrupt, how do they play a role in preventing shock hazards? 20.7 Nerve Conduction–Electrocardiograms 30. Note that in Figure 20.31, both the concentration gradient and the Coulomb force tend to move Na+ prevents this? ions into the cell. What 31. Define depolarization, repolarization, and the action potential. 32. Explain the properties of myelinated nerves in terms of the insulating properties of myelin. 906 Chapter 20 \| Electric Current, Resistance, and Ohm's Law Problems & Exercises 20.1 Current 1. What is the current in milliamperes produced by the solar cells of a pocket calculator through which 4.00 C of charge passes in 4.00 h? 2. A total of 600 C of charge passes through a flashlight in 0.500 h. What is the average current? 3. What is the current when a typical static charge of 0.250 C moves from your finger to a metal doorknob in 1.00 s ? 4. Find the current when 2.00 nC jumps between your comb and hair over a 0.500 - s time interval. 5. A large lightning bolt had a 20,000-A current and moved 30.0 C of charge. What was its duration? 6. The 200-A current through a spark plug moves 0.300 mC of charge. How long does the spark last? 7. (a) A defibrillator sends a 6.00-A current through the chest of a patient by applying a 10,000-V potential as in the figure below. What is the resista
nce of the path? (b) The defibrillator paddles make contact with the patient through a conducting gel that greatly reduces the path resistance. Discuss the difficulties that would ensue if a larger voltage were used to produce the same current through the patient, but with the path having perhaps 50 times the resistance. (Hint: The current must be about the same, so a higher voltage would imply greater power. Use this equation for power: = 2 .) Figure 20.41 The capacitor in a defibrillation unit drives a current through the heart of a patient. 8. During open-heart surgery, a defibrillator can be used to bring a patient out of cardiac arrest. The resistance of the path is 500 Ω and a 10.0-mA current is needed. What voltage should be applied? 9. (a) A defibrillator passes 12.0 A of current through the torso of a person for 0.0100 s. How much charge moves? (b) How many electrons pass through the wires connected to the patient? (See figure two problems earlier.) 10. A clock battery wears out after moving 10,000 C of charge through the clock at a rate of 0.500 mA. (a) How long did the clock run? (b) How many electrons per second flowed? This content is available for free at http://cnx.org/content/col11844/1.13 11. The batteries of a submerged non-nuclear submarine supply 1000 A at full speed ahead. How long does it take to move Avogadro's number ( 6.021023 rate? ) of electrons at this 12. Electron guns are used in X-ray tubes. The electrons are accelerated through a relatively large voltage and directed onto a metal target, producing X-rays. (a) How many electrons per second strike the target if the current is 0.500 mA? (b) What charge strikes the target in 0.750 s? 13. A large cyclotron directs a beam of He++ target with a beam current of 0.250 mA. (a) How many He++ nuclei per second is this? (b) How long does it take for 1.00 C to strike the target? (c) How long before 1.00 mol of He++ nuclei strike the target? nuclei onto a 14. Repeat the above example on Example 20.3, but for a wire made of silver and given there is one free electron per silver atom. 15. Using the results of the above example on Example 20.3, find the drift velocity in a copper wire of twice the diameter and carrying 20.0 A. 16. A 14-gauge copper wire has a diameter of 1.628 mm. What magnitude current flows when the drift velocity is 1.00 mm/s? (See above example on Example 20.3 for useful information.) 17. SPEAR, a storage ring about 72.0 m in diameter at the Stanford Linear Accelerator (closed in 2009), has a 20.0-A circulating beam of electrons that are moving at nearly the speed of light. (See Figure 20.42.) How many electrons are in the beam? Figure 20.42 Electrons circulating in the storage ring called SPEAR constitute a 20.0-A current. Because they travel close to the speed of light, each electron completes many orbits in each second. 20.2 Ohm’s Law: Resistance and Simple Circuits 18. What current flows through the bulb of a 3.00-V flashlight when its hot resistance is 3.60 Ω ? 19. Calculate the effective resistance of a pocket calculator that has a 1.35-V battery and through which 0.200 mA flows. 20. What is the effective resistance of a car's starter motor when 150 A flows through it as the car battery applies 11.0 V to the motor? 21. How many volts are supplied to operate an indicator light on a DVD player that has a resistance of 140 Ω , given that 25.0 mA passes through it? 22. (a) Find the voltage drop in an extension cord having a 0.0600- Ω resistance and through which 5.00 A is flowing. (b) A cheaper cord utilizes thinner wire and has a resistance of 0.300 Ω . What is the voltage drop in it when 5.00 A Chapter 20 \| Electric Current, Resistance, and Ohm's Law 907 flows? (c) Why is the voltage to whatever appliance is being used reduced by this amount? What is the effect on the appliance? 23. A power transmission line is hung from metal towers with glass insulators having a resistance of 1.00109 Ω . What current flows through the insulator if the voltage is 200 kV? (Some high-voltage lines are DC.) low temperatures. Discuss why and whether this is the case here. (Hint: Resistance can't become negative.) 38. Integrated Concepts (a) Redo Exercise 20.25 taking into account the thermal expansion of the tungsten filament. You may assume a thermal expansion coefficient of 1210−6 / ºC . (b) By what percentage does your answer differ from that in the example? 20.3 Resistance and Resistivity 39. Unreasonable Results 24. What is the resistance of a 20.0-m-long piece of 12-gauge copper wire having a 2.053-mm diameter? 25. The diameter of 0-gauge copper wire is 8.252 mm. Find the resistance of a 1.00-km length of such wire used for power transmission. 26. If the 0.100-mm diameter tungsten filament in a light bulb is to have a resistance of 0.200 Ω at 20.0ºC , how long should it be? 27. Find the ratio of the diameter of aluminum to copper wire, if they have the same resistance per unit length (as they might in household wiring). 28. What current flows through a 2.54-cm-diameter rod of pure silicon that is 20.0 cm long, when 1.00 × 103 V is applied to it? (Such a rod may be used to make nuclearparticle detectors, for example.) 29. (a) To what temperature must you raise a copper wire, originally at 20.0ºC , to double its resistance, neglecting any changes in dimensions? (b) Does this happen in household wiring under ordinary circumstances? 30. A resistor made of Nichrome wire is used in an application where its resistance cannot change more than 1.00% from its value at 20.0ºC . Over what temperature range can it be used? 31. Of what material is a resistor made if its resistance is 40.0% greater at 100ºC than at 20.0ºC ? 32. An electronic device designed to operate at any temperature in the range from –10.0ºC to 55.0ºC contains pure carbon resistors. By what factor does their resistance increase over this range? 33. (a) Of what material is a wire made, if it is 25.0 m long with a 0.100 mm diameter and has a resistance of 77.7 Ω at 20.0ºC ? (b) What is its resistance at 150ºC ? 34. Assuming a constant temperature coefficient of resistivity, what is the maximum percent decrease in the resistance of a constantan wire starting at 20.0ºC ? 35. A wire is drawn through a die, stretching it to four times its original length. By what factor does its resistance increase? 36. A copper wire has a resistance of 0.500 Ω at 20.0ºC , and an iron wire has a resistance of 0.525 Ω at the same temperature. At what temperature are their resistances equal? 37. (a) Digital medical thermometers determine temperature by measuring the resistance of a semiconductor device called a thermistor (which has = – 0.0600 / ºC ) when it is at the same temperature as the patient. What is a patient's temperature if the thermistor's resistance at that temperature is 82.0% of its value at 37.0ºC (normal body temperature)? (b) The negative value for may not be maintained for very (a) To what temperature must you raise a resistor made of constantan to double its resistance, assuming a constant temperature coefficient of resistivity? (b) To cut it in half? (c) What is unreasonable about these results? (d) Which assumptions are unreasonable, or which premises are inconsistent? 20.4 Electric Power and Energy 40. What is the power of a 1.00×102 MV lightning bolt having a current of 2.00 × 104 A ? 41. What power is supplied to the starter motor of a large truck that draws 250 A of current from a 24.0-V battery hookup? 42. A charge of 4.00 C of charge passes through a pocket calculator's solar cells in 4.00 h. What is the power output, given the calculator's voltage output is 3.00 V? (See Figure 20.43.) Figure 20.43 The strip of solar cells just above the keys of this calculator convert light to electricity to supply its energy needs. (credit: Evan-Amos, Wikimedia Commons) 43. How many watts does a flashlight that has 6.00×102 C pass through it in 0.500 h use if its voltage is 3.00 V? 44. Find the power dissipated in each of these extension cords: (a) an extension cord having a 0.0600 - Ω resistance and through which 5.00 A is flowing; (b) a cheaper cord utilizing thinner wire and with a resistance of 0.300 Ω . 45. Verify that the units of a volt-ampere are watts, as implied by the equation = . 46. Show that the units 1 V2 / Ω = 1W , as implied by the equation = 2 / . 47. Show that the units 1 A2 ⋅ Ω = 1 W , as implied by the equation = 2 . 908 Chapter 20 \| Electric Current, Resistance, and Ohm's Law minute? (b) How much water must you put into the vaporizer for 8.00 h of overnight operation? (See Figure 20.45.) 48. Verify the energy unit equivalence that 1 kW ⋅ h = 3.60106 J . 49. Electrons in an X-ray tube are accelerated through 1.00×102 kV and directed toward a target to produce Xrays. Calculate the power of the electron beam in this tube if it has a current of 15.0 mA. 50. An electric water heater consumes 5.00 kW for 2.00 h per day. What is the cost of running it for one year if electricity costs 12.0 cents/kW ⋅ h ? See Figure 20.44. Figure 20.44 On-demand electric hot water heater. Heat is supplied to water only when needed. (credit: aviddavid, Flickr) 60. Integrated Concepts Figure 20.45 This cold vaporizer passes current directly through water, vaporizing it directly with relatively little temperature increase. (a) What energy is dissipated by a lightning bolt having a 20,000-A current, a voltage of 1.00×102 MV , and a length of 1.00 ms? (b) What mass of tree sap could be raised from 18.0ºC to its boiling point and then evaporated by this energy, assuming sap has the same thermal characteristics as water? 61. Integrated Concepts What current must be produced by a 12.0-V battery-operated bottle warmer in order to heat 75.0 g of glass, 250 g of baby formula, and 3.00×102 g of aluminum from 20.0ºC to 90.0ºC in 5.00 min? 62. Integrated Concepts How much time is needed for a surgical cauterizer to raise the temperature of 1.00 g of tissue from 37.0ºC to 100ºC and then boil away 0.500 g of
water, if it puts out 2.00 mA at 15.0 kV? Ignore heat transfer to the surroundings. 63. Integrated Concepts Hydroelectric generators (see Figure 20.46) at Hoover Dam produce a maximum current of 8.00×103 A at 250 kV. (a) What is the power output? (b) The water that powers the generators enters and leaves the system at low speed (thus its kinetic energy does not change) but loses 160 m in altitude. How many cubic meters per second are needed, assuming 85.0% efficiency? 51. With a 1200-W toaster, how much electrical energy is needed to make a slice of toast (cooking time = 1 minute)? At 9.0 cents/kW · h , how much does this cost? 52. What would be the maximum cost of a CFL such that the total cost (investment plus operating) would be the same for both CFL and incandescent 60-W bulbs? Assume the cost of the incandescent bulb is 25 cents and that electricity costs 10 cents/kWh . Calculate the cost for 1000 hours, as in the cost effectiveness of CFL example. 53. Some makes of older cars have 6.00-V electrical systems. (a) What is the hot resistance of a 30.0-W headlight in such a car? (b) What current flows through it? 54. Alkaline batteries have the advantage of putting out constant voltage until very nearly the end of their life. How long will an alkaline battery rated at 1.00 A ⋅ h and 1.58 V keep a 1.00-W flashlight bulb burning? 55. A cauterizer, used to stop bleeding in surgery, puts out 2.00 mA at 15.0 kV. (a) What is its power output? (b) What is the resistance of the path? 56. The average television is said to be on 6 hours per day. Estimate the yearly cost of electricity to operate 100 million TVs, assuming their power consumption averages 150 W and the cost of electricity averages 12.0 cents/kW ⋅ h . 57. An old lightbulb draws only 50.0 W, rather than its original 60.0 W, due to evaporative thinning of its filament. By what factor is its diameter reduced, assuming uniform thinning along its length? Neglect any effects caused by temperature differences. 58. 00-gauge copper wire has a diameter of 9.266 mm. Calculate the power loss in a kilometer of such wire when it carries 1.00×102 A . 59. Integrated Concepts Cold vaporizers pass a current through water, evaporating it with only a small increase in temperature. One such home device is rated at 3.50 A and utilizes 120 V AC with 95.0% efficiency. (a) What is the vaporization rate in grams per This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 20 \| Electric Current, Resistance, and Ohm's Law 909 (a) An immersion heater utilizing 120 V can raise the temperature of a 1.00×102 -g aluminum cup containing 350 g of water from 20.0ºC to 95.0ºC in 2.00 min. Find its resistance, assuming it is constant during the process. (b) A lower resistance would shorten the heating time. Discuss the practical limits to speeding the heating by lowering the resistance. 68. Integrated Concepts (a) What is the cost of heating a hot tub containing 1500 kg of water from 10.0ºC to 40.0ºC , assuming 75.0% efficiency to account for heat transfer to the surroundings? The cost of electricity is 9 cents/kW ⋅ h . (b) What current was used by the 220-V AC electric heater, if this took 4.00 h? 69. Unreasonable Results (a) What current is needed to transmit 1.00×102 MW of power at 480 V? (b) What power is dissipated by the transmission lines if they have a 1.00 - Ω resistance? (c) What is unreasonable about this result? (d) Which assumptions are unreasonable, or which premises are inconsistent? 70. Unreasonable Results (a) What current is needed to transmit 1.00×102 MW of power at 10.0 kV? (b) Find the resistance of 1.00 km of wire that would cause a 0.0100% power loss. (c) What is the diameter of a 1.00-km-long copper wire having this resistance? (d) What is unreasonable about these results? (e) Which assumptions are unreasonable, or which premises are inconsistent? 71. Construct Your Own Problem Consider an electric immersion heater used to heat a cup of water to make tea. Construct a problem in which you calculate the needed resistance of the heater so that it increases the temperature of the water and cup in a reasonable amount of time. Also calculate the cost of the electrical energy used in your process. Among the things to be considered are the voltage used, the masses and heat capacities involved, heat losses, and the time over which the heating takes place. Your instructor may wish for you to consider a thermal safety switch (perhaps bimetallic) that will halt the process before damaging temperatures are reached in the immersion unit. 20.5 Alternating Current versus Direct Current 72. (a) What is the hot resistance of a 25-W light bulb that runs on 120-V AC? (b) If the bulb's operating temperature is 2700ºC , what is its resistance at 2600ºC ? 73. Certain heavy industrial equipment uses AC power that has a peak voltage of 679 V. What is the rms voltage? 74. A certain circuit breaker trips when the rms current is 15.0 A. What is the corresponding peak current? 75. Military aircraft use 400-Hz AC power, because it is possible to design lighter-weight equipment at this higher frequency. What is the time for one complete cycle of this power? 76. A North American tourist takes his 25.0-W, 120-V AC razor to Europe, finds a special adapter, and plugs it into 240 V AC. Assuming constant resistance, what power does the razor consume as it is ruined? Figure 20.46 Hydroelectric generators at the Hoover dam. (credit: Jon Sullivan) 64. Integrated Concepts (a) Assuming 95.0% efficiency for the conversion of electrical power by the motor, what current must the 12.0-V batteries of a 750-kg electric car be able to supply: (a) To accelerate from rest to 25.0 m/s in 1.00 min? (b) To climb a 2.00×102 -m high hill in 2.00 min at a constant 25.0-m/s speed while exerting 5.00×102 N of force to overcome air resistance and friction? (c) To travel at a constant 25.0-m/s speed, exerting a 5.00×102 N force to overcome air resistance and friction? See Figure 20.47. Figure 20.47 This REVAi, an electric car, gets recharged on a street in London. (credit: Frank Hebbert) 65. Integrated Concepts A light-rail commuter train draws 630 A of 650-V DC electricity when accelerating. (a) What is its power consumption rate in kilowatts? (b) How long does it take to reach 20.0 m/s starting from rest if its loaded mass is 5.30104 kg , assuming 95.0% efficiency and constant power? (c) Find its average acceleration. (d) Discuss how the acceleration you found for the light-rail train compares to what might be typical for an automobile. 66. Integrated Concepts (a) An aluminum power transmission line has a resistance of 0.0580 Ω / km . What is its mass per kilometer? (b) What is the mass per kilometer of a copper line having the same resistance? A lower resistance would shorten the heating time. Discuss the practical limits to speeding the heating by lowering the resistance. 67. Integrated Concepts 910 Chapter 20 \| Electric Current, Resistance, and Ohm's Law 90. (a) During surgery, a current as small as 20.0 μA applied directly to the heart may cause ventricular fibrillation. If the resistance of the exposed heart is 300 Ω , what is the smallest voltage that poses this danger? (b) Does your answer imply that special electrical safety precautions are needed? 91. (a) What is the resistance of a 220-V AC short circuit that generates a peak power of 96.8 kW? (b) What would the average power be if the voltage was 120 V AC? 92. A heart defibrillator passes 10.0 A through a patient's torso for 5.00 ms in an attempt to restore normal beating. (a) How much charge passed? (b) What voltage was applied if 500 J of energy was dissipated? (c) What was the path's resistance? (d) Find the temperature increase caused in the 8.00 kg of affected tissue. 93. Integrated Concepts A short circuit in a 120-V appliance cord has a 0.500- Ω resistance. Calculate the temperature rise of the 2.00 g of surrounding materials, assuming their specific heat capacity is 0.200 cal/g⋅ºC and that it takes 0.0500 s for a circuit breaker to interrupt the current. Is this likely to be damaging? 94. Construct Your Own Problem Consider a person working in an environment where electric currents might pass through her body. Construct a problem in which you calculate the resistance of insulation needed to protect the person from harm. Among the things to be considered are the voltage to which the person might be exposed, likely body resistance (dry, wet, …), and acceptable currents (safe but sensed, safe and unfelt, …). 20.7 Nerve Conduction–Electrocardiograms 95. Integrated Concepts Use the ECG in Figure 20.37 to determine the heart rate in beats per minute assuming a constant time between beats. 96. Integrated Concepts (a) Referring to Figure 20.37, find the time systolic pressure lags behind the middle of the QRS complex. (b) Discuss the reasons for the time lag. 77. In this problem, you will verify statements made at the end of the power losses for Example 20.10. (a) What current is needed to transmit 100 MW of power at a voltage of 25.0 kV? (b) Find the power loss in a 1.00 - Ω transmission line. (c) What percent loss does this represent? 78. A small office-building air conditioner operates on 408-V AC and consumes 50.0 kW. (a) What is its effective resistance? (b) What is the cost of running the air conditioner during a hot summer month when it is on 8.00 h per day for 30 days and electricity costs 9.00 cents/kW ⋅ h ? 79. What is the peak power consumption of a 120-V AC microwave oven that draws 10.0 A? 80. What is the peak current through a 500-W room heater that operates on 120-V AC power? 81. Two different electrical devices have the same power consumption, but one is meant to be operated on 120-V AC and the other on 240-V AC. (a) What is the ratio of their resistances? (b) What is the ratio of their currents? (c) Assuming its resistance is unaffected, by what factor will the power increase if a 120-V AC device is connected to 240-V AC?
82. Nichrome wire is used in some radiative heaters. (a) Find the resistance needed if the average power output is to be 1.00 kW utilizing 120-V AC. (b) What length of Nichrome wire, having a cross-sectional area of 5.00mm2 , is needed if the operating temperature is 500º C ? (c) What power will it draw when first switched on? 83. Find the time after = 0 when the instantaneous voltage of 60-Hz AC first reaches the following values: (a) 0 / 2 (b) 0 (c) 0. 84. (a) At what two times in the first period following = 0 does the instantaneous voltage in 60-Hz AC equal rms ? (b) −rms ? 20.6 Electric Hazards and the Human Body 85. (a) How much power is dissipated in a short circuit of 240-V AC through a resistance of 0.250 Ω ? (b) What current flows? 86. What voltage is involved in a 1.44-kW short circuit through a 0.100 - Ω resistance? 87. Find the current through a person and identify the likely effect on her if she touches a 120-V AC source: (a) if she is standing on a rubber mat and offers a total resistance of 300 k Ω ; (b) if she is standing barefoot on wet grass and has a resistance of only 4000 k Ω . 88. While taking a bath, a person touches the metal case of a radio. The path through the person to the drainpipe and ground has a resistance of 4000 Ω . What is the smallest voltage on the case of the radio that could cause ventricular fibrillation? 89. Foolishly trying to fish a burning piece of bread from a toaster with a metal butter knife, a man comes into contact with 120-V AC. He does not even feel it since, luckily, he is wearing rubber-soled shoes. What is the minimum resistance of the path the current follows through the person? This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 20 \| Electric Current, Resistance, and Ohm's Law 911 Test Prep for AP® Courses 20.1 Current 1. Which of the following can be explained on the basis of conservation of charge in a closed circuit consisting of a battery, resistor, and metal wires? a. The number of electrons leaving the battery will be equal to the number of electrons entering the battery. b. The number of electrons leaving the battery will be less than the number of electrons entering the battery. c. The number of protons leaving the battery will be equal to the number of protons entering the battery. d. The number of protons leaving the battery will be less than the number of protons entering the battery. 2. When a battery is connected to a bulb, there is 2.5 A of current in the circuit. What amount of charge will flow though the circuit in a time of 0.5 s? a. 0.5 C b. 1 C c. 1.25 C d. 1.5 C 3. If 0.625 × 1020 electrons flow through a circuit each second, what is the current in the circuit? 4. Two students calculate the charge flowing through a circuit. The first student concludes that 300 C of charge flows in 1 minute. The second student concludes that 3.125 × 1019 electrons flow per second. If the current measured in the circuit is 5 A, which of the two students (if any) have performed the calculations correctly? 20.2 Ohm’s Law: Resistance and Simple Circuits 5. If the voltage across a fixed resistance is doubled, what happens to the current? It doubles. It halves. It stays the same. a. b. c. d. The current cannot be determined. Figure 20.48 If the four wires are made from the same material, which of the following is true? Select two answers. a. Resistance of Wire 3 > Resistance of Wire 2 b. Resistance of Wire 1 > Resistance of Wire 2 c. Resistance of Wire 1 < Resistance of Wire 4 d. Resistance of Wire 4 < Resistance of Wire 3 10. Suppose the resistance of a wire is R Ω. What will be the resistance of another wire of the same material having the same length but double the diameter? a. R/2 b. 2R c. R/4 d. 4R 11. The resistances of two wires having the same lengths and cross section areas are 3 Ω and 11 Ω. If the resistivity of the 3 Ω wire is 2.65 × 10−8 Ω·m, find the resistivity of the 1 Ω wire. 12. The lengths and diameters of three wires are given below. If they all have the same resistance, find the ratio of their resistivities. Table 20.5 Wire Length Diameter Wire 1 2 m 1 cm Wire 2 1 m 0.5 cm Wire 3 1 m 1 cm 13. Suppose the resistance of a wire is 2 Ω. If the wire is stretched to three times its length, what will be its resistance? Assume that the volume does not change. 20.4 Electric Power and Energy 6. The table below gives the voltages and currents recorded across a resistor. 14. Table 20.4 Voltage (V) 2.50 5.00 7.50 10.00 12.50 Current (A) 0.69 1.38 2.09 2.76 3.49 a. Plot the graph and comment on the shape. b. Calculate the value of the resistor. 7. What is the resistance of a bulb if the current in it is 1.25 A when a 4 V voltage supply is connected to it? If the voltage supply is increased to 7 V, what will be the current in the bulb? 20.3 Resistance and Resistivity 8. Which of the following affect the resistivity of a wire? length a. b. area of cross section c. material d. all of the above 9. The lengths and diameters of four wires are given as shown. Figure 20.49 The circuit shown contains a resistor R connected to a voltage supply. The graph shows the total energy E dissipated by the resistance as a function of time. Which of the following shows the corresponding graph for double resistance, i.e., if R is replaced by 2R? 912 Chapter 20 \| Electric Current, Resistance, and Ohm's Law a. Figure 20.50 b. Figure 20.51 c. Figure 20.52 d. Figure 20.53 15. What will be the ratio of the resistance of a 120 W, 220 V lamp to that of a 100 W, 110 V lamp? This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments 913 21 CIRCUITS, BIOELECTRICITY, AND DC INSTRUMENTS Figure 21.1 Electric circuits in a computer allow large amounts of data to be quickly and accurately analyzed.. (credit: Airman 1st Class Mike Meares, United States Air Force) Chapter Outline 21.1. Resistors in Series and Parallel 21.2. Electromotive Force: Terminal Voltage 21.3. Kirchhoff’s Rules 21.4. DC Voltmeters and Ammeters 21.5. Null Measurements 21.6. DC Circuits Containing Resistors and Capacitors Connection for AP® Courses Electric circuits are commonplace in our everyday lives. Some circuits are simple, such as those in flashlights while others are extremely complex, such as those used in supercomputers. This chapter takes the topic of electric circuits a step beyond simple circuits by addressing both changes that result from interactions between systems (Big Idea 4) and constraints on such changes due to laws of conservation (Big Idea 5). When the circuit is purely resistive, everything in this chapter applies to both DC and AC. However, matters become more complex when capacitance is involved. We do consider what happens when capacitors are connected to DC voltage sources, but the interaction of capacitors (and other nonresistive devices) with AC sources is left for a later chapter. In addition, a number of important DC instruments, such as meters that measure voltage and current, are covered in this chapter. Information and examples presented in the chapter examine cause-effect relationships inherent in interactions involving electrical systems. The electrical properties of an electric circuit can change due to other systems (Enduring Understanding 4.E). More specifically, values of currents and potential differences in electric circuits depend on arrangements of individual circuit components (Essential Knowledge 4.E.5). In this chapter several series and parallel combinations of resistors are discussed and their effects on currents and potential differences are analyzed. In electric circuits the total energy (Enduring Understanding 5.B) and the total electric charge (Enduring Understanding 5.C) are conserved. Kirchoff’s rules describe both, energy conservation (Essential Knowledge 5.B.9) and charge conservation (Essential 914 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments Knowledge 5.C.3). Energy conservation is discussed in terms of the loop rule which specifies that the potential around any closed circuit path must be zero. Charge conservation is applied as conservation of current by equating the sum of all currents entering a junction to the sum of all currents leaving the junction (also known as the junction rule). Kirchoff’s rules are used to calculate currents and potential differences in circuits that combine resistors in series and parallel, and resistors and capacitors. The concepts in this chapter support: Big Idea 4 Interactions between systems can result in changes in those systems. Enduring Understanding 4.E The electric and magnetic properties of a system can change in response to the presence of, or changes in, other objects or systems. Essential Knowledge 4.E.5 The values of currents and electric potential differences in an electric circuit are determined by the properties and arrangement of the individual circuit elements such as sources of emf, resistors, and capacitors. Big Idea 5 Changes that occur as a result of interactions are constrained by conservation laws. Enduring Understanding 5.B The energy of a system is conserved. Essential Knowledge 5.B.9 Kirchhoff’s loop rule describes conservation of energy in electrical circuits. Enduring Understanding 5.C The electric charge of a system is conserved. Essential Knowledge 5.C.3 Kirchhoff’s junction rule describes the conservation of electric charge in electrical circuits. Since charge is conserved, current must be conserved at each junction in the circuit. Examples should include circuits that combine resistors in series and parallel. 21.1 Resistors in Series and Parallel By the end of this section, you will be able to: Learning Objectives • Draw a circuit with resistors in parallel and in series. • Use Ohm’s law to calculate the voltage drop across a resistor when current passes through it. • Contrast the way total resistance is calculated for resistors in series and in parallel. • Explain why total resis
tance of a parallel circuit is less than the smallest resistance of any of the resistors in that circuit. • Calculate total resistance of a circuit that contains a mixture of resistors connected in series and in parallel. The information presented in this section supports the following AP® learning objectives and science practices: • 4.E.5.1 The student is able to make and justify a quantitative prediction of the effect of a change in values or arrangements of one or two circuit elements on the currents and potential differences in a circuit containing a small number of sources of emf, resistors, capacitors, and switches in series and/or parallel. (S.P. 2.2, 6.4) • 4.E.5.2 The student is able to make and justify a qualitative prediction of the effect of a change in values or arrangements of one or two circuit elements on currents and potential differences in a circuit containing a small number of sources of emf, resistors, capacitors, and switches in series and/or parallel. (S.P. 6.1, 6.4) • 4.E.5.3 The student is able to plan data collection strategies and perform data analysis to examine the values of currents and potential differences in an electric circuit that is modified by changing or rearranging circuit elements, including sources of emf, resistors, and capacitors. (S.P. 2.2, 4.2, 5.1) • 5.B.9.3 The student is able to apply conservation of energy (Kirchhoff’s loop rule) in calculations involving the total electric potential difference for complete circuit loops with only a single battery and resistors in series and/or in, at most, one parallel branch. (S.P. 2.2, 6.4, 7.2) Most circuits have more than one component, called a resistor that limits the flow of charge in the circuit. A measure of this limit on charge flow is called resistance. The simplest combinations of resistors are the series and parallel connections illustrated in Figure 21.2. The total resistance of a combination of resistors depends on both their individual values and how they are connected. Figure 21.2 (a) A series connection of resistors. (b) A parallel connection of resistors. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments 915 Resistors in Series When are resistors in series? Resistors are in series whenever the flow of charge, called the current, must flow through devices sequentially. For example, if current flows through a person holding a screwdriver and into the Earth, then 1 in Figure 21.2(a) could be the resistance of the screwdriver’s shaft, 2 the resistance of its handle, 3 the person’s body resistance, and 4 the resistance of her shoes. Figure 21.3 shows resistors in series connected to a voltage source. It seems reasonable that the total resistance is the sum of the individual resistances, considering that the current has to pass through each resistor in sequence. (This fact would be an advantage to a person wishing to avoid an electrical shock, who could reduce the current by wearing high-resistance rubbersoled shoes. It could be a disadvantage if one of the resistances were a faulty high-resistance cord to an appliance that would reduce the operating current.) Figure 21.3 Three resistors connected in series to a battery (left) and the equivalent single or series resistance (right). To verify that resistances in series do indeed add, let us consider the loss of electrical power, called a voltage drop, in each resistor in Figure 21.3. According to Ohm’s law, the voltage drop, , across a resistor when a current flows through it is calculated using the equation = , where equals the current in amps (A) and is the resistance in ohms ( Ω ) . Another way to think of this is that is the voltage necessary to make a current flow through a resistance . So the voltage drop across 1 is 1 = 1 , that across 2 is 2 = 2 , and that across 3 is 3 = 3 . The sum of these voltages equals the voltage output of the source; that is, = 1 + 2 + 3. (21.1) This equation is based on the conservation of energy and conservation of charge. Electrical potential energy can be described by the equation = , where is the electric charge and is the voltage. Thus the energy supplied by the source is , while that dissipated by the resistors is 1 + 2 + 3. (21.2) Connections: Conservation Laws The derivations of the expressions for series and parallel resistance are based on the laws of conservation of energy and conservation of charge, which state that total charge and total energy are constant in any process. These two laws are directly involved in all electrical phenomena and will be invoked repeatedly to explain both specific effects and the general behavior of electricity. These energies must be equal, because there is no other source and no other destination for energy in the circuit. Thus, = 1 + 2 + 3 . The charge cancels, yielding = 1 + 2 + 3 , as stated. (Note that the same amount of charge passes through the battery and each resistor in a given amount of time, since there is no capacitance to store charge, there is no place for charge to leak, and charge is conserved.) Now substituting the values for the individual voltages gives = 1 + 2 + 3 = (1 + 2 + 3). Note that for the equivalent single series resistance s , we have = s. This implies that the total or equivalent series resistance s of three resistors is s = 1 + 2 + 3 . This logic is valid in general for any number of resistors in series; thus, the total resistance s of a series connection is s = 1 + 2 + 3 + ..., (21.3) (21.4) (21.5) 916 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments as proposed. Since all of the current must pass through each resistor, it experiences the resistance of each, and resistances in series simply add up. Example 21.1 Calculating Resistance, Current, Voltage Drop, and Power Dissipation: Analysis of a Series Circuit Suppose the voltage output of the battery in Figure 21.3 is 12.0 V , and the resistances are 1 = 1.00 Ω , 2 = 6.00 Ω , and 3 = 13.0 Ω . (a) What is the total resistance? (b) Find the current. (c) Calculate the voltage drop in each resistor, and show these add to equal the voltage output of the source. (d) Calculate the power dissipated by each resistor. (e) Find the power output of the source, and show that it equals the total power dissipated by the resistors. Strategy and Solution for (a) The total resistance is simply the sum of the individual resistances, as given by this equation.00 Ω + 6.00 Ω + 13.0 Ω = 20.0 Ω. (21.6) Strategy and Solution for (b) The current is found using Ohm’s law, = . Entering the value of the applied voltage and the total resistance yields the current for the circuit: = s = 12.0 V 20.0 Ω = 0.600 A. (21.7) Strategy and Solution for (c) The voltage—or drop—in a resistor is given by Ohm’s law. Entering the current and the value of the first resistance yields Similarly, and 1 = 1 = (0.600 A)(1.0 Ω ) = 0.600 V. 2 = 2 = (0.600 A)(6.0 Ω ) = 3.60 V 3 = 3 = (0.600 A)(13.0 Ω ) = 7.80 V. Discussion for (c) The three drops add to 12.0 V , as predicted: 1 + 2 + 3 = (0.600 + 3.60 + 7.80) V = 12.0 V. (21.8) (21.9) (21.10) (21.11) Strategy and Solution for (d) The easiest way to calculate power in watts (W) dissipated by a resistor in a DC circuit is to use Joule’s law, = , where is electric power. In this case, each resistor has the same full current flowing through it. By substituting Ohm’s law = into Joule’s law, we get the power dissipated by the first resistor as Similarly, and 1 = 21 = (0.600 A)2(1.00 Ω ) = 0.360 W. 2 = 22 = (0.600 A)2(6.00 Ω ) = 2.16 W 3 = 23 = (0.600 A)2(13.0 Ω ) = 4.68 W. (21.12) (21.13) (21.14) Discussion for (d) Power can also be calculated using either = or = 2 full voltage of the source). The same values will be obtained. Strategy and Solution for (e) , where is the voltage drop across the resistor (not the This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments 917 The easiest way to calculate power output of the source is to use = , where is the source voltage. This gives = (0.600 A)(12.0 V) = 7.20 W. (21.15) Discussion for (e) Note, coincidentally, that the total power dissipated by the resistors is also 7.20 W, the same as the power put out by the source. That is, 1 + 2 + 3 = (0.360 + 2.16 + 4.68) W = 7.20 W. (21.16) Power is energy per unit time (watts), and so conservation of energy requires the power output of the source to be equal to the total power dissipated by the resistors. Major Features of Resistors in Series 1. Series resistances add: s = 1 + 2 + 3 + .... 2. The same current flows through each resistor in series. 3. Individual resistors in series do not get the total source voltage, but divide it. Resistors in Parallel Figure 21.4 shows resistors in parallel, wired to a voltage source. Resistors are in parallel when each resistor is connected directly to the voltage source by connecting wires having negligible resistance. Each resistor thus has the full voltage of the source applied to it. Each resistor draws the same current it would if it alone were connected to the voltage source (provided the voltage source is not overloaded). For example, an automobile’s headlights, radio, and so on, are wired in parallel, so that they utilize the full voltage of the source and can operate completely independently. The same is true in your house, or any building. (See Figure 21.4(b).) 918 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments Figure 21.4 (a) Three resistors connected in parallel to a battery and the equivalent single or parallel resistance. (b) Electrical power setup in a house. (credit: Dmitry G, Wikimedia Commons) To find an expression for the equivalent parallel resistance p , let us consider the currents that flow and how they are related to resistance. Since each resistor in the circuit has the full voltage, the currents flowing through the individual resistors are 1 = 1 of these currents: . Conservation of charge impl
ies that the total current produced by the source is the sum , and 3 = 3 , 2 = 2 Substituting the expressions for the individual currents gives = 1 + 2 + 3 . Note that Ohm’s law for the equivalent single resistance gives = p = 1 p . (21.17) (21.18) (21.19) The terms inside the parentheses in the last two equations must be equal. Generalizing to any number of resistors, the total resistance p of a parallel connection is related to the individual resistances by 3 + .... (21.20) This relationship results in a total resistance p that is less than the smallest of the individual resistances. (This is seen in the next example.) When resistors are connected in parallel, more current flows from the source than would flow for any of them individually, and so the total resistance is lower. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments 919 Example 21.2 Calculating Resistance, Current, Power Dissipation, and Power Output: Analysis of a Parallel Circuit Let the voltage output of the battery and resistances in the parallel connection in Figure 21.4 be the same as the previously considered series connection: = 12.0 V , 1 = 1.00 Ω , 2 = 6.00 Ω , and 3 = 13.0 Ω . (a) What is the total resistance? (b) Find the total current. (c) Calculate the currents in each resistor, and show these add to equal the total current output of the source. (d) Calculate the power dissipated by each resistor. (e) Find the power output of the source, and show that it equals the total power dissipated by the resistors. Strategy and Solution for (a) The total resistance for a parallel combination of resistors is found using the equation below. Entering known values gives .00 Ω + 1 6.00 Ω + 1 13.0 Ω . Thus, 1 p = 1.00 Ω + 0.1667 Ω + 0.07692 Ω = 1.2436 Ω . (Note that in these calculations, each intermediate answer is shown with an extra digit.) We must invert this to find the total resistance p . This yields The total resistance with the correct number of significant digits is p = 0.804 Ω . p = 1 1.2436 Ω = 0.8041 Ω . Discussion for (a) p is, as predicted, less than the smallest individual resistance. Strategy and Solution for (b) The total current can be found from Ohm’s law, substituting p for the total resistance. This gives = p = 12.0 V 0.8041 Ω = 14.92 A. (21.21) (21.22) (21.23) (21.24) Discussion for (b) Current for each device is much larger than for the same devices connected in series (see the previous example). A circuit with parallel connections has a smaller total resistance than the resistors connected in series. Strategy and Solution for (c) The individual currents are easily calculated from Ohm’s law, since each resistor gets the full voltage. Thus, Similarly, and 1 = 1 = 12.0 V 1.00 Ω = 12.0 A. 2 = 2 = 12.0 V 6.00 Ω = 2.00 A 3 = 3 = 12.0 V 13.0 Ω = 0.92 A. Discussion for (c) The total current is the sum of the individual currents: 1 + 2 + 3 = 14.92 A. This is consistent with conservation of charge. Strategy and Solution for (d) The power dissipated by each resistor can be found using any of the equations relating power to current, voltage, and resistance, since all three are known. Let us use = 2 , since each resistor gets full voltage. Thus, (21.25) (21.26) (21.27) (21.28) 920 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments 1 = 2 1 = (12.0 V)2 1.00 Ω = 144 W. 2 = 2 2 = (12.0 V)2 6.00 Ω = 24.0 W 3 = 2 3 = (12.0 V)2 13.0 Ω = 11.1 W. Similarly, and Discussion for (d) (21.29) (21.30) (21.31) The power dissipated by each resistor is considerably higher in parallel than when connected in series to the same voltage source. Strategy and Solution for (e) The total power can also be calculated in several ways. Choosing = , and entering the total current, yields = = (14.92 A)(12.0 V) = 179 W. Discussion for (e) Total power dissipated by the resistors is also 179 W: 1 + 2 + 3 = 144 W + 24.0 W + 11.1 W = 179 W. This is consistent with the law of conservation of energy. Overall Discussion Note that both the currents and powers in parallel connections are greater than for the same devices in series. (21.32) (21.33) Major Features of Resistors in Parallel 1. Parallel resistance is found from combination. + ... , and it is smaller than any individual resistance in the 2. Each resistor in parallel has the same full voltage of the source applied to it. (Power distribution systems most often use parallel connections to supply the myriad devices served with the same voltage and to allow them to operate independently.) 3. Parallel resistors do not each get the total current; they divide it. Combinations of Series and Parallel More complex connections of resistors are sometimes just combinations of series and parallel. These are commonly encountered, especially when wire resistance is considered. In that case, wire resistance is in series with other resistances that are in parallel. Combinations of series and parallel can be reduced to a single equivalent resistance using the technique illustrated in Figure 21.5. Various parts are identified as either series or parallel, reduced to their equivalents, and further reduced until a single resistance is left. The process is more time consuming than difficult. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments 921 Figure 21.5 This combination of seven resistors has both series and parallel parts. Each is identified and reduced to an equivalent resistance, and these are further reduced until a single equivalent resistance is reached. The simplest combination of series and parallel resistance, shown in Figure 21.6, is also the most instructive, since it is found in many applications. For example, 1 could be the resistance of wires from a car battery to its electrical devices, which are in parallel. 2 and 3 could be the starter motor and a passenger compartment light. We have previously assumed that wire resistance is negligible, but, when it is not, it has important effects, as the next example indicates. Example 21.3 Calculating Resistance, IR Drop, Current, and Power Dissipation: Combining Series and Parallel Circuits Figure 21.6 shows the resistors from the previous two examples wired in a different way—a combination of series and parallel. We can consider 1 to be the resistance of wires leading to 2 and 3 . (a) Find the total resistance. (b) What is the drop in 1 ? (c) Find the current 2 through 2 . (d) What power is dissipated by 2 ? Figure 21.6 These three resistors are connected to a voltage source so that 2 and 3 are in parallel with one another and that combination is in series with 1 . Strategy and Solution for (a) To find the total resistance, we note that 2 and 3 are in parallel and their combination p is in series with 1 . Thus the total (equivalent) resistance of this combination is tot = 1 + p. First, we find p using the equation for resistors in parallel and entering known values.00 Ω + 1 13.0 Ω = 0.2436 Ω . (21.34) (21.35) 922 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments Inverting gives So the total resistance is Discussion for (a) p = 1 0.2436 Ω = 4.11 Ω . tot = 1 + p = 1.00 Ω + 4.11 Ω = 5.11 Ω . (21.36) (21.37) The total resistance of this combination is intermediate between the pure series and pure parallel values ( 20.0 Ω and 0.804 Ω , respectively) found for the same resistors in the two previous examples. Strategy and Solution for (b) To find the drop in 1 , we note that the full current flows through 1 . Thus its drop is We must find before we can calculate 1 . The total current is found using Ohm’s law for the circuit. That is, 1 = . = tot = 12.0 V 5.11 Ω = 2.35 A. Entering this into the expression above, we get 1 = = (2.35 A)(1.00 Ω ) = 2.35 V. Discussion for (b) (21.38) (21.39) (21.40) The voltage applied to 2 and 3 is less than the total voltage by an amount 1 . When wire resistance is large, it can significantly affect the operation of the devices represented by 2 and 3 . Strategy and Solution for (c) To find the current through 2 , we must first find the voltage applied to it. We call this voltage p , because it is applied to a parallel combination of resistors. The voltage applied to both 2 and 3 is reduced by the amount 1 , and so it is Now the current 2 through resistance 2 is found using Ohm’s law: p = − 1 = 12.0 V − 2.35 V = 9.65 V. 2 = p 2 = 9.65 V 6.00 Ω = 1.61 A. (21.41) (21.42) Discussion for (c) The current is less than the 2.00 A that flowed through 2 when it was connected in parallel to the battery in the previous parallel circuit example. Strategy and Solution for (d) The power dissipated by 2 is given by 2 = (2 )22 = (1.61 A)2(6.00 Ω ) = 15.5 W. (21.43) Discussion for (d) The power is less than the 24.0 W this resistor dissipated when connected in parallel to the 12.0-V source. Applying the Science Practices: Circuit Construction Kit (DC only) Plan an experiment to analyze the effect on currents and potential differences due to rearrangement of resistors and variations in voltage sources. Your experimental investigation should include data collection for at least five different scenarios of rearranged resistors (i.e., several combinations of series and parallel) and three scenarios of different voltage sources. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments 923 Practical Implications One implication of this last example is that resistance in wires reduces the current and power delivered to a resistor. If wire resistance is relatively large, as in a worn (or a very long) extension cord, then this loss can be significant. If a large current is drawn, the drop in the wires can also be significant. For example, when you are rummaging in the refrigerator and the motor comes on, the refrigerator light dims momentarily. Similarly, you can see the passenger compartment light dim when you start the engine of yo
ur car (although this may be due to resistance inside the battery itself). What is happening in these high-current situations is illustrated in Figure 21.7. The device represented by 3 has a very low resistance, and so when it is switched on, a large current flows. This increased current causes a larger drop in the wires represented by 1 , reducing the voltage across the light bulb (which is 2 ), which then dims noticeably. Figure 21.7 Why do lights dim when a large appliance is switched on? The answer is that the large current the appliance motor draws causes a significant drop in the wires and reduces the voltage across the light. Check Your Understanding Can any arbitrary combination of resistors be broken down into series and parallel combinations? See if you can draw a circuit diagram of resistors that cannot be broken down into combinations of series and parallel. Solution No, there are many ways to connect resistors that are not combinations of series and parallel, including loops and junctions. In such cases Kirchhoff’s rules, to be introduced in Kirchhoff’s Rules, will allow you to analyze the circuit. Problem-Solving Strategies for Series and Parallel Resistors 1. Draw a clear circuit diagram, labeling all resistors and voltage sources. This step includes a list of the knowns for the problem, since they are labeled in your circuit diagram. 2. Identify exactly what needs to be determined in the problem (identify the unknowns). A written list is useful. 3. Determine whether resistors are in series, parallel, or a combination of both series and parallel. Examine the circuit diagram to make this assessment. Resistors are in series if the same current must pass sequentially through them. 4. Use the appropriate list of major features for series or parallel connections to solve for the unknowns. There is one list for series and another for parallel. If your problem has a combination of series and parallel, reduce it in steps by considering individual groups of series or parallel connections, as done in this module and the examples. Special note: When finding , the reciprocal must be taken with care. 5. Check to see whether the answers are reasonable and consistent. Units and numerical results must be reasonable. Total series resistance should be greater, whereas total parallel resistance should be smaller, for example. Power should be greater for the same devices in parallel compared with series, and so on. 21.2 Electromotive Force: Terminal Voltage By the end of this section, you will be able to: Learning Objectives 924 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments • Compare and contrast the voltage and the electromagnetic force of an electric power source. • Describe what happens to the terminal voltage, current, and power delivered to a load as internal resistance of the voltage source increases. • Explain why it is beneficial to use more than one voltage source connected in parallel. The information presented in this section supports the following AP® learning objectives and science practices: • 5.B.9.7 The student is able to refine and analyze a scientific question for an experiment using Kirchhoff’s loop rule for circuits that includes determination of internal resistance of the battery and analysis of a nonohmic resistor. (S.P. 4.1, 4.2, 5.1, 5.3) When you forget to turn off your car lights, they slowly dim as the battery runs down. Why don’t they simply blink off when the battery’s energy is gone? Their gradual dimming implies that battery output voltage decreases as the battery is depleted. Furthermore, if you connect an excessive number of 12-V lights in parallel to a car battery, they will be dim even when the battery is fresh and even if the wires to the lights have very low resistance. This implies that the battery’s output voltage is reduced by the overload. The reason for the decrease in output voltage for depleted or overloaded batteries is that all voltage sources have two fundamental parts—a source of electrical energy and an internal resistance. Let us examine both. Electromotive Force You can think of many different types of voltage sources. Batteries themselves come in many varieties. There are many types of mechanical/electrical generators, driven by many different energy sources, ranging from nuclear to wind. Solar cells create voltages directly from light, while thermoelectric devices create voltage from temperature differences. A few voltage sources are shown in Figure 21.8. All such devices create a potential difference and can supply current if connected to a resistance. On the small scale, the potential difference creates an electric field that exerts force on charges, causing current. We thus use the name electromotive force, abbreviated emf. Emf is not a force at all; it is a special type of potential difference. To be precise, the electromotive force (emf) is the potential difference of a source when no current is flowing. Units of emf are volts. Figure 21.8 A variety of voltage sources (clockwise from top left): the Brazos Wind Farm in Fluvanna, Texas (credit: Leaflet, Wikimedia Commons); the Krasnoyarsk Dam in Russia (credit: Alex Polezhaev); a solar farm (credit: U.S. Department of Energy); and a group of nickel metal hydride batteries (credit: Tiaa Monto). The voltage output of each depends on its construction and load, and equals emf only if there is no load. Electromotive force is directly related to the source of potential difference, such as the particular combination of chemicals in a battery. However, emf differs from the voltage output of the device when current flows. The voltage across the terminals of a battery, for example, is less than the emf when the battery supplies current, and it declines further as the battery is depleted or loaded down. However, if the device’s output voltage can be measured without drawing current, then output voltage will equal emf (even for a very depleted battery). Internal Resistance As noted before, a 12-V truck battery is physically larger, contains more charge and energy, and can deliver a larger current than a 12-V motorcycle battery. Both are lead-acid batteries with identical emf, but, because of its size, the truck battery has a smaller internal resistance . Internal resistance is the inherent resistance to the flow of current within the source itself. Figure 21.9 is a schematic representation of the two fundamental parts of any voltage source. The emf (represented by a script E in the figure) and internal resistance are in series. The smaller the internal resistance for a given emf, the more current and the more power the source can supply. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments 925 Figure 21.9 Any voltage source (in this case, a carbon-zinc dry cell) has an emf related to its source of potential difference, and an internal resistance related to its construction. (Note that the script E stands for emf.). Also shown are the output terminals across which the terminal voltage is measured. Since = emf − , terminal voltage equals emf only if there is no current flowing. The internal resistance can behave in complex ways. As noted, increases as a battery is depleted. But internal resistance may also depend on the magnitude and direction of the current through a voltage source, its temperature, and even its history. The internal resistance of rechargeable nickel-cadmium cells, for example, depends on how many times and how deeply they have been depleted. Things Great and Small: The Submicroscopic Origin of Battery Potential Various types of batteries are available, with emfs determined by the combination of chemicals involved. We can view this as a molecular reaction (what much of chemistry is about) that separates charge. The lead-acid battery used in cars and other vehicles is one of the most common types. A single cell (one of six) of this battery is seen in Figure 21.10. The cathode (positive) terminal of the cell is connected to a lead oxide plate, while the anode (negative) terminal is connected to a lead plate. Both plates are immersed in sulfuric acid, the electrolyte for the system. Figure 21.10 Artist’s conception of a lead-acid cell. Chemical reactions in a lead-acid cell separate charge, sending negative charge to the anode, which is connected to the lead plates. The lead oxide plates are connected to the positive or cathode terminal of the cell. Sulfuric acid conducts the charge as well as participating in the chemical reaction. The details of the chemical reaction are left to the reader to pursue in a chemistry text, but their results at the molecular level help explain the potential created by the battery. Figure 21.11 shows the result of a single chemical reaction. Two electrons are placed on the anode, making it negative, provided that the cathode supplied two electrons. This leaves the cathode positively charged, because it has lost two electrons. In short, a separation of charge has been driven by a chemical reaction. Note that the reaction will not take place unless there is a complete circuit to allow two electrons to be supplied to the cathode. Under many circumstances, these electrons come from the anode, flow through a resistance, and return to the cathode. Note also that since the chemical reactions involve substances with resistance, it is not possible to create the emf without an internal resistance. 926 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments Figure 21.11 Artist’s conception of two electrons being forced onto the anode of a cell and two electrons being removed from the cathode of the cell. The chemical reaction in a lead-acid battery places two electrons on the anode and removes two from the cathode. It requires a closed circuit to proceed, since the two electrons must be supplied to the cathode. Why are the chemicals able to produce a unique potential difference? Quantum mechanica
l descriptions of molecules, which take into account the types of atoms and numbers of electrons in them, are able to predict the energy states they can have and the energies of reactions between them. In the case of a lead-acid battery, an energy of 2 eV is given to each electron sent to the anode. Voltage is defined as the electrical potential energy divided by charge: = E . An electron volt is the energy given to a single electron by a voltage of 1 V. So the voltage here is 2 V, since 2 eV is given to each electron. It is the energy produced in each molecular reaction that produces the voltage. A different reaction produces a different energy and, hence, a different voltage. Terminal Voltage The voltage output of a device is measured across its terminals and, thus, is called its terminal voltage . Terminal voltage is given by = emf − , (21.44) where is the internal resistance and is the current flowing at the time of the measurement. is positive if current flows away from the positive terminal, as shown in Figure 21.9. You can see that the larger the current, the smaller the terminal voltage. And it is likewise true that the larger the internal resistance, the smaller the terminal voltage. Suppose a load resistance load is connected to a voltage source, as in Figure 21.12. Since the resistances are in series, the total resistance in the circuit is load + . Thus the current is given by Ohm’s law to be = emf load + . (21.45) Figure 21.12 Schematic of a voltage source and its load load . Since the internal resistance is in series with the load, it can significantly affect the terminal voltage and current delivered to the load. (Note that the script E stands for emf.) We see from this expression that the smaller the internal resistance , the greater the current the voltage source supplies to its load load . As batteries are depleted, increases. If becomes a significant fraction of the load resistance, then the current is significantly reduced, as the following example illustrates. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments 927 Example 21.4 Calculating Terminal Voltage, Power Dissipation, Current, and Resistance: Terminal Voltage and Load A certain battery has a 12.0-V emf and an internal resistance of 0.100 Ω . (a) Calculate its terminal voltage when connected to a 10.0- Ω load. (b) What is the terminal voltage when connected to a 0.500- Ω load? (c) What power does the 0.500- Ω load dissipate? (d) If the internal resistance grows to 0.500 Ω , find the current, terminal voltage, and power dissipated by a 0.500- Ω load. Strategy The analysis above gave an expression for current when internal resistance is taken into account. Once the current is found, the terminal voltage can be calculated using the equation = emf − . Once current is found, the power dissipated by a resistor can also be found. Solution for (a) Entering the given values for the emf, load resistance, and internal resistance into the expression above yields = emf load + = 12.0 V 10.1 Ω = 1.188 A. Enter the known values into the equation = emf − to get the terminal voltage: = emf − = 12.0 V − (1.188 A)(0.100 Ω) = 11.9 V. Discussion for (a) (21.46) (21.47) The terminal voltage here is only slightly lower than the emf, implying that 10.0 Ω is a light load for this particular battery. Solution for (b) Similarly, with load = 0.500 Ω , the current is = emf load + = 12.0 V 0.600 Ω = 20.0 A. = emf − = 12.0 V − (20.0 A)(0.100 Ω) = 10.0 V. The terminal voltage is now Discussion for (b) (21.48) (21.49) This terminal voltage exhibits a more significant reduction compared with emf, implying 0.500 Ω is a heavy load for this battery. Solution for (c) The power dissipated by the 0.500 - Ω load can be found using the formula = 2 . Entering the known values gives load = 2load = (20.0 A)2(0.500 Ω) = 2.00×102 W. (21.50) Discussion for (c) Note that this power can also be obtained using the expressions 2 this case). Solution for (d) or , where is the terminal voltage (10.0 V in Here the internal resistance has increased, perhaps due to the depletion of the battery, to the point where it is as great as the load resistance. As before, we first find the current by entering the known values into the expression, yielding Now the terminal voltage is = emf load + = 12.0 V 1.00 Ω = 12.0 A. = emf − = 12.0 V − (12.0 A)(0.500 Ω) = 6.00 V, and the power dissipated by the load is (21.51) (21.52) 928 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments load = 2load = (12.0 A)2(0.500 Ω ) = 72.0 W. (21.53) Discussion for (d) We see that the increased internal resistance has significantly decreased terminal voltage, current, and power delivered to a load. Applying the Science Practices: Internal Resistance The internal resistance of a battery can be estimated using a simple activity. The circuit shown in the figure below includes a resistor R in series with a battery along with an ammeter and voltmeter to measure the current and voltage respectively. Figure 21.13 The currents and voltages measured for several R values are shown in the table below. Using the data given in the table and applying graphical analysis, determine the emf and internal resistance of the battery. Your response should clearly explain the method used to obtain the result. Table 21.1 Resistance Current (A) Voltage (V) R1 R2 R3 R4 3.53 2.07 1.46 1.13 4.24 4.97 5.27 5.43 Answer Plot the measured currents and voltages on a graph. The terminal voltage of a battery is equal to the emf of the battery minus the voltage drop across the internal resistance of the battery or V = emf – Ir. Using this linear relationship and the plotted graph, the internal resistance and emf of the battery can be found. The graph for this case is shown below. The equation is V = -0.50I + 6.0 and hence the internal resistance will be equal to 0.5 Ω and emf will be equal to 6 V. Figure 21.14 Battery testers, such as those in Figure 21.15, use small load resistors to intentionally draw current to determine whether the terminal voltage drops below an acceptable level. They really test the internal resistance of the battery. If internal resistance is high, the battery is weak, as evidenced by its low terminal voltage. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments 929 Figure 21.15 These two battery testers measure terminal voltage under a load to determine the condition of a battery. The large device is being used by a U.S. Navy electronics technician to test large batteries aboard the aircraft carrier USS Nimitz and has a small resistance that can dissipate large amounts of power. (credit: U.S. Navy photo by Photographer’s Mate Airman Jason A. Johnston) The small device is used on small batteries and has a digital display to indicate the acceptability of their terminal voltage. (credit: Keith Williamson) Some batteries can be recharged by passing a current through them in the direction opposite to the current they supply to a resistance. This is done routinely in cars and batteries for small electrical appliances and electronic devices, and is represented pictorially in Figure 21.16. The voltage output of the battery charger must be greater than the emf of the battery to reverse current through it. This will cause the terminal voltage of the battery to be greater than the emf, since = emf − , and is now negative. Figure 21.16 A car battery charger reverses the normal direction of current through a battery, reversing its chemical reaction and replenishing its chemical potential. Multiple Voltage Sources There are two voltage sources when a battery charger is used. Voltage sources connected in series are relatively simple. When voltage sources are in series, their internal resistances add and their emfs add algebraically. (See Figure 21.17.) Series connections of voltage sources are common—for example, in flashlights, toys, and other appliances. Usually, the cells are in series in order to produce a larger total emf. But if the cells oppose one another, such as when one is put into an appliance backward, the total emf is less, since it is the algebraic sum of the individual emfs. A battery is a multiple connection of voltaic cells, as shown in Figure 21.18. The disadvantage of series connections of cells is that their internal resistances add. One of the authors once owned a 1957 MGA that had two 6-V batteries in series, rather than a single 12-V battery. This arrangement produced a large internal resistance that caused him many problems in starting the engine. Figure 21.17 A series connection of two voltage sources. The emfs (each labeled with a script E) and internal resistances add, giving a total emf of emf1 + emf2 and a total internal resistance of 1 + 2 . 930 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments Figure 21.18 Batteries are multiple connections of individual cells, as shown in this modern rendition of an old print. Single cells, such as AA or C cells, are commonly called batteries, although this is technically incorrect. If the series connection of two voltage sources is made into a complete circuit with the emfs in opposition, then a current of magnitude = emf1 – emf2 1 + 2 flows. See Figure 21.19, for example, which shows a circuit exactly analogous to the battery charger discussed above. If two voltage sources in series with emfs in the same sense are connected to a load load , as in Figure 21.20, then = emf1 + emf2 1 + 2 + load flows. Figure 21.19 These two voltage sources are connected in series with their emfs in opposition. Current flows in the direction of the greater emf and is limited to = emf1 − emf2 1 + 2 by the sum of the internal resistances. (Note that each emf is represented by script E in the figure.) A battery charger connected to a battery is an example of such a connection. The charger must have a larger emf than the battery to
reverse current through it. Figure 21.20 This schematic represents a flashlight with two cells (voltage sources) and a single bulb (load resistance) in series. The current that flows is = emf1 + emf2 1 + 2 + load . (Note that each emf is represented by script E in the figure.) Take-Home Experiment: Flashlight Batteries Find a flashlight that uses several batteries and find new and old batteries. Based on the discussions in this module, predict the brightness of the flashlight when different combinations of batteries are used. Do your predictions match what you observe? Now place new batteries in the flashlight and leave the flashlight switched on for several hours. Is the flashlight still quite bright? Do the same with the old batteries. Is the flashlight as bright when left on for the same length of time with old and new batteries? What does this say for the case when you are limited in the number of available new batteries? Figure 21.21 shows two voltage sources with identical emfs in parallel and connected to a load resistance. In this simple case, the total emf is the same as the individual emfs. But the total internal resistance is reduced, since the internal resistances are in parallel. The parallel connection thus can produce a larger current. Here, = emf tot + load flows through the load, and tot is less than those of the individual batteries. For example, some diesel-powered cars use two 12-V batteries in parallel; they produce a total emf of 12 V but can deliver the larger current needed to start a diesel engine. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments 931 Figure 21.21 Two voltage sources with identical emfs (each labeled by script E) connected in parallel produce the same emf but have a smaller total internal resistance than the individual sources. Parallel combinations are often used to deliver more current. Here = flows emf tot + load through the load. Animals as Electrical Detectors A number of animals both produce and detect electrical signals. Fish, sharks, platypuses, and echidnas (spiny anteaters) all detect electric fields generated by nerve activity in prey. Electric eels produce their own emf through biological cells (electric organs) called electroplaques, which are arranged in both series and parallel as a set of batteries. Electroplaques are flat, disk-like cells; those of the electric eel have a voltage of 0.15 V across each one. These cells are usually located toward the head or tail of the animal, although in the case of the electric eel, they are found along the entire body. The electroplaques in the South American eel are arranged in 140 rows, with each row stretching horizontally along the body and containing 5,000 electroplaques. This can yield an emf of approximately 600 V, and a current of 1 A—deadly. The mechanism for detection of external electric fields is similar to that for producing nerve signals in the cell through depolarization and repolarization—the movement of ions across the cell membrane. Within the fish, weak electric fields in the water produce a current in a gel-filled canal that runs from the skin to sensing cells, producing a nerve signal. The Australian platypus, one of the very few mammals that lay eggs, can detect fields of 30 mV m , while sharks have been found to be able to sense a field in their snouts as small as 100 mV m (Figure 21.22). Electric eels use their own electric fields produced by the electroplaques to stun their prey or enemies. Figure 21.22 Sand tiger sharks (Carcharias taurus), like this one at the Minnesota Zoo, use electroreceptors in their snouts to locate prey. (credit: Jim Winstead, Flickr) Solar Cell Arrays Another example dealing with multiple voltage sources is that of combinations of solar cells—wired in both series and parallel combinations to yield a desired voltage and current. Photovoltaic generation (PV), the conversion of sunlight directly into 932 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments electricity, is based upon the photoelectric effect, in which photons hitting the surface of a solar cell create an electric current in the cell. Most solar cells are made from pure silicon—either as single-crystal silicon, or as a thin film of silicon deposited upon a glass or metal backing. Most single cells have a voltage output of about 0.5 V, while the current output is a function of the amount of sunlight upon the cell (the incident solar radiation—the insolation). Under bright noon sunlight, a current of about 100 mA/cm2 of cell surface area is produced by typical single-crystal cells. Individual solar cells are connected electrically in modules to meet electrical-energy needs. They can be wired together in series or in parallel—connected like the batteries discussed earlier. A solar-cell array or module usually consists of between 36 and 72 cells, with a power output of 50 W to 140 W. The output of the solar cells is direct current. For most uses in a home, AC is required, so a device called an inverter must be used to convert the DC to AC. Any extra output can then be passed on to the outside electrical grid for sale to the utility. Take-Home Experiment: Virtual Solar Cells One can assemble a “virtual” solar cell array by using playing cards, or business or index cards, to represent a solar cell. Combinations of these cards in series and/or parallel can model the required array output. Assume each card has an output of 0.5 V and a current (under bright light) of 2 A. Using your cards, how would you arrange them to produce an output of 6 A at 3 V (18 W)? Suppose you were told that you needed only 18 W (but no required voltage). Would you need more cards to make this arrangement? 21.3 Kirchhoff’s Rules By the end of this section, you will be able to: Learning Objectives • Analyze a complex circuit using Kirchhoff’s rules, applying the conventions for determining the correct signs of various terms. The information presented in this section supports the following AP® learning objectives and science practices: • 5.B.9.1 The student is able to construct or interpret a graph of the energy changes within an electrical circuit with only a single battery and resistors in series and/or in, at most, one parallel branch as an application of the conservation of energy (Kirchhoff’s loop rule). (S.P. 1.1, 1.4) • 5.B.9.2 The student is able to apply conservation of energy concepts to the design of an experiment that will demonstrate the validity of Kirchhoff’s loop rule in a circuit with only a battery and resistors either in series or in, at most, one pair of parallel branches. (S.P. 4.2, 6.4, 7.2) • 5.B.9.3 The student is able to apply conservation of energy (Kirchhoff’s loop rule) in calculations involving the total electric potential difference for complete circuit loops with only a single battery and resistors in series and/or in, at most, one parallel branch. (S.P. 2.2, 6.4, 7.2) • 5.B.9.4 The student is able to analyze experimental data including an analysis of experimental uncertainty that will demonstrate the validity of Kirchhoff’s loop rule. (S.P. 5.1) • 5.B.9.5 The student is able to use conservation of energy principles (Kirchhoff’s loop rule) to describe and make predictions regarding electrical potential difference, charge, and current in steady-state circuits composed of various combinations of resistors and capacitors. (S.P. 6.4) • 5.C.3.1 The student is able to apply conservation of electric charge (Kirchhoff’s junction rule) to the comparison of electric current in various segments of an electrical circuit with a single battery and resistors in series and in, at most, one parallel branch and predict how those values would change if configurations of the circuit are changed. (S.P. 6.4, 7.2) • 5.C.3.2 The student is able to design an investigation of an electrical circuit with one or more resistors in which evidence of conservation of electric charge can be collected and analyzed. (S.P. 4.1, 4.2, 5.1) • 5.C.3.3 The student is able to use a description or schematic diagram of an electrical circuit to calculate unknown values of current in various segments or branches of the circuit. (S.P. 1.4, 2.2) • 5.C.3.4 The student is able to predict or explain current values in series and parallel arrangements of resistors and other branching circuits using Kirchhoff’s junction rule and relate the rule to the law of charge conservation. (S.P. 6.4, 7.2) • 5.C.3.5 The student is able to determine missing values and direction of electric current in branches of a circuit with resistors and NO capacitors from values and directions of current in other branches of the circuit through appropriate selection of nodes and application of the junction rule. (S.P. 1.4, 2.2) Many complex circuits, such as the one in Figure 21.23, cannot be analyzed with the series-parallel techniques developed in Resistors in Series and Parallel and Electromotive Force: Terminal Voltage. There are, however, two circuit analysis rules that can be used to analyze any circuit, simple or complex. These rules are special cases of the laws of conservation of charge and conservation of energy. The rules are known as Kirchhoff’s rules, after their inventor Gustav Kirchhoff (1824–1887). This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments 933 Figure 21.23 This circuit cannot be reduced to a combination of series and parallel connections. Kirchhoff’s rules, special applications of the laws of conservation of charge and energy, can be used to analyze it. (Note: The script E in the figure represents electromotive force, emf.) Kirchhoff’s Rules • Kirchhoff’s first rule—the junction rule. The sum of all currents entering a junction must equal the sum of all currents leaving the junction. • Kirchhoff’s second rule—the loop rule. The algebraic sum of changes in potential around any closed circuit path
(loop) must be zero. Explanations of the two rules will now be given, followed by problem-solving hints for applying Kirchhoff’s rules, and a worked example that uses them. Kirchhoff’s First Rule Kirchhoff’s first rule (the junction rule) is an application of the conservation of charge to a junction; it is illustrated in Figure 21.24. Current is the flow of charge, and charge is conserved; thus, whatever charge flows into the junction must flow out. Kirchhoff’s first rule requires that 1 = 2 + 3 (see figure). Equations like this can and will be used to analyze circuits and to solve circuit problems. Making Connections: Conservation Laws Kirchhoff’s rules for circuit analysis are applications of conservation laws to circuits. The first rule is the application of conservation of charge, while the second rule is the application of conservation of energy. Conservation laws, even used in a specific application, such as circuit analysis, are so basic as to form the foundation of that application. Figure 21.24 The junction rule. The diagram shows an example of Kirchhoff’s first rule where the sum of the currents into a junction equals the sum of the currents out of a junction. In this case, the current going into the junction splits and comes out as two currents, so that 1 = 2 + 3 . Here 1 must be 11 A, since 2 is 7 A and 3 is 4 A. 934 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments Kirchhoff’s Second Rule Kirchhoff’s second rule (the loop rule) is an application of conservation of energy. The loop rule is stated in terms of potential, , rather than potential energy, but the two are related since PEelec = . Recall that emf is the potential difference of a source when no current is flowing. In a closed loop, whatever energy is supplied by emf must be transferred into other forms by devices in the loop, since there are no other ways in which energy can be transferred into or out of the circuit. Figure 21.25 illustrates the changes in potential in a simple series circuit loop. Kirchhoff’s second rule requires emf − − 1 − 2 = 0 . Rearranged, this is emf = + 1 + 2 , which means the emf equals the sum of the (voltage) drops in the loop. Figure 21.25 The loop rule. An example of Kirchhoff’s second rule where the sum of the changes in potential around a closed loop must be zero. (a) In this standard schematic of a simple series circuit, the emf supplies 18 V, which is reduced to zero by the resistances, with 1 V across the internal resistance, and 12 V and 5 V across the two load resistances, for a total of 18 V. (b) This perspective view represents the potential as something like a roller coaster, where charge is raised in potential by the emf and lowered by the resistances. (Note that the script E stands for emf.) Applying Kirchhoff’s Rules By applying Kirchhoff’s rules, we generate equations that allow us to find the unknowns in circuits. The unknowns may be currents, emfs, or resistances. Each time a rule is applied, an equation is produced. If there are as many independent equations as unknowns, then the problem can be solved. There are two decisions you must make when applying Kirchhoff’s rules. These decisions determine the signs of various quantities in the equations you obtain from applying the rules. 1. When applying Kirchhoff’s first rule, the junction rule, you must label the current in each branch and decide in what direction it is going. For example, in Figure 21.23, Figure 21.24, and Figure 21.25, currents are labeled 1 , 2 , 3 , and , and arrows indicate their directions. There is no risk here, for if you choose the wrong direction, the current will be of the correct magnitude but negative. 2. When applying Kirchhoff’s second rule, the loop rule, you must identify a closed loop and decide in which direction to go around it, clockwise or counterclockwise. For example, in Figure 21.25 the loop was traversed in the same direction as the current (clockwise). Again, there is no risk; going around the circuit in the opposite direction reverses the sign of every term in the equation, which is like multiplying both sides of the equation by –1. Figure 21.26 and the following points will help you get the plus or minus signs right when applying the loop rule. Note that the resistors and emfs are traversed by going from a to b. In many circuits, it will be necessary to construct more than one loop. In traversing each loop, one needs to be consistent for the sign of the change in potential. (See Example 21.5.) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments 935 Figure 21.26 Each of these resistors and voltage sources is traversed from a to b. The potential changes are shown beneath each element and are explained in the text. (Note that the script E stands for emf.) • When a resistor is traversed in the same direction as the current, the change in potential is − . (See Figure 21.26.) • When a resistor is traversed in the direction opposite to the current, the change in potential is + . (See Figure 21.26.) • When an emf is traversed from – to + (the same direction it moves positive charge), the change in potential is +emf. (See Figure 21.26.) • When an emf is traversed from + to – (opposite to the direction it moves positive charge), the change in potential is − emf. (See Figure 21.26.) Example 21.5 Calculating Current: Using Kirchhoff’s Rules Find the currents flowing in the circuit in Figure 21.27. Figure 21.27 This circuit is similar to that in Figure 21.23, but the resistances and emfs are specified. (Each emf is denoted by script E.) The currents in each branch are labeled and assumed to move in the directions shown. This example uses Kirchhoff’s rules to find the currents. Strategy This circuit is sufficiently complex that the currents cannot be found using Ohm’s law and the series-parallel techniques—it is necessary to use Kirchhoff’s rules. Currents have been labeled 1 , 2 , and 3 in the figure and assumptions have been made about their directions. Locations on the diagram have been labeled with letters a through h. In the solution we will apply the junction and loop rules, seeking three independent equations to allow us to solve for the three unknown currents. Solution We begin by applying Kirchhoff’s first or junction rule at point a. This gives 1 = 2 + 3, (21.54) since 1 flows into the junction, while 2 and 3 flow out. Applying the junction rule at e produces exactly the same equation, so that no new information is obtained. This is a single equation with three unknowns—three independent equations are needed, and so the loop rule must be applied. 936 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments Now we consider the loop abcdea. Going from a to b, we traverse 2 in the same (assumed) direction of the current 2 , and so the change in potential is −22 . Then going from b to c, we go from – to +, so that the change in potential is +emf1 . Traversing the internal resistance 1 from c to d gives −21 . Completing the loop by going from d to a again traverses a resistor in the same direction as its current, giving a change in potential of −11 . The loop rule states that the changes in potential sum to zero. Thus, −22 + emf1 − 21 − 11 = −2(2 + 1) + emf1 − 11 = 0. Substituting values from the circuit diagram for the resistances and emf, and canceling the ampere unit gives Now applying the loop rule to aefgha (we could have chosen abcdefgha as well) similarly gives + 11 + 33 + 32 − emf2= +1 1 + 3 3 + 2 − emf2 = 0. −3 2 + 18 − 6 1 = 0. (21.55) (21.56) (21.57) Note that the signs are reversed compared with the other loop, because elements are traversed in the opposite direction. With values entered, this becomes These three equations are sufficient to solve for the three unknown currents. First, solve the second equation for 2 : + 6 1 + 2 3 − 45 = 0. Now solve the third equation for 3 : 2 = 6 − 2 1. 3 = 22.5 − 3 1. Substituting these two new equations into the first one allows us to find a value for 1 : 1 = 2 + 3 = (6 − 2 1) + (22.5 − 3 1) = 28.5 − 5 1. Combining terms gives 6 1 = 28.5, and 1 = 4.75 A. Substituting this value for 1 back into the fourth equation gives .50 2 = −3.50 A. The minus sign means 2 flows in the direction opposite to that assumed in Figure 21.27. Finally, substituting the value for 1 into the fifth equation gives 3 = 22.5−3 1 = 22.5 − 14.25 3 = 8.25 A. (21.58) (21.59) (21.60) (21.61) (21.62) (21.63) (21.64) (21.65) (21.66) (21.67) Discussion Just as a check, we note that indeed 1 = 2 + 3 . The results could also have been checked by entering all of the values into the equation for the abcdefgha loop. Problem-Solving Strategies for Kirchhoff’s Rules 1. Make certain there is a clear circuit diagram on which you can label all known and unknown resistances, emfs, and currents. If a current is unknown, you must assign it a direction. This is necessary for determining the signs of potential changes. If you assign the direction incorrectly, the current will be found to have a negative value—no harm done. 2. Apply the junction rule to any junction in the circuit. Each time the junction rule is applied, you should get an equation with a current that does not appear in a previous application—if not, then the equation is redundant. 3. Apply the loop rule to as many loops as needed to solve for the unknowns in the problem. (There must be as many independent equations as unknowns.) To apply the loop rule, you must choose a direction to go around the loop. Then This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments 937 carefully and consistently determine the signs of the potential changes for each element using the four bulleted points discussed above in conjunction with Figure 21.26. 4. Solve the simultaneous equations for the unknowns. This may involve many algebraic steps, requiring careful checking and rechecking. 5. Check to see
whether the answers are reasonable and consistent. The numbers should be of the correct order of magnitude, neither exceedingly large nor vanishingly small. The signs should be reasonable—for example, no resistance should be negative. Check to see that the values obtained satisfy the various equations obtained from applying the rules. The currents should satisfy the junction rule, for example. The material in this section is correct in theory. We should be able to verify it by making measurements of current and voltage. In fact, some of the devices used to make such measurements are straightforward applications of the principles covered so far and are explored in the next modules. As we shall see, a very basic, even profound, fact results—making a measurement alters the quantity being measured. Check Your Understanding Can Kirchhoff’s rules be applied to simple series and parallel circuits or are they restricted for use in more complicated circuits that are not combinations of series and parallel? Solution Kirchhoff's rules can be applied to any circuit since they are applications to circuits of two conservation laws. Conservation laws are the most broadly applicable principles in physics. It is usually mathematically simpler to use the rules for series and parallel in simpler circuits so we emphasize Kirchhoff’s rules for use in more complicated situations. But the rules for series and parallel can be derived from Kirchhoff’s rules. Moreover, Kirchhoff’s rules can be expanded to devices other than resistors and emfs, such as capacitors, and are one of the basic analysis devices in circuit analysis. Making Connections: Parallel Resistors A simple circuit shown below – with two parallel resistors and a voltage source – is implemented in a laboratory experiment with ɛ= 6.00 ± 0.02 V and R1 = 4.8 ± 0.1 Ω and R2 = 9.6 ± 0.1 Ω. The values include an allowance for experimental uncertainties as they cannot be measured with perfect certainty. For example if you measure the value for a resistor a few times, you may get slightly different results. Hence values are expressed with some level of uncertainty. Figure 21.28 In the laboratory experiment the currents measured in the two resistors are I1 = 1.27 A and I2 = 0.62 A respectively. Let us examine these values using Kirchhoff’s laws. For the two loops, E - I1R1 = 0 or I1 = E/R1 E - I2R2 = 0 or I2 = E/R2 Converting the given uncertainties for voltage and resistances into percentages, we get E = 6.00 V ± 0.33% R1 = 4.8 Ω ± 2.08% R2 = 9.6 Ω ± 1.04% We now find the currents for the two loops. While the voltage is divided by the resistance to find the current, uncertainties in voltage and resistance are directly added to find the uncertainty in the current value. I1 = (6.00/4.8) ± (0.33%+2.08%) = 1.25 ± 2.4% = 1.25 ± 0.03 A I2 = (6.00/9.6) ± (0.33%+1.04%) = 0.63 ± 1.4% = 0.63 ± 0.01 A Finally you can check that the two measured values in this case are within the uncertainty ranges found for the currents. However there can also be additional experimental uncertainty in the measurements of currents. 938 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments 21.4 DC Voltmeters and Ammeters By the end of this section, you will be able to: Learning Objectives • Explain why a voltmeter must be connected in parallel with the circuit. • Draw a diagram showing an ammeter correctly connected in a circuit. • Describe how a galvanometer can be used as either a voltmeter or an ammeter. • Find the resistance that must be placed in series with a galvanometer to allow it to be used as a voltmeter with a given reading. • Explain why measuring the voltage or current in a circuit can never be exact. Voltmeters measure voltage, whereas ammeters measure current. Some of the meters in automobile dashboards, digital cameras, cell phones, and tuner-amplifiers are voltmeters or ammeters. (See Figure 21.29.) The internal construction of the simplest of these meters and how they are connected to the system they monitor give further insight into applications of series and parallel connections. Figure 21.29 The fuel and temperature gauges (far right and far left, respectively) in this 1996 Volkswagen are voltmeters that register the voltage output of “sender” units, which are hopefully proportional to the amount of gasoline in the tank and the engine temperature. (credit: Christian Giersing) Voltmeters are connected in parallel with whatever device’s voltage is to be measured. A parallel connection is used because objects in parallel experience the same potential difference. (See Figure 21.30, where the voltmeter is represented by the symbol V.) Ammeters are connected in series with whatever device’s current is to be measured. A series connection is used because objects in series have the same current passing through them. (See Figure 21.31, where the ammeter is represented by the symbol A.) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments 939 Figure 21.30 (a) To measure potential differences in this series circuit, the voltmeter (V) is placed in parallel with the voltage source or either of the resistors. Note that terminal voltage is measured between points a and b. It is not possible to connect the voltmeter directly across the emf without including its internal resistance, . (b) A digital voltmeter in use. (credit: Messtechniker, Wikimedia Commons) Figure 21.31 An ammeter (A) is placed in series to measure current. All of the current in this circuit flows through the meter. The ammeter would have the same reading if located between points d and e or between points f and a as it does in the position shown. (Note that the script capital E stands for emf, and stands for the internal resistance of the source of potential difference.) Analog Meters: Galvanometers Analog meters have a needle that swivels to point at numbers on a scale, as opposed to digital meters, which have numerical readouts similar to a hand-held calculator. The heart of most analog meters is a device called a galvanometer, denoted by G. Current flow through a galvanometer, G , produces a proportional needle deflection. (This deflection is due to the force of a magnetic field upon a current-carrying wire.) The two crucial characteristics of a given galvanometer are its resistance and current sensitivity. Current sensitivity is the current that gives a full-scale deflection of the galvanometer’s needle, the maximum current that the instrument can measure. For example, a galvanometer with a current sensitivity of 50 μA has a maximum deflection of its needle when 50 μA flows through it, reads half-scale when 25 μA flows through it, and so on. 940 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments If such a galvanometer has a 25- Ω resistance, then a voltage of only = = full-scale reading. By connecting resistors to this galvanometer in different ways, you can use it as either a voltmeter or ammeter that can measure a broad range of voltages or currents. (25 Ω) = 1.25 mV produces a 50 μA Galvanometer as Voltmeter Figure 21.32 shows how a galvanometer can be used as a voltmeter by connecting it in series with a large resistance, . The value of the resistance is determined by the maximum voltage to be measured. Suppose you want 10 V to produce a fullscale deflection of a voltmeter containing a 25-Ω galvanometer with a 50-μA sensitivity. Then 10 V applied to the meter must produce a current of 50 μA . The total resistance must be tot = + = = 10 V 50 μA = 200 kΩ, or = tot − = 200 kΩ − 25 Ω ≈ 200 k Ω . (21.68) (21.69) ( is so large that the galvanometer resistance, , is nearly negligible.) Note that 5 V applied to this voltmeter produces a halfscale deflection by producing a 25-μA current through the meter, and so the voltmeter’s reading is proportional to voltage as desired. This voltmeter would not be useful for voltages less than about half a volt, because the meter deflection would be small and difficult to read accurately. For other voltage ranges, other resistances are placed in series with the galvanometer. Many meters have a choice of scales. That choice involves switching an appropriate resistance into series with the galvanometer. Figure 21.32 A large resistance placed in series with a galvanometer G produces a voltmeter, the full-scale deflection of which depends on the choice of . The larger the voltage to be measured, the larger must be. (Note that represents the internal resistance of the galvanometer.) Galvanometer as Ammeter The same galvanometer can also be made into an ammeter by placing it in parallel with a small resistance , often called the shunt resistance, as shown in Figure 21.33. Since the shunt resistance is small, most of the current passes through it, allowing an ammeter to measure currents much greater than those producing a full-scale deflection of the galvanometer. Suppose, for example, an ammeter is needed that gives a full-scale deflection for 1.0 A, and contains the same 25- Ω galvanometer with its 50-μA sensitivity. Since and are in parallel, the voltage across them is the same. These drops are = G so that = G = . Solving for , and noting that G is 50 μA and is 0.999950 A, we have = G = (25 Ω ) 50 μA 0.999950 A = 1.2510−3 Ω . (21.70) Figure 21.33 A small shunt resistance placed in parallel with a galvanometer G produces an ammeter, the full-scale deflection of which depends on the choice of . The larger the current to be measured, the smaller must be. Most of the current ( ) flowing through the meter is shunted through to protect the galvanometer. (Note that represents the internal resistance of the galvanometer.) Ammeters may also have multiple scales for greater flexibility in application. The various scales are achieved by switching various shunt resistances in parallel with the galvanometer—the greater the maximum current to be measured, the smaller the shunt resistance must be. Taking Measurements Alters the
Circuit When you use a voltmeter or ammeter, you are connecting another resistor to an existing circuit and, thus, altering the circuit. Ideally, voltmeters and ammeters do not appreciably affect the circuit, but it is instructive to examine the circumstances under which they do or do not interfere. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments 941 First, consider the voltmeter, which is always placed in parallel with the device being measured. Very little current flows through the voltmeter if its resistance is a few orders of magnitude greater than the device, and so the circuit is not appreciably affected. (See Figure 21.34(a).) (A large resistance in parallel with a small one has a combined resistance essentially equal to the small one.) If, however, the voltmeter’s resistance is comparable to that of the device being measured, then the two in parallel have a smaller resistance, appreciably affecting the circuit. (See Figure 21.34(b).) The voltage across the device is not the same as when the voltmeter is out of the circuit. Figure 21.34 (a) A voltmeter having a resistance much larger than the device ( Voltmeter >> ) with which it is in parallel produces a parallel resistance essentially the same as the device and does not appreciably affect the circuit being measured. (b) Here the voltmeter has the same resistance as the device ( Voltmeter ≅ ), so that the parallel resistance is half of what it is when the voltmeter is not connected. This is an example of a significant alteration of the circuit and is to be avoided. An ammeter is placed in series in the branch of the circuit being measured, so that its resistance adds to that branch. Normally, the ammeter’s resistance is very small compared with the resistances of the devices in the circuit, and so the extra resistance is negligible. (See Figure 21.35(a).) However, if very small load resistances are involved, or if the ammeter is not as low in resistance as it should be, then the total series resistance is significantly greater, and the current in the branch being measured is reduced. (See Figure 21.35(b).) A practical problem can occur if the ammeter is connected incorrectly. If it was put in parallel with the resistor to measure the current in it, you could possibly damage the meter; the low resistance of the ammeter would allow most of the current in the circuit to go through the galvanometer, and this current would be larger since the effective resistance is smaller. Figure 21.35 (a) An ammeter normally has such a small resistance that the total series resistance in the branch being measured is not appreciably increased. The circuit is essentially unaltered compared with when the ammeter is absent. (b) Here the ammeter’s resistance is the same as that of the branch, so that the total resistance is doubled and the current is half what it is without the ammeter. This significant alteration of the circuit is to be avoided. One solution to the problem of voltmeters and ammeters interfering with the circuits being measured is to use galvanometers with greater sensitivity. This allows construction of voltmeters with greater resistance and ammeters with smaller resistance than when less sensitive galvanometers are used. There are practical limits to galvanometer sensitivity, but it is possible to get analog meters that make measurements accurate to a few percent. Note that the inaccuracy comes from altering the circuit, not from a fault in the meter. Connections: Limits to Knowledge Making a measurement alters the system being measured in a manner that produces uncertainty in the measurement. For macroscopic systems, such as the circuits discussed in this module, the alteration can usually be made negligibly small, but it cannot be eliminated entirely. For submicroscopic systems, such as atoms, nuclei, and smaller particles, measurement alters the system in a manner that cannot be made arbitrarily small. This actually limits knowledge of the system—even limiting what nature can know about itself. We shall see profound implications of this when the Heisenberg uncertainty principle is discussed in the modules on quantum mechanics. There is another measurement technique based on drawing no current at all and, hence, not altering the circuit at all. These are called null measurements and are the topic of Null Measurements. Digital meters that employ solid-state electronics and null measurements can attain accuracies of one part in 106 . 950 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments heart defibrillator is slightly more complex than the one in Figure 21.42, to compensate for magnetic and AC effects that will be covered in Magnetism. Check Your Understanding When is the potential difference across a capacitor an emf? Solution Only when the current being drawn from or put into the capacitor is zero. Capacitors, like batteries, have internal resistance, so their output voltage is not an emf unless current is zero. This is difficult to measure in practice so we refer to a capacitor’s voltage rather than its emf. But the source of potential difference in a capacitor is fundamental and it is an emf. PhET Explorations: Circuit Construction Kit (DC only) An electronics kit in your computer! Build circuits with resistors, light bulbs, batteries, and switches. Take measurements with the realistic ammeter and voltmeter. View the circuit as a schematic diagram, or switch to a life-like view. Figure 21.45 Circuit Construction Kit (DC only) (http://cnx.org/content/m55370/1.3/circuit-construction-kit-dc_en.jar) Glossary ammeter: an instrument that measures current analog meter: a measuring instrument that gives a readout in the form of a needle movement over a marked gauge bridge device: a device that forms a bridge between two branches of a circuit; some bridge devices are used to make null measurements in circuits capacitance: the maximum amount of electric potential energy that can be stored (or separated) for a given electric potential capacitor: an electrical component used to store energy by separating electric charge on two opposing plates conservation laws: require that energy and charge be conserved in a system current: the flow of charge through an electric circuit past a given point of measurement current sensitivity: the maximum current that a galvanometer can read digital meter: a measuring instrument that gives a readout in a digital form electromotive force (emf): the potential difference of a source of electricity when no current is flowing; measured in volts full-scale deflection: the maximum deflection of a galvanometer needle, also known as current sensitivity; a galvanometer with a full-scale deflection of 50 μA has a maximum deflection of its needle when 50 μA flows through it galvanometer: an analog measuring device, denoted by G, that measures current flow using a needle deflection caused by a magnetic field force acting upon a current-carrying wire internal resistance: the amount of resistance within the voltage source Joule’s law: the relationship between potential electrical power, voltage, and resistance in an electrical circuit, given by: = junction rule: Kirchhoff’s first rule, which applies the conservation of charge to a junction; current is the flow of charge; thus, whatever charge flows into the junction must flow out; the rule can be stated 1 = 2 + 3 Kirchhoff’s rules: a set of two rules, based on conservation of charge and energy, governing current and changes in potential in an electric circuit loop rule: Kirchhoff’s second rule, which states that in a closed loop, whatever energy is supplied by emf must be transferred into other forms by devices in the loop, since there are no other ways in which energy can be transferred into or out of This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments 951 the circuit. Thus, the emf equals the sum of the (voltage) drops in the loop and can be stated: emf = + 1 + 2 null measurements: methods of measuring current and voltage more accurately by balancing the circuit so that no current flows through the measurement device ohmmeter: an instrument that applies a voltage to a resistance, measures the current, calculates the resistance using Ohm’s law, and provides a readout of this calculated resistance Ohm’s law: the relationship between current, voltage, and resistance within an electrical circuit: = parallel: the wiring of resistors or other components in an electrical circuit such that each component receives an equal voltage from the power source; often pictured in a ladder-shaped diagram, with each component on a rung of the ladder potential difference: the difference in electric potential between two points in an electric circuit, measured in volts potentiometer: a null measurement device for measuring potentials (voltages) RC circuit: a circuit that contains both a resistor and a capacitor resistance: causing a loss of electrical power in a circuit resistor: a component that provides resistance to the current flowing through an electrical circuit series: a sequence of resistors or other components wired into a circuit one after the other shunt resistance: a small resistance placed in parallel with a galvanometer G to produce an ammeter; the larger the current to be measured, the smaller must be; most of the current flowing through the meter is shunted through to protect the galvanometer terminal voltage: the voltage measured across the terminals of a source of potential difference voltage: the electrical potential energy per unit charge; electric pressure created by a power source, such as a battery voltage drop: the loss of electrical power as a current travels through a resistor, wire or other component voltmeter: an instrument that measures voltage Wheatstone bridge: a null measurement device for calculating resistance b
y balancing potential drops in a circuit Section Summary 21.1 Resistors in Series and Parallel • The total resistance of an electrical circuit with resistors wired in a series is the sum of the individual resistances: s = 1 + 2 + 3 + .... • Each resistor in a series circuit has the same amount of current flowing through it. • The voltage drop, or power dissipation, across each individual resistor in a series is different, and their combined total adds up to the power source input. • The total resistance of an electrical circuit with resistors wired in parallel is less than the lowest resistance of any of the components and can be determined using the formula + .... • Each resistor in a parallel circuit has the same full voltage of the source applied to it. • The current flowing through each resistor in a parallel circuit is different, depending on the resistance. • If a more complex connection of resistors is a combination of series and parallel, it can be reduced to a single equivalent resistance by identifying its various parts as series or parallel, reducing each to its equivalent, and continuing until a single resistance is eventually reached. 21.2 Electromotive Force: Terminal Voltage • All voltage sources have two fundamental parts—a source of electrical energy that has a characteristic electromotive force (emf), and an internal resistance . • The emf is the potential difference of a source when no current is flowing. • The numerical value of the emf depends on the source of potential difference. • The internal resistance of a voltage source affects the output voltage when a current flows. • The voltage output of a device is called its terminal voltage and is given by = emf − , where is the electric current and is positive when flowing away from the positive terminal of the voltage source. 952 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments • When multiple voltage sources are in series, their internal resistances add and their emfs add algebraically. • Solar cells can be wired in series or parallel to provide increased voltage or current, respectively. 21.3 Kirchhoff’s Rules • Kirchhoff’s rules can be used to analyze any circuit, simple or complex. • Kirchhoff’s first rule—the junction rule: The sum of all currents entering a junction must equal the sum of all currents leaving the junction. • Kirchhoff’s second rule—the loop rule: The algebraic sum of changes in potential around any closed circuit path (loop) must be zero. • The two rules are based, respectively, on the laws of conservation of charge and energy. • When calculating potential and current using Kirchhoff’s rules, a set of conventions must be followed for determining the correct signs of various terms. • The simpler series and parallel rules are special cases of Kirchhoff’s rules. 21.4 DC Voltmeters and Ammeters • Voltmeters measure voltage, and ammeters measure current. • A voltmeter is placed in parallel with the voltage source to receive full voltage and must have a large resistance to limit its effect on the circuit. • An ammeter is placed in series to get the full current flowing through a branch and must have a small resistance to limit its effect on the circuit. • Both can be based on the combination of a resistor and a galvanometer, a device that gives an analog reading of current. • Standard voltmeters and ammeters alter the circuit being measured and are thus limited in accuracy. 21.5 Null Measurements • Null measurement techniques achieve greater accuracy by balancing a circuit so that no current flows through the measuring device. • One such device, for determining voltage, is a potentiometer. • Another null measurement device, for determining resistance, is the Wheatstone bridge. • Other physical quantities can also be measured with null measurement techniques. 21.6 DC Circuits Containing Resistors and Capacitors • An circuit is one that has both a resistor and a capacitor. • The time constant for an circuit is = . • When an initially uncharged ( 0 = 0 at = 0 ) capacitor in series with a resistor is charged by a DC voltage source, the voltage rises, asymptotically approaching the emf of the voltage source; as a function of time, = emf(1 − − / )(charging). • Within the span of each time constant , the voltage rises by 0.632 of the remaining value, approaching the final voltage asymptotically. If a capacitor with an initial voltage 0 is discharged through a resistor starting at = 0 , then its voltage decreases exponentially as given by In each time constant , the voltage falls by 0.368 of its remaining initial value, approaching zero asymptotically. = 0− / (discharging). • • Conceptual Questions 21.1 Resistors in Series and Parallel 1. A switch has a variable resistance that is nearly zero when closed and extremely large when open, and it is placed in series with the device it controls. Explain the effect the switch in Figure 21.46 has on current when open and when closed. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments 953 Figure 21.46 A switch is ordinarily in series with a resistance and voltage source. Ideally, the switch has nearly zero resistance when closed but has an extremely large resistance when open. (Note that in this diagram, the script E represents the voltage (or electromotive force) of the battery.) 2. What is the voltage across the open switch in Figure 21.46? 3. There is a voltage across an open switch, such as in Figure 21.46. Why, then, is the power dissipated by the open switch small? 4. Why is the power dissipated by a closed switch, such as in Figure 21.46, small? 5. A student in a physics lab mistakenly wired a light bulb, battery, and switch as shown in Figure 21.47. Explain why the bulb is on when the switch is open, and off when the switch is closed. (Do not try this—it is hard on the battery!) Figure 21.47 A wiring mistake put this switch in parallel with the device represented by . (Note that in this diagram, the script E represents the voltage (or electromotive force) of the battery.) 6. Knowing that the severity of a shock depends on the magnitude of the current through your body, would you prefer to be in series or parallel with a resistance, such as the heating element of a toaster, if shocked by it? Explain. 7. Would your headlights dim when you start your car’s engine if the wires in your automobile were superconductors? (Do not neglect the battery’s internal resistance.) Explain. 8. Some strings of holiday lights are wired in series to save wiring costs. An old version utilized bulbs that break the electrical connection, like an open switch, when they burn out. If one such bulb burns out, what happens to the others? If such a string operates on 120 V and has 40 identical bulbs, what is the normal operating voltage of each? Newer versions use bulbs that short circuit, like a closed switch, when they burn out. If one such bulb burns out, what happens to the others? If such a string operates on 120 V and has 39 remaining identical bulbs, what is then the operating voltage of each? 9. If two household lightbulbs rated 60 W and 100 W are connected in series to household power, which will be brighter? Explain. 10. Suppose you are doing a physics lab that asks you to put a resistor into a circuit, but all the resistors supplied have a larger resistance than the requested value. How would you connect the available resistances to attempt to get the smaller value asked for? 11. Before World War II, some radios got power through a “resistance cord” that had a significant resistance. Such a resistance cord reduces the voltage to a desired level for the radio’s tubes and the like, and it saves the expense of a transformer. Explain why resistance cords become warm and waste energy when the radio is on. 12. Some light bulbs have three power settings (not including zero), obtained from multiple filaments that are individually switched and wired in parallel. What is the minimum number of filaments needed for three power settings? 21.2 Electromotive Force: Terminal Voltage 13. Is every emf a potential difference? Is every potential difference an emf? Explain. 14. Explain which battery is doing the charging and which is being charged in Figure 21.48. 954 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments Figure 21.48 15. Given a battery, an assortment of resistors, and a variety of voltage and current measuring devices, describe how you would determine the internal resistance of the battery. 16. Two different 12-V automobile batteries on a store shelf are rated at 600 and 850 “cold cranking amps.” Which has the smallest internal resistance? 17. What are the advantages and disadvantages of connecting batteries in series? In parallel? 18. Semitractor trucks use four large 12-V batteries. The starter system requires 24 V, while normal operation of the truck’s other electrical components utilizes 12 V. How could the four batteries be connected to produce 24 V? To produce 12 V? Why is 24 V better than 12 V for starting the truck’s engine (a very heavy load)? 21.3 Kirchhoff’s Rules 19. Can all of the currents going into the junction in Figure 21.49 be positive? Explain. Figure 21.49 20. Apply the junction rule to junction b in Figure 21.50. Is any new information gained by applying the junction rule at e? (In the figure, each emf is represented by script E.) Figure 21.50 21. (a) What is the potential difference going from point a to point b in Figure 21.50? (b) What is the potential difference going from c to b? (c) From e to g? (d) From e to d? 22. Apply the loop rule to loop afedcba in Figure 21.50. 23. Apply the loop rule to loops abgefa and cbgedc in Figure 21.50. 21.4 DC Voltmeters and Ammeters This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments 955 24. Why should you not connect an ammeter directly across a voltage source as shown i
n Figure 21.51? (Note that script E in the figure stands for emf.) Figure 21.51 25. Suppose you are using a multimeter (one designed to measure a range of voltages, currents, and resistances) to measure current in a circuit and you inadvertently leave it in a voltmeter mode. What effect will the meter have on the circuit? What would happen if you were measuring voltage but accidentally put the meter in the ammeter mode? 26. Specify the points to which you could connect a voltmeter to measure the following potential differences in Figure 21.52: (a) the potential difference of the voltage source; (b) the potential difference across 1 ; (c) across 2 ; (d) across 3 ; (e) across 2 and 3 . Note that there may be more than one answer to each part. Figure 21.52 27. To measure currents in Figure 21.52, you would replace a wire between two points with an ammeter. Specify the points between which you would place an ammeter to measure the following: (a) the total current; (b) the current flowing through 1 ; (c) through 2 ; (d) through 3 . Note that there may be more than one answer to each part. 21.5 Null Measurements 28. Why can a null measurement be more accurate than one using standard voltmeters and ammeters? What factors limit the accuracy of null measurements? 29. If a potentiometer is used to measure cell emfs on the order of a few volts, why is it most accurate for the standard emfs to be the same order of magnitude and the resistances to be in the range of a few ohms? 21.6 DC Circuits Containing Resistors and Capacitors 30. Regarding the units involved in the relationship = , verify that the units of resistance times capacitance are time, that is, Ω ⋅ F = s . 31. The time constant in heart defibrillation is crucial to limiting the time the current flows. If the capacitance in the defibrillation unit is fixed, how would you manipulate resistance in the circuit to adjust the constant ? Would an adjustment of the applied voltage also be needed to ensure that the current delivered has an appropriate value? 32. When making an ECG measurement, it is important to measure voltage variations over small time intervals. The time is limited by the constant of the circuit—it is not possible to measure time variations shorter than . How would you manipulate and in the circuit to allow the necessary measurements? 33. Draw two graphs of charge versus time on a capacitor. Draw one for charging an initially uncharged capacitor in series with a resistor, as in the circuit in Figure 21.41, starting from t = 0 . Draw the other for discharging a capacitor through a resistor, as in the circuit in Figure 21.42, starting at t = 0 , with an initial charge 0 . Show at least two intervals of . 34. When charging a capacitor, as discussed in conjunction with Figure 21.41, how long does it take for the voltage on the capacitor to reach emf? Is this a problem? 956 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments 35. When discharging a capacitor, as discussed in conjunction with Figure 21.42, how long does it take for the voltage on the capacitor to reach zero? Is this a problem? 36. Referring to Figure 21.41, draw a graph of potential difference across the resistor versus time, showing at least two intervals of . Also draw a graph of current versus time for this situation. 37. A long, inexpensive extension cord is connected from inside the house to a refrigerator outside. The refrigerator doesn’t run as it should. What might be the problem? 38. In Figure 21.44, does the graph indicate the time constant is shorter for discharging than for charging? Would you expect ionized gas to have low resistance? How would you adjust to get a longer time between flashes? Would adjusting affect the discharge time? 39. An electronic apparatus may have large capacitors at high voltage in the power supply section, presenting a shock hazard even when the apparatus is switched off. A “bleeder resistor” is therefore placed across such a capacitor, as shown schematically in Figure 21.53, to bleed the charge from it after the apparatus is off. Why must the bleeder resistance be much greater than the effective resistance of the rest of the circuit? How does this affect the time constant for discharging the capacitor? Figure 21.53 A bleeder resistor bl discharges the capacitor in this electronic device once it is switched off. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments 957 Problems & Exercises 21.1 Resistors in Series and Parallel Note: Data taken from figures can be assumed to be accurate to three significant digits. 1. (a) What is the resistance of ten 275-Ω resistors connected in series? (b) In parallel? 2. (a) What is the resistance of a 1.00×102 -Ω , a 2.50-kΩ , and a 4.00-k Ω resistor connected in series? (b) In parallel? 3. What are the largest and smallest resistances you can obtain by connecting a 36.0-Ω , a 50.0-Ω , and a 700-Ω resistor together? 4. An 1800-W toaster, a 1400-W electric frying pan, and a 75-W lamp are plugged into the same outlet in a 15-A, 120-V circuit. (The three devices are in parallel when plugged into the same socket.). (a) What current is drawn by each device? (b) Will this combination blow the 15-A fuse? 5. Your car’s 30.0-W headlight and 2.40-kW starter are ordinarily connected in parallel in a 12.0-V system. What power would one headlight and the starter consume if connected in series to a 12.0-V battery? (Neglect any other resistance in the circuit and any change in resistance in the two devices.) 6. (a) Given a 48.0-V battery and 24.0-Ω and 96.0-Ω resistors, find the current and power for each when connected in series. (b) Repeat when the resistances are in parallel. 7. Referring to the example combining series and parallel circuits and Figure 21.6, calculate 3 in the following two different ways: (a) from the known values of and 2 ; (b) using Ohm’s law for 3 . In both parts explicitly show how you follow the steps in the Problem-Solving Strategies for Series and Parallel Resistors. 8. Referring to Figure 21.6: (a) Calculate 3 and note how it compares with 3 found in the first two example problems in this module. (b) Find the total power supplied by the source and compare it with the sum of the powers dissipated by the resistors. 9. Refer to Figure 21.7 and the discussion of lights dimming when a heavy appliance comes on. (a) Given the voltage source is 120 V, the wire resistance is 0.400 Ω , and the bulb is nominally 75.0 W, what power will the bulb dissipate if a total of 15.0 A passes through the wires when the motor comes on? Assume negligible change in bulb resistance. (b) What power is consumed by the motor? 10. A 240-kV power transmission line carrying 5.00×102 A is hung from grounded metal towers by ceramic insulators, each having a 1.00109 -Ω resistance. Figure 21.54. (a) What is the resistance to ground of 100 of these insulators? (b) Calculate the power dissipated by 100 of them. (c) What fraction of the power carried by the line is this? Explicitly show how you follow the steps in the Problem-Solving Strategies for Series and Parallel Resistors. Figure 21.54 High-voltage (240-kV) transmission line carrying 5.00×102 A is hung from a grounded metal transmission tower. The row of ceramic insulators provide 1.00×109 Ω of resistance each. 11. Show that if two resistors 1 and 2 are combined and one is much greater than the other ( 1 >>2 ): (a) Their series resistance is very nearly equal to the greater resistance 1 . (b) Their parallel resistance is very nearly equal to smaller resistance 2 . 12. Unreasonable Results Two resistors, one having a resistance of 145 Ω , are connected in parallel to produce a total resistance of 150 Ω . (a) What is the value of the second resistance? (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent? 13. Unreasonable Results Two resistors, one having a resistance of 900 kΩ , are connected in series to produce a total resistance of 0.500 MΩ . (a) What is the value of the second resistance? (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent? 21.2 Electromotive Force: Terminal Voltage 14. Standard automobile batteries have six lead-acid cells in series, creating a total emf of 12.0 V. What is the emf of an individual lead-acid cell? 15. Carbon-zinc dry cells (sometimes referred to as nonalkaline cells) have an emf of 1.54 V, and they are produced as single cells or in various combinations to form other voltages. (a) How many 1.54-V cells are needed to make the common 9-V battery used in many small electronic devices? (b) What is the actual emf of the approximately 9-V battery? (c) Discuss how internal resistance in the series connection of cells will affect the terminal voltage of this approximately 9-V battery. 16. What is the output voltage of a 3.0000-V lithium cell in a digital wristwatch that draws 0.300 mA, if the cell’s internal resistance is 2.00 Ω ? 17. (a) What is the terminal voltage of a large 1.54-V carbonzinc dry cell used in a physics lab to supply 2.00 A to a circuit, if the cell’s internal resistance is 0.100 Ω ? (b) How much 958 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments represent the situation. (b) If the internal resistance of the power supply is 2000 Ω , what is the current through his body? (c) What is the power dissipated in his body? (d) If the power supply is to be made safe by increasing its internal resistance, what should the internal resistance be for the maximum current in this situation to be 1.00 mA or less? (e) Will this modification compromise the effectiveness of the power supply for driving low-resistance devices? Explain your reasoning. 27. Electric fish generate current with biological cells called electroplaques, which are physiological emf devices. The electroplaques in the South American eel are arranged in 140 rows, each row stretching horizontally al
ong the body and each containing 5000 electroplaques. Each electroplaque has an emf of 0.15 V and internal resistance of 0.25 Ω . If the water surrounding the fish has resistance of 800 Ω , how much current can the eel produce in water from near its head to near its tail? 28. Integrated Concepts A 12.0-V emf automobile battery has a terminal voltage of 16.0 V when being charged by a current of 10.0 A. (a) What is the battery’s internal resistance? (b) What power is dissipated inside the battery? (c) At what rate (in ºC/min ) will its temperature increase if its mass is 20.0 kg and it has a specific heat of 0.300 kcal/kg ⋅ ºC , assuming no heat escapes? 29. Unreasonable Results A 1.58-V alkaline cell with a 0.200-Ω internal resistance is supplying 8.50 A to a load. (a) What is its terminal voltage? (b) What is the value of the load resistance? (c) What is unreasonable about these results? (d) Which assumptions are unreasonable or inconsistent? 30. Unreasonable Results (a) What is the internal resistance of a 1.54-V dry cell that supplies 1.00 W of power to a 15.0-Ω bulb? (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent? 21.3 Kirchhoff’s Rules 31. Apply the loop rule to loop abcdefgha in Figure 21.27. 32. Apply the loop rule to loop aedcba in Figure 21.27. 33. Verify the second equation in Example 21.5 by substituting the values found for the currents 1 and 2 . 34. Verify the third equation in Example 21.5 by substituting the values found for the currents 1 and 3 . 35. Apply the junction rule at point a in Figure 21.55. electrical power does the cell produce? (c) What power goes to its load? 18. What is the internal resistance of an automobile battery that has an emf of 12.0 V and a terminal voltage of 15.0 V while a current of 8.00 A is charging it? 19. (a) Find the terminal voltage of a 12.0-V motorcycle battery having a 0.600-Ω internal resistance, if it is being charged by a current of 10.0 A. (b) What is the output voltage of the battery charger? 20. A car battery with a 12-V emf and an internal resistance of 0.050 Ω is being charged with a current of 60 A. Note that in this process the battery is being charged. (a) What is the potential difference across its terminals? (b) At what rate is thermal energy being dissipated in the battery? (c) At what rate is electric energy being converted to chemical energy? (d) What are the answers to (a) and (b) when the battery is used to supply 60 A to the starter motor? 21. The hot resistance of a flashlight bulb is 2.30 Ω , and it is run by a 1.58-V alkaline cell having a 0.100-Ω internal resistance. (a) What current flows? (b) Calculate the power supplied to the bulb using 2 bulb . (c) Is this power the same as calculated using 2 bulb ? 22. The label on a portable radio recommends the use of rechargeable nickel-cadmium cells (nicads), although they have a 1.25-V emf while alkaline cells have a 1.58-V emf. The radio has a 3.20-Ω resistance. (a) Draw a circuit diagram of the radio and its batteries. Now, calculate the power delivered to the radio. (b) When using Nicad cells each having an internal resistance of 0.0400 Ω . (c) When using alkaline cells each having an internal resistance of 0.200 Ω . (d) Does this difference seem significant, considering that the radio’s effective resistance is lowered when its volume is turned up? 23. An automobile starter motor has an equivalent resistance of 0.0500 Ω and is supplied by a 12.0-V battery with a 0.0100-Ω internal resistance. (a) What is the current to the motor? (b) What voltage is applied to it? (c) What power is supplied to the motor? (d) Repeat these calculations for when the battery connections are corroded and add 0.0900 Ω to the circuit. (Significant problems are caused by even small amounts of unwanted resistance in low-voltage, high-current applications.) 24. A child’s electronic toy is supplied by three 1.58-V alkaline cells having internal resistances of 0.0200 Ω in series with a 1.53-V carbon-zinc dry cell having a 0.100-Ω internal resistance. The load resistance is 10.0 Ω . (a) Draw a circuit diagram of the toy and its batteries. (b) What current flows? (c) How much power is supplied to the load? (d) What is the internal resistance of the dry cell if it goes bad, resulting in only 0.500 W being supplied to the load? 25. (a) What is the internal resistance of a voltage source if its terminal voltage drops by 2.00 V when the current supplied increases by 5.00 A? (b) Can the emf of the voltage source be found with the information supplied? 26. A person with body resistance between his hands of 10.0 k Ω accidentally grasps the terminals of a 20.0-kV power supply. (Do NOT do this!) (a) Draw a circuit diagram to This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments 959 44. Find the resistance that must be placed in series with a 25.0-Ω galvanometer having a 50.0-μA sensitivity (the same as the one discussed in the text) to allow it to be used as a voltmeter with a 0.100-V full-scale reading. 45. Find the resistance that must be placed in series with a 25.0-Ω galvanometer having a 50.0-μA sensitivity (the same as the one discussed in the text) to allow it to be used as a voltmeter with a 3000-V full-scale reading. Include a circuit diagram with your solution. 46. Find the resistance that must be placed in parallel with a 25.0-Ω galvanometer having a 50.0-μA sensitivity (the same as the one discussed in the text) to allow it to be used as an ammeter with a 10.0-A full-scale reading. Include a circuit diagram with your solution. 47. Find the resistance that must be placed in parallel with a 25.0-Ω galvanometer having a 50.0-μA sensitivity (the same as the one discussed in the text) to allow it to be used as an ammeter with a 300-mA full-scale reading. 48. Find the resistance that must be placed in series with a 10.0-Ω galvanometer having a 100-μA sensitivity to allow it to be used as a voltmeter with: (a) a 300-V full-scale reading, and (b) a 0.300-V full-scale reading. 49. Find the resistance that must be placed in parallel with a 10.0-Ω galvanometer having a 100-μA sensitivity to allow it to be used as an ammeter with: (a) a 20.0-A full-scale reading, and (b) a 100-mA full-scale reading. 50. Suppose you measure the terminal voltage of a 1.585-V alkaline cell having an internal resistance of 0.100 Ω by placing a 1.00-k Ω voltmeter across its terminals. (See Figure 21.57.) (a) What current flows? (b) Find the terminal voltage. (c) To see how close the measured terminal voltage is to the emf, calculate their ratio. Figure 21.57 51. Suppose you measure the terminal voltage of a 3.200-V lithium cell having an internal resistance of 5.00 Ω by placing a 1.00-k Ω voltmeter across its terminals. (a) What current flows? (b) Find the terminal voltage. (c) To see how close the measured terminal voltage is to the emf, calculate their ratio. 52. A certain ammeter has a resistance of 5.0010−5 Ω on its 3.00-A scale and contains a 10.0-Ω galvanometer. What is the sensitivity of the galvanometer? 53. A 1.00-MΩ voltmeter is placed in parallel with a 75.0-k Ω resistor in a circuit. (a) Draw a circuit diagram of the connection. (b) What is the resistance of the combination? (c) If the voltage across the combination is kept the same as it was across the 75.0-k Ω resistor alone, what is the percent increase in current? (d) If the current through the combination is kept the same as it was through the 75.0-k Ω resistor alone, what is the percentage decrease in voltage? (e) Are the changes found in parts (c) and (d) significant? Discuss. Figure 21.55 36. Apply the loop rule to loop abcdefghija in Figure 21.55. 37. Apply the loop rule to loop akledcba in Figure 21.55. 38. Find the currents flowing in the circuit in Figure 21.55. Explicitly show how you follow the steps in the ProblemSolving Strategies for Series and Parallel Resistors. 39. Solve Example 21.5, but use loop abcdefgha instead of loop akledcba. Explicitly show how you follow the steps in the Problem-Solving Strategies for Series and Parallel Resistors. 40. Find the currents flowing in the circuit in Figure 21.50. 41. Unreasonable Results Consider the circuit in Figure 21.56, and suppose that the emfs are unknown and the currents are given to be 1 = 5.00 A , 2 = 3.0 A , and 3 = –2.00 A . (a) Could you find the emfs? (b) What is wrong with the assumptions? Figure 21.56 21.4 DC Voltmeters and Ammeters 42. What is the sensitivity of the galvanometer (that is, what current gives a full-scale deflection) inside a voltmeter that has a 1.00-M Ω resistance on its 30.0-V scale? 43. What is the sensitivity of the galvanometer (that is, what current gives a full-scale deflection) inside a voltmeter that has a 25.0-k Ω resistance on its 100-V scale? 960 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments 54. A 0.0200-Ω ammeter is placed in series with a 10.00-Ω resistor in a circuit. (a) Draw a circuit diagram of the connection. (b) Calculate the resistance of the combination. (c) If the voltage is kept the same across the combination as it was through the 10.00-Ω resistor alone, what is the percent decrease in current? (d) If the current is kept the same through the combination as it was through the 10.00-Ω resistor alone, what is the percent increase in voltage? (e) Are the changes found in parts (c) and (d) significant? Discuss. 55. Unreasonable Results Suppose you have a 40.0-Ω galvanometer with a 25.0-μA sensitivity. (a) What resistance would you put in series with it to allow it to be used as a voltmeter that has a full-scale deflection for 0.500 mV? (b) What is unreasonable about this result? (c) Which assumptions are responsible? 56. Unreasonable Results (a) What resistance would you put in parallel with a 40.0-Ω galvanometer having a 25.0-μA sensitivity to allow it to be used as an ammeter that has a full-scale deflection for 10.0-μA ? (b) What is unr
easonable about this result? (c) Which assumptions are responsible? 21.5 Null Measurements 57. What is the emfx of a cell being measured in a potentiometer, if the standard cell’s emf is 12.0 V and the potentiometer balances for x = 5.000 Ω and s = 2.500 Ω ? 58. Calculate the emfx of a dry cell for which a potentiometer is balanced when x = 1.200 Ω , while an alkaline standard cell with an emf of 1.600 V requires s = 1.247 Ω to balance the potentiometer. 59. When an unknown resistance x is placed in a Wheatstone bridge, it is possible to balance the bridge by adjusting 3 to be 2500 Ω . What is x if 2 1 = 0.625 ? 60. To what value must you adjust 3 to balance a Wheatstone bridge, if the unknown resistance x is 100 Ω , 1 is 50.0 Ω , and 2 is 175 Ω ? 61. (a) What is the unknown emfx in a potentiometer that balances when x is 10.0 Ω , and balances when s is 15.0 Ω for a standard 3.000-V emf? (b) The same emfx is placed in the same potentiometer, which now balances when s is 15.0 Ω for a standard emf of 3.100 V. At what resistance x will the potentiometer balance? 62. Suppose you want to measure resistances in the range from 10.0 Ω to 10.0 kΩ using a Wheatstone bridge that = 2.000 . Over what range should 3 be has 2 1 adjustable? This content is available for free at http://cnx.org/content/col11844/1.13 21.6 DC Circuits Containing Resistors and Capacitors 63. The timing device in an automobile’s intermittent wiper system is based on an time constant and utilizes a 0.500-μF capacitor and a variable resistor. Over what range must be made to vary to achieve time constants from 2.00 to 15.0 s? 64. A heart pacemaker fires 72 times a minute, each time a 25.0-nF capacitor is charged (by a battery in series with a resistor) to 0.632 of its full voltage. What is the value of the resistance? 65. The duration of a photographic flash is related to an time constant, which is 0.100 μs for a certain camera. (a) If the resistance of the flash lamp is 0.0400 Ω during discharge, what is the size of the capacitor supplying its energy? (b) What is the time constant for charging the capacitor, if the charging resistance is 800 kΩ ? 66. A 2.00- and a 7.50-μF capacitor can be connected in series or parallel, as can a 25.0- and a 100-kΩ resistor. Calculate the four time constants possible from connecting the resulting capacitance and resistance in series. 67. After two time constants, what percentage of the final voltage, emf, is on an initially uncharged capacitor , charged through a resistance ? 68. A 500-Ω resistor, an uncharged 1.50-μF capacitor, and a 6.16-V emf are connected in series. (a) What is the initial current? (b) What is the time constant? (c) What is the current after one time constant? (d) What is the voltage on the capacitor after one time constant? 69. A heart defibrillator being used on a patient has an time constant of 10.0 ms due to the resistance of the patient and the capacitance of the defibrillator. (a) If the defibrillator has an 8.00-μF capacitance, what is the resistance of the path through the patient? (You may neglect the capacitance of the patient and the resistance of the defibrillator.) (b) If the initial voltage is 12.0 kV, how long does it take to decline to 6.00×102 V ? 70. An ECG monitor must have an time constant less than 1.00×102 μs to be able to measure variations in voltage over small time intervals. (a) If the resistance of the circuit (due mostly to that of the patient’s chest) is 1.00 kΩ , what is the maximum capacitance of the circuit? (b) Would it be difficult in practice to limit the capacitance to less than the value found in (a)? 71. Figure 21.58 shows how a bleeder resistor is used to discharge a capacitor after an electronic device is shut off, allowing a person to work on the electronics with less risk of shock. (a) What is the time constant? (b) How long will it take to reduce the voltage on the capacitor to 0.250% (5% of 5%) of its full value once discharge begins? (c) If the capacitor is charged to a voltage 0 through a 100-Ω resistance, calculate the time it takes to rise to 0.8650 (This is about two time constants.) Chapter 21 \| Circuits, Bioelectricity, and DC Instruments 961 Consider a rechargeable lithium cell that is to be used to power a camcorder. Construct a problem in which you calculate the internal resistance of the cell during normal operation. Also, calculate the minimum voltage output of a battery charger to be used to recharge your lithium cell. Among the things to be considered are the emf and useful terminal voltage of a lithium cell and the current it should be able to supply to a camcorder. Figure 21.58 72. Using the exact exponential treatment, find how much time is required to discharge a 250-μF capacitor through a 500-Ω resistor down to 1.00% of its original voltage. 73. Using the exact exponential treatment, find how much time is required to charge an initially uncharged 100-pF capacitor through a 75.0-M Ω resistor to 90.0% of its final voltage. 74. Integrated Concepts If you wish to take a picture of a bullet traveling at 500 m/s, then a very brief flash of light produced by an discharge through a flash tube can limit blurring. Assuming 1.00 mm of motion during one constant is acceptable, and given that the flash is driven by a 600-μF capacitor, what is the resistance in the flash tube? 75. Integrated Concepts A flashing lamp in a Christmas earring is based on an discharge of a capacitor through its resistance. The effective duration of the flash is 0.250 s, during which it produces an average 0.500 W from an average 3.00 V. (a) What energy does it dissipate? (b) How much charge moves through the lamp? (c) Find the capacitance. (d) What is the resistance of the lamp? 76. Integrated Concepts A 160-μF capacitor charged to 450 V is discharged through a 31.2-k Ω resistor. (a) Find the time constant. (b) Calculate the temperature increase of the resistor, given that its mass is 2.50 g and its specific heat is 1.67 kJ kg ⋅ ºC , noting that most of the thermal energy is retained in the short time of the discharge. (c) Calculate the new resistance, assuming it is pure carbon. (d) Does this change in resistance seem significant? 77. Unreasonable Results (a) Calculate the capacitance needed to get an time constant of 1.00×103 s with a 0.100-Ω resistor. (b) What is unreasonable about this result? (c) Which assumptions are responsible? 78. Construct Your Own Problem Consider a camera’s flash unit. Construct a problem in which you calculate the size of the capacitor that stores energy for the flash lamp. Among the things to be considered are the voltage applied to the capacitor, the energy needed in the flash and the associated charge needed on the capacitor, the resistance of the flash lamp during discharge, and the desired time constant. 79. Construct Your Own Problem 962 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments Test Prep for AP® Courses 21.1 Resistors in Series and Parallel 1. Figure 21.59 The figure above shows a circuit containing two batteries and three identical resistors with resistance R. Which of the following changes to the circuit will result in an increase in the current at point P? Select two answers. a. Reversing the connections to the 14 V battery. b. Removing the 2 V battery and connecting the wires to close the left loop. c. Rearranging the resistors so all three are in series. d. Removing the branch containing resistor Z. 2. In a circuit, a parallel combination of six 1.6-kΩ resistors is connected in series with a parallel combination of four 2.4-kΩ resistors. If the source voltage is 24 V, what will be the percentage of total current in one of the 2.4-kΩ resistors? a. 10% b. 12% c. 20% d. 25% 3. If the circuit in the previous question is modified by removing some of the 1.6 kΩ resistors, the total current in the circuit is 24 mA. How many resistors were removed? a. 1 b. 2 c. 3 d. 4 4. Figure 21.60 Two resistors, with resistances R and 2R are connected to a voltage source as shown in this figure. If the power dissipated in R is 10 W, what is the power dissipated in 2R? a. 1 W b. 2.5 W c. 5 W d. 10 W 5. In a circuit, a parallel combination of two 20-Ω and one 10-Ω resistors is connected in series with a 4-Ω resistor. The source voltage is 36 V. a. Find the resistor(s) with the maximum current. b. Find the resistor(s) with the maximum voltage drop. c. Find the power dissipated in each resistor and hence the total power dissipated in all the resistors. Also find the power output of the source. Are they equal or not? Justify your answer. d. Will the answers for questions (a) and (b) differ if a 3 Ω resistor is added in series to the 4 Ω resistor? If yes, repeat the question(s) for the new resistor combination. This content is available for free at http://cnx.org/content/col11844/1.13 e. If the values of all the resistors and the source voltage are doubled, what will be the effect on the current? 21.2 Electromotive Force: Terminal Voltage 6. Suppose there are two voltage sources – Sources A and B – with the same emfs but different internal resistances, i.e., the internal resistance of Source A is lower than Source B. If they both supply the same current in their circuits, which of the following statements is true? a. External resistance in Source A’s circuit is more than Source B’s circuit. b. External resistance in Source A’s circuit is less than Source B’s circuit. c. External resistance in Source A’s circuit is the same as Source B’s circuit. d. The relationship between external resistances in the two circuits can’t be determined. 7. Calculate the internal resistance of a voltage source if the terminal voltage of the source increases by 1 V when the current supplied decreases by 4 A? Suppose this source is connected in series (in the same direction) to another source with a different voltage but same internal resistance. What will be the total internal resistance? How will the total internal resistance change if the sources are connecte
d in the opposite direction? 21.3 Kirchhoff’s Rules 8. An experiment was set up with the circuit diagram shown. Assume R1 = 10 Ω, R2 = R3 = 5 Ω, r = 0 Ω and E = 6 V. Figure 21.61 a. One of the steps to examine the set-up is to test points with the same potential. Which of the following points can be tested? a. Points b, c and d. b. Points d, e and f. c. Points f, h and j. d. Points a, h and i. b. At which three points should the currents be measured so that Kirchhoff’s junction rule can be directly confirmed? a. Points b, c and d. b. Points d, e and f. c. Points f, h and j. d. Points a, h and i. c. If the current in the branch with the voltage source is upward and currents in the other two branches are downward, i.e. Ia = Ii + Ic, identify which of the following can be true? Select two answers. a. b. c. d. Ii = Ij - If Ie = Ih - Ii Ic = Ij - Ia Id = Ih - Ij Chapter 21 \| Circuits, Bioelectricity, and DC Instruments 963 d. The measurements reveal that the current through R1 is 0.5 A and R3 is 0.6 A. Based on your knowledge of Kirchoff’s laws, confirm which of the following statements are true. a. The measured current for R1 is correct but for R3 is incorrect. b. The measured current for R3 is correct but for R1 is incorrect. c. Both the measured currents are correct. d. Both the measured currents are incorrect. e. The graph shown in the following figure is the energy dissipated at R1 as a function of time. Figure 21.62 Which of the following shows the graph for energy dissipated at R2 as a function of time? c. Figure 21.65 d. Figure 21.66 9. For this question, consider the circuit shown in the following figure. Figure 21.67 a. Assuming that none of the three currents (I1, I2, and I3) are equal to zero, which of the following statements is false? a. b. c. The current through R3 is equal to the current I3 = I1 + I2 at point a. I2 = I3 - I1 at point e. through R5. d. The current through R1 is equal to the current through R5. b. Which of the following statements is true? a. E1 + E2 + I1R1 - I2R2 + I1r1 - I2r2 + I1R5 = 0 b. - E1 + E2 + I1R1 - I2R2 + I1r1 - I2r2 - I1R5 = 0 c. E1 - E2 - I1R1 + I2R2 - I1r1 + I2r2 - I1R5 = 0 d. E1 + E2 - I1R1 + I2R2 - I1r1 + I2r2 + I1R5 = 0 c. If I1 = 5 A and I3 = -2 A, which of the following statements is false? a. The current through R1 will flow from a to b and will be equal to 5 A. b. The current through R3 will flow from a to j and will be equal to 2 A. c. The current through R5 will flow from d to e and will be equal to 5 A. d. None of the above. d. If I1 = 5 A and I3 = -2 A, I2 will be equal to a. 3 A -3 A b. c. 7 A -7 A d. 10. a. Figure 21.63 b. Figure 21.64 964 Chapter 21 \| Circuits, Bioelectricity, and DC Instruments Figure 21.68 In an experiment this circuit is set up. Three ammeters are used to record the currents in the three vertical branches (with R1, R2, and E). The readings of the ammeters in the resistor branches (i.e. currents in R1 and R2) are 2 A and 3 A respectively. a. Find the equation obtained by applying Kirchhoff’s loop rule in the loop involving R1 and R2. b. What will be the reading of the third ammeter (i.e. the branch with E)? If E were replaced by 3E, how would this reading change? If the original circuit is modified by adding another voltage source (as shown in the following circuit), find the readings of the three ammeters. c. Figure 21.69 11. 0.19 A. According to the results calculated in part (d) identify the resistor(s). Justify any difference in measured and calculated value. 21.6 DC Circuits Containing Resistors and Capacitors 12. A battery is connected to a resistor and an uncharged capacitor. The switch for the circuit is closed at t = 0 s. a. While the capacitor is being charged, which of the following is true? a. Current through and voltage across the resistor increase. b. Current through and voltage across the resistor decrease. c. Current through and voltage across the resistor first increase and then decrease. d. Current through and voltage across the resistor first decrease and then increase. b. When the capacitor is fully charged, which of the following is NOT zero? a. Current in the resistor. b. Voltage across the resistor. c. Current in the capacitor. d. None of the above. 13. An uncharged capacitor C is connected in series (with a switch) to a resistor R1 and a voltage source E. Assume E = 24 V, R1 = 1.2 kΩ and C = 1 mF. a. What will be the current through the circuit as the switch is closed? Draw a circuit diagram and show the direction of current after the switch is closed. How long will it take for the capacitor to be 99% charged? b. After full charging, this capacitor is connected in series to another resistor, R2 = 1 kΩ. What will be the current in the circuit as soon as it’s connected? Draw a circuit diagram and show the direction of current. How long will it take for the capacitor voltage to reach 3.24 V? Figure 21.70 In this circuit, assume the currents through R1, R2 and R3 are I1, I2 and I3 respectively and all are flowing in the clockwise direction. a. Find the equation obtained by applying Kirchhoff’s junction rule at point A. b. Find the equations obtained by applying Kirchhoff’s loop rule in the upper and lower loops. c. Assume R1 = R2 = 6 Ω, R3 = 12 Ω, r1 = r2 = 0 Ω, E1 = 6 V and E2 = 4 V. Calculate I1, I2 and I3. d. For the situation in which E2 is replaced by a closed switch, repeat parts (a) and (b). Using the values for R1, R2, R3, r1 and E1 from part (c) calculate the currents through the three resistors. e. For the circuit in part (d) calculate the output power of the voltage source and across all the resistors. Examine if energy is conserved in the circuit. f. A student implemented the circuit of part (d) in the lab and measured the current though one of the resistors as This content is available for free at http://cnx.org/content/col11844/1.13 Appendix A 1531 A ATOMIC MASSES Table A1 Atomic Masses Atomic Number 10 11 Name neutron Hydrogen Deuterium Tritium Helium Lithium Beryllium Boron Carbon Nitrogen Oxygen Fluorine Neon Sodium Atomic Mass Number, A Symbol 10 11 11 12 13 14 13 14 15 15 16 18 18 19 20 22 22 1H 2H or D 3H or T 3He 4He 6Li 7Li 7Be 9Be 10B 11B 11C 12C 13C 14C 13N 14N 15N 15O 16O 18O 18F 19F 20Ne 22Ne 22Na 3.016 050 3.016 030 4.002 603 6.015 121 7.016 003 7.016 928 9.012 182 10.012 937 11.009 305 11.011 432 12.000 000 13.003 355 14.003 241 13.005 738 14.003 074 15.000 108 15.003 065 15.994 915 17.999 160 18.000 937 Atomic Mass (u) 1.008 665 Percent Abundance or Decay Mode − Half-life, t1/2 10.37 min 1.007 825 99.985% 2.014 102 0.015% − 12.33 y 1.38×10−4% ≈100% 7.5% 92.5% EC 100% 19.9% 80.1% EC, + 98.90% 1.10% − + 99.63% 0.37% EC, + 99.76% 0.200% EC, + 53.29 d 5730 y 9.96 min 122 s 1.83 h 2.602 y 18.998 403 100% 19.992 435 90.51% 21.991 383 21.994 434 9.22% + 1532 Appendix A Atomic Number, Z Name Atomic Mass Number, A Symbol Atomic Mass (u) Percent Abundance or Decay Mode Half-life, t1/2 12 13 14 Magnesium Aluminum Silicon 15 Phosphorus 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 Sulfur Chlorine Argon Potassium Calcium Scandium Titanium Vanadium Chromium Manganese Iron Cobalt Nickel Copper Zinc Gallium 23 24 24 27 28 31 31 32 32 35 35 37 40 39 40 40 45 48 51 52 55 56 59 60 58 60 63 65 64 66 69 23Na 24Na 24Mg 27Al 28Si 31Si 31P 32P 32S 35S 35Cl 37Cl 40Ar 39K 40K 40Ca 45Sc 48Ti 51V 52Cr 22.989 767 23.990 961 100% − 23.985 042 78.99% 26.981 539 100% 14.96 h 27.976 927 92.23% 2.62h 30.975 362 30.973 762 31.973 907 − 100% − 31.972 070 95.02% 14.28 d 87.4 d 34.969 031 34.968 852 36.965 903 39.962 384 38.963 707 39.963 999 44.955 910 47.947 947 50.943 962 51.940 509 − 75.77% 24.23% 99.60% 93.26% 100% 73.8% 99.75% 83.79% 100% 0.0117%, EC, − 1.28×109 y 39.962 591 96.94% 55Mn 54.938 047 5.271 y 56Fe 59Co 60Co 58Ni 60Ni 63Cu 65Cu 64Zn 66Zn 69Ga 55.934 939 91.72% 58.933 198 59.933 819 57.935 346 59.930 788 62.939 598 64.927 793 63.929 145 65.926 034 68.925 580 100% − 68.27% 26.10% 69.17% 30.83% 48.6% 27.9% 60.1% This content is available for free at http://cnx.org/content/col11844/1.13 Appendix A 1533 Atomic Number, Z Name Atomic Mass Number, A Symbol Atomic Mass (u) Percent Abundance or Decay Mode Half-life, t1/2 71.922 079 73.921 177 74.921 594 79.916 520 78.918 336 83.911 507 84.911 794 85.909 267 87.905 619 89.907 738 88.905 849 89.907 152 27.4% 36.5% 100% 49.7% 50.69% 57.0% 72.17% 9.86% 82.58% − 100% − 89.904 703 51.45% 92.906 377 100% 97.905 406 24.13% 97.907 215 101.904 348 102.905 500 105.903 478 106.905 092 108.904 757 113.903 357 114.903 880 − 31.6% 100% 27.33% 51.84% 48.16% 28.73% 32 Germanium 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 Arsenic Selenium Bromine Krypton Rubidium Strontium Yttrium Zirconium Niobium Molybdenum Technetium Ruthenium Rhodium Palladium Silver Cadmium Indium Tin Antimony Tellurium Iodine Xenon 72 74 75 80 79 84 85 86 88 90 89 90 90 93 98 98 102 103 106 107 109 114 115 120 121 130 127 131 132 136 72Ge 74Ge 75As 80Se 79Br 84Kr 85Rb 86Sr 88Sr 90Sr 89Y 90Y 90Zr 93Nb 98Mo 98Tc 102Ru 103Rh 106Pd 107Ag 109Ag 114Cd 115In 120Sn 121Sb 130Te 127I 131I 132Xe 136Xe 95.7%, − 4.4×1014y 119.902 200 32.59% 120.903 821 57.3% 129.906 229 126.904 473 130.906 114 131.904 144 135.907 214 33.8%, − 2.5×1021y 100% − 26.9% 8.9% 8.040 d 28.8 y 64.1 h 4.2×106y 1534 Atomic Number, Z 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 Name Cesium Barium Lanthanum Cerium Praseodymium Neodymium Promethium Samarium Europium Gadolinium Terbium Dysprosium Holmium Erbium Thulium Ytterbium Lutecium Hafnium Tantalum Tungsten Rhenium Osmium Iridium Platinum Gold Mercury Atomic Mass Number, A Symbol Atomic Mass (u) Percent Abundance or Decay Mode Half-life, t1/2 Appendix A 133 134 137 138 139 140 141 142 145 152 153 158 159 164 165 166 169 174 175 180 181 184 187 191 192 191 193 195 197 198 199 133Cs 134Cs 137Ba 138Ba 139La 140Ce 141Pr 142Nd 145Pm 152Sm 153Eu 158Gd 159Tb 164Dy 165Ho 166Er 132.905 429 133.906 696 136.905 812 137.905 232 138.906 346 139.905 433 140.9
07 647 100% EC, − 11.23% 71.70% 99.91% 88.48% 100% 141.907 719 27.13% 144.912 743 151.919 729 152.921 225 EC, 26.7% 52.2% 157.924 099 24.84% 2.06 y 17.7 y 158.925 342 163.929 171 164.930 319 165.930 290 169Tm 168.934 212 174Yb 175Lu 180Hf 181Ta 173.938 859 174.940 770 179.946 545 180.947 992 184W 183.950 928 100% 28.2% 100% 33.6% 100% 31.8% 97.41% 35.10% 99.98% 30.67% 187Re 191Os 192Os 191Ir 193Ir 195Pt 197Au 198Au 199Hg 186.955 744 190.960 920 191.961 467 190.960 584 192.962 917 194.964 766 196.966 543 197.968 217 62.6%, − 4.6×1010y − 41.0% 37.3% 62.7% 33.8% 100% − 15.4 d 2.696 d 198.968 253 16.87% This content is available for free at http://cnx.org/content/col11844/1.13 Appendix A 1535 Atomic Number, Z Name Atomic Mass Number, A Symbol Atomic Mass (u) Percent Abundance or Decay Mode Half-life, t1/2 202Hg 201.970 617 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 Thallium Lead Bismuth Polonium Astatine Radon Francium Radium Actinium Thorium Protactinium Uranium Neptunium Plutonium Americium Curium Berkelium 202 205 206 207 208 210 211 212 209 211 210 218 222 223 226 227 228 232 231 233 235 236 238 239 239 239 243 245 247 205Tl 206Pb 207Pb 208Pb 210Pb 211Pb 212Pb 209Bi 211Bi 210Po 218At 222Rn 223Fr 226Ra 227Ac 228Th 232Th 231Pa 233U 235U 236U 238U 239U 239Np 239Pu 204.974 401 205.974 440 206.975 872 207.976 627 209.984 163 210.988 735 211.991 871 208.980 374 210.987 255 209.982 848 218.008 684 222.017 570 223.019 733 226.025 402 227.027 750 228.028 715 29.86% 70.48% 24.1% 22.1% 52.4% − − − 100% − − − − 22.3 y 36.1 min 10.64 h 2.14 min 138.38 d 1.6 s 3.82 d 21.8 min 1.60×103y 21.8 y 1.91 y 232.038 054 100%, 1.41×1010y 231.035 880 233.039 628 235.043 924 0.720%, 236.045 562 238.050 784 99.2745%, 239.054 289 239.052 933 239.052 157 − − 243Am 243.061 375 fission 245Cm 245.065 483 247Bk 247.070 300 3.28×104y 1.59×103y 7.04×108y 2.34×107y 4.47×109y 23.5 min 2.355 d 2.41×104y 7.37×103y 8.50×103y 1.38×103y 1536 Appendix A Atomic Number, Z Name Atomic Mass Number, A Symbol Atomic Mass (u) Percent Abundance or Decay Mode Half-life, t1/2 98 99 100 101 102 103 104 105 106 107 108 109 Californium Einsteinium Fermium Mendelevium Nobelium Lawrencium Rutherfordium Dubnium Seaborgium Bohrium Hassium Meitnerium 249 254 253 255 255 257 261 262 263 262 264 266 249Cf 254Es 253Fm 255Md 255No 257Lr 261Rf 262Db 263Sg 262Bh 264Hs 266Mt 249.074 844 254.088 019 253.085 173 255.091 081 255.093 260 257.099 480 261.108 690 262.113 760 263.11 86 262.123 1 264.128 5 266.137 8 − EC, EC, EC, EC, fission fission 351 y 276 d 3.00 d 27 min 3.1 min 0.646 s 1.08 min 34 s 0.8 s 0.102 s 0.08 ms 3.4 ms This content is available for free at http://cnx.org/content/col11844/1.13 Appendix B 1537 B SELECTED RADIOACTIVE ISOTOPES Decay modes are , − , + would + are roughly one-half the maxima. decay. IT is a transition from a metastable excited state. Energies for ± , electron capture (EC) and isomeric transition (IT). EC results in the same daughter nucleus as decays are the maxima; average energies Table B1 Selected Radioactive Isotopes Isotope t 1/2 DecayMode(s) Energy(MeV) Percent γ -Ray Energy(MeV) Percent 100% 100% 100% 90% 1.27 100% 0.0186 0.156 1.20 0.55 1.71 0.167 0.710 1.31 0.827 100% 100% 100% 89% 87% 0.257 100% 3.69 28% 1.80 43% 0.273 0.466 0.318 45% 55% 100% 3H 14C 13N 22Na 32P 35S 36Cl 40K 43K 45Ca 51Cr 52Mn 52Fe 59Fe 60Co 65Zn 67Ga 12.33 y 5730 y 9.96 min 2.602 y 14.28 d 87.4 d 3.00×105y 1.28×109y 22.3 h 165 d 27.70 d 5.59d 8.27 h 44.6 d 5.271 y 244.1 d 78.3 h − − + + − − − − − − EC + + − s − EC EC 75Se 118.5 d EC s s s s s s 0.373 0.618 0.320 1.33 1.43 0.169 0.378 1.10 1.29 1.17 1.33 1.12 0.0933 0.185 0.300 others 0.121 0.136 0.265 87% 87% 10% 28% 28% 43% 43% 57% 43% 100% 100% 51% 70% 35% 19% 20% 65% 68% 1538 Isotope t 1/2 DecayMode(s) Energy(MeV) Percent γ -Ray Energy(MeV) Appendix B Percent 20% 9% 0.280 others 1.08 0.514 100% 0.142 0.392 0.159 0.364 others 0.0400 0.372 0.411 others 0.662 0.030 0.044 0.537 others 0.412 100% 100% ≈100% 85% 35% 32% 25% 95% 25% 65% 24% ≈100% 0.0733 100% 0.186 100% ≈100% s numerous <0.400% s s 23% 77% 11% 15% 73% 0.050 23% numerous <0.250% 7.5×10−5 0.013 0.052 73% 15% 10% 86Rb 18.8 d 85Sr 90Sr 90Y 99mTc 113mIn 123I 131I 64.8 d 28.8 y 64.1 h 6.02 h 99.5 min 13.0 h 8.040 d − s EC − − IT IT EC − s 129Cs 32.3 h EC 137Cs 30.17 y 140Ba 12.79 d 198Au 197Hg 210Po 226Ra 2.696 d 64.1 h 138.38 d 1.60×103y 235U 238U 7.038×108y 4.468×109y 237Np 2.14×106y 239Pu 2.41×104y − s − − EC s s s s 0.69 1.77 0.546 2.28 0.248 0.607 others 0.511 1.17 1.035 s s 9% 91% 100% 100% 7% 93% 95% 5% ≈100% s 1.161 ≈100% 100% 5% 95% 5.41 4.68 4.87 4.68 4.22 4.27 numerous 4.96 (max.) 5.19 5.23 5.24 This content is available for free at http://cnx.org/content/col11844/1.13 Appendix B 1539 Isotope t 1/2 DecayMode(s) Energy(MeV) Percent γ -Ray Energy(MeV) Percent 243Am 7.37×103y s Max. 5.44 s 5.37 5.32 others 88% 11% others 0.075 others 1540 Appendix B This content is available for free at http://cnx.org/content/col11844/1.13 Appendix C 1541 C USEFUL INFORMATION This appendix is broken into several tables. • Table C1, Important Constants • Table C2, Submicroscopic Masses • Table C3, Solar System Data • Table C4, Metric Prefixes for Powers of Ten and Their Symbols • Table C5, The Greek Alphabet • Table C6, SI units • Table C7, Selected British Units • Table C8, Other Units • Table C9, Useful Formulae Table C1 Important Constants [1] Symbol Meaning Best Value Approximate Value σ ε0 μ0 Speed of light in vacuum Gravitational constant Avogadro’s number Boltzmann’s constant Gas constant StefanBoltzmann constant Coulomb force constant Charge on electron Permittivity of free space Permeability of free space Planck’s constant 2.99792458 × 108 m / s 3.00 × 108 m / s 6.67384(80) × 10−11 N ⋅ m2 / kg2 6.67 × 10−11 N ⋅ m2 / kg2 6.02214129(27) × 1023 6.02 × 1023 1.3806488(13) × 10−23 J / K 1.38 × 10−23 J / K 8.3144621(75) J / mol ⋅ K 8.31 J / mol ⋅ K = 1.99 cal / mol ⋅ K = 0.0821atm ⋅ L / mol ⋅ K 5.670373(21) × 10−8 W / m2 ⋅ K 5.67 × 10−8 W / m2 ⋅ K 8.987551788... × 109 N ⋅ m2 / C2 8.99 × 109 N ⋅ m2 / C2 −1.602176565(35) × 10−19 C −1.60 × 10−19 C 8.854187817... × 10−12 C2 / N ⋅ m2 8.85 × 10−12 C2 / N ⋅ m2 4π × 10−7 T ⋅ m / A 1.26 × 10−6 T ⋅ m / A 6.62606957(29) × 10−34 J ⋅ s 6.63 × 10−34 J ⋅ s Table C2 Submicroscopic Masses [2] Symbol Meaning Best Value Approximate Value Electron mass 9.10938291(40)×10−31kg 9.11×10−31kg Proton mass 1.672621777(74)×10−27kg 1.6726×10−27kg 1. Stated values are according to the National Institute of Standards and Technology Reference on Constants, Units, and Uncertainty, www.physics.nist.gov/cuu (http://www.physics.nist.gov/cuu) (accessed May 18, 2012). Values in parentheses are the uncertainties in the last digits. Numbers without uncertainties are exact as defined. 2. Stated values are according to the National Institute of Standards and Technology Reference on Constants, Units, and Uncertainty, www.physics.nist.gov/cuu (http://www.physics.nist.gov/cuu) (accessed May 18, 2012). Values in parentheses are the uncertainties in the last digits. Numbers without uncertainties are exact as defined. 1542 Appendix C Symbol Meaning Best Value Approximate Value u Neutron mass 1.674927351(74)×10−27kg 1.6749×10−27kg Atomic mass unit 1.660538921(73)×10−27kg 1.6605×10−27kg Table C3 Solar System Data Sun mass average radius Earth-sun distance (average) Earth mass average radius orbital period Moon mass average radius orbital period (average) 1.99×1030kg 6.96×108m 1.496×1011m 5.9736×1024kg 6.376×106m 3.16×107s 7.35×1022kg 1.74×106m 2.36×106s Earth-moon distance (average) 3.84×108m Table C4 Metric Prefixes for Powers of Ten and Their Symbols Prefix Symbol Value Prefix Symbol Value tera giga mega kilo hecto deka T G M k h da 1012 109 106 103 102 101 deci centi milli micro nano pico — — 100( = 1) femto d c m n p f 10−1 10−2 10−3 10−6 10−9 10−12 10−15 Table C5 The Greek Alphabet Alpha Α Eta Η Nu Ν Tau Τ Beta Β Theta Θ Xi Ξ Upsilon Υ Gamma Γ Iota Ι Omicron Ο Phi Φ Delta Δ Kappa Κ Pi Epsilon Ε Lambda Λ Rho Π Chi Ρ Psi Χ Ψ Zeta Ζ Mu Μ Sigma Σ Omega Ω This content is available for free at http://cnx.org/content/col11844/1.13 Appendix C 1543 Table C6 SI Units Entity Abbreviation Name Fundamental units Length Mass Time Current Supplementary unit Angle Derived units Force Energy Power Pressure m kg s A rad meter kilogram second ampere radian N = kg ⋅ m / s2 newton J = kg ⋅ m2 / s2 joule W = J / s Pa = N / m2 watt pascal hertz volt farad coulomb Frequency Hz = 1 / s Electronic potential V = J / C Capacitance Charge Resistance Ω = V / A ohm Magnetic field T = N / (A ⋅ m) tesla Nuclear decay rate Bq = 1 / s becquerel Table C7 Selected British Units Length 1 inch (in.) = 2.54 cm (exactly) 1 foot (ft) = 0.3048 m 1 mile (mi) = 1.609 km Force 1 pound (lb) = 4.448 N Energy 1 British thermal unit (Btu) = 1.055×103 J Power 1 horsepower (hp) = 746 W Pressure 1 lb / in2 = 6.895×103 Pa Table C8 Other Units Length 1 light year (ly) = 9.46×1015 m 1 astronomical unit (au) = 1.50×1011 m 1 nautical mile = 1.852 km 1 angstrom(Å) = 10−10 m Area 1 acre (ac) = 4.05×103 m2 1 square foot (ft2) = 9.29×10−2 m2 1 barn () = 10−28 m2 Volume 1 liter () = 10−3 m3 1544 Appendix C 1 U.S. gallon (gal) = 3.785×10−3 m3 Mass 1 solar mass = 1.99×1030 kg Time Speed Angle 1 metric ton = 103 kg 1 atomic mass unit () = 1.6605×10−27 kg 1 year () = 3.16×107 s 1 day () = 86400 s 1 mile per hour (mph) = 1.609 km / h 1 nautical mile per hour (naut) = 1.852 km / h 1 degree () = 1.745×10−2 rad 1 minute of arc (') = 1 / 60 degree 1 second of arc ('') = 1 / 60 minute of arc 1 grad = 1.571×10−2 rad Energy 1 kiloton TNT (kT) = 4.2×1012 J 1 kilowatt hour (kW ⋅ ) = 3.60×106 J 1 food calorie (kcal) = 4186 J 1 calorie (cal) = 4.186 J 1 electron volt (eV) = 1.60×10−19 J Pressure 1 atmosphere (atm) = 1.013×105 Pa 1 millimeter of mercury (mm Hg) = 133.3 Pa 1 torricelli (torr) = 1 mm Hg = 133.3 Pa Nuclear decay rate 1 curie (Ci) = 3.70×1010 Bq Table C9 Useful Formulae Circumference of a circle with radius or diamete
r = 4 / 3) 3 Area of a circle with radius or diameter Area of a sphere with radius Volume of a sphere with radius This content is available for free at http://cnx.org/content/col11844/1.13 Appendix D 1545 D GLOSSARY OF KEY SYMBOLS AND NOTATION In this glossary, key symbols and notation are briefly defined. Table D1 Symbol Definition any symbol ¯ average (indicated by a bar over a symbol—e.g., is average velocity) C F // ⊥ ∝ ± 0 − + Celsius degree Fahrenheit degree parallel perpendicular proportional to plus or minus zero as a subscript denotes an initial value alpha rays angular acceleration temperature coefficient(s) of resistivity beta rays sound level volume coefficient of expansion electron emitted in nuclear beta decay positron decay gamma rays surface tension = 1 / 1 − 2 / 2 a constant used in relativity Δ change in whatever quantity follows uncertainty in whatever quantity follows change in energy between the initial and final orbits of an electron in an atom uncertainty in energy difference in mass between initial and final products number of decays that occur change in momentum uncertainty in momentum PEg change in gravitational potential energy 1546 Appendix D Symbol Definition 0 0 0 k s + − 0 c fl ¯ obj / w Υ rotation angle distance traveled along a circular path uncertainty in time proper time as measured by an observer at rest relative to the process potential difference uncertainty in position permittivity of free space viscosity angle between the force vector and the displacement vector angle between two lines contact angle direction of the resultant Brewster's angle critical angle dielectric constant decay constant of a nuclide wavelength wavelength in a medium permeability of free space coefficient of kinetic friction coefficient of static friction electron neutrino positive pion negative pion neutral pion density critical density, the density needed to just halt universal expansion fluid density average density of an object specific gravity characteristic time constant for a resistance and inductance () or resistance and capacitance () circuit characteristic time for a resistor and capacitor () circuit torque upsilon meson This content is available for free at http://cnx.org/content/col11844/1.13 Appendix D 1547 Symbol Definition Φ Ω A B c t AC AM atm ¯ Bint Borb BE BE / Bq p s CG CM magnetic flux phase angle ohm (unit) angular velocity ampere (current unit) area cross-sectional area total number of nucleons acceleration Bohr radius centripetal acceleration tangential acceleration alternating current amplitude modulation atmosphere baryon number blue quark color antiblue (yellow) antiquark color quark flavor bottom or beauty bulk modulus magnetic field strength electron’s intrinsic magnetic field orbital magnetic field binding energy of a nucleus—it is the energy required to completely disassemble it into separate protons and neutrons binding energy per nucleon becquerel—one decay per second capacitance (amount of charge stored per volt) coulomb (a fundamental SI unit of charge) total capacitance in parallel total capacitance in series center of gravity center of mass quark flavor charm specific heat speed of light 1548 Appendix D Symbol Definition Cal cal hp ref cos cot csc dB i o DC emf 0 0 EC cap in ind out + eV F F kilocalorie calorie heat pump’s coefficient of performance coefficient of performance for refrigerators and air conditioners cosine cotangent cosecant diffusion constant displacement quark flavor down decibel distance of an image from the center of a lens distance of an object from the center of a lens direct current electric field strength emf (voltage) or Hall electromotive force electromotive force energy of a single photon nuclear reaction energy relativistic total energy total energy ground state energy for hydrogen rest energy electron capture energy stored in a capacitor efficiency—the useful work output divided by the energy input Carnot efficiency energy consumed (food digested in humans) energy stored in an inductor energy output emissivity of an object antielectron or positron electron volt farad (unit of capacitance, a coulomb per volt) focal point of a lens This content is available for free at http://cnx.org/content/col11844/1.13 Appendix D 1549 Symbol Definition F B c i Fnet o FM force magnitude of a force restoring force buoyant force centripetal force force input net force force output frequency modulation focal length frequency resonant frequency of a resistance, inductance, and capacitance () series circuit threshold frequency for a particular material (photoelectric effect) fundamental first overtone second overtone beat frequency magnitude of kinetic friction magnitude of static friction gravitational constant green quark color antigreen (magenta) antiquark color acceleration due to gravity gluons (carrier particles for strong nuclear force) change in vertical position height above some reference point maximum height of a projectile Planck's constant photon energy height of the image height of the object electric current intensity intensity of a transmitted wave 1550 Appendix D Symbol Definition 0 ave rms J / Ψ K kcal KE moment of inertia (also called rotational inertia) intensity of a polarized wave before passing through a filter average intensity for a continuous sinusoidal electromagnetic wave average current joule Joules/psi meson kelvin Boltzmann constant force constant of a spring x rays created when an electron falls into an = 1 shell vacancy from the = 3 shell x rays created when an electron falls into an = 2 shell vacancy from the = 3 shell kilocalorie translational kinetic energy KE + PE mechanical energy KE KErel KErot KE kg L f kinetic energy of an ejected electron relativistic kinetic energy rotational kinetic energy thermal energy kilogram (a fundamental SI unit of mass) angular momentum liter magnitude of angular momentum self-inductance angular momentum quantum number x rays created when an electron falls into an = 2 shell from the = 3 shell electron total family number muon family total number tau family total number heat of fusion f and v latent heat coefficients Lorb s v orbital angular momentum heat of sublimation heat of vaporization z - component of the angular momentum This content is available for free at http://cnx.org/content/col11844/1.13 Appendix D 1551 Symbol Definition m m X MA e o mol OE p angular magnification mutual inductance indicates metastable state magnification mass mass of an object as measured by a person at rest relative to the object meter (a fundamental SI unit of length) order of interference overall magnification (product of the individual magnifications) atomic mass of a nuclide mechanical advantage magnification of the eyepiece mass of the electron angular momentum projection quantum number mass of a neutron magnification of the objective lens mole mass of a proton spin projection quantum number magnitude of the normal force newton normal force number of neutrons index of refraction number of free charges per unit volume Avogadro's number Reynolds number newton-meter (work-energy unit) newtons times meters (SI unit of torque) other energy power power of a lens pressure momentum momentum magnitude relativistic momentum 1552 Appendix D Symbol ptot Definition total momentum ptot abs atm atm PE PEel PEelec PEs g in out + − QF ¯ R ⊥ total momentum some time later absolute pressure atmospheric pressure standard atmospheric pressure potential energy elastic potential energy electric potential energy potential energy of a spring gauge pressure power consumption or input useful power output going into useful work or a desired, form of energy latent heat net heat transferred into a system flow rate—volume per unit time flowing past a point positive charge negative charge electron charge charge of a proton test charge quality factor activity, the rate of decay radius of curvature of a spherical mirror red quark color antired (cyan) quark color resistance resultant or total displacement Rydberg constant universal gas constant distance from pivot point to the point where a force is applied internal resistance perpendicular lever arm radius of a nucleus radius of curvature This content is available for free at http://cnx.org/content/col11844/1.13 Appendix D 1553 Symbol Definition r or rad rem rad RBE rms p s s S s s sec sin c T 1 / 2 tan u u resistivity radiation dose unit roentgen equivalent man radian relative biological effectiveness resistor and capacitor circuit root mean square radius of the nth H-atom orbit total resistance of a parallel connection total resistance of a series connection Schwarzschild radius entropy intrinsic spin (intrinsic angular momentum) magnitude of the intrinsic (internal) spin angular momentum shear modulus strangeness quantum number quark flavor strange second (fundamental SI unit of time) spin quantum number total displacement secant sine z-component of spin angular momentum period—time to complete one oscillation temperature critical temperature—temperature below which a material becomes a superconductor tension tesla (magnetic field strength B) quark flavor top or truth time half-life—the time in which half of the original nuclei decay tangent internal energy quark flavor up unified atomic mass unit velocity of an object relative to an observer 1554 Appendix D Symbol Definition u' V v v v¯ velocity relative to another observer electric potential terminal voltage volt (unit) volume relative velocity between two observers speed of light in a material velocity average fluid velocity B − A change in potential vd p rms s vtot w vw W fl c nc out X C L rms drift velocity transformer input voltage rms voltage transformer output voltage total velocity propagation speed of sound or other wave wave velocity work net work done by a system watt weight weight of the fluid displaced by an object total work done by all conservative forces total work done by all nonconservative forces useful work output
amplitude symbol for an element notation for a particular nuclide deformation or displacement from equilibrium displacement of a spring from its undeformed position horizontal axis capacitive reactance inductive reactance root mean square diffusion distance vertical axis This content is available for free at http://cnx.org/content/col11844/1.13 Appendix D 1555 Symbol Definition elastic modulus or Young's modulus atomic number (number of protons in a nucleus) impedance 1556 Appendix D This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key 1557 ANSWER KEY Chapter 1 Problems & Exercises 1 a. 27.8 m/s b. 62.1 mph 3 s × 3600 s 1 hr 1.0 m s = 1.0 m = 3.6 km/h . × 1 km 1000 m 5 length: 377 ft ; 4.53×103 in. width: 280 ft ; 3.3×103 in . 7 8.847 km 9 (a) 1.3×10−9 m (b) 40 km/My 11 2 kg 13 a. 85.5 to 94.5 km/h b. 53.1 to 58.7 mi/h 15 (a) 7.6×107 beats (b) 7.57×107 beats (c) 7.57×107 beats 17 a. 3 b. 3 c. 3 19 a) 2.2% (b) 59 to 61 km/h 21 80 ± 3 beats/min 23 2.8 h 25 11 ± 1 cm3 27 12.06 ± 0.04 m2 1558 Answer Key 29 Sample answer: 2×109 heartbeats 31 Sample answer: 2×1031 if an average human lifetime is taken to be about 70 years. 33 Sample answer: 50 atoms 35 Sample answers: (a) 1012 cells/hummingbird (b) 1016 cells/human Chapter 2 Problems & Exercises 1 (a) 7 m (b) 7 m (c) +7 m 3 (a) 13 m (b) 9 m (c) +9 m 5 (a) 3.0×104 m/s (b) 0 m/s 7 2×107 years 9 34.689 m/s = 124.88 km/h 11 (a) 40.0 km/h (b) 34.3 km/h, 25º S of E. (c) average speed = 3.20 km/h, - = 0. 13 384,000 km 15 (a) 6.61×1015 rev/s (b) 0 m/s 16 4.29 m/s2 18 (a) 1.43 s (b) −2.50 m/s2 20 (a) 10.8 m/s This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key (b) 1559 Figure 2.48. 21 38.9 m/s (about 87 miles per hour) 23 (a) 16.5 s (b) 13.5 s (c) −2.68 m/s2 25 (a) 20.0 m (b) −1.00 m/s (c) This result does not really make sense. If the runner starts at 9.00 m/s and decelerates at 2.00 m/s2 , then she will have stopped after 4.50 s. If she continues to decelerate, she will be running backwards. 27 0.799 m 29 (a) 28.0 m/s (b) 50.9 s (c) 7.68 km to accelerate and 713 m to decelerate 31 (a) 51.4 m (b) 17.1 s 33 (a) −80.4 m/s2 (b) 9.33×10−2 s 35 (a) 7.7 m/s (b) −15×102 m/s2 . This is about 3 times the deceleration of the pilots, who were falling from thousands of meters high! 37 (a) 32.6 m/s2 (b) 162 m/s (c) > max , because the assumption of constant acceleration is not valid for a dragster. A dragster changes gears, and would have a greater acceleration in first gear than second gear than third gear, etc. The acceleration would be 1560 Answer Key greatest at the beginning, so it would not be accelerating at 32.6 m/s2 during the last few meters, but substantially less, and the final velocity would be less than 162 m/s. 39 104 s 40 (a) = 12.2 m/s ; = 4.07 m/s2 (b) = 11.2 m/s 41 (a) 1 = 6.28 m ; 1 = 10.1 m/s (b) 2 = 10.1 m ; 2 = 5.20 m/s (c) 3 = 11.5 m ; 3 = 0.300 m/s (d) 4 = 10.4 m ; 4 = −4.60 m/s 43 0 = 4.95 m/s 45 (a) = −9.80 m/s2 ; 0 = 13.0 m/s ; 0 = 0 m (b) = 0m/s . Unknown is distance to top of trajectory, where velocity is zero. Use equation 2 = 0 because it contains all known values except for , so we can solve for . Solving for gives 2 + 2( − 0) 2 = 2( − 00 m/s)2 − (13.0 m/s)2 −9.80 m/s2 2 = 8.62 m (2.100) Dolphins measure about 2 meters long and can jump several times their length out of the water, so this is a reasonable result. (c) 2.65 s 47 Figure 2.57. (a) 8.26 m (b) 0.717 s 49 1.91 s 51 (a) 94.0 m (b) 3.13 s This content is available for free at http://cnx.org/content/col11844/1.13 1561 = (11.7 − 6.95)×103 m (40.0 – 20.0) s = 238 m/s (2.114) Answer Key 53 (a) -70.0 m/s (downward) (b) 6.10 s 55 (a) 19.6 m (b) 18.5 m 57 (a) 305 m (b) 262 m, -29.2 m/s (c) 8.91 s 59 (a) 115 m/s (b) 5.0 m/s2 61 63 Figure 2.63. 65 (a) 6 m/s (b) 12 m/s (c) 3 m/s2 (d) 10 s Test Prep for AP® Courses 1 (a) 3 a. Use tape to mark off two distances on the track — one for cart A before the collision and one for the combined carts after the collision. Push cart A to give it an initial speed. Use a stopwatch to measure the time it takes for the cart(s) to cross the marked distances. The speeds are the distances divided by the times. If the measurement errors are of the same magnitude, they will have a greater effect after the collision. The speed of the combined carts will be less than the initial speed of cart A. As a result, these errors will be a greater percentage of the actual velocity value after the collision occurs. (Note: Other arguments could properly be made for ‘more error before the collision' and error that ‘equally affects both sets of measurement.') b. 5 1562 Answer Key The position vs. time graph should be represented with a positively sloped line whose slope steadily decreases to zero. The y-intercept of the graph may be any value. The line on the velocity vs. time graph should have a positive yintercept and a negative slope. Because the final velocity of the book is zero, the line should finish on the x-axis. The position vs. time graph should be represented with a negatively sloped line whose slope steadily decreases to zero. The y-intercept of the graph may be any value. The line on the velocity vs. time graph should have a negative yintercept and a positive slope. Because the final velocity of the book is zero, the line should finish on the x-axis.] 7 (c) Chapter 3 Problems & Exercises 1 (a) 480 m (b) 379 m , 18.4° east of north 3 north component 3.21 km, east component 3.83 km 5 19.5 m , 4.65° south of west 7 (a) 26.6 m , 65.1° north of east (b) 26.6 m , 65.1° south of west 9 52.9 m , 90.1° with respect to the x-axis. 11 x-component 4.41 m/s y-component 5.07 m/s 13 (a) 1.56 km (b) 120 m east 15 North-component 87.0 km, east-component 87.0 km 17 30.8 m, 35.8 west of north 19 (a) 30.8 m , 54.2º south of west (b) 30.8 m , 54.2º north of east 21 18.4 km south, then 26.2 km west(b) 31.5 km at 45.0º south of west, then 5.56 km at 45.0º west of north 23 7.34 km , 63.5º south of east 25 = 1.30 m×102 = 30.9 m. 27 (a) 3.50 s (b) 28.6 m/s (c) 34.3 m/s (d) 44.7 m/s, 50.2° below horizontal 29 This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key (a) 18.4° (b) The arrow will go over the branch. 31 = 0 sin2θ0 For = 45°, = 0 1563 = 91.8 m for 0 = 30 m/s ; = 163 m for 0 = 40 m/s ; = 255 m for 0 = 50 m/s . 33 (a) 560 m/s (b) 8.00×103 m (c) 80.0 m. This error is not significant because it is only 1% of the answer in part (b). 35 1.50 m, assuming launch angle of 45° 37 = 6.1° yes, the ball lands at 5.3 m from the net 39 (a) −0.486 m (b) The larger the muzzle velocity, the smaller the deviation in the vertical direction, because the time of flight would be smaller. Air resistance would have the effect of decreasing the time of flight, therefore increasing the vertical deviation. 41 4.23 m. No, the owl is not lucky; he misses the nest. 43 No, the maximum range (neglecting air resistance) is about 92 m. 45 15.0 m/s 47 (a) 24.2 m/s (b) The ball travels a total of 57.4 m with the brief gust of wind. 49 − 0 = 0 = 0 − 1 22 = (0 sin ) − 1 22 , so that = 2(0 sin ) − 0 = 0 = (0 cos ) = and substituting for gives: = 0 cos 20 sin = 20 2 sin cos since 2 sin cos = sin 2θ the range is: = 0 2 sin 2θ . 52 (a) 35.8 km , 45º south of east (b) 5.53 m/s , 45º south of east 1564 Answer Key (c) 56.1 km , 45º south of east 54 (a) 0.70 m/s faster (b) Second runner wins (c) 4.17 m 56 17.0 m/s , 22.1º 58 (a) 230 m/s , 8.0º south of west (b) The wind should make the plane travel slower and more to the south, which is what was calculated. 60 (a) 63.5 m/s (b) 29.6 m/s 62 6.68 m/s , 53.3º south of west 64 (a) average = 14.9km/s Mly (b) 20.2 billion years 66 1.72 m/s , 42.3º north of east Test Prep for AP® Courses 1 (d) 3 We would need to know the horizontal and vertical positions of each ball at several times. From that data, we could deduce the velocities over several time intervals and also the accelerations (both horizontal and vertical) for each ball over several time intervals. 5 The graph of the ball's vertical velocity over time should begin at 4.90 m/s during the time interval 0 - 0.1 sec (there should be a data point at t = 0.05 sec, v = 4.90 m/s). It should then have a slope of -9.8 m/s2, crossing through v = 0 at t = 0.55 sec and ending at v = -0.98 m/s at t = 0.65 sec. The graph of the ball's horizontal velocity would be a constant positive value, a flat horizontal line at some positive velocity from t = 0 until t = 0.7 sec. Chapter 4 Problems & Exercises 1 265 N 3 13.3 m/s2 7 (a) 12 m/s2 . (b) The acceleration is not one-fourth of what it was with all rockets burning because the frictional force is still as large as it was with all rockets burning. 9 (a) The system is the child in the wagon plus the wagon. (b This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key 1565 Figure 4.10. (c) = 0.130 m/s2 in the direction of the second child’s push. (d) = 0.00 m/s2 11 (a) 3.68×103 N . This force is 5.00 times greater than his weight. (b) 3750 N; 11.3º above horizontal 13 1.5×103 N, 150 kg, 150 kg 15 Force on shell: 2.64×107 N Force exerted on ship = −2.64×107 N , by Newton’s third law 17 a. b. 0.11 m/s2 1.2×104 N 19 (a) 7.84×10-4 N (b) 1.89×10–3 N . This is 2.41 times the tension in the vertical strand. 21 Newton’s second law applied in vertical direction gives = − 2 sin = 0 = 2 sin = . 2 sin () () () 23 1566 Answer Key Figure 4.26. Using the free-body diagram: net = − − = , so that = − − = 1.250×107 N − 4.50×106 − (5.00×105 kg)(9.80 m/s2) 5.00×105 kg = 6.20 m/s2 . 25 1. Use Newton’s laws of motion. Figure 4.26. 2. Given : = 4.00 = (4.00)(9.80 m/s2 ) = 39.2 m/s2 ; = 70.0 kg , Find: . 3. ∑ =+ − = , so that = + = + = ( + ) . = (70.0 kg)[(39.2 m/s2 ) + (9.80 m/s2)] down on the ground, but is up from the ground and makes him jump. = 3.43×103N . The force exerted by the high-jumper is actually 4. This result is reasonable, since it is quite possible for a person to exert a force of the magnitude o
f 103 N . This content is available for free at http://cnx.org/content/col11844/1.13 1567 Answer Key 27 (a) 4.41×105 N (b) 1.50×105 N 29 (a) 910 N (b) 1.11×103 N 31 = 0.139 m/s , = 12.4º north of east 33 1. Use Newton’s laws since we are looking for forces. 2. Draw a free-body diagram: Figure 4.29. 3. The tension is given as = 25.0 N. Find app . Using Newton’s laws gives: = 0, so that applied force is due to the y-components of the two tensions: = 2 sin = 2(25.0 N)sin The x-components of the tension cancel. ∑ = 0 . 15º = 12.9 N 4. This seems reasonable, since the applied tensions should be greater than the force applied to the tooth. 40 10.2 m/s2 , 4.67º from vertical 42 1568 Answer Key Figure 4.35. 1 = 736 N 2 = 194 N 44 (a) 7.43 m/s (b) 2.97 m 46 (a) 4.20 m/s (b) 29.4 m/s2 (c) 4.31×103 N 48 (a) 47.1 m/s (b) 2.47×103 m/s2 (c) 6.18×103 N . The average force is 252 times the shell’s weight. 52 (a) 1×10−13 (b) 1×10−11 54 102 Test Prep for AP® Courses 1 This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key 1569 Figure 4.4. Car X is shown on the left, and Car Y is shown on the right. i. Car X takes longer to accelerate and does not spend any time traveling at top speed. Car Y accelerates over a shorter time and spends time going at top speed. So Car Y must cover the straightaways in a shorter time. Curves take the same time, so Car Y must overall take a shorter time. ii. The only difference in the calculations for the time of one segment of linear acceleration is the difference in distances. That shows that Car X takes longer to accelerate. The equation = corresponds to Car Y traveling for a time at 4 top speed. Substituting = 1 into the displacement equation in part (b) ii gives = 3 2 1 . This shows that a car takes less time to reach its maximum speed when it accelerates over a shorter distance. Therefore, Car Y reaches its maximum speed more quickly, and spends more time at its maximum speed than Car X does, as argued in part (b) i. 3 A body cannot exert a force on itself. The hawk may accelerate as a result of several forces. The hawk may accelerate toward Earth as a result of the force due to gravity. The hawk may accelerate as a result of the additional force exerted on it by wind. The hawk may accelerate as a result of orienting its body to create less air resistance, thus increasing the net force forward. 5 (a) A soccer player, gravity, air, and friction commonly exert forces on a soccer ball being kicked. (b) Gravity and the surrounding water commonly exert forces on a dolphin jumping. (The dolphin moves its muscles to exert a force on the water. The water exerts an equal force on the dolphin, resulting in the dolphin’s motion.) (c) Gravity and air exert forces on a parachutist drifting to Earth. 7 (c) 9 Figure 4.14. The diagram consists of a black dot in the center and two small red arrows pointing up (Fb) and down (Fg) and two long red arrows pointing right (Fc = 9.0 N) and left (Fw=13.0 N). In the diagram, Fg represents the force due to gravity on the balloon, and Fb represents the buoyant force. These two forces are equal in magnitude and opposite in direction. Fc represents the force of the current. Fw represents the force of the wind. The net force on the balloon will be − = 4.0 N and the balloon will accelerate in the direction the wind is blowing. 11 1570 Answer Key Since = / , the parachutist has a mass of 539 N/9.8 km/s2 = 55 kg . For the first 2 s, the parachutist accelerates at 9.8 m/s2. • 2s = = 9.8 m s2 = 17.6m s Her speed after 2 s is 19.6 m/s. From 2 s to 10 s, the net force on the parachutist is 539 N – 615 N, or 76 N upward. = = −76 N 55 kg = −1.4 m s2 Since = 0 + , = 17.6 m/s2 + ( − 1.4 m/s2)(8s) = 6.5 m/s2 . At 10 s, the parachutist is falling to Earth at 8.4 m/s. 13 The system includes the gardener and the wheelbarrow with its contents. The following forces are important to include: the weight of the wheelbarrow, the weight of the gardener, the normal force for the wheelbarrow and the gardener, the force of the gardener pushing against the ground and the equal force of the ground pushing back against the gardener, and any friction in the wheelbarrow’s wheels. 15 The system undergoing acceleration is the two figure skaters together. Net force = 120 N – 5.0 N = 115 N . Total mass = 40 kg + 50 kg = 90 kg . Using Newton’s second law, we have that = = 115 N 90 kg = 1.28 m s2 The pair accelerates forward at 1.28 m/s2. 17 The force of tension must equal the force of gravity plus the force necessary to accelerate the mass. = can be used to calculate the first, and = can be used to calculate the second. For gravity: = = (120.0 kg)(9.8 m/s2) = 1205.4 N For acceleration: = = (120.0 kg)(1.3 m/s2) = 159.9 N The total force of tension in the cable is 1176 N + 156 N = 1332 N. 19 (b) This content is available for free at http://cnx.org/content/col11844/1.13 1571 Answer Key 21 Figure 4.24. The diagram has a black dot and three solid red arrows pointing away from the dot. Arrow Ft is long and pointing to the left and slightly down. Arrow Fw is also long and is a bit below a diagonal line halfway between pointing up and pointing to the right. A short arrow Fg is pointing down. Fg is the force on the kite due to gravity. Fw is the force exerted on the kite by the wind. Ft is the force of tension in the string holding the kite. It must balance the vector sum of the other two forces for the kite to float stationary in the air. 23 (b) 25 (d) 27 A free-body diagram would show a northward force of 64 N and a westward force of 38 N. The net force is equal to the sum of the two applied forces. It can be found using the Pythagorean theorem: net = 2 + = (38 N)2 + (64 N)2 = 74.4 N Since = , = 74.4 N 825 kg = 0.09 m/s2 The boulder will accelerate at 0.09 m/s2. 29 (b) 31 (b) 33 (d) Chapter 5 Problems & Exercises 1 5.00 N 4 (a) 588 N (b) 1.96 m/s2 Answer Key (5.30) (5.58) (5.59) 1572 6 (a) 3.29 m/s2 (b) 3.52 m/s2 (c) 980 N; 945 N 10 1.83 m/s2 14 (a) 4.20 m/s2 (b) 2.74 m/s2 (c) –0.195 m/s2 16 (a) 1.03×106 N (b) 3.48×105 N 18 (a) 51.0 N (b) 0.720 m/s2 20 115 m/s; 414 km/hr 22 25 m/s; 9.9 m/s 24 2.9 26 28 0.76 kg/m ⋅ s 29 [] = s [][] = kg ⋅ m/s2 m ⋅ m/s = kg m ⋅ s 1.90×10−3 cm 31 (a)1 mm (b) This does seem reasonable, since the lead does seem to shrink a little when you push on it. 33 (a)9 cm (b)This seems reasonable for nylon climbing rope, since it is not supposed to stretch that much. 35 8.59 mm 37 1.49×10−7 m 39 (a) 3.99×10−7 m (b) 9.67×10−8 m 41 4×106 N/m2 . This is about 36 atm, greater than a typical jar can withstand. 43 1.4 cm This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key 1573 Test Prep for AP® Courses 1 (b) 3 (c) Chapter 6 Problems & Exercises 1 723 km 3 5×107 rotations 5 117 rad/s 7 76.2 rad/s 728 rpm 8 (a) 33.3 rad/s (b) 500 N (c) 40.8 m 10 12.9 rev/min 12 4×1021 m 14 a) 3.47×104 m / s2 , 3.55×103 b) 51.1 m / s 16 a) 31.4 rad/s b) 118 m/s c) 384 m/s d)The centripetal acceleration felt by Olympic skaters is 12 times larger than the acceleration due to gravity. That's quite a lot of acceleration in itself. The centripetal acceleration felt by Button's nose was 39.2 times larger than the acceleration due to gravity. It is no wonder that he ruptured small blood vessels in his spins. 18 a) 0.524 km/s b) 29.7 km/s 20 (a) 1.35×103 rpm (b) 8.47×103 m/s2 (c) 8.47×10–12 N (d) 865 21 (a) 16.6 m/s Answer Key 1574 (b) 19.6 m / s2 (c) Figure 6.10. (d) 1.76×103 N or 3.00 , that is, the normal force (upward) is three times her weight. (e) This answer seems reasonable, since she feels like she's being forced into the chair MUCH stronger than just by gravity. 22 a) 40.5 m / s2 b) 905 N c) The force in part (b) is very large. The acceleration in part (a) is too much, about 4 g. d) The speed of the swing is too large. At the given velocity at the bottom of the swing, there is enough kinetic energy to send the child all the way over the top, ignoring friction. 23 a) 483 N b) 17.4 N c) 2.24 times her weight, 0.0807 times her weight 25 4.14º 27 a) 24.6 m b) 36.6 m / s2 c) c = 3.73 This does not seem too large, but it is clear that bobsledders feel a lot of force on them going through sharply banked turns. 29 a) 2.56 rad/s b) 5.71º 30 a) 16.2 m/s b) 0.234 32 a) 1.84 b) A coefficient of friction this much greater than 1 is unreasonable . c) The assumed speed is too great for the tight curve. 33 a) 5.979×1024 kg b) This is identical to the best value to three significant figures. This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key 35 a) 1.62 m / s2 b) 3.75 m / s2 37 a) 3.42×10–5 m / s2 b) 3.34×10–5 m / s2 1575 The values are nearly identical. One would expect the gravitational force to be the same as the centripetal force at the core of the system. 39 a) 7.01×10–7 N b) 1.35×10–6 N , 0.521 41 a) 1.66×10–10 m / s2 b) 2.17×105 m/s 42 a) 2.94×1017 kg b) 4.92×10–8 of the Earth's mass. c) The mass of the mountain and its fraction of the Earth's mass are too great. d) The gravitational force assumed to be exerted by the mountain is too great. 44 1.98×1030 kg 46 48 a) 7.4×103 m/s = 316 b) 1.05×103 m/s c) 2.86×10−7 s d) 1.84×107 N e) 2.76×104 J 49 a) 5.08×103 km b) This radius is unreasonable because it is less than the radius of earth. c) The premise of a one-hour orbit is inconsistent with the known radius of the earth. Test Prep for AP® Courses 1 (a) 3 (b) Answer Key (7.8) (7.9) 3.00 J = 7.17×10−4 kcal 3.14×103 J 1576 5 (b) Chapter 7 Problems & Exercises 1 3 (a) 5.92×105 J (b) −5.88×105 J (c) The net force is zero. 5 7 (a) −700 J (b) 0 (c) 700 J (d) 38.6 N (e) 0 9 1 / 250 11 1.1×1010 J 13 2.8×103 N 15 102 N 16 (a) 1.961016 J (b) The ratio of gravitational potential energy in the lake to the energy stored in the bomb is 0.52. That is, the energy stored in the lake is approximately half that in a 9-megaton fusion bomb. 18 (a) 1.8 J (b) 8.6 J 20 22 = 2 + 0 2 = 2(9.80 m/s2)( − 0.180 m) + (2.00 m/s)2 = 0.687 m/s 7.81105 N/m (7.45) (7.60) 2
4 9.46 m/s 26 4104 molecules 27 Equating ΔPEg and ΔKE , we obtain = 2 + 0 29 (a) 25×106 years 2 = 2(9.80 m/s2)(20.0 m) + (15.0 m/s)2 = 24.8 m/s This content is available for free at http://cnx.org/content/col11844/1.13 1577 (7.81) Answer Key (b) This is much, much longer than human time scales. 210−10 30 32 (a) 40 (b) 8 million 34 $149 36 (a) 208 W (b) 141 s 38 (a) 3.20 s (b) 4.04 s 40 (a) 9.46107 J (b) 2.54 y 42 Identify knowns: = 950 kg , slope angle = 2.00º , = 3.00 m/s , = 600 N Identify unknowns: power of the car, force that car applies to road Solve for unknown: = = = = where is parallel to the incline and must oppose the resistive forces and the force of gravity: = + = 600 N + sin Insert this into the expression for power and solve: = = + sin 600 N + = 2.77×104 W 950 kg 9.80 m/s2 sin 2º (30.0 m/s) About 28 kW (or about 37 hp) is reasonable for a car to climb a gentle incline. 44 (a) 9.5 min (b) 69 flights of stairs 46 641 W, 0.860 hp 48 31 g 50 14.3% 52 (a) 3.21104 N (b) 2.35103 N (c) Ratio of net force to weight of person is 41.0 in part (a); 3.00 in part (b) 54 Answer Key 1578 (a) 108 kJ (b) 599 W 56 (a) 144 J (b) 288 W 58 (a) 2.501012 J (b) 2.52% (c) 1.4104 kg (14 metric tons) 60 (a) 294 N (b) 118 J (c) 49.0 W 62 (a) 0.500 m/s2 (b) 62.5 N (c) Assuming the acceleration of the swimmer decreases linearly with time over the 5.00 s interval, the frictional force must therefore be increasing linearly with time, since = − . If the acceleration decreases linearly with time, the velocity will contain a term dependent on time squared ( 2 ). Therefore, the water resistance will not depend linearly on the velocity. 64 (a) 16.1×103 N (b) 3.22×105 J (c) 5.66 m/s (d) 4.00 kJ 66 (a) 4.65×103 kcal (b) 38.8 kcal/min (c) This power output is higher than the highest value on Table 7.5, which is about 35 kcal/min (corresponding to 2415 watts) for sprinting. (d) It would be impossible to maintain this power output for 2 hours (imagine sprinting for 2 hours!). 69 (a) 4.32 m/s (b) 3.47×103 N (c) 8.93 kW Test Prep for AP® Courses 1 (b) 3 (d) 5 (a) 7 This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key 1579 The kinetic energy should change in the form of –cos, with an initial value of 0 or slightly above, and ending at the same level. 9 Any force acting perpendicular will have no effect on kinetic energy. Obvious examples are gravity and the normal force, but others include wind directly from the side and rain or other precipitation falling straight down. 11 Note that the wind is pushing from behind and one side, so your KE will increase. The net force has components of 1400 N in the direction of travel and 212 N perpendicular to the direction of travel. So the net force is 1420 N at 8.5 degrees from the direction of travel. 13 Gravity has a component perpendicular to the cannon (and to displacement, so it is irrelevant) and has a component parallel to the cannon. The latter is equal to 9.8 N. Thus the net force in the direction of the displacement is 50 N − 9.8 N, and the kinetic energy is 60 J. 15 The potato cannon (and many other projectile launchers) above is an option, with a force launching the projectile, friction, potentially gravity depending on the direction it is pointed, etc. A drag (or other) car accelerating is another possibility. 17 The kinetic energy of the rear wagon increases. The front wagon does not, until the rear wagon collides with it. The total system may be treated by its center of mass, halfway between the wagons, and its energy increases by the same amount as the sum of the two individual wagons. 19 (d) 21 0.049 J; 0.041 m, 0.25 m 23 20 m high, 20 m/s. 25 (a) 27 (d) 29 (c) 31 (b) 33 (c) 35 (c) 37 (c), (d) 39 (a) 41 (b) Chapter 8 Problems & Exercises 1 (a) 1.50×104 kg ⋅ m/s (b) 625 to 1 (c) 6.66×102 kg ⋅ m/s 3 (a) 8.00×104 m/s (b) 1.20×106 kg · m/s 1580 Answer Key (c) Because the momentum of the airplane is 3 orders of magnitude smaller than of the ship, the ship will not recoil very much. The recoil would be −0.0100 m/s , which is probably not noticeable. 5 54 s 7 9.00×103 N 9 a) 2.40×103 N toward the leg b) The force on each hand would have the same magnitude as that found in part (a) (but in opposite directions by Newton’s third law) because the change in momentum and the time interval are the same. 11 a) 800 kg ⋅ m/s away from the wall b) 1.20 m/s away from the wall 13 (a) 1.50×106 N away from the dashboard (b) 1.00×105 N away from the dashboard 15 4.69×105 N in the boat’s original direction of motion 17 2.10×103 N away from the wall 19 = 2 p = v ⇒ 2 = 22 ⇒ 2 ⇒ 2 = 1 2 = 2 2 22 = (8.35) 21 60.0 g 23 0.122 m/s 25 In a collision with an identical car, momentum is conserved. Afterwards f = 0 for both cars. The change in momentum will be the same as in the crash with the tree. However, the force on the body is not determined since the time is not known. A padded stop will reduce injurious force on body. 27 22.4 m/s in the same direction as the original motion 29 0.250 m/s 31 (a) 86.4 N perpendicularly away from the bumper (b) 0.389 J (c) 64.0% 33 (a) 8.06 m/s (b) -56.0 J (c)(i) 7.88 m/s; (ii) -223 J 35 (a) 0.163 m/s in the direction of motion of the more massive satellite This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key 1581 (b) 81.6 J (c) 8.70×10−2 m/s in the direction of motion of the less massive satellite, 81.5 J. Because there are no external forces, the velocity of the center of mass of the two-satellite system is unchanged by the collision. The two velocities calculated above are the velocity of the center of mass in each of the two different individual reference frames. The loss in KE is the same in both reference frames because the KE lost to internal forces (heat, friction, etc.) is the same regardless of the coordinate system chosen. 37 0.704 m/s –2.25 m/s 38 (a) 4.58 m/s away from the bullet (b) 31.5 J (c) –0.491 m/s (d) 3.38 J 40 (a) 1.02×10−6 m/s (b) 5.63×1020 J (almost all KE lost) (c) Recoil speed is 6.79×10−17 m/s , energy lost is 6.25×109 J . The plume will not affect the momentum result because the plume is still part of the Moon system. The plume may affect the kinetic energy result because a significant part of the initial kinetic energy may be transferred to the kinetic energy of the plume particles. 42 24.8 m/s 44 (a) 4.00 kg (b) 210 J (c) The clown does work to throw the barbell, so the kinetic energy comes from the muscles of the clown. The muscles convert the chemical potential energy of ATP into kinetic energy. 45 (a) 3.00 m/s, 60º below -axis (b) Find speed of first puck after collision: 0 = ′1 sin 30º−′2 sin 60º ⇒ ′1 = 2 sin 60º sin 30º = 5.196 m/s KE = 1 KE = 1 21 2′1 2 = 18 J 2 + 1 2′2 KE KE′ 2 = 18 J = 1.00 Verify that ratio of initial to final KE equals one: 47 (a) −2.26 m/s (b) 7.63×103 J (c) The ground will exert a normal force to oppose recoil of the cannon in the vertical direction. The momentum in the vertical direction is transferred to the earth. The energy is transferred into the ground, making a dent where the cannon is. After long barrages, cannon have erratic aim because the ground is full of divots. 49 (a) 5.36×105 m/s at −29.5º (b) 7.52×10−13 J 51 1582 Answer Key We are given that 1 = 2 ≡ . The given equations then become: and Square each equation to get 1 = 1 cos 1 + 2 cos 2 0 = ′1 sin 1 + ′2 sin 2. 1 0 2 = ′1 = ′1 2 cos2 1 + ′2 2 sin2 1 + ′2 2 cos2 2 + 2′1 ′2 cos 1 cos 2 2 sin2 2 + 2′1 ′2 sin 1 sin 2 . 1 Add these two equations and simplify: 2 + ′2 2 + ′2 2 + ′2 2 = ′1 = ′1 2 + 2′1 ′2 cos 1 cos 2 + sin 1 sin 2 1 2 + 2′1 ′2 2 2 + 2′1 ′2 cos 1 cos cos . 1 + 2 + 1 2 cos 1 − 2 − 1 2 cos 1 + 2 (8.107) (8.108) (8.109) (8.110) Multiply the entire equation by 1 2 to recover the kinetic energy: 2 + 1 2 = 1 2′1 2′2 1 21 2 + ′1 ′2 cos 1 − 2 (8.111) 53 39.2 m/s2 55 4.16×103 m/s 57 The force needed to give a small mass Δ an acceleration Δ is = ΔΔ . To accelerate this mass in the small time interval Δ at a speed e requires e = ΔΔ , so = e Δ in magnitude to the thrust force acting on the rocket, so thrust = e Δ Newton’s second law to the rocket gives thrust − = ⇒ = e and unburnt fuel. 60 2.63×103 kg 61 − , where is the mass of the rocket , where all quantities are positive. Applying Δ Δ Δ Δ . By Newton’s third law, this force is equal (a) 0.421 m/s away from the ejected fluid. (b) 0.237 J . Test Prep for AP® Courses 1 (b) 3 (b) 5 (a) 7 (c) (based on calculation of = Δ Δ ) 9 (c) 11 (d) 13 (b) This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key 1583 15 (d) 17 (b) 19 (c) 21 (b) 23 (c) 25 (b) 27 (a) 29 (c) 31 (b) 33 (a) 35 (b) 37 (a) 39 (a) 41 (d) 43 (c). Because of conservation of momentum, the final velocity of the combined mass must be 4.286 m/s. The initial kinetic energy is (0.5)(2.0)(15)2 = 225 J . The final kinetic energy is (0.5)(7.0)(4.286)2 = 64 J , so the difference is −161 J. 45 (a) 47 (d) 49 (c) 51 (b) Chapter 9 Problems & Exercises 1 a) 46.8 N·m b) It does not matter at what height you push. The torque depends on only the magnitude of the force applied and the perpendicular distance of the force's application from the hinges. (Children don't have a tougher time opening a door because they push lower than adults, they have a tougher time because they don't push far enough from the hinges.) 3 23.3 N 5 Given: 1 = 26.0 kg, 2 = 32.0 kg, s = 12.0 kg, 1 = 1.60 m, s = 0.160 m, find (a) 2, (b) p (9.26) a) Since children are balancing: Answer Key (9.27) (9.28) (9.29) (9.30) 1584 So, solving for 2 gives: net cw = – net ccw ⇒ 1 1 + s s = 22 26.0 kg)(1.60 m) + (12.0 kg)(0.160 m) 32.0 kg b) Since the children are not moving: = 1.36 m So that net = 26.0 kg + 32.0 kg + 12.0 kg)(9.80 m / s2) = 686 N 6 wall = 1.43×103 N 8 a) 2.55×103 N, 16.3º to the left of vertical (i.e., toward the wall) b) 0.292 10 B = 2.12×104 N 12 a) 0.167, or about one-sixth of the weight is supported by the opposite shore. b) = 2.0×104 N , straight up. 14 a) 21.6 N b) 21.6 N 16 350 N directly upwar
ds 19 25 50 N 21 a) MA = 18.5 b) i = 29.1 N c) 510 N downward 23 1.3×103 N 25 a) = 299 N b) 897 N upward 26 This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key 1585 B = 470 N; 1 = 4.00 cm; a = 2.50 kg; 2 = 16.0 cm;b = 4.00 kg; 3 = 38.0 cm 16.0 cm 9.80 m / s2 4.0 cm 38.0 cm 9.80 m / s2 4.00 cm – 1 – 1 = 2.50 kg + 4.00 kg = 407 N 28 1.1×103 N = 190º ccw from positive axis 30 V = 97 N, = 59º 32 (a) 25 N downward (b) 75 N upward 33 (a) A = 2.21×103 N upward (b) B = 2.94×103 N downward 35 (a) teeth on bullet = 1.2×102 N upward (b) J = 84 N downward 37 (a) 147 N downward (b) 1680 N, 3.4 times her weight (c) 118 J (d) 49.0 W 39 a) 2 = 2.33 m b) The seesaw is 3.0 m long, and hence, there is only 1.50 m of board on the other side of the pivot. The second child is off the board. c) The position of the first child must be shortened, i.e. brought closer to the pivot. Test Prep for AP® Courses 1 (a) 3 Both objects are in equilibrium. However, they will respond differently if a force is applied to their sides. If the cone placed on its base is displaced to the side, its center of gravity will remain over its base and it will return to its original position. When the traffic cone placed on its tip is displaced to the side, its center of gravity will drift from its base, causing a torque that will accelerate it to the ground. 5 (d) 7 a. FL = 7350 N, FR = 2450 N b. As the car moves to the right side of the bridge, FL will decrease and FR will increase. (At exactly halfway across the bridge, FL and FR will both be 4900 N.) 9 1586 Answer Key The student should mention that the guiding principle behind simple machines is the second condition of equilibrium. Though the torque leaving a machine must be equivalent to torque entering a machine, the same requirement does not exist for forces. As a result, by decreasing the lever arm to the existing force, the size of the existing force will be increased. The mechanical advantage will be equivalent to the ratio of the forces exiting and entering the machine. 11 a. The force placed on your bicep muscle will be greater than the force placed on the dumbbell. The bicep muscle is closer to your elbow than the downward force placed on your hand from the dumbbell. Because the elbow is the pivot point of the system, this results in a decreased lever arm for the bicep. As a result, the force on the bicep must be greater than that placed on the dumbbell. (How much greater? The ratio between the bicep and dumbbell forces is equal to the inverted ratio of their distances from the elbow. If the dumbbell is ten times further from the elbow than the bicep, the force on the bicep will be 200 pounds!) b. The force placed on your bicep muscle will decrease. As the forearm lifts the dumbbell, it will get closer to the elbow. As a result, the torque placed on the arm from the weight will decrease and the countering torque created by the bicep muscle will do so as well. Chapter 10 Problems & Exercises 1 = 0.737 rev/s 3 (a) −0.26 rad/s2 (b) 27 rev 5 (a) 80 rad/s2 (b) 1.0 rev 7 (a) 45.7 s (b) 116 rev 9 a) 600 rad/s2 b) 450 rad/s c) 21.0 m/s 10 (a) 0.338 s (b) 0.0403 rev (c) 0.313 s 12 0.50 kg ⋅ m2 14 (a) 50.4 N ⋅ m (b) 17.1 rad/s2 (c) 17.0 rad/s2 16 3.96×1018 s or 1.26×1011 y This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key 18 1587 2 = + 42 = 1 2 32 − 1 42 = 1 122 Thus, = − 1 19 (a) 2.0 ms (b) The time interval is too short. (c) The moment of inertia is much too small, by one to two orders of magnitude. A torque of 500 N ⋅ m is reasonable. 20 (a) 17,500 rpm (b) This angular velocity is very high for a disk of this size and mass. The radial acceleration at the edge of the disk is > 50,000 gs. (c) Flywheel mass and radius should both be much greater, allowing for a lower spin rate (angular velocity). KErot = 434 J (10.104) 21 (a) 185 J (b) 0.0785 rev (c) = 9.81 N 23 (a) 2.57×1029 J (b) KErot = 2.65×1033 J 25 27 (a) 128 rad/s (b) 19.9 m 29 (a) 10.4 rad/s2 (b) net = 6.11 J 34 (a) 1.49 kJ (b) 2.52×104 N 36 (a) 2.66×1040 kg ⋅ m2/s (b) 7.07×1033 kg ⋅ m2/s The angular momentum of the Earth in its orbit around the Sun is 3.77×106 times larger than the angular momentum of the Earth around its axis. 38 22.5 kg ⋅ m2/s 40 25.3 rpm 43 Answer Key 1588 (a) 0.156 rad/s (b) 1.17×10−2 J (c) 0.188 kg ⋅ m/s 45 (a) 3.13 rad/s (b) Initial KE = 438 J, final KE = 438 J 47 (a) 1.70 rad/s (b) Initial KE = 22.5 J, final KE = 2.04 J (c) 1.50 kg ⋅ m/s 48 (a) 5.64×1033 kg ⋅ m2 /s (b) 1.39×1022 N ⋅ m (c) 2.17×1015 N Test Prep for AP® Courses 1 (b) 3 (d) 5 (d) You are given a thin rod of length 1.0 m and mass 2.0 kg, a small lead weight of 0.50 kg, and a not-so-small lead weight of 1.0 kg. The rod has three holes, one in each end and one through the middle, which may either hold a pivot point or one of the small lead weights. 7 (a) 9 (c) 11 (a) 13 (a) 15 (b) 17 (c) 19 (b) 21 (b) 23 (c) 25 (d) 27 A door on hinges is a rotational system. When you push or pull on the door handle, the angular momentum of the system changes. If a weight is hung on the door handle, then pushing on the door with the same force will cause a different increase in angular momentum. If you push or pull near the hinges with the same force, the resulting angular momentum of the system will also be different. 29 Since the globe is stationary to start with, This content is available for free at http://cnx.org/content/col11844/1.13 1589 Answer Key = Δ Δ ⋅ Δ = Δ By substituting, 120 N•m • 1.2 s = 144 N•m•s. The angular momentum of the globe after 1.2 s is 144 N•m•s. Chapter 11 Problems & Exercises 1 1.610 cm3 3 (a) 2.58 g (b) The volume of your body increases by the volume of air you inhale. The average density of your body decreases when you take a deep breath, because the density of air is substantially smaller than the average density of the body before you took the deep breath. 4 2.70 g/cm3 6 (a) 0.163 m (b) Equivalent to 19.4 gallons, which is reasonable 8 7.9×102 kg/m3 9 15.6 g/cm3 10 (a) 1018 kg/m3 (b) 2×104 m 11 3.59×106 Pa ; or 521 lb/in2 13 2.36×103 N 14 0.760 m 16 units = (m) = kg/m3 kg ⋅ m/s2 m/s2 1/m2 = kg ⋅ m2 / m3 ⋅ s2 (11.30) 18 (a) 20.5 mm Hg = N/m2 (b) The range of pressures in the eye is 12–24 mm Hg, so the result in part (a) is within that range 20 1.09×103 N/m2 22 24.0 N 24 1590 Answer Key 2.55×107 Pa ; or 251 atm 26 5.76×103 N extra force 28 (a) = ii = oo ⇒ o = i i o . Now, using equation: Finally = oo = i o i i i o = ii = i. (11.32) (11.33) In other words, the work output equals the work input. (b) If the system is not moving, friction would not play a role. With friction, we know there are losses, so that out = in − f ; therefore, the work output is less than the work input. In other words, with friction, you need to push harder on the input piston than was calculated for the nonfriction case. 29 Balloon: = 5.00 cm H2 O, g abs = 1.035×103 cm H2 O. Jar: = −50.0 mm Hg, g abs = 710 mm Hg. 31 4.08 m 33 Δ = 38.7 mm Hg, Leg blood pressure = 159 119 . 35 22.4 cm2 36 91.7% 38 815 kg/m3 40 (a) 41.4 g (b) 41.4 cm3 (c) 1.09 g/cm3 42 (a) 39.5 g (b) 50 cm3 (c) 0.79 g/cm3 It is ethyl alcohol. This content is available for free at http://cnx.org/content/col11844/1.13 1591 Answer Key 44 8.21 N 46 (a) 960 kg/m3 (b) 6.34% She indeed floats more in seawater. 48 (a) 0.24 (b) 0.68 (c) Yes, the cork will float because obj < ethyl alcohol(0.678 g/cm3 < 0.79 g/cm3) 50 The difference is 0.006%. 52 net = fl − 1fl 2 − 1 = 2 − 1 fl where fl = density of fluid. Therefore, net = (2 − 1)fl = flfl = fl = fl where is fl the weight of the fluid displaced. 54 592 N/m2 56 2.23×10−2 mm Hg 58 (a) 1.65×10−3 m (b) 3.71×10–4 m 60 6.32×10−2 N/m Based on the values in table, the fluid is probably glycerin. 62 w = 14.6 N/m2 a = 4.46 N/m2 sw = 7.40 N/m2 . Alcohol forms the most stable bubble, since the absolute pressure inside is closest to atmospheric pressure. 64 5.1º This is near the value of = 0º for most organic liquids. 66 −2.78 The ratio is negative because water is raised whereas mercury is lowered. Answer Key 1592 68 479 N 70 1.96 N 71 −63.0 cm H2 O 73 (a) 3.81×103 N/m2 (b) 28.7 mm Hg , which is sufficient to trigger micturition reflex 75 (a) 13.6 m water (b) 76.5 cm water 77 (a) 3.98×106 Pa (b) 2.1×10−3 cm 79 (a) 2.97 cm (b) 3.39×10−6 J (c) Work is done by the surface tension force through an effective distance / 2 to raise the column of water. 81 (a) 2.01×104 N (b) 1.17×10−3 m (c) 2.56×1010 N/m2 83 (a) 1.38×104 N (b) 2.81×107 N/m2 (c) 283 N 85 (a) 867 N (b) This is too much force to exert with a hand pump. (c) The assumed radius of the pump is too large; it would be nearly two inches in diameter—too large for a pump or even a master cylinder. The pressure is reasonable for bicycle tires. Test Prep for AP® Courses 1 (e) 3 (a) 100 kg/m3 (b) 60% (c) yes; yes (76% will be submerged) (d) answers vary 5 (d) Chapter 12 Problems & Exercises 1 This content is available for free at http://cnx.org/content/col11844/1.13 1593 Answer Key 2.78 cm3 /s 3 27 cm/s 5 (a) 0.75 m/s (b) 0.13 m/s 7 (a) 40.0 cm2 (b) 5.09×107 9 (a) 22 h (b) 0.016 s 11 (a) 12.6 m/s (b) 0.0800 m3 /s (c) No, independent of density. 13 (a) 0.402 L/s (b) 0.584 cm 15 (a) 127 cm3 /s (b) 0.890 cm 17 = Force Area () units = N/m2 = N ⋅ m/m3 = J/m3 = energy/volume 19 184 mm Hg 21 2.54×105 N 23 (a) 1.58×106 N/m2 (b) 163 m 25 (a) 9.56×108 W (b) 1.4 27 1.26 W 29 (a) 3.02×10−3 N (b) 1.03×10−3 31 1.60 cm3 /min 1594 33 8.7×10−11 m3 /s 35 0.316 37 (a) 1.52 Answer Key (b) Turbulence will decrease the flow rate of the blood, which would require an even larger increase in the pressure difference, leading to higher blood pressure. 225 mPa ⋅ s 0.138 Pa ⋅ s, (12.98) (12.99) 39 41 or Olive oil. 43 (a) 1.62×104 N/m2 (b) 0.111 cm3 /s (c)10.6 cm 45 1.59 47 2.95×106 N/m2 (gauge pressure) 51 R = 1.99×102 < 2000 53 (a) nozzle: 1.27×105 , not laminar (b) hose: 3.51×104 , not laminar. 55 2.54 << 2000, laminar. 57 1.02 m/s 1.28×10–2 L/s 59 (a) ≥ 13.0 m (b) 2.68×10−6 N/m2 61 (a) 23.7 atm o
r 344 lb/in2 (b) The pressure is much too high. (c) The assumed flow rate is very high for a garden hose. (d) 5.27×106 > > 3000, turbulent, contrary to the assumption of laminar flow when using this equation. 62 1.41×10−3 m 64 1.3×102 s 66 This content is available for free at http://cnx.org/content/col11844/1.13 1595 Answer Key 0.391 s Test Prep for AP® Courses 1 (c) 3 (a) 5 (a) 7 (a) 9 (d) Chapter 13 Problems & Exercises 1 102ºF 3 20.0ºC and 25.6ºC 5 9890ºF 7 (a) 22.2ºC Δ(ºF) = 2 (ºF) − 1(ºF) (bºC) + 32.0º − 5 = 9 5 2 (ºC) − 1(ºC) 1 (ºC) + 32.0º Δ(ºC) 9 169.98 m 11 5.4×10−6 m 13 Because the area gets smaller, the price of the land DECREASES by ~17000. 15 = 0 + Δ = 0(1 + Δ) (13.25) = (60.00 L) 1 + 950×10−6 / ºC (35.0ºC − 15.0ºC) = 61.1 L 17 (a) 9.35 mL (b) 7.56 mL 19 0.832 mm 21 We know how the length changes with temperature: Δ = 0Δ . Also we know that the volume of a cube is related to its length by = 3 , so the final volume is then = 0 + Δ = 0 + Δ 3 . Substituting for Δ gives = 0 + 0Δ 3 = 0 3(1 + Δ)3 . Now, because Δ is small, we can use the binomial expansion: ≈ 0 3(1 + 3αΔT) = 0 3 + 3α0 3Δ. (13.26) (13.27) 1596 So writing the length terms in terms of volumes gives = 0 + Δ ≈ 0 + 3α 0Δ and so Δ = 0Δ ≈ 3α 0Δ or ≈ 3α. Answer Key (13.28) 22 1.62 atm 24 (a) 0.136 atm (b) 0.135 atm. The difference between this value and the value from part (a) is negligible. 26 (a) = (mol)(J/mol ⋅ K)(K) = J (b) = (mol)(cal/mol ⋅ K)(K) = cal (c) = (mol)(L ⋅ atm/mol ⋅ K)(K) = L ⋅ atm = (m3)(N/m2) = N ⋅ m = J 28 7.86×10−2 mol 30 (a) 6.02×105 km3 (b) 6.02×108 km 32 −73.9ºC 34 (a) 9.14×106 N/m2 (b) 8.23×106 N/m2 (c) 2.16 K (d) No. The final temperature needed is much too low to be easily achieved for a large object. 36 41 km 38 (a) 3.7×10−17 Pa (b) 6.0×1017 m3 (c) 8.4×102 km 39 1.25×103 m/s 41 (a) 1.20×10−19 J (b) 1.24×10−17 J 43 458 K 45 1.95×107 K 47 6.09×105 m/s 49 This content is available for free at http://cnx.org/content/col11844/1.13 1597 Answer Key 7.89×104 Pa 51 (a) 1.99×105 Pa (b) 0.97 atm 53 3.12×104 Pa 55 78.3% 57 (a) 2.12×104 Pa (b) 1.06 % 59 (a) 8.80×10−2 g (b) 6.30×103 Pa ; the two values are nearly identical. 61 82.3% 63 4.77ºC 65 38.3 m 67 B / Cu B / Cu circumstances. 69 (a) 4.41×1010 mol/m3 = 1.02 . The buoyant force supports nearly the exact same amount of force on the copper block in both (b) It’s unreasonably large. (c) At high pressures such as these, the ideal gas law can no longer be applied. As a result, unreasonable answers come up when it is used. 71 (a) 7.03×108 m/s (b) The velocity is too high—it’s greater than the speed of light. (c) The assumption that hydrogen inside a supernova behaves as an idea gas is responsible, because of the great temperature and density in the core of a star. Furthermore, when a velocity greater than the speed of light is obtained, classical physics must be replaced by relativity, a subject not yet covered. Test Prep for AP® Courses 1 (a), (c) 3 (d) 5 (b) 7 (a) 7.29 × 10-21 J; (b) 352K or 79ºC Chapter 14 Problems & Exercises 5.02×108 J 3.07×103 J 0.171ºC Answer Key (14.18) (14.19) (14.20) 1598 1 3 5 7 10.8 9 617 W 11 35.9 kcal 13 (a) 591 kcal (b) 4.94×103 s 15 13.5 W 17 (a) 148 kcal (b) 0.418 s, 3.34 s, 4.19 s, 22.6 s, 0.456 s 19 33.0 g 20 (a) 9.67 L (b) Crude oil is less dense than water, so it floats on top of the water, thereby exposing it to the oxygen in the air, which it uses to burn. Also, if the water is under the oil, it is less efficient in absorbing the heat generated by the oil. 22 a) 319 kcal b) 2.00ºC 24 20.6ºC 26 4.38 kg 28 (a) 1.57×104 kcal (b) 18.3 kW ⋅ h (c) 1.29×104 kcal 30 (a) 1.01×103 W (b) One 32 84.0 W 34 2.59 kg 36 (a) 39.7 W (b) 820 kcal 38 35 to 1, window to wall This content is available for free at http://cnx.org/content/col11844/1.13 1599 Answer Key 40 1.05×103 K 42 (a) 83 W (b) 24 times that of a double pane window. 44 20.0 W, 17.2% of 2400 kcal per day 45 10 m/s 47 85.7ºC 49 1.48 kg 51 2×104 MW 53 (a) 97.2 J (b) 29.2 W (c) 9.49 W (d) The total rate of heat loss would be 29.2 W + 9.49 W = 38.7 W . While sleeping, our body consumes 83 W of power, while sitting it consumes 120 to 210 W. Therefore, the total rate of heat loss from breathing will not be a major form of heat loss for this person. 55 −21.7 kW Note that the negative answer implies heat loss to the surroundings. 57 −266 kW 59 −36.0 W 61 (a) 1.31% (b) 20.5% 63 (a) −15.0 kW (b) 4.2 cm 65 (a) 48.5ºC (b) A pure white object reflects more of the radiant energy that hits it, so a white tent would prevent more of the sunlight from heating up the inside of the tent, and the white tunic would prevent that heat which entered the tent from heating the rider. Therefore, with a white tent, the temperature would be lower than 48.5ºC , and the rate of radiant heat transferred to the rider would be less than 20.0 W. 67 (a) 3×1017 J (b) 1×1013 kg (c) When a large meteor hits the ocean, it causes great tidal waves, dissipating large amount of its energy in the form of kinetic energy of the water. 69 (a) 3.44×105 m3 /s 1600 Answer Key (b) This is equivalent to 12 million cubic feet of air per second. That is tremendous. This is too large to be dissipated by heating the air by only 5ºC . Many of these cooling towers use the circulation of cooler air over warmer water to increase the rate of evaporation. This would allow much smaller amounts of air necessary to remove such a large amount of heat because evaporation removes larger quantities of heat than was considered in part (a). 71 20.9 min 73 (a) 3.96×10-2 g (b) 96.2 J (c) 16.0 W 75 (a) 1.102 (b) 2.79×104 J (c) 12.6 J. This will not cause a significant cooling of the air because it is much less than the energy found in part (b), which is the energy required to warm the air from 20.0ºC to 50.0ºC . 76 (a) 36ºC (b) Any temperature increase greater than about 3ºC would be unreasonably large. In this case the final temperature of the person would rise to 73ºC (163ºF) . (c) The assumption of 95% heat retention is unreasonable. 78 (a) 1.46 kW (b) Very high power loss through a window. An electric heater of this power can keep an entire room warm. (c) The surface temperatures of the window do not differ by as great an amount as assumed. The inner surface will be warmer, and the outer surface will be cooler. Test Prep for AP® Courses 1 (c) 3 (a) 5 (b) 7 (a) 9 (d) Chapter 15 Problems & Exercises 1 1.6×109 J 3 -9.30108 J 5 (a) −1.0×104 J , or −2.39 kcal (b) 5.00% 7 (a) 122 W This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key (b) 2.10×106 J 1601 (c) Work done by the motor is 1.61×107 J ;thus the motor produces 7.67 times the work done by the man 9 (a) 492 kJ (b) This amount of heat is consistent with the fact that you warm quickly when exercising. Since the body is inefficient, the excess heat produced must be dissipated through sweating, breathing, etc. 10 6.77×103 J 12 (a) = Δ = 1.76×105 J (b) = = 1.76×105 J . Yes, the answer is the same. 14 = 4.5×103 J 16 is not equal to the difference between the heat input and the heat output. 20 (a) 18.5 kJ (b) 54.1% 22 (a) 1.32 × 109 J (b) 4.68 × 109 J 24 (a) 3.80 × 109 J (b) 0.667 barrels 26 (a) 8.30 × 1012 J , which is 3.32% of 2.50 × 1014 J . (b) –8.30 × 1012 J , where the negative sign indicates a reduction in heat transfer to the environment. 28 403ºC 30 (a) 244ºC (b) 477ºC (c)Yes, since automobiles engines cannot get too hot without overheating, their efficiency is limited. 32 (a) (b) 1 = 1 − c,1 h,1 = 1 − 543 K 723 K = 0.249 or 24.9% 2 = 1 − 423 K 543 K = 0.221 or 22.1% 1602 (c) 1 = 1 − c,1 h,1 ⇒ c,1 = h,1 1, − , similarly, c,2 = h,2 1 − 1 2 using h,2 = c,1 in above equation gives c,2 = h,,1 2 1 − overall 1 − 1 overall = 1 − (1 − 0.249)(1 − 0.221) = 41.5% 2 Answer Key 1 − overall (d) overall = 1 − 423 K 723 K = 0.415 or 41.5% 34 The heat transfer to the cold reservoir is c = h − = 25 kJ − 12 kJ = 13 kJ , so the efficiency is = 1 − c h = 1 − 13 kJ 25 kJ = 0.48 . The Carnot efficiency is C = 1 − c h = 1 − 300 K 600 K = 0.50 . The actual efficiency is 96% of the Carnot efficiency, which is much higher than the best-ever achieved of about 70%, so her scheme is likely to be fraudulent. 36 (a) –56.3ºC (b) The temperature is too cold for the output of a steam engine (the local environment). It is below the freezing point of water. (c) The assumed efficiency is too high. 37 4.82 39 0.311 41 (a) 4.61 (b) 1.66×108 J or 3.97×104 kcal (c) To transfer 1.66×108 J , heat pump costs $1.00, natural gas costs $1.34. 43 27.6ºC 45 (a) 1.44×107 J (b) 40 cents (c) This cost seems quite realistic; it says that running an air conditioner all day would cost $9.59 (if it ran continuously). 47 (a) 9.78×104 J/K (b) In order to gain more energy, we must generate it from things within the house, like a heat pump, human bodies, and other appliances. As you know, we use a lot of energy to keep our houses warm in the winter because of the loss of heat to the outside. 49 8.01×105 J 51 (a) 1.04×1031 J/K (b) 3.28×1031 J 53 199 J/K 55 This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key (a) 2.47×1014 J (b) 1.60×1014 J (c) 2.85×1010 J/K (d) 8.29×1012 J 1603 57 It should happen twice in every 1.27×1030 s or once in every 6.35×1029 s 1 y 6.35×1029 s 365.25 d 1 h 3600 s 1 d 24 h 2.0×1022 y = 59 (a) 3.0×1029 (b) 24% 61 (a) -2.3810 – 23 J/K (b) 5.6 times more likely (c) If you were betting on two heads and 8 tails, the odds of breaking even are 252 to 45, so on average you would break even. So, no, you wouldn't bet on odds of 252 to 45. Test Prep for AP® Courses 1 (d) 3 (a) 5 (b) 7 (c) 9 (d) 11 (a) 13 (c) 15 (b) Chapter 16 Problems & Exercises 1 (a) 1.23×103 N/m (b) 6.88 kg (c) 4.00 mm 3 (a) 889 N/m (b) 133 N 5 (a) 6.53×103 N/m Answer Key 1604 (b) Yes 7 16.7 ms 8 0.400 s / beats 9 400 Hz 10 12,500 Hz 11 1.50 kHz 12 (a) 93.8 m/s (b) 11.3×103 rev/min 13 2.37 N/m 15 0.389 kg 18 94.7 kg 21 1.94 s 22 6.21 cm 24 2.01 s 26 2.23 Hz 28 (a) 2.99541 s (b) Since the period is related to the square root of
the acceleration of gravity, when the acceleration changes by 1% the period changes by (0.01)2 = 0.01% so it is necessary to have at least 4 digits after the decimal to see the changes. 30 (a) Period increases by a factor of 1.41 ( 2 ) (b) Period decreases to 97.5% of old period 32 Slow by a factor of 2.45 34 length must increase by 0.0116%. 35 (a) 1.99 Hz (b) 50.2 cm (c) 1.41 Hz, 0.710 m 36 (a) 3.95×106 N/m (b) 7.90×106 J 37 a). 0.266 m/s b). 3.00 J 39 This content is available for free at http://cnx.org/content/col11844/1.13 1605 (16.75) (16.76) (16.77) (16.78) (16.79) Answer Key ± 3 2 42 384 J 44 (a). 0.123 m (b). −0.600 J (c). 0.300 J. The rest of the energy may go into heat caused by friction and other damping forces. = 9.26 d = 40.0 Hz w = 16.0 m/s = 700 m = 34.0 cm 46 (a) 5.00×105 J (b) 1.20×103 s 47 49 51 53 55 57 = 4 Hz 59 462 Hz, 4 Hz 61 (a) 3.33 m/s (b) 1.25 Hz 63 0.225 W 65 7.07 67 16.0 d 68 2.50 kW 70 3.38×10–5 W/m2 Test Prep for AP® Courses 1 (d) 3 (b) 5 The frequency is given by = 1 = 50 30 = 1.66 Time period is: 1606 Answer Key = 1 = 1 1.66 = 0.6 s 7 (c) 9 The energy of the particle at the center of the oscillation is given by = 1 ×0.2 kg×(5 m·s−1)2 = 2.5 J 22 = 1 2 11 (b) 13 19.7 J 15 (c) 17 = 2 2 − 2 = 50 N ⋅ m−1 = 0.06 = 0.5kg = 50N ⋅ m−1 2×0.06×9.8m ⋅ s−2 (0.2)2 − 0.06×0.5kg×9.8m ⋅ s−2) (50N ⋅ m−1)2 2 = 1.698 m 19 The waves coming from a tuning fork are mechanical waves that are longitudinal in nature, whereas electromagnetic waves are transverse in nature. 21 The sound energy coming out of an instrument depends on its size. The sound waves produced are relative to the size of the musical instrument. A smaller instrument such as a tambourine will produce a high-pitched sound (higher frequency, shorter wavelength), whereas a larger instrument such as a drum will produce a deeper sound (lower frequency, longer wavelength). 23 2 m 25 The student explains the principle of superposition and then shows two waves adding up to form a bigger wave when a crest adds with a crest and a trough with another trough. Also the student shows a wave getting cancelled out when a crest meets a trough and vice versa. 27 The student must note that the shape of the wave remains the same and there is first an overlap and then receding of the waves. 29 (c) Chapter 17 Problems & Exercises 1 0.288 m 3 332 m/s 5 7 0.223 9 (a) 7.70 m w = (331 m/s) 273 K = (331 m/s) 293 K 273 K = 343 m/s (17.12) This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key 1607 (b) This means that sonar is good for spotting and locating large objects, but it isn’t able to resolve smaller objects, or detect the detailed shapes of objects. Objects like ships or large pieces of airplanes can be found by sonar, while smaller pieces must be found by other means. 11 (a) 18.0 ms, 17.1 ms (b) 5.00% (c) This uncertainty could definitely cause difficulties for the bat, if it didn’t continue to use sound as it closed in on its prey. A 5% uncertainty could be the difference between catching the prey around the neck or around the chest, which means that it could miss grabbing its prey. (17.23) (17.24) (17.25) (17.36) 3.16×10–4 W/m2 3.04×10–4 W/m2 1.45×10–3 J 3.79×103 Hz 12 14 16 106 dB 18 (a) 93 dB (b) 83 dB 20 (a) 50.1 (b) 5.01×10–3 or 1 200 22 70.0 dB 24 100 26 28 28.2 dB 30 (a) 878 Hz (b) 735 Hz 32 34 (a) 12.9 m/s (b) 193 Hz 36 First eagle hears 4.23×103 Hz Second eagle hears 3.56×103 Hz 38 0.7 Hz 40 0.3 Hz, 0.2 Hz, 0.5 Hz 42 (a) 256 Hz (b) 512 Hz 44 180 Hz, 270 Hz, 360 Hz Answer Key 1608 46 1.56 m 48 (a) 0.334 m (b) 259 Hz 50 3.39 to 4.90 kHz 52 (a) 367 Hz (b) 1.07 kHz 54 (a) = 47.6 Hz , = 1, 3, 5,..., 419 (b) = 95.3 Hz , = 1, 2, 3,..., 210 55 1×106 km (17.49) 57 498.5 or 501.5 Hz 59 82 dB 61 approximately 48, 9, 0, –7, and 20 dB, respectively 63 (a) 23 dB (b) 70 dB 65 Five factors of 10 67 (a) 2×10−10 W/m2 (b) 2×10−13 W/m2 69 2.5 71 1.26 72 170 dB 74 103 dB 76 (a) 1.00 (b) 0.823 (c) Gel is used to facilitate the transmission of the ultrasound between the transducer and the patient’s body. 78 (a) 77.0 μm (b) Effective penetration depth = 3.85 cm, which is enough to examine the eye. (c) 16.6 μm 80 (a) 5.78×10–4 m (b) 2.67×106 Hz This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key 82 1609 (a) w = 1540 m/s = = 1540 m/s 100×103 Hz = 0.0154 m < 3.50 m. Because the wavelength is much shorter than ⇒ the distance in question, the wavelength is not the limiting factor. (b) 4.55 ms 84 974 Hz (Note: extra digits were retained in order to show the difference.) Test Prep for AP® Courses 1 (b) 3 (e) 5 (c) 7 Answers vary. Students could include a sketch showing an increased amplitude when two waves occupy the same location. Students could also cite conceptual evidence such as sound waves passing through each other. 9 (d) 11 (c) 13 (a) 15 (c) 17 (b) 19 (a), (b) 21 (c) Chapter 18 Problems & Exercises 1 (a) 1.25×1010 (b) 3.13×1012 3 -600 C 5 1.03×1012 7 9.09×10−13 9 1.48×108 C 15 (a) = 1.00 cm = − ∞ (b) 2.12×105 N/C (c) one charge of + 1610 17 (a) 0.252 N to the left (b) = 6.07 cm 19 Answer Key (a)The electric field at the center of the square will be straight up, since and are positive and and are negative and all have the same magnitude. (b) 2.04×107 N/C (upward) 21 0.102 N in the − 23 direction = 4.36×103 N/C 35.0º , below the horizontal. → (a) (b) No 25 (a) 0.263 N (b) If the charges are distributed over some area, there will be a concentration of charge along the side closest to the oppositely charged object. This effect will increase the net force. 27 The separation decreased by a factor of 5. 31 = \|1 2\| 2 = = 2 ⇒ 2 9.00×109 N ⋅ m2/ C2 = 1.60×10–19 m 2 1.67×10–27 kg 2.00×10–9 m = 3.45×1016 m/s2 2 32 (a) 3.2 (b) If the distance increases by 3.2, then the force will decrease by a factor of 10 ; if the distance decreases by 3.2, then the force will increase by a factor of 10. Either way, the force changes by a factor of 10. 34 (a) 1.04×10−9 C (b) This charge is approximately 1 nC, which is consistent with the magnitude of charge typical for static electricity 37 1.02×10−11 39 a. 0.859 m beyond negative charge on line connecting two charges b. 0.109 m from lesser charge on line connecting two charges 42 8.75×10−4 N 44 (a) 6.94×10−8 C (b) 6.25 N/C 46 (a) 300 N/C (east) This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key 1611 (b) 4.80×10−17 N (east) 52 (a) 5.58×10−11 N/C (b)the coulomb force is extraordinarily stronger than gravity 54 (a) −6.76×105 C (b) 2.63×1013 m/s2 (upward) (c) 2.45×10−18 kg 56 The charge 2 is 9 times greater than 1 . Test Prep for AP® Courses 1 (b) 3 (c) 5 (a) 7 (b) 9 (a) -0.1 C, (b) 1.1 C, (c) Both charges will be equal to 1 C, law of conservation of charge, (d) 0.9 C 11 W is negative, X is positive, Y is negative, Z is neutral. 13 (c) 15 (c) 17 (b) 19 a) Ball 1 will have positive charge and Ball 2 will have negative charge. b) The negatively charged rod attracts positive charge of Ball 1. The electrons of Ball 1 are transferred to Ball 2, making it negatively charged. c) If Ball 2 is grounded while the rod is still there, it will lose its negative charge to the ground. d) Yes, Ball 1 will be positively charged and Ball 2 will be negatively charge. 21 (c) 23 decrease by 77.78%. 25 (a) 27 (d) 29 (a) 3.60×1010 N, (b) It will become 1/4 of the original value; hence it will be equal to 8.99×109 N 31 (c) 33 (a) 35 (b) 37 (a) 350 N/C, (b) west, (c) 5.6×10−17 N, (d) west. 39 (b) 1612 41 Answer Key (a) i) Field vectors near objects point toward negatively charged objects and away from positively charged objects. (a) ii) The vectors closest to R and T are about the same length and start at about the same distance. We have that / 2 = / 2 , so the charge on R is about the same as the charge on T. The closest vectors around S are about the same length as those around R and T. The vectors near S start at about 6 units away, while vectors near R and T start at about 4 units. We have that / 2 = / 2 , so / = 2/ 2 = 36/ 16 = 2.25 , and so the charge on S is about twice that on R and T. (b) Figure 18.35. A vector diagram. (c) = − ( + )2 + 2 ()2 + ( − )2 (d) The statement is not true. The vector diagram shows field vectors in this region with nonzero length, and the vectors not shown have even greater lengths. The equation in part (c) shows that, when 0 < < , the denominator of the negative term is always greater than the denominator of the third term, but the numerator is the same. So the negative term always has a smaller magnitude than the third term and since the second term is positive the sum of the terms is always positive. Chapter 19 Problems & Exercises 1 42.8 4 1.00105 K 6 (a) 4104 W (b) A defibrillator does not cause serious burns because the skin conducts electricity well at high voltages, like those used in defibrillators. The gel used aids in the transfer of energy to the body, and the skin doesn’t absorb the energy, but rather lets it pass through to the heart. 8 (a) 7.40103 C (b) 1.541020 electrons per second 9 3.89106 C 11 (a) 1.44×1012 V (b) This voltage is very high. A 10.0 cm diameter sphere could never maintain this voltage; it would discharge. (c) An 8.00 C charge is more charge than can reasonably be accumulated on a sphere of that size. 15 This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key (a) 3.00 kV (b) 750 V 1613 17 (a) No. The electric field strength between the plates is 2.5×106 V/m, which is lower than the breakdown strength for air ( 3.0×106 V/m ). (b) 1.7 mm 19 44.0 mV 21 15 kV 23 (a) 800 KeV (b) 25.0 km 24 144 V 26 (a) 1.80 km (b) A charge of 1 C is a very large amount of charge; a sphere of radius 1.80 km is not practical. 28 –2.22×10 – 13 C 30 (a) 3.31×106 V (b) 152 MeV 32 (a) 2.78×10-7 C (b) 2.00×10-10 C 35 (a) 2.96×109 m/s (b) This velocity is far too great. It is faster than the speed of light. (c) The assumption that the speed of the electron is far less than that of light and that the problem do
es not require a relativistic treatment produces an answer greater than the speed of light. 46 21.6 mC 48 80.0 mC 50 20.0 kV 52 667 pF 54 (a) 4.4 µF (b) 4.0×10 – 5 C 56 (a) 14.2 kV (b) The voltage is unreasonably large, more than 100 times the breakdown voltage of nylon. 1614 Answer Key (c) The assumed charge is unreasonably large and cannot be stored in a capacitor of these dimensions. 57 0.293 μF 59 3.08 µF in series combination, 13.0 µF in parallel combination 60 2.79 µF 62 (a) –3.00 µF (b) You cannot have a negative value of capacitance. (c) The assumption that the capacitors were hooked up in parallel, rather than in series, was incorrect. A parallel connection always produces a greater capacitance, while here a smaller capacitance was assumed. This could happen only if the capacitors are connected in series. 63 (a) 405 J (b) 90.0 mC 64 (a) 3.16 kV (b) 25.3 mC 66 (a) 1.42×10−5 C , 6.38×10−5 J (b) 8.46×10−5 C , 3.81×10−4 J 67 (a) 4.43×10 – 12 F (b) 452 V (c) 4.52×10 – 7 J 70 (a) 133 F (b) Such a capacitor would be too large to carry with a truck. The size of the capacitor would be enormous. (c) It is unreasonable to assume that a capacitor can store the amount of energy needed. Test Prep for AP® Courses 1 (a) 3 (b) 5 (c) 7 (a) 9 (b) 11 (b) 13 (a) 15 (c) 17 This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key 1615 (b) 19 (a) 21 (d) 23 (d) 25 (a) 27 (b) 29 (c) 31 (d) 33 (a) 35 (c) 37 (c) 39 (b) 41 (a) 43 (d) Chapter 20 Problems & Exercises 1 0.278 mA 3 0.250 A 5 1.50ms 7 (a) 1.67k Ω (b) If a 50 times larger resistance existed, keeping the current about the same, the power would be increased by a factor of about 50 (based on the equation = 2 ), causing much more energy to be transferred to the skin, which could cause serious burns. The gel used reduces the resistance, and therefore reduces the power transferred to the skin. 9 (a) 0.120 C (b) 7.501017 electrons 11 96.3 s 13 (a) 7.81 × 1014 He++ nuclei/s (b) 4.00 × 103 s (c) 7.71 × 108 s 15 −1.1310−4 m/s 1616 17 9.421013 electrons 18 0.833 A 20 7.3310−2 Ω 22 (a) 0.300 V (b) 1.50 V Answer Key (c) The voltage supplied to whatever appliance is being used is reduced because the total voltage drop from the wall to the final output of the appliance is fixed. Thus, if the voltage drop across the extension cord is large, the voltage drop across the appliance is significantly decreased, so the power output by the appliance can be significantly decreased, reducing the ability of the appliance to work properly. 24 0.104 Ω 26 2.810−2 m 28 1.1010−3 A 30 −5ºC to 45ºC 32 1.03 34 0.06% 36 −17ºC 38 (a) 4.7 Ω (total) (b) 3.0% decrease 40 2.001012 W 44 (a) 1.50 W (b) 7.50 W = V2 V/A 46 V2 Ω 48 1 kW ⋅ h= = AV = ×103 J 1 s (1 h) 3600 s 1 h = 3.60×106 J 50 $438/y 52 $6.25 54 1.58 h 56 $3.94 billion/year 58 25.5 W 60 (a) 2.00109 J This content is available for free at http://cnx.org/content/col11844/1.13 1617 Answer Key (b) 769 kg 62 45.0 s 64 (a) 343 A (b) 2.17103 A (c) 1.10103 A 66 (a) 1.23×103 kg (b) 2.64×103 kg 69 (a) 2.08×105 A (b) 4.33×104 MW (c) The transmission lines dissipate more power than they are supposed to transmit. (d) A voltage of 480 V is unreasonably low for a transmission voltage. Long-distance transmission lines are kept at much higher voltages (often hundreds of kilovolts) to reduce power losses. 73 480 V 75 2.50 ms 77 (a) 4.00 kA (b) 16.0 MW (c) 16.0% 79 2.40 kW 81 (a) 4.0 (b) 0.50 (c) 4.0 83 (a) 1.39 ms (b) 4.17 ms (c) 8.33 ms 85 (a) 230 kW (b) 960 A 87 (a) 0.400 mA, no effect (b) 26.7 mA, muscular contraction for duration of the shock (can't let go) 89 1.20105 Ω 91 (a) 1.00 Ω 1618 (b) 14.4 kW 93 Temperature increases 860º C . It is very likely to be damaging. 95 80 beats/minute Test Prep for AP® Courses Answer Key 1 (a) 3 10 A 5 (a) 7 3.2 Ω, 2.19 A 9 (b), (d) 11 9.72 × 10−8 Ω·m 13 18 Ω 15 10:3 or 3.33 Chapter 21 Problems & Exercises 1 (a) 2.75 k Ω (b) 27.5 Ω 3 (a) 786 Ω (b) 20.3 Ω 5 29.6 W 7 (a) 0.74 A (b) 0.742 A 9 (a) 60.8 W (b) 3.18 kW 11 (a >>2 (b , so that >>2 . 13 This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key (a) -400 k Ω (b) Resistance cannot be negative. 1619 (c) Series resistance is said to be less than one of the resistors, but it must be greater than any of the resistors. 14 2.00 V 16 2.9994 V 18 0.375 Ω 21 (a) 0.658 A (b) 0.997 W (c) 0.997 W; yes 23 (a) 200 A (b) 10.0 V (c) 2.00 kW (d) 0.1000 Ω 80.0 A, 4.0 V, 320 W 25 (a) 0.400 Ω (b) No, there is only one independent equation, so only can be found. 29 (a) –0.120 V (b) -1.4110−2 Ω (c) Negative terminal voltage; negative load resistance. (d) The assumption that such a cell could provide 8.50 A is inconsistent with its internal resistance. −22 + emf1 − 21 + 33 + 32 - emf2 = 0 3 = 1 + 2 emf2 - 22 - 22 + 15 + 11 - emf1 + 11 = 0 (21.69) (21.70) (21.71) 31 35 37 39 (a) I1 = 4.75 A (b) I2 = -3.5 A (c) I3 = 8.25 A 41 (a) No, you would get inconsistent equations to solve. (b) 1 ≠ 2 + 3 . The assumed currents violate the junction rule. 42 30 44 1.98 k Ω 46 1.2510-4 Ω (21.75) Answer Key 1620 48 (a) 3.00 M Ω (b) 2.99 k Ω 50 (a) 1.58 mA (b) 1.5848 V (need four digits to see the difference) (c) 0.99990 (need five digits to see the difference from unity) 52 15.0 μA 54 (a) Figure 21.39. (b) 10.02 Ω (c) 0.9980, or a 2.0×10–1 percent decrease (d) 1.002, or a 2.0×10–1 percent increase (e) Not significant. 56 (a) −66.7 Ω (b) You can’t have negative resistance. (c) It is unreasonable that G is greater than tot (see Figure 21.36). You cannot achieve a full-scale deflection using a current less than the sensitivity of the galvanometer. Range = 5.00 Ω to 5.00 k Ω (21.82) 57 24.0 V 59 1.56 k Ω 61 (a) 2.00 V (b) 9.68 Ω 62 63 range 4.00 to 30.0 M Ω 65 (a) 2.50 μF (b) 2.00 s 67 86.5% 69 (a) 1.25 k Ω (b) 30.0 ms 71 (a) 20.0 s This content is available for free at http://cnx.org/content/col11844/1.13 1621 Answer Key (b) 120 s (c) 16.0 ms 73 1.73×10−2 s 74 3.3310−3 Ω 76 (a) 4.99 s (b) 3.87ºC (c) 31.1 k Ω (d) No Test Prep for AP® Courses 1 (a), (b) 3 (b) 5 (a) 4-Ω resistor; (b) combination of 20-Ω, 20-Ω, and 10-Ω resistors; (c) 20 W in each 20-Ω resistor, 40 W in 10-Ω resistor, 64 W in 4-Ω resistor, total 144W total in resistors, output power is 144 W, yes they are equal (law of conservation of energy); (d) 4 Ω and 3 Ω for part (a) and no change for part (b); (e) no effect, it will remain the same. 7 0.25 Ω, 0.50 Ω, no change 9 a. (c) b. (c) c. (d) d. (d) 11 a. I1 + I3 = I2 b. E1 - I1R1 - I2R2 - I1r1 = 0; - E2 + I1R1 - I3R3 - I3r2 = 0 c. d. I1 = 8/15 A, I2 = 7/15 A and I3 = -1/15 A I1 = 2/5 A, I2 = 3/5 A and I3 = 1/5 A e. PE1 = 18/5 W and PR1 = 24/25 W, PR2 = 54/25 W, PR3 = 12/25 W. Yes, PE1 = PR1+ PR2 + PR3 f. R3, losses in the circuit 13 (a) 20 mA, Figure 21.44, 5.5 s; (b) 24 mA, Figure 21.35, 2 s Chapter 22 Problems & Exercises 1 (a) Left (West) (b) Into the page (c) Up (North) (d) No force (e) Right (East) (f) Down (South) 3 (a) East (right) 1622 (b) Into page (c) South (down) 5 (a) Into page (b) West (left) (c) Out of page Answer Key 7 7.50×10−7 N perpendicular to both the magnetic field lines and the velocity 9 (a) 3.01×10−5 T (b) This is slightly less then the magnetic field strength of 5×10−5 T at the surface of the Earth, so it is consistent. 11 (a) 6.67×10−10 C (taking the Earth’s field to be 5.00×10−5 T ) (b) Less than typical static, therefore difficult 12 4.27 m 14 (a) 0.261 T (b) This strength is definitely obtainable with today’s technology. Magnetic field strengths of 0.500 T are obtainable with permanent magnets. 16 4.36×10−4 m 18 (a) 3.00 kV/m (b) 30.0 V 20 0.173 m 22 7.50×10−4 V 24 (a) 1.18 × 10 3 m/s (b) Once established, the Hall emf pushes charges one direction and the magnetic force acts in the opposite direction resulting in no net force on the charges. Therefore, no current flows in the direction of the Hall emf. This is the same as in a current-carrying conductor—current does not flow in the direction of the Hall emf. 26 11.3 mV 28 1.16 μV 30 2.00 T 31 (a) west (left) (b) into page (c) north (up) (d) no force (e) east (right) (f) south (down) This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key 33 (a) into page (b) west (left) (c) out of page 35 (a) 2.50 N (b) This is about half a pound of force per 100 m of wire, which is much less than the weight of the wire itself. Therefore, it does not cause any special concerns. 1623 37 1.80 T 39 (a) 30º (b) 4.80 N 41 (a) τ decreases by 5.00% if B decreases by 5.00% (b) 5.26% increase 43 10.0 A 45 A ⋅ m2 ⋅ T = A ⋅ m2 47 3.48×10−26 N ⋅ m 49 (a) 0.666 N ⋅ m west b) This is not a very significant torque, so practical use would be limited. Also, the current would need to be alternated to make the loop rotate (otherwise it would oscillate). 50 (a) 8.53 N, repulsive (b) This force is repulsive and therefore there is never a risk that the two wires will touch and short circuit. 52 400 A in the opposite direction 54 (a) 1.67×10−3 N/m (b) 3.33×10−3 N/m (c) Repulsive (d) No, these are very small forces 56 (a) Top wire: 2.65×10−4 N/m s, 10.9º to left of up (b) Lower left wire: 3.61×10−4 N/m , 13.9º down from right (c) Lower right wire: 3.46×10−4 N/m , 30.0º down from left 58 (a) right-into page, left-out of page (b) right-out of page, left-into page (c) right-out of page, left-into page Answer Key 1624 60 (a) clockwise (b) clockwise as seen from the left (c) clockwise as seen from the right 61 1.01×1013 T 63 (a) 4.80×10−4 T (b) Zero (c) If the wires are not paired, the field is about 10 times stronger than Earth’s magnetic field and so could severely disrupt the use of a compass. 65 39.8 A 67 (a) 3.14×10−5 T (b) 0.314 T 69 7.55×10−5 T , 23.4º 71 10.0 A 73 (a) 9.09×10−7 N upward (b) 3.03×10−5 m/s2 75 60.2 cm 77 (a) 1.02×103 N/m2 (b) Not a significant fraction of an atmosphere 79 17.0×10−4%/ºC 81 18.3 MHz 83 (a) Straight up (b) 6.00×10−4 N/m (c) 94.1 μm (d)2.47 Ω/m, 49.4 V/m 85 (a) 571 C (b) Impossible to have such a large separated charge on such a small object. (c) The 1
.00-N force is much too great to be realistic in the Earth’s field. 87 (a) 2.40×106 m/s (b) The speed is too high to be practical ≤ 1% speed of light This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key 1625 (c) The assumption that you could reasonably generate such a voltage with a single wire in the Earth’s field is unreasonable 89 (a) 25.0 kA (b) This current is unreasonably high. It implies a total power delivery in the line of 50.0x10^9 W, which is much too high for standard transmission lines. (c)100 meters is a long distance to obtain the required field strength. Also coaxial cables are used for transmission lines so that there is virtually no field for DC power lines, because of cancellation from opposing currents. The surveyor’s concerns are not a problem for his magnetic field measurements. Test Prep for AP® Courses 1 (a) 3 (b) 5 (b) 7 (a) 9 (b) 11 (e) 13 (c) 15 (c) Chapter 23 Problems & Exercises 1 Zero 3 (a) CCW (b) CW (c) No current induced 5 (a) 1 CCW, 2 CCW, 3 CW (b) 1, 2, and 3 no current induced (c) 1 CW, 2 CW, 3 CCW 9 (a) 3.04 mV (b) As a lower limit on the ring, estimate R = 1.00 mΩ. The heat transferred will be 2.31 mJ. This is not a significant amount of heat. 11 0.157 V 13 proportional to 1 17 (a) 0.630 V (b) No, this is a very small emf. 19 2.22 m/s 25 1626 (a) 10.0 N (b) 2.81×108 J (c) 0.36 m/s Answer Key (d) For a week-long mission (168 hours), the change in velocity will be 60 m/s, or approximately 1%. In general, a decrease in velocity would cause the orbit to start spiraling inward because the velocity would no longer be sufficient to keep the circular orbit. The long-term consequences are that the shuttle would require a little more fuel to maintain the desired speed, otherwise the orbit would spiral slightly inward. 28 474 V 30 0.247 V 32 (a) 50 (b) yes 34 (a) 0.477 T (b) This field strength is small enough that it can be obtained using either a permanent magnet or an electromagnet. 36 (a) 5.89 V (b) At t=0 (c) 0.393 s (d) 0.785 s 38 (a) 1.92×106 rad/s (b) This angular velocity is unreasonably high, higher than can be obtained for any mechanical system. (c) The assumption that a voltage as great as 12.0 kV could be obtained is unreasonable. 39 (a) 12.00 Ω (b) 1.67 A 41 72.0 V 43 0.100 Ω 44 (a) 30.0 (b) 9.75×10−2 A 46 (a) 20.0 mA (b) 2.40 W (c) Yes, this amount of power is quite reasonable for a small appliance. 48 (a) 0.063 A (b) Greater input current needed. 50 (a) 2.2 (b) 0.45 This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key (c) 0.20, or 20.0% 52 (a) 335 MV (b) way too high, well beyond the breakdown voltage of air over reasonable distances (c) input voltage is too high 1627 54 (a) 15.0 V (b) 75.0 A (c) yes 55 1.80 mH 57 3.60 V 61 (a) 31.3 kV (b) 125 kJ (c) 1.56 MW (d) No, it is not surprising since this power is very high. 63 (a) 1.39 mH (b) 3.33 V (c) Zero 65 60.0 mH 67 (a) 200 H (b) 5.00ºC 69 500 H 71 50.0 Ω 73 1.00×10–18 s to 0.100 s 75 95.0% 77 (a) 24.6 ms (b) 26.7 ms (c) 9% difference, which is greater than the inherent uncertainty in the given parameters. 79 531 Hz 81 1.33 nF 83 (a) 2.55 A (b) 1.53 mA 85 63.7 µH 87 1628 (a) 21.2 mH (b) 8.00 Ω 89 (a) 3.18 mF (b) 16.7 Ω 92 Answer Key (a) 40.02 Ω at 60.0 Hz, 193 Ω at 10.0 kHz (b) At 60 Hz, with a capacitor, Z=531 Ω , over 13 times as high as without the capacitor. The capacitor makes a large difference at low frequencies. At 10 kHz, with a capacitor Z=190 Ω , about the same as without the capacitor. The capacitor has a smaller effect at high frequencies. 94 (a) 529 Ω at 60.0 Hz, 185 Ω at 10.0 kHz (b) These values are close to those obtained in Example 23.12 because at low frequency the capacitor dominates and at high frequency the inductor dominates. So in both cases the resistor makes little contribution to the total impedance. 96 9.30 nF to 101 nF 98 3.17 pF 100 (a) 1.31 μH (b) 1.66 pF 102 (a) 12.8 kΩ (b) 1.31 kΩ (c) 31.9 mA at 500 Hz, 312 mA at 7.50 kHz (d) 82.2 kHz (e) 0.408 A 104 (a) 0.159 (b) 80.9º (c) 26.4 W (d) 166 W 106 16.0 W Test Prep for AP® Courses 1 (c) This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key 1629 Figure 23.6. 3 (c) 5 (a), (d) 7 (c) Chapter 24 Problems & Exercises 3 150 kV/m 6 (a) 33.3 cm (900 MHz) 11.7 cm (2560 MHz) (b) The microwave oven with the smaller wavelength would produce smaller hot spots in foods, corresponding to the one with the frequency 2560 MHz. 8 26.96 MHz 10 5.0×1014 Hz 12 14 0.600 m 16 = = 3.00108 m/s 1.201015 Hz = 2.5010 – 7 m () (a) = = 3.00108 m/s 110-10 m = 31018 Hz (b) X-rays 19 (a) 6.00×106 m (b) 4.33×10−5 T 21 (a) 1.50 × 10 6 Hz, AM band (b) The resonance of currents on an antenna that is 1/4 their wavelength is analogous to the fundamental resonant mode of an air column closed at one end, since the tube also has a length equal to 1/4 the wavelength of the fundamental oscillation. 23 (a) 1.55×1015 Hz (b) The shortest wavelength of visible light is 380 nm, so that 1630 Answer Key () visible UV = 380 nm 193 nm = 1.97. In other words, the UV radiation is 97% more accurate than the shortest wavelength of visible light, or almost twice as accurate! 25 3.90×108 m 27 (a) 1.50×1011 m (b) 0.500 s (c) 66.7 ns 29 (a) −3.5102 W/m2 (b) 88% (c) 1.7 T 30 = = 2 0 0 2 () 3.00108 m/s 8.8510–12 C2 /N ⋅ m2 (125 V/m)2 2 = 20.7 W/m2 32 (a) = = 2 = 0.25010−3 W 0.50010−3 m 2 = 318 W/m2 ave = 2 0 2μ0 ⇒ 0 = 2μ0 1 / 2 = 410−7 T ⋅ m/A 2 318.3 W/m2 3.00108 m/s 1 / 2 = 1.6310−6 T 0 = 0 = 3.00108 m/s 1.63310−6 T = 4.90102 V/m (b) (c) 34 (a) 89.2 cm (b) 27.4 V/m 36 (a) 333 T (b) 1.331019 W/m2 (c) 13.3 kJ 38 This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key (a) = = 4π 2 ∝ 1 2 1631 (b 40 13.5 pF 42 (a) 4.07 kW/m2 (b) 1.75 kV/m (c) 5.84 T (d) 2 min 19 s 44 (a) 5.00103 W/m2 (b) 3.88×10−6 N (c) 5.18×10−12 N 46 (a) = 0 (b) 7.50×10−10 s (c) 1.00×10−9 s 48 (a) 1.01×106 W/m2 (b) Much too great for an oven. (c) The assumed magnetic field is unreasonably large. 50 (a) 2.53×10−20 H (b) L is much too small. (c) The wavelength is unreasonably small. Test Prep for AP® Courses 1 (b) 3 (a) 5 (d) 7 (d) 9 (d) 11 (a) Chapter 25 Problems & Exercises 1 1632 Answer Key Top 1.715 m from floor, bottom 0.825 m from floor. Height of mirror is 0.890 m , or precisely one-half the height of the person. 5 2.25×108 m/s in water 2.04×108 m/s in glycerine 7 1.490 , polystyrene 9 1.28 s 11 1.03 ns 13 = 1.46 , fused quartz 17 (a) 0.898 (b) Can’t have < 1.00 since this would imply a speed greater than . (c) Refracted angle is too big relative to the angle of incidence. 19 (a) 5.00 (b) Speed of light too slow, since index is much greater than that of diamond. (c) Angle of refraction is unreasonable relative to the angle of incidence. 22 66.3º 24 > 1.414 26 1.50, benzene 29 46.5º, red; 46.0º, violet 31 (a) 0.043º (b) 1.33 m 33 71.3º 35 53.5º, red; 55.2º, violet 37 5.00 to 12.5 D 39 −0.222 m 41 (a) 3.43 m (b) 0.800 by 1.20 m 42 (a) −1.35 m (on the object side of the lens). (b) +10.0 (c) 5.00 cm This content is available for free at http://cnx.org/content/col11844/1.13 1633 Answer Key 43 44.4 cm 45 (a) 6.60 cm (b) –0.333 47 (a) +7.50 cm (b) 13.3 D (c) Much greater 49 (a) +6.67 (b) +20.0 (c) The magnification increases without limit (to infinity) as the object distance increases to the limit of the focal distance. 51 −0.933 mm 53 +0.667 m 55 (a) –1.5×10–2 m (b) –66.7 D 57 +0.360 m (concave) 59 (a) +0.111 (b) -0.334 cm (behind “mirror”) (c) 0.752cm 61 63 6.82 kW/m2 = i o = − i o = − −25.61) Test Prep for AP® Courses 1 (c) 3 (c) 5 (a) 7 Since light bends toward the normal upon entering a medium with a higher index of refraction, the upper path is a more accurate representation of a light ray moving from A to B. 9 First, measure the angle of incidence and the angle of refraction for light entering the plastic from air. Since the two angles can be measured and the index of refraction of air is known, the student can solve for the index of refraction of the plastic. Next, measure the angle of incidence and the angle of refraction for light entering the gas from the plastic. Since the two angles can be measured and the index of refraction of the plastic is known, the student can solve for the index of refraction of the gas. 11 1634 Answer Key The speed of light in a medium is simply c/n, so the speed of light in water is 2.25 × 108 m/s. From Snell’s law, the angle of incidence is 44°. 13 (d) 15 (a) 17 (a) 19 (b) Chapter 26 Problems & Exercises 1 52.0 D 3 (a) −0.233 mm (b) The size of the rods and the cones is smaller than the image height, so we can distinguish letters on a page. 5 (a) +62.5 D (b) –0.250 mm (c) –0.0800 mm 6 2.00 m 8 (a) ±0.45 D (b) The person was nearsighted because the patient was myopic and the power was reduced. 10 0.143 m 12 1.00 m 14 20.0 cm 16 –5.00 D 18 25.0 cm 20 –0.198 D 22 30.8 cm 24 –0.444 D 26 (a) 4.00 (b) 1600 28 (a) 0.501 cm (b) Eyepiece should be 204 cm behind the objective lens. 30 (a) +18.3 cm (on the eyepiece side of the objective lens) This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key (b) -60.0 (c) -11.3 cm (on the objective side of the eyepiece) 1635 (d) +6.67 (e) -400 33 −40.0 35 −1.67 37 +10.0 cm 39 (a) 0.251 μm (b) Yes, this thickness implies that the shape of the cornea can be very finely controlled, producing normal distant vision in more than 90% of patients. Test Prep for AP® Courses 1 (a) 3 (c) 5 (a) 7 (b) 9 (d) 11 (c) Chapter 27 Problems & Exercises 1 1 / 1.333 = 0.750 3 1.49, Polystyrene 5 0.877 glass to water 6 0.516º 8 1.22×10−6 m 10 600 nm 12 2.06º 14 1200 nm (not visible) 16 (a) 760 nm (b) 1520 nm 18 For small angles sin − tan ≈ (in radians) . For two adjacent fringes we have, 1636 and Subtracting these equations gives sin m = sin m + 1 = ( + 1) ( + 1) − sin m + 1 − sin tan = = Answer Key (27.11) (27.12) (27.13) 20 450 nm 21 5.97º 23 8.99×103 25 707 nm 27 (a) 11.8º, 12.5º, 14.1º, 19.2º (b) 24.2º, 25.7º, 29.1º, 41.0º (c) Decreasi
ng the number of lines per centimeter by a factor of x means that the angle for the x‐order maximum is the same as the original angle for the first- order maximum. 29 589.1 nm and 589.6 nm 31 28.7º 33 43.2º 35 90.0º 37 (a) The longest wavelength is 333.3 nm, which is not visible. (b) 333 nm (UV) (c) 6.58×103 cm 39 1.13×10−2 m 41 (a) 42.3 nm (b) Not a visible wavelength The number of slits in this diffraction grating is too large. Etching in integrated circuits can be done to a resolution of 50 nm, so slit separations of 400 nm are at the limit of what we can do today. This line spacing is too small to produce diffraction of light. 43 (a) 33.4º (b) No 45 (a) 1.35×10−6 m This content is available for free at http://cnx.org/content/col11844/1.13 1637 Answer Key (b) 69.9º 47 750 nm 49 (a) 9.04º (b) 12 51 (a) 0.0150º (b) 0.262 mm (c) This distance is not easily measured by human eye, but under a microscope or magnifying glass it is quite easily measurable. 53 (a) 30.1º (b) 48.7º (c) No (d) 2θ1 = (2)(14.5º) = 29º 2 − 1 = 30.05º−14.5º=15.56º . Thus, 29º ≈ (2)(15.56º) = 31.1º . 55 23.6º and 53.1º 57 (a) 1.63×10−4 rad (b) 326 ly 59 1.46×10−5 rad 61 (a) 3.04×10−7 rad (b) Diameter of 235 m 63 5.15 cm 65 (a) Yes. Should easily be able to discern. (b) The fact that it is just barely possible to discern that these are separate bodies indicates the severity of atmospheric aberrations. 70 532 nm (green) 72 83.9 nm 74 620 nm (orange) 76 380 nm 78 33.9 nm 80 4.42×10−5 m 82 The oil film will appear black, since the reflected light is not in the visible part of the spectrum. 84 45.0º 86 Answer Key 1638 45.7 mW/m2 88 90.0% 90 0 92 48.8º 94 41.2º 96 (a) 1.92, not diamond (Zircon) (b) 55.2º 98 2 = 0.707 1 100 (a) 2.07×10-2 °C/s (b) Yes, the polarizing filters get hot because they absorb some of the lost energy from the sunlight. Test Prep for AP® Courses 1 (b) 3 (b) and (c) 5 (b) 7 (b) 9 (b) 11 (d) 13 (b) 15 (d) 17 (b) Chapter 28 Problems & Exercises 1 (a) 1.0328 (b) 1.15 3 5.96×10−8 s 5 0.800 7 0.140 9 (a) 0.745 (b) 0.99995 (to five digits to show effect) This content is available for free at http://cnx.org/content/col11844/1.13 1639 Answer Key 11 (a) 0.996 (b) cannot be less than 1. (c) Assumption that time is longer in moving ship is unreasonable. 12 48.6 m 14 (a) 1.387 km = 1.39 km (b) 0.433 km (c) = Thus, the distances in parts (a) and (b) are related when = 3.20 . 16 (a) 4.303 y (to four digits to show any effect) (b) 0.1434 y (c) Δt = γΔt0 ⇒ = Δt Δt0 = 4.303 y 0.1434 y = 30.0 Thus, the two times are related when 30.00 . 18 (a) 0.250 (b) must be ≥1 (c) The Earth-bound observer must measure a shorter length, so it is unreasonable to assume a longer length. 20 (a) 0.909 (b) 0.400 22 0.198 24 a) 658 nm b) red c) / 9.92×10−5 (negligible) 26 0.991 28 −0.696 30 0.01324 32 ′ = , so = 1 + ( 2) = 1 + ( / 2) = 1 + ( / ) = () = 34 a) 0.99947 1640 b) 1.2064×1011 y c) 1.2058×1011 y (all to sufficient digits to show effects) 35 4.09×10–19 kg ⋅ m/s 37 (a) 3.000000015×1013 kg ⋅ m/s . Answer Key (b) Ratio of relativistic to classical momenta equals 1.000000005 (extra digits to show small effects) 39 2.9957×108 m/s 41 (a) 1.121×10–8 m/s (b) The small speed tells us that the mass of a proton is substantially smaller than that of even a tiny amount of macroscopic matter! 43 8.20×10−14 J 0.512 MeV 45 2.3×10−30 kg 47 (a) 1.11×1027 kg (b) 5.56×10−5 49 7.1×10−3 kg 7.1×10−3 The ratio is greater for hydrogen. 51 208 0.999988 53 6.92×105 J 1.54 55 (a) 0.914 (b) The rest mass energy of an electron is 0.511 MeV, so the kinetic energy is approximately 150% of the rest mass energy. The electron should be traveling close to the speed of light. 57 90.0 MeV 59 This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key 1641 (a) 2 − 1 2 = 22 + 24 = 22 4 so that 2 4 , and therefore 2 2 = ()2 2 2 = 2 − 1 (b) yes 61 1.07×103 63 6.56×10−8 kg 4.37×10−10 65 0.314 0.99995 67 (a) 1.00 kg (b) This much mass would be measurable, but probably not observable just by looking because it is 0.01% of the total mass. 69 (a) 6.3×1011 kg/s (b) 4.5×1010 y (c) 4.44×109 kg (d) 0.32% Test Prep for AP® Courses 1 (a) 3 The relativistic Doppler effect takes into account the special relativity concept of time dilation and also does not require a medium of propagation to be used as a point of reference (light does not require a medium for propagation). 5 Relativistic kinetic energy is given as KErel = ( − 1)2 where = 1 1 − 2 2 Classical kinetic energy is given as KEclass = 1 22 At low velocities = 0 , a binomial expansion and subsequent approximation of gives: = 1 + 12 22 or − 1 = 12 22 Substituting − 1 in the expression for KErel gives KErel = 12 22 2 = 1 22 = KEclass 1642 Answer Key Hence, relativistic kinetic energy becomes classical kinetic energy when ≪ . Chapter 29 Problems & Exercises 1 (a) 0.070 eV (b) 14 3 (a) 2.21×1034 J (b) 2.26×1034 (c) No 4 263 nm 6 3.69 eV 8 0.483 eV 10 2.25 eV 12 (a) 264 nm (b) Ultraviolet 14 1.95×106 m/s 16 (a) 4.02×1015 /s (b) 0.256 mW 18 (a) –1.90 eV (b) Negative kinetic energy (c) That the electrons would be knocked free. 20 6.34×10−9 eV , 1.01×10−27 J 22 2.42×1020 Hz 24 = 6.62607×10−34 J ⋅ s 2.99792×108 m/s 109 nm 1 m 1.00000 eV 1.60218×10−19 J (29.22) = 1239.84 eV ⋅ nm ≈ 1240 eV ⋅ nm 26 (a) 0.0829 eV (b) 121 (c) 1.24 MeV (d) 1.24×105 28 This content is available for free at http://cnx.org/content/col11844/1.13 1643 Answer Key (a) 25.0×103 eV (b) 6.04×1018 Hz 30 (a) 2.69 (b) 0.371 32 (a) 1.25×1013 photons/s (b) 997 km 34 8.33×1013 photons/s 36 181 km 38 (a) 1.66×10−32 kg ⋅ m/s (b) The wavelength of microwave photons is large, so the momentum they carry is very small. 40 (a) 13.3 μm (b) 9.38×10-2 eV 42 (a) 2.65×10−28 kg ⋅ m/s (b) 291 m/s (c) electron 3.86×10−26 J , photon 7.96×10−20 J , ratio 2.06×106 44 (a) 1.32×10−13 m (b) 9.39 MeV (c) 4.70×10−2 MeV 46 = 2 and = , so = 2 = 2 . (29.35) As the mass of particle approaches zero, its velocity will approach , so that the ratio of energy to momentum in this limit is lim→0 = 2 = (29.36) which is consistent with the equation for photon energy. 48 (a) 3.00×106 W (b) Headlights are way too bright. (c) Force is too large. 49 7.28×10–4 m Answer Key 15.1 keV (29.42) 1644 51 6.62×107 m/s 53 1.32×10–13 m 55 (a) 6.62×107 m/s (b) 22.9 MeV 57 59 (a) 5.29 fm (b) 4.70×10−12 J (c) 29.4 MV 61 (a) 7.28×1012 m/s (b) This is thousands of times the speed of light (an impossibility). (c) The assumption that the electron is non-relativistic is unreasonable at this wavelength. 62 (a) 57.9 m/s (b) 9.55×10−9 eV (c) From Table 29.1, we see that typical molecular binding energies range from about 1eV to 10 eV, therefore the result in part (b) is approximately 9 orders of magnitude smaller than typical molecular binding energies. 64 29 nm, 290 times greater 66 1.10×10−13 eV 68 3.3×10−22 s 70 2.66×10−46 kg 72 0.395 nm 74 (a) 1.3×10−19 J (b) 2.1×1023 (c) 1.4×102 s 76 (a) 3.35×105 J (b) 1.12×10–3 kg ⋅ m/s (c) 1.12×10–3 m/s (d) 6.23×10–7 J This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key 78 (a) 1.06×103 (b) 5.33×10−16 kg ⋅ m/s (c) 1.24×10−18 m 80 (a) 1.62×103 m/s 1645 (b) 4.42×10−19 J for photon, 1.19×10−24 J for electron, photon energy is 3.71×105 times greater (c) The light is easier to make because 450-nm light is blue light and therefore easy to make. Creating electrons with 7.43 μeV of energy would not be difficult, but would require a vacuum. 81 (a) 2.30×10−6 m (b) 3.20×10−12 m 83 3.69×10−4 ºC 85 (a) 2.00 kJ (b) 1.33×10−5 kg ⋅ m/s (c) 1.33×10−5 N (d) yes Test Prep for AP® Courses 1 (b) 3 (c) 5 (b) 7 (c) 9 (c) 11 (a) 13 (a) 15 (c) 17 (d) 19 (d) Chapter 30 Problems & Exercises 1 1.84×103 3 50 km 1646 4 6×1020 kg/m3 6 (a) 10.0 μm Answer Key (b) It isn’t hard to make one of approximately this size. It would be harder to make it exactly 10.0 μm . i ⋅ f)2 2 2 − f i i = 2, f = 1, so that = m 1.097×107 (2×1)2 22 − 12 = 1.22×10−7 m = 122 nm , which is UV radiation6.626×10−34 J·s)2 4 2(9.109×10−31 kg)(8.988×109 N·m2 / C2)(1)(1.602×10−19 C)2 = 0.529×10−10 m 11 0.850 eV 13 2.12×10–10 m 15 365 nm It is in the ultraviolet. 17 No overlap 365 nm 122 nm 19 7 21 (a) 2 (b) 54.4 eV = 2 23 2 2 velocity, giving: = 2 , so that = so that = 2 2 2 . From the equation = 2 , we can substitute for the 2 2 4 2 = 2 B where B = 2 2 4π2 . 25 (a) 0.248×10−10 m (b) 50.0 keV (c) The photon energy is simply the applied voltage times the electron charge, so the value of the voltage in volts is the same as the value of the energy in electron volts. 27 (a) 100×103 eV , 1.60×10−14 J (b) 0.124×10−10 m 29 (a) 8.00 keV This content is available for free at http://cnx.org/content/col11844/1.13 1647 Answer Key (b) 9.48 keV 30 (a) 1.96 eV (b) (1240 eV·nm) / (1.96 eV) = 633 nm (c) 60.0 nm 32 693 nm 34 (a) 590 nm (b) (1240 eV·nm) / (1.17 eV) = 1.06 μm 35 = 4, 3 are possible since < and ∣ ∣ ≤ . 37 = 4 ⇒ = 3, 2, 1, 0 ⇒ = ±3, ± 2, ± 1, 0 are possible. 39 (a) 1.49×10−34 J ⋅ s (b) 1.06×10−34 J ⋅ s 41 (a) 3.66×10−34 J ⋅ s (b) = 9.13×10−35 J ⋅ s (c) = 12 3 / 4 = 4 43 = 54.7º, 125.3º 44 (a) 32. (b) 2 in 6 in 10 in and 14 in , for a total of 32. 46 (a) 2 (b) 3 9 48 (b) ≥ is violated, (c) cannot have 3 electrons in subshell since 3 > (2 + 1) = 2 (d) cannot have 7 electrons in subshell since 7 > (2 + 1) = 2(2 + 1) = 6 50 (a) The number of different values of overall factor of 2 since each can have equal to either +1 / 2 or −1 / 2 ⇒ 2(2 + 1) . is ± ± ( − 1), ...,0 for each > 0 and one for = 0 ⇒ (2 + 1). Also an (b) for each value of , you get 2(2 + 1) = 0, 1, 2, ...,(–1) ⇒ 2 (2)(0) + 1 + (2)(1) + 1 + .... + (2)( − 1) + 1 = 2 1 + 3 + ... + (2 − 3) + (2 − 1) ⏟ terms to see that the expression in the box is = 2 imagine taking ( − 1) from the last term and adding it to first term 1648 Answer Key = 2 1 + (–1) + 3 + ... + (2 − 3) + (2 − 1)( − 1) = 2 . Now take ( − 3) from penultimate term and add to the second term 2[ + + ... + + ] + 3 + .... + (2 − 3) + = 22 . ⏟ terms 52 The electric force on the electron is up (toward the positively c
harged plate). The magnetic force is down (by the RHR). 54 401 nm 56 (a) 6.54×10−16 kg (b) 5.54×10−7 m 58 1.76×1011 C/kg , which agrees with the known value of 1.759×1011 C/kg to within the precision of the measurement 60 (a) 2.78 fm (b) 0.37 of the nuclear radius. 62 (a) 1.34×1023 (b) 2.52 MW 64 (a) 6.42 eV (b) 7.27×10−20 J/molecule (c) 0.454 eV, 14.1 times less than a single UV photon. Therefore, each photon will evaporate approximately 14 molecules of tissue. This gives the surgeon a rather precise method of removing corneal tissue from the surface of the eye. 66 91.18 nm to 91.22 nm 68 (a) 1.24×1011 V (b) The voltage is extremely large compared with any practical value. (c) The assumption of such a short wavelength by this method is unreasonable. Test Prep for AP® Courses 1 (a), (d) 3 (a) 5 (a) 7 (b) 9 (a) 11 (d) 13 (d) 15 (a), (c) This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key 1649 Chapter 31 Problems & Exercises 1 1.67×104 5 = = 3 ⇒ = 1 3 1/3 = 2.3×1017 kg 1000 kg/m3 = 61×103 m = 61 km 7 1.9 fm 9 (a) 4.6 fm (b) 0.61 to 1 11 85.4 to 1 13 12.4 GeV 15 19.3 to 1 17 19 21 23 25 27 29 3 H2 → 2 1 ¯ 3 He1 + − + 50 25 → 24 25 50 Cr26 + + + 7 Be3 + − → 3 4 7 Li4 + 210 Po126 → 82 84 206 Pb124 + 2 4He2 137 Cs82 → 56 55 ¯ 137 Ba81 + − + 232 Th142 → 88 90 228 Ra140 + 2 4He2 (31.17) (31.47) (31.48) (31.49) (31.50) (31.51) (31.52) (a) charge:(+1) + (−1) = 0 electron family number: (+1) + (−1b) 0.511 MeV (c) The two rays must travel in exactly opposite directions in order to conserve momentum, since initially there is zero momentum if the center of mass is initially at rest. 31 33 35 = ( + 1) − 1; = ; efn : 0 = (+1) + (−1) - 1 = − 1; = ; efn :(+1) = (+1) (31.53) (31.54) (a) 88 226 Ra138 → 86 222 Rn136 + 2 4He2 (b) 4.87 MeV 37 1650 ¯ (a) n → p + − + (b) ) 0.783 MeV 39 1.82 MeV 41 (a) 4.274 MeV (b) 1.927×10−5 Answer Key (c) Since U-238 is a slowly decaying substance, only a very small number of nuclei decay on human timescales; therefore, although those nuclei that decay lose a noticeable fraction of their mass, the change in the total mass of the sample is not detectable for a macroscopic sample. 43 (a) 8 15 O7 + − → 7 15 N8 + (b) 2.754 MeV 44 57,300 y 46 (a) 0.988 Ci (b) The half-life of 226 Ra is now better known. 48 1.22×103 Bq 50 (a) 16.0 mg (b) 0.0114% 52 1.48×1017 y 54 5.6×104 y 56 2.71 y 58 (a) 1.56 mg (b) 11.3 Ci 60 (a) 1.23×10−3 (b) Only part of the emitted radiation goes in the direction of the detector. Only a fraction of that causes a response in the detector. Some of the emitted radiation (mostly particles) is observed within the source. Some is absorbed within the source, some is absorbed by the detector, and some does not penetrate the detector. 62 (a) 1.68×10 – 5 Ci (b) 8.65×1010 J (c) $ 2.9×103 64 (a) 6.97×1015 Bq This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key (b) 6.24 kW (c) 5.67 kW 68 (a) 84.5 Ci 1651 (b) An extremely large activity, many orders of magnitude greater than permitted for home use. (c) The assumption of 1.00 μA is unreasonably large. Other methods can detect much smaller decay rates. 69 1.112 MeV, consistent with graph 71 7.848 MeV, consistent with graph 73 (a) 7.680 MeV, consistent with graph (b) 7.520 MeV, consistent with graph. Not significantly different from value for 12 C , but sufficiently lower to allow decay into another nuclide that is more tightly bound. 75 (a) 1.46×10−8 u vs. 1.007825 u for 1 H (b) 0.000549 u (c) 2.66×10−5 76 (a) –9.315 MeV (b) The negative binding energy implies an unbound system. (c) This assumption that it is two bound neutrons is incorrect. 78 22.8 cm 79 (a) 92 235 U143 → 90 231 Th141 + 2 4 He2 (b) 4.679 MeV (c) 4.599 MeV 81 a) 2.4×108 u (b) The greatest known atomic masses are about 260. This result found in (a) is extremely large. (c) The assumed radius is much too large to be reasonable. 82 (a) –1.805 MeV (b) Negative energy implies energy input is necessary and the reaction cannot be spontaneous. (c) Although all conversation laws are obeyed, energy must be supplied, so the assumption of spontaneous decay is incorrect. Test Prep for AP® Courses 1 (c) 3 (a) 5 When 95 7 241 Am undergoes α decay, it loses 2 neutrons and 2 protons. The resulting nucleus is therefore 93 237 Np . 1652 Answer Key During this process, the nucleus emits a particle with -1 charge. In order for the overall charge of the system to remain constant, the charge of the nucleus must therefore increase by +1. 9 a. No. Nucleon number is conserved (238 = 234 + 4), but the atomic number or charge is NOT conserved (92 ≠ 88+2). b. Yes. Nucleon number is conserved (223 = 209 + 14), and atomic number is conserved (88 = 82 + 6). c. Yes. Nucleon number is conserved (14 = 14), and charge is conserved if the electron’s charge is properly counted (6 = 7 + (-1)). d. No. Nucleon number is not conserved (24 ≠ 23). The positron released counts as a charge to conserve charge, but it doesn’t count as a nucleon. 11 This must be alpha decay since 4 nucleons (2 positive charges) are lost from the parent nucleus. The number remaining is found from: −0.693 N() = N0 1 2 = 3.4×1017 −(0.693)(0.035) 0.00173 N() = 4.1×1011 nuclei Chapter 32 Problems & Exercises 1 5.701 MeV 3 99 Mo57 → 43 42 5 1.43×10−9 g 7 ¯ 99 Tc56 + − + (a) 6.958 MeV (b) 5.7×10−10 g 8 (a) 100 mSv (b) 80 mSv (c) ~30 mSv 10 ~2 Gy 12 1.69 mm 14 1.24 MeV 16 7.44×108 18 4.92×10–4 Sv 20 4.43 g 22 0.010 g 24 95% 26 This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key 1653 (a) =1+1=2 , =1+1=1+1 , efn = 0 = −1 + 1 (b) =1+2=3 , =1+1=2 , efn=0=0 (c) =3+3=4+1+1 , =2+2=2+1+1 , efn=0=0 28 = (i − f)2 4 − 4(1.007825) − 4.002603 4 He = 1 H 2 = = 26.73 MeV (931.5 MeV) 30 3.12×105 kg (about 200 tons) 32 = (i − f)2 1 = (1.008665 + 3.016030 − 4.002603)(931.5 MeV) = 20.58 MeV 2 = (1.008665 + 1.007825 − 2.014102)(931.5 MeV) = 2.224 MeV 4 He is more tightly bound, since this reaction gives off more energy per nucleon. 34 1.19×104 kg 36 2− + 41 H → 4 He + 7γ + 2 38 (a) =12+1=13 , =6+1=7 , efn = 0 = 0 (b) =13=13 , =7=6+1 , efn = 0 = −1 + 1 (c) =13 + 1=14 , =6+1=7 , efn = 0 = 0 (d) =14 + 1=15 , =7+1=8 , efn = 0 = 0 (e) =15=15 , =8=7+1 , efn = 0 = −1 + 1 (f) =15 + 1=12 + 4 , =7+1=6 + 2 , efn = 0 = 0 40 = 20.6 MeV 4 He = 5.68×10-2 MeV 42 (a) 3×109 y (b) This is approximately half the lifetime of the Earth. 43 (a) 177.1 MeV (b) Because the gain of an external neutron yields about 6 MeV, which is the average BE/ for heavy nuclei. (c) = 1 + 238 = 96 + 140 + 1 + 1 + 1, = 92 = 38 + 53 efn = 0 = 0 45 (a) 180.6 MeV 1654 Answer Key (b) = 1 + 239 = 96 + 140 + 1 + 1 + 1 + 1, = 94 = 38 + 56 efn = 0 = 0 47 238 U + → 239 U + 4.81 MeV 239 U → 239 Np + − + 0.753 MeV Np → Pu + − + 0.211 MeV 49 (a) 2.57×103 MW (b) 8.03×1019 fission/s (c) 991 kg 51 0.56 g 53 4.781 MeV 55 (a) Blast yields 2.1×1012 J to 8.4×1011 J , or 2.5 to 1, conventional to radiation enhanced. (b) Prompt radiation yields 6.3×1011 J to 2.1×1011 J , or 3 to 1, radiation enhanced to conventional. 57 (a) 1.1×1025 fissions , 4.4 kg (b) 3.2×1026 fusions , 2.7 kg (c) The nuclear fuel totals only 6 kg, so it is quite reasonable that some missiles carry 10 overheads. The mass of the fuel would only be 60 kg and therefore the mass of the 10 warheads, weighing about 10 times the nuclear fuel, would be only 1500 lbs. If the fuel for the missiles weighs 5 times the total weight of the warheads, the missile would weigh about 9000 lbs or 4.5 tons. This is not an unreasonable weight for a missile. 59 7×104 g 61 (a) 4.86×109 W (b) 11.0 y Test Prep for AP® Courses 1 (b) 3 (c) 5 (d) 7 (d) 9 (b) Chapter 33 Problems & Exercises 1 3×10−39 s 3 This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key 1655 1.99×10−16 m (0.2 fm) 4 (a) 10−11 to 1, weak to EM (b) 1 to 1 6 (a) 2.09×10−5 s (b) 4.77×104 Hz 8 78.0 cm 10 1.40×106 12 100 GeV 13 67.5 MeV 15 (a) 1×1014 (b) 2×1017 17 (a) 1671 MeV (b) = 1. = − 1; ′ = − 1; = 0; ′ = − 1 + 1 = 0 ¯ − → −+ + ¯ ⇒ − antiparticle of +; of ¯ ; of (c) 19 (a) 3.9 eV (b) 2.9×10−8 21 (a) The composition is the same as for a proton. (b) 3.3×10−24 s (c) Strong (short lifetime) 23 a) Δ++() = ) 1656 Answer Key Figure 33.20. 25 (a) +1 (b) = 1 = 1 + 0, = = 0 + ( − 1) , all lepton numbers are 0 before and after (c) () → () + ( ) 27 (ab) 277.9 MeV (c) 547.9 MeV 29 No. Charge = −1 is conserved. i 31 = 0 ≠ f = 2 is not conserved. = 1 is conserved. (a)Yes. = −1 = 0 + ( − 1) , = 1 = 1 + 0 , all lepton family numbers are 0 before and after, spontaneous since mass greater before reaction. (b) → + 33 (a) 216 (b) There are more baryons observed because we have the 6 antiquarks and various mixtures of quarks (as for the π-meson) as well. = −1, − 1 3 35 Ω. 37 = 1, (a)803 MeV This content is available for free at http://cnx.org/content/col11844/1.13 Answer Key (b) 938.8 MeV (c) The annihilation energy of an extra electron is included in the total energy. 1657 39 ¯ 41 a)The antiproton - → 0 + − b) 43 (a) 5×1010 (b) 5×104 particles/m2 45 2.5×10−17 m 47 (a) 33.9 MeV (b) Muon antineutrino 29.8 MeV, muon 4.1 MeV (kinetic energy) 49 (a) 7.2×105 kg (b) 7.2×102 m3 (c) 100 months Test Prep for AP® Courses 1 (d) 3 (d) 5 (b) 7 (a) 9 (c), though this comes from Einstein's special relativity 11 (a) 13 (d) 15 (b) 17 (b) Chapter 34 Problems & Exercises 1 3×1041 kg 3 (a) 3×1052 kg (b) 2×1079 1658 (c) 4×1088 5 0.30 Gly 7 (a) 2.0×105 km/s (b) 0.67 Answer Key 9 2.7×105 m/s 11 6×10−11 (an overestimate, since some of the light from Andromeda is blocked by gas and dust within that galaxy) 13 (a) 2×10−8 kg (b) 1×1019 15 (a) 30km/s ⋅ Mly (b) 15km/s ⋅ Mly 17 960 rev/s 19 89.999773º (many digits are used to show the difference between 90º ) 22 23.6 km 24 (a) 2.95×1012 m (b) 3.12×10−4 ly 26 (a) 1×1020 (b) 10 times greater 27 29 31 1.5×1015 0.6 m−3 0.30 Ω (34.6) (34.7) (34.8) This content is available for free at http://cnx.org/content/col11844/1.13 Index Index Symbols (peak) emf, 1030 RC circuit, 945 A aberration, 1176 aberrations,
1175 absolute pressure, 454, 478 absolute zero, 532, 565 AC current, 887, 901 AC voltage, 887, 901 acceleration, 43, 82, 146, 178 acceleration due to gravity, 68, 82 accommodation, 1155, 1176 Accuracy, 22 accuracy, 29 acoustic impedance , 755, 761 active transport, 518, 519 activity, 1400, 1410 adaptive optics, 1174, 1176 adhesive forces, 465, 478 adiabatic, 632 adiabatic process, 661 air resistance, 113, 128 alpha, 1379 alpha decay, 1391, 1410 alpha rays, 1410 Alternating current, 886 alternating current, 901 ammeter, 950 ammeters, 938 ampere, 869, 901 Ampere’s law, 991, 999 amplitude, 681, 711, 1074, 1090, 1092 amplitude modulation, 1080 amplitude modulation (AM), 1092 analog meter, 950 Analog meters, 939 analytical method, 128 Analytical methods, 107 Anger camera, 1427, 1453 angular acceleration, 391, 425 angular magnification, 1172, 1176 angular momentum, 413, 425 angular momentum quantum number, 1353, 1364 angular velocity, 222, 249 antielectron, 1395, 1410 antimatter, 1394, 1410 antinode, 706, 711, 743, 761 approximation, 29 approximations, 27 arc length, 221, 249 Archimedes' principle, 459, 478 astigmatism, 1161, 1176 atom, 1318, 1364 atomic de-excitation, 1340, 1364 atomic excitation, 1340, 1364 atomic mass, 1386, 1410 atomic number, 1358, 1364, 1386, 1410 atomic spectra, 1279, 1306 Average Acceleration, 43 average acceleration, 84, 82 Average speed, 41 average speed, 82 Average velocity, 40 average velocity, 82 Avogadro’s number, 546, 565 axions, 1518, 1524 axis of a polarizing filter, 1213, 1224 B B-field, 973, 999 back emf, 1031, 1056 banked curve, 249 banked curves, 230 barrier penetration, 1408, 1410 baryon number, 1478, 1492 Baryons, 1478 baryons, 1492 basal metabolic rate, 296, 301 beat frequency, 707, 711 becquerel, 1400, 1410 Bernoulli's equation, 496, 519 Bernoulli's principle, 497, 519 beta, 1379 beta decay, 1393, 1410 beta rays, 1410 Big Bang, 1504, 1524 binding energy, 1281, 1306, 1403, 1410 binding energy per nucleon, 1405, 1410 bioelectricity, 895, 901 Biot-Savart law, 991, 999 birefringent, 1221, 1224 Black holes, 1511 black holes, 1524 blackbodies, 1278 blackbody, 1306 blackbody radiation, 1278, 1306 Bohr radius, 1331, 1364 Boltzmann constant, 544, 565 boson, 1476, 1492 bottom, 1486, 1492 bow wake, 739, 761 break-even, 1442, 1453 breeder reactors, 1448, 1454 breeding, 1448, 1454 bremsstrahlung, 1287, 1306 Brewster’s angle, 1216, 1224 Brewster’s law, 1216, 1224 bridge device, 950 bridge devices, 944 Brownian motion, 1319, 1364 buoyant force, 458, 478 C capacitance, 843, 855, 945, 950 capacitive reactance, 1049, 1056 capacitor, 841, 855, 945, 950 capillary action, 470, 478 carbon-14 dating, 1398, 1410 Carnot cycle, 640, 661 Carnot efficiency, 640, 661 Carnot engine, 640, 661 carrier particle, 178 carrier particles, 177 carrier wave, 1080, 1092 cathode-ray tube, 1364 cathode-ray tubes, 1320 Celsius, 531 1659 Celsius scale, 565 center of gravity, 364, 380 center of mass, 237, 249 centrifugal force, 233, 249 centrifuge, 226 centripetal acceleration, 225, 249 centripetal force, 228, 249 change in angular velocity, 391, 425 change in entropy, 650, 661 change in momentum, 319, 343 Chaos, 1520 chaos, 1524 characteristic time constant, 1045, 1057 characteristic x rays, 1287, 1306 charm, 1486, 1492 chart of the nuclides, 1389, 1410 chemical energy, 287, 301 classical physics, 12, 29 Classical relativity, 125 classical relativity, 128 classical velocity addition, 1267 coefficient of linear expansion, 537, 565 coefficient of performance, 648, 661 coefficient of volume expansion, 539, 565 coherent, 1191, 1224 cohesive forces, 465, 478 Colliding beams, 1473 colliding beams, 1492 color, 1487, 1492 color constancy, 1164, 1176 commutative, 104, 129, 128 complexity, 1519, 1524 component (of a 2-d vector), 128 components , 106 compound microscope, 1165, 1176 Compton effect, 1291, 1306 Conduction, 590 conduction, 606 conductor, 782, 805 Conductors, 786 confocal microscopes, 1223, 1224 conservation laws, 933, 950 conservation of mechanical energy, 279, 301 conservation of momentum principle, 324, 343 Conservation of total , 1478 conservation of total baryon number, 1478, 1492 conservation of total electron family number, 1492 conservation of total , 1478 conservation of total muon family number, 1492 conservative force, 277, 301 constructive interference, 704, 711 constructive interference for a diffraction grating, 1197, 1224 constructive interference for a double slit, 1193, 1224 contact angle, 470, 478 Contrast, 1222 contrast, 1224 Convection, 590 convection, 607 1660 Index converging (or convex) lens, 1122 converging lens, 1142 converging mirror, 1142 conversion factor, 19, 29 Coriolis force, 234, 249 corner reflector, 1116, 1142 correspondence principle, 1277, 1306 cosmic microwave background, 1505, 1524 cosmological constant, 1516, 1525 cosmological red shift, 1504, 1525 Cosmology, 1502 cosmology, 1525 Coulomb force, 792, 805 Coulomb forces, 791 Coulomb interaction, 799, 805 Coulomb's law, 790, 805 critical angle, 1112, 1142 Critical damping, 694 critical damping, 711 critical density, 1516, 1525 critical mass, 1446, 1454 critical point, 557, 565 Critical pressure, 557 critical pressure, 565 critical temperature, 557, 565, 1521, 1525 criticality, 1447, 1454 curie, 1400, 1411 Curie temperature, 970, 999 current, 915, 950 Current sensitivity, 939 current sensitivity, 950 cyclical process, 636, 661 cyclotron, 1472, 1492 D Dalton’s law of partial pressures, 560, 565 dark matter, 1515, 1525 daughter, 1411 daughters, 1390 de Broglie wavelength, 1296, 1306 decay, 1379, 1390, 1411 decay constant, 1398, 1411 decay equation, 1392, 1396, 1411 decay series, 1390, 1411 deceleration, 82 defibrillator, 853, 855 deformation, 203, 212, 675, 711 degree Celsius, 531, 565 degree Fahrenheit, 531, 565 Density, 441 density, 478 dependent variable, 75, 82 derived units, 16, 29 destructive interference, 704, 711 destructive interference for a double slit, 1193, 1225 destructive interference for a single slit, 1202, 1225 dew point, 561, 565 dialysis, 518, 519 diastolic pressure, 455, 478 Diastolic pressure, 474 dielectric, 846, 855 dielectric strength, 855 dielectric strengths, 846 diffraction, 1190, 1225 diffraction grating, 1196, 1225 Diffusion, 515 diffusion, 519 digital meter, 950 digital meters, 939 dipole, 799, 805 Direct current, 886 direct current, 901 direction, 102 direction (of a vector), 128 direction of magnetic field lines, 973, 999 direction of polarization, 1213, 1225 Dispersion, 1118 dispersion, 1142 displacement, 34, 82 Distance, 36 distance, 82 Distance traveled, 36 distance traveled, 82 diverging lens, 1124, 1142 diverging mirror, 1142 domains, 969, 999 Doppler effect, 736, 762 Doppler shift, 736, 762 Doppler-shifted ultrasound, 759, 762 Double-slit interference, 1329 double-slit interference, 1364 down, 1481, 1492 drag force, 198, 212 drift velocity , 872, 901 dynamic equilibrium, 358, 380 Dynamics, 141, 144, 179 dynamics, 178 E eddy current, 1024, 1057 efficiency, 290, 301 Elapsed time, 39 elapsed time, 83 elastic collision, 328, 343 elastic potential energy, 677, 711 electric and magnetic fields, 1090 electric charge, 777, 805 electric current, 869, 901 electric field, 786, 805, 1074, 1092 electric field lines, 805, 1092 Electric field lines, 1071 electric field strength, 793, 1092 electric fields, 794 electric generator, 1057 Electric generators, 1028 electric potential, 824, 855 electric power, 883, 901 Electrical energy, 287 electrical energy, 301 electrocardiogram (ECG), 900, 901 electromagnet, 999 electromagnetic force, 806 electromagnetic induction, 1018, 1057 electromagnetic spectrum, 1092 electromagnetic waves, 1073, 1090, 1092 Electromagnetism, 970 electromagnetism, 999 electromagnets, 970 electromotive force, 924 This content is available for free at http://cnx.org/content/col11844/1.13 electromotive force (emf), 950, 1092 electron, 806 Electron capture, 1396 electron capture, 1411 electron capture equation, 1396, 1411 electron family number , 1478, 1492 electron volt, 828, 856 electrons, 777 electron’s antineutrino, 1394, 1411 electron’s neutrino, 1396, 1411 electrostatic equilibrium, 786, 806 electrostatic force, 790, 806 electrostatic precipitators, 803, 806 Electrostatic repulsion, 783 electrostatic repulsion, 806 electrostatics, 800, 806 electroweak epoch, 1508, 1525 electroweak theory, 1488, 1493 emf, 934 emf induced in a generator coil, 1029, 1057 emissivity, 603, 607 endoscope, 1114 energies of hydrogen-like atoms, 1332, 1364 energy, 301 energy stored in an inductor , 1044, 1057 energy-level diagram, 1330, 1364 English units, 15, 29 entropy, 649, 661 equipotential line, 856 equipotential lines, 838 escape velocity, 1511, 1525 event horizon, 1511, 1525 external force, 146, 178 external forces, 144 External forces, 179 Extremely low frequency (ELF), 1079 extremely low frequency (ELF), 1092 eyepiece, 1166, 1176 F Fahrenheit, 531 Fahrenheit scale, 565 far point, 1158, 1176 Faraday cage, 788, 806 Faraday’s law of induction, 1019, 1057 Farsightedness, 1158 farsightedness, 1176 fermion, 1476, 1493 ferromagnetic, 969, 999 Feynman diagram, 1470, 1493 Fiber optics, 1114 fiber optics, 1142 fictitious force, 233, 249 field, 806 fine structure, 1351, 1364 first law of thermodynamics, 621, 661 first postulate of special relativity, 1239, 1267 fission fragments, 1445, 1454 flat (zero curvature) universe, 1516, 1525 flavors, 1481, 1493 Flow rate, 490 flow rate, 519 fluid dynamics, 519 Index 1661 fluids, 440, 478 Fluorescence, 1340 fluorescence, 1364 focal length, 1122, 1142 focal point, 1122, 1142 Food irradiation, 1437 food irradiation, 1454 force, 144, 178 Force, 179 force constant, 676, 711 force field, 175, 176, 178, 792 fossil fuels, 298, 301 free charge, 806 free charges, 786 free electron, 806 free electrons, 782 free radicals, 1438, 1454 free-body diagram, 144, 179, 167, 178 free-fall, 68, 83, 149, 178 Frequency, 680 frequency, 712, 1074, 1092 frequency modulation, 1080 frequency modulation (
FM), 1092 friction, 147, 178, 212, 301 Friction, 192, 282 full-scale deflection, 939, 950 fundamental, 744, 762 fundamental frequency, 706, 712 fundamental particle, 1480, 1493 fundamental units, 16, 29 G galvanometer, 939, 950 gamma, 1379 gamma camera, 1427, 1454 Gamma decay, 1397 gamma decay, 1411 gamma ray, 1088, 1092, 1306 Gamma rays, 1284 gamma rays, 1411 gauge boson, 1493 gauge bosons, 1476 gauge pressure, 454, 478 gauss, 975, 999 Geiger tube, 1383, 1411 general relativity, 1509, 1525 geometric optics, 1103, 1142 glaucoma, 475, 478 gluons, 1470, 1493 Gluons, 1489 grand unified theory, 1493 Grand Unified Theory (GUT), 1488 gravitational constant, 237 gravitational constant, G, 249 gravitational potential energy, 272, 301 Gravitational waves, 1512 gravitational waves, 1525 gray (Gy), 1429, 1454 greenhouse effect, 605, 607 grounded, 801, 806 grounding, 839, 856 GUT epoch, 1508, 1525 H Hadrons, 1476 hadrons, 1493 half-life, 1397, 1411 Hall effect, 982, 999 Hall emf, 982, 999 harmonics, 744, 762 head, 100 head (of a vector), 128 head-to-tail method, 100, 129, 128 Hearing, 724, 749 hearing, 762 heat, 576, 607 heat engine, 627, 661 heat of sublimation, 589, 607 heat pump, 661 heat pump's coefficient of performance, 646 Heisenberg uncertainty principle, 1302 Heisenberg’s uncertainty principle , 1303, 1306 henry, 1041, 1057 hertz, 1092 Higgs boson, 1490, 1493 high dose, 1430, 1454 hologram, 1347, 1364 Holography, 1347 holography, 1364 Hooke's law, 203, 212 horizontally polarized, 1213, 1225 Hormesis, 1431 hormesis, 1454 horsepower, 293, 301 Hubble constant, 1504, 1525 hues, 1162, 1176 Human metabolism, 625 human metabolism, 661 Huygens’s principle, 1188, 1225 Hydrogen spectrum wavelength, 1329 hydrogen spectrum wavelengths, 1364 hydrogen-like atom, 1331, 1364 hydrogen-spectrum wavelengths, 1328 hyperopia, 1158, 1176 I ideal angle, 249 ideal banking, 230, 249 ideal gas law, 544, 565 ideal speed, 249 Ignition, 1442 ignition, 1454 Image distance , 1128 impedance, 1051, 1057 impulse, 319, 343 Incoherent, 1191 incoherent, 1225 independent variable, 75, 83 index of refraction, 1108, 1142 inductance, 1040, 1057 induction, 783, 806, 1057 inductive reactance, 1047, 1057 inductor, 1042, 1057 inelastic collision, 332, 343 inertia, 146, 179 Inertia, 179 inertial confinement, 1442, 1454 inertial frame of reference, 165, 179, 1239, 1267 inflationary scenario, 1509, 1525 Infrared radiation, 1083 infrared radiation, 1306 infrared radiation (IR), 1092 Infrared radiation (IR), 1289 infrasound, 749, 762 ink jet printer, 802 ink-jet printer, 806 Instantaneous acceleration, 49 instantaneous acceleration, 83 Instantaneous speed, 41 instantaneous speed, 83 Instantaneous velocity, 40 instantaneous velocity, 83 insulator, 806 insulators, 782 intensity, 709, 712, 732, 762, 1091, 1092 intensity reflection coefficient, 756, 762 Interference microscopes, 1222 interference microscopes, 1225 internal energy, 622, 661 Internal kinetic energy, 328 internal kinetic energy, 343 internal resistance, 924, 950 intraocular pressure, 475, 478 intrinsic magnetic field , 1351, 1364 intrinsic spin, 1351, 1364 ionizing radiation, 1284, 1306, 1381, 1411 ionosphere, 787, 806 irreversible process, 635, 661 isobaric process, 628, 661 isochoric, 630 isochoric process, 661 isolated system, 325, 343 isothermal, 632 isothermal process, 661 isotopes, 1387, 1411 J joule, 265, 301 Joule’s law, 916, 950 junction rule, 933, 950 K Kelvin, 532 Kelvin scale, 565 kilocalorie, 576, 607 kilogram, 16, 29 kilowatt-hour, 301 kilowatt-hours, 295 kinematics, 83, 114, 128 kinematics of rotational motion, 395, 425 kinetic energy, 268, 301 kinetic friction, 192, 212 Kirchhoff’s rules, 932, 950 L Laminar, 504 laminar, 519 laser, 1343, 1364 laser printer, 806 Laser printers, 802 Laser vision correction, 1161 laser vision correction, 1176 latent heat coefficient, 607 latent heat coefficients, 585 law, 11, 29 law of conservation of angular momentum, 416, 425 law of conservation of charge, 780, 806 1662 Index law of conservation of energy, 287, 301 law of inertia, 146, 179, 179 law of reflection, 1142 law of refraction, 1110 Length contraction , 1248 length contraction, 1267 Lenz’s law, 1019, 1057 leptons, 1476, 1493 linear accelerator, 1474, 1493 linear hypothesis, 1431, 1454 Linear momentum, 316 linear momentum, 343 liquid drop model, 1445, 1454 liter, 490, 519 longitudinal wave, 702, 712 loop rule, 934, 950 Lorentz force, 975, 999 loudness, 749, 762 low dose, 1430, 1454 M MACHOs, 1518, 1525 macrostate, 656, 661 magic numbers, 1389, 1411 magnetic confinement, 1442, 1454 magnetic damping, 1024, 1057 magnetic field, 973, 999, 1074, 1092 magnetic field lines, 973, 1000, 1093 Magnetic field lines, 1072 magnetic field strength, 1093 magnetic field strength (magnitude) produced by a long straight currentcarrying wire, 990, 1000 magnetic field strength at the center of a circular loop, 991, 1000 magnetic field strength inside a solenoid, 992, 1000 magnetic flux, 1018, 1057 magnetic force, 975, 1000 magnetic monopoles, 972, 1000 Magnetic resonance imaging (MRI), 998 magnetic resonance imaging (MRI), 1000 magnetized, 969, 1000 magnetocardiogram (MCG), 999, 1000 magnetoencephalogram (MEG), 999, 1000 magnification, 1128, 1143 magnitude, 102 magnitude (of a vector), 128 magnitude of kinetic friction, 212 magnitude of kinetic friction fk , 193 magnitude of static friction, 212 magnitude of static friction fs , 193 magnitude of the intrinsic (internal) spin angular momentum, 1355, 1364 mass, 146, 179 Mass, 179 mass number, 1386, 1411 massive compact halo objects, 1518 maximum field strength, 1090, 1093 Maxwell’s equations, 991, 1000, 1071, 1093 mechanical advantage, 371, 380 mechanical energy, 279, 301, 856 Mechanical energy, 829 mechanical equivalent of heat, 577, 607 meson, 1468, 1493 Mesons, 1478 metabolic rate, 296, 301 metastable, 1342, 1364 meter, 16, 29, 1000 Meters, 988 method of adding percents, 24, 29 metric system, 17, 29 Michelson-Morley experiment, 1240, 1267 Microgravity, 242 microgravity, 249 microlensing, 1518, 1525 microshock sensitive, 894, 901 microstate, 656, 661 Microwaves, 1082, 1289 microwaves, 1093, 1306 micturition reflex, 477, 478 mirror, 1143 model, 11, 29, 41, 83 moderate dose, 1430, 1454 Modern physics, 14 modern physics, 29 mole, 546, 565 moment of inertia, 401, 401, 425 motion, 113, 128 motor, 1000 Motors, 986 muon family number, 1478, 1493 Mutual inductance, 1040 mutual inductance, 1057 myopia, 1158, 1176 N natural frequency, 698, 712 near point, 1158, 1176 Nearsightedness, 1158 nearsightedness, 1176 negatively curved, 1516, 1525 Nerve conduction, 895 nerve conduction, 901 net external force, 147, 179 net rate of heat transfer by radiation, 604, 607 net work, 267, 301 neutral equilibrium, 366, 380 neutralinos, 1518, 1525 neutrino, 1393, 1411 neutrino oscillations, 1518, 1525 neutron, 1386, 1411 Neutron stars, 1512 neutron stars, 1525 Neutron-induced fission, 1445 neutron-induced fission, 1454 newton, 149 Newton's universal law of gravitation, 236, 249 newton-meters, 265 Newton’s first law of motion, 145, 179, 179 Newton’s second law of motion, 146, 179 Newton’s third law of motion, 153, 180, 179 node, 743, 762 Nodes, 705 This content is available for free at http://cnx.org/content/col11844/1.13 nodes, 712 non-inertial frame of reference, 233, 249 nonconservative force, 282, 301 normal force, 158, 179 north magnetic pole, 967, 1000 note, 762 notes, 749 Nuclear energy, 287 nuclear energy, 301 Nuclear fission, 1444 nuclear fission, 1454 Nuclear fusion, 1439 nuclear fusion, 1454 nuclear magnetic resonance (NMR), 998, 1000 nuclear radiation, 1379, 1411 nuclear reaction energy, 1392, 1411 nucleons, 1386, 1411 nucleus, 1411 nuclide, 1386, 1411 Null measurements, 942 null measurements, 951 numerical aperture, 1176 numerical aperture , 1167 O objective lens, 1166, 1177 ohm, 875, 901 Ohm's law, 874, 901 ohmic, 875, 901 ohmmeter, 951 ohmmeters, 943 Ohm’s law, 915, 951 optically active, 1220, 1225 orbital angular momentum, 1350, 1364 orbital magnetic field, 1350, 1364 order, 1193, 1225 order of magnitude, 17, 29 oscillate, 712, 1093 Osmosis, 518 osmosis, 519 osmotic pressure, 518, 519 Otto cycle, 638, 661 over damping, 712 overdamped, 695 overtones, 706, 712, 744, 762 P parallel, 917, 951 parallel plate capacitor, 842, 856 parent, 1390, 1411 Partial pressure, 560 partial pressure, 565 particle physics, 1493 particle-wave duality, 1295, 1306 Particle-wave duality, 1304 Pascal's principle, 451 Pascal's Principle, 478 Pauli exclusion principle, 1358, 1365 peak emf, 1057 percent relative humidity, 563, 565 percent uncertainty, 24, 29 perfectly inelastic collision, 332, 343 period, 680, 712 periodic motion, 680, 712 permeability of free space, 990, 1000 perpendicular lever arm, 360, 380 Index 1663 phase angle, 1054, 1057 phase diagram, 565 phase diagrams, 558 phase-contrast microscope, 1223, 1225 phon, 750, 762 Phosphorescence, 1342 phosphorescence, 1365 photoconductor, 801, 806 photoelectric effect, 1280, 1306 photomultiplier, 1384, 1411 photon, 1280, 1290, 1306 photon energy, 1280, 1306 photon momentum, 1291, 1307 physical quantity, 15, 29 Physics, 8 physics, 29 pion, 1467, 1493 pit, 221, 249 pitch, 726, 749, 762 Planck’s constant, 1278, 1307 planetary model of the atom, 1326, 1365 point charge, 792, 806 point masses, 336, 343 Poiseuille's law, 507, 519 Poiseuille's law for resistance, 506, 519 polar molecule, 799, 806, 847, 856 polarization, 783, 806, 1225 Polarization, 1213 polarization microscope, 1223, 1225 polarized, 786, 806, 1213, 1225 population inversion, 1343, 1365 position, 34, 83 positively curved, 1516, 1525 positron, 1396, 1411 positron decay, 1395, 1411 positron emission tomography (PET), 1427, 1454 potential difference, 824, 924, 951 potential difference (or voltage), 856 potential energy, 277, 279, 301 potential energy of a spring, 278, 301 potentiometer, 943, 951 power, 292, 301, 1123, 1143 power factor, 1054, 1057 precision, 23, 29 presbyopia, 1158, 1177 pressure
, 444, 447, 478 Pressure, 451 probability distribution, 1300, 1307 projectile, 113, 128 Projectile motion, 113 projectile motion, 128 Proper length , 1248 proper length, 1267 Proper time , 1243 proper time, 1267 proton, 806 proton-proton cycle, 1440, 1454 protons, 777, 1386, 1411 PV diagram, 557, 565 Q quality factor, 1429, 1454 quantized, 1277, 1307 quantum chromodynamics, 1487, 1490, 1493 quantum electrodynamics, 1470, 1493 Quantum gravity, 1509, 1525 quantum mechanical tunneling, 1408, 1411 Quantum mechanics, 14 quantum mechanics, 29, 1277, 1307 quantum numbers, 1353, 1365 quark, 343, 1493 quarks, 327, 1480 quasars, 1512, 1525 R R factor, 607 factor, 594 rad, 1429, 1454 Radar, 1082 radar, 1093 radians, 221, 250 radiant energy, 287, 302 radiation, 590, 602, 607 radiation detector, 1383, 1411 radio waves, 1070, 1079, 1093 radioactive, 1379, 1411 Radioactive dating, 1398 radioactive dating, 1411 radioactivity, 1379, 1412 radiolytic products, 1438, 1454 radiopharmaceutical, 1425, 1454 radiotherapy, 1435, 1454 radius of a nucleus, 1387, 1412 radius of curvature, 221, 250 rainbow, 1143 range, 119, 128 range of radiation, 1381, 1412 rate of conductive heat transfer, 593, 607 rate of decay, 1400, 1412 ray, 1102, 1143 Ray tracing, 1125 Rayleigh criterion, 1204, 1225 RC circuit, 951 real image, 1127, 1143 reflected light is completely polarized, 1216 reflected light that is completely polarized, 1225 refraction, 1106, 1143 relative biological effectiveness, 1429 relative biological effectiveness (RBE), 1454 relative humidity, 561, 565 relative osmotic pressure, 518, 519 relative velocities, 125 relative velocity, 128 relativistic Doppler effects, 1269, 1267 Relativistic kinetic energy, 1263 relativistic kinetic energy, 1267 Relativistic momentum, 1257 relativistic momentum, 1267 relativistic velocity addition, 1253, 1267 Relativity, 14 relativity, 29, 125, 128, 1239, 1267 Renewable forms of energy, 298 renewable forms of energy, 302 resistance, 874, 901, 914, 951 resistivity, 878, 902 resistor, 914, 945, 951 resonance, 698, 712 resonant, 1075, 1093 resonant frequency, 1053, 1057 resonate, 698, 712 Rest energy, 1259 rest energy, 1267 rest mass, 1257, 1267 restoring force, 675, 712 resultant, 101, 128 resultant vector, 101, 128 retinex, 1177 retinex theory of color vision, 1165, 1177 retinexes, 1165 reverse dialysis, 518, 519 Reverse osmosis, 518 reverse osmosis, 519 reversible process, 633, 661 Reynolds number, 512, 519 right hand rule 1, 975 right hand rule 1 (RHR-1), 1000 right hand rule 2, 989 right hand rule 2 (RHR-2), 1000 right-hand rule, 423, 425 RLC circuit, 1093 rms current, 888, 902 rms voltage, 888, 902 rods and cones, 1162, 1177 roentgen equivalent man, 1429 roentgen equivalent man (rem), 1454 rotation angle, 221, 250 rotational inertia, 401, 425 rotational kinetic energy, 405, 425 Rydberg constant, 1328, 1333, 1365 S saturation, 561, 565 scalar, 37, 83, 106, 128, 830, 856 Schwarzschild radius, 1511, 1525 scientific method, 12, 29 scintillators, 1384, 1412 screening, 799, 806 second, 16, 29 second law of motion, 317, 344 second law of thermodynamics, 635, 636, 640, 661 second law of thermodynamics stated in terms of entropy, 661 second postulate of special relativity, 1240, 1268 Self-inductance, 1042 self-inductance, 1057 semipermeable, 517, 519, 895, 902 series, 915, 951 shear deformation, 209, 212 shell, 1359, 1365 shielding, 1432, 1454 shock hazard, 890, 902, 1036, 1057 short circuit, 891, 902 shunt resistance, 940, 951 SI unit of torque, 361 SI units, 15, 29 SI units of torque, 380 sievert, 1430, 1454 significant figures, 25, 29 simple circuit, 875, 902 Simple Harmonic Motion, 681 simple harmonic motion, 712 simple harmonic oscillator, 681, 712 1664 Index simple pendulum, 686, 712 simplified theory of color vision, 1163, 1177 single-photon-emission computed tomography (SPECT), 1454 single-photon-emission computed tomography(SPECT), 1427 slope, 75, 83 solenoid, 992, 1000 Solid-state radiation detectors, 1385 solid-state radiation detectors, 1412 sonic boom, 739, 762 sound, 724, 762 sound intensity level, 733, 762 sound pressure level, 735, 762 south magnetic pole, 967, 1000 space quantization, 1351, 1365 special relativity, 1268 special relativity., 1239 specific gravity, 461, 478 specific heat, 579, 607 speed of light, 1093 spin projection quantum number, 1355, 1365 spin quantum number, 1355, 1365 spontaneous symmetry breaking, 1509, 1525 stable equilibrium, 365, 380 Standard Model, 1490 standard model, 1493 standing wave, 705, 1075, 1093 static electricity, 806 static equilibrium, 358, 368, 380 static friction, 192, 212 statistical analysis, 658, 661 Stefan-Boltzmann law of radiation, 603, 607 step-down transformer, 1034, 1057 step-up transformer, 1034, 1057 Stimulated emission, 1343 stimulated emission, 1365 Stokes' law, 202, 212 strain, 208, 212 strange, 1481, 1493 strangeness, 1478, 1493 stress, 208, 212 sublimation, 558, 566, 607 Sublimation, 589 subshell, 1359, 1365 Superconductors, 1521, 1525 supercriticality, 1447, 1454 superforce, 1508, 1525 superposition, 704, 712 Superstring theory, 1491, 1515, 1525 superstring theory, 1493 surface tension, 465, 478 synchrotron, 1472, 1493 synchrotron radiation, 1472, 1493 system, 146, 179 systolic pressure, 455, 478 Systolic pressure, 474 T tagged, 1425, 1455 tail, 100, 128 tangential acceleration, 392, 425 tau family number, 1493 Television, 1081 Temperature, 530 temperature, 566 temperature coefficient of resistivity, 880, 902 tensile strength, 205, 212 tension, 161, 179 terminal speed, 514, 519 terminal voltage, 926, 951 tesla, 975, 1000 test charge , 792, 806 the second law of thermodynamics stated in terms of entropy, 652 theory, 11, 29 theory of quark confinement, 1487, 1493 therapeutic ratio, 1435, 1455 thermal agitation, 1082, 1093 thermal conductivity, 593, 607 thermal energy, 282, 287, 302, 552, 566 thermal equilibrium, 536, 566 thermal expansion, 537, 566 thermal hazard, 890, 902, 1036, 1057 Thermal stress, 541 thermal stress, 566 thin film interference, 1208, 1225 thin lens, 1125 thin lens equations, 1128 thought experiment, 1509, 1525 three-wire system, 1036, 1057 thrust, 154, 180, 179 timbre, 750, 762 time, 39, 83 Time dilation, 1242 time dilation, 1268 TOE epoch, 1508, 1525 tone, 750, 762 top, 1486, 1493 Torque, 360 torque, 380, 400, 425 Total energy , 1259 total energy, 1268 total internal reflection, 1112 trajectory, 113, 128 transformer, 1057 transformer equation, 1034, 1058 Transformers, 1032 transverse wave, 702, 712, 1075, 1093 triple point, 559, 566 Tunneling, 1409 tunneling, 1412 turbulence, 504, 519 TV, 1093 twin paradox, 1268 U ultra high frequency, 1081 ultra-high frequency (UHF), 1093 ultracentrifuge, 227, 250 ultrasound, 749, 762 Ultraviolet (UV) microscopes, 1222 ultraviolet (UV) microscopes, 1225 Ultraviolet radiation, 1287 ultraviolet radiation, 1307 ultraviolet radiation (UV), 1085, 1093 uncertainty, 23, 29 uncertainty in energy, 1303, 1307 uncertainty in momentum, 1301, 1307 This content is available for free at http://cnx.org/content/col11844/1.13 uncertainty in position, 1301, 1307 uncertainty in time, 1303, 1307 under damping, 712 underdamped, 695 uniform circular motion, 250 units, 15, 29 unpolarized, 1213, 1225 unstable equilibrium, 365, 380 up, 1481, 1493 useful work, 296, 302 V Van de Graaff, 1472, 1493 Van de Graaff generator, 806 Van de Graaff generators, 801 vapor, 559, 566 Vapor pressure, 560 vapor pressure, 566 vector, 37, 83, 99, 128, 806, 830, 856 vector addition, 122, 128, 796, 806 vectors, 97, 794 velocity, 122, 128 vertically polarized, 1213, 1225 very high frequency, 1081 very high frequency (VHF), 1093 virtual image, 1130, 1143 virtual particles, 1467, 1494 viscosity, 506, 519 viscous drag, 513, 519 Visible light, 1084 visible light, 1093, 1288, 1307 voltage, 824, 915, 951 voltage drop, 915, 951 voltmeter, 951 Voltmeters, 938 W watt, 292, 302 wave, 700, 712 wave velocity, 700, 712 wavelength, 701, 712, 1074, 1093 wavelength in a medium, 1187, 1225 weakly interacting massive particles, 1518 weight, 149, 179 Weight, 157 Wheatstone bridge, 944, 951 WIMPs, 1518, 1525 work, 263, 302 work-energy theorem, 268, 302, 406, 427, 425 X x ray, 1307 x rays, 1285, 1365 X rays, 1334 X-ray, 1088, 1093 x-ray diffraction, 1338, 1365 xerography, 801, 806 Y y-intercept, 75, 83 Z z-component of spin angular momentum, 1365 z-component of the angular momentum, 1365 Index 1665 -component of spin angular momentum, 1355 -component of the angular momentum, 1353 Zeeman effect, 1350, 1365 zeroth law of thermodynamics, 536, 566 zircon, 1143
ent with greater accuracy than is possible with the means at our disposal. The reasons will be apparent after studying later paragraphs (page 15). 28.7 (c) Approximate Numbers In all measurement, you must first decide what will be the limits of accuracy of your work and then realize that a possible error exists in the result. In order to make this small, the possible error is not permitted to exceed one-half of the smallest unit employed in the operaThus, in the previous examples, tion. if 1/8 inch were the smallest unit on the measuring instrument used, a possible error of 1/16 inch more or less is found in the answer. Accordingly the length of the room in part (a) would be between 14 feet, 15/16 inch and 15 feet, 1/16 inch and the width of the door would be between 29 15/16 and 30 1/16 Similarly, if the present room inches. temperature is given as 21.5°C., and the thermometer is marked off into 0.1 degree units, the possible error will be .05 centigrade degrees and the actual reading may be between 21.45°C. and 21.55°C. The reading 21.5°C. represents the temperature as accurately as the means of measurement at our disposal will permit. Other examples follow: Measurement 55.3 cm. 719 ft. 19.0 ft. 0.0032 in. From 55.25 cm. Limits of Accuracy up to but not including 55.35 cm. a “ “ 718.5 ft. 18.95 ft. 0.00315 in. “ “ “ “ “ “ “ 719.5 ft. 19.05 ft. 0.00325 in. To allay the fear that naturally arises over the presence of this possible error, let us calculate what per cent of the whole it represents. The percentage error is possible error divided by the accepted number times Using the first measurement in the preceding table we get: 100. The percentage error =: From this, we see that approximate numbers contain an extremely small error. The more precise the measuring device is, the smaller the unit of measurement will be. The smaller the unit of measurement is, the smaller the possible error will be. (d) How to Deal with Approximate Numbers 55.3 - 55.25 55.3 - 55.35 X 100 or 55.3 = +_X100or- — X.oo 55.3 .05 .05 = -f .09% or - .09% X 100 1. Significant Digits When performing a measurement, only those numbers in the measurement that are certain, like the 15 feet, 30 inches, 21.5°C. are recorded. The digits in these numbers paragraphs, previous of 13 Chap. 2 MECHANICS are called the significant digits in the result, there being two in the first two numbers above, and three in the third. Mathematicians have laid down certain rules to guide us in working with them. They are as follows: 1. All the digits from 1 to 9 including any zeros between them or after them are significant digits. 2. The position of the decimal is disregarded in determining the number of significant digits. 3. Rounding off Approximate Numbers A number like the value of pi (tt), 3.14159, is correct to 6 significant digits. It may be made consistent with measurements having fewer than six significant digits by the process of rounding-off. This means dropping necessary number of digits off the end of a number and adding one to the last remaining digit if the next digit was five or more. the 4. Examples 3. The zeros preceding the first digit Rounding-off Accuracy are not significant. Significant Digits IN Numbers Number 604 60.4 6.04 0.604 0.0604 0.06004 604.0 60.40 Significant Digits 3 3 3 3 3 4 4 4 2. Representing Significant Digits in Large Numbers digit, Where a number like 600 is correct it should be to one significant written as 6 X 10^. To express it as 600 indicates digits. Similarly, the distance from the earth to the sun is correct to only two signiwritten ficant 93 X 10® rather than 93,000,000 miles. digits and should significant three be 3.14159 3.142 3.1 6 4 2 significant digits a a Calculations With Significant Digits How to carry out mathematical operations with approximate numbers is indicated by the following rules: 1. Addition and Subtraction In operations involving addition and subtraction of approximate numbers of which the least precise has N places of decimals, round off the other numbers where possible to N + 1 places and the answer to N places. 2. Multiplication and Division In operations involving multiplication and division of approximate numbers of which the least accurate has N significant digits, round off the others where possible to N + 1 digits and the answer to N digits. Examples 1. Add 2.0149 3.02864 1.239 1.97 14 Method Round off all to three decimals where possible. Round off answer to 2 decimals. Result 2.015 3.029 1.239 1.97 Total = 8.253 Proper answer = 8.25 MEASUREMENT Method Result 2. Subtract 21.347 As for addition from 32.5 Sec. 1:3 32.5 21.35 3. Multiply 2.1 by 2.56 and by 9.547 Round oflF all numbers to 3 significant digits. Keep 3 significant digits in all partial products. Round off answer to 2 significant digits. 4. Divide 96.568 As for multiplication by 7.02 Difference = 11.15 Proper answer = 11.2 2.1 X 2.56 = 5.376 Proper product = 5.38 5.38 X 9.55 = 51.3790 Proper answer = 51. 96.57 = 13.75 7.02 = 13.75 Quotient Proper answer =13.8 On page 13 we multiplied 7.54 by getting 28.7274 for an answer. 3.81, However, as the multiplier and multiplicand are correct to three significant digits only, 28.7 is the proper answer. You may wonder about the usefulness of the above material, but be assured that the method is used daily by scientists and mathematicians whose work has contributed so much to each modern development and invention. (e) Applying Approximate Numbers in Elementary Physics The diameter of a cylindrical solid is 2.50 cm. and it is 10.04 cm. long. (7r = 3.1416). Its mass is 280.76 gm. Calculate its density. Radius = 1.25 cm. Length = 10.04 cm. =3.1416 = 3.142 TT Area of the end of the cylinder = tt R^ = 3.142 X 1.25 X 1.25 = 4.910 sq. cm. Volume of the cylinder = area of end X length = 4.910 X 10.04 = 49.30 c.c. The mass given = 280.76 gm. The proper mass = 280.8 gm. 49.30 c.c. of solid weigh 280.8 gm. • . . 1 1 1-^ c.c. of solid weighs r • u 280.8 49.30 = 5.695 gm. The density = 5.70 gm. per c.c. 15 Chap. 2 : 4 I MECHANICS QUESTIONS 1. 2. (a) Why Is measurement necessary? (b) Why was it necessary to establish standard units? (a) Name the two systems of measurement and name the fundamental units for each. (b) What advantages has the metric system over the British system of measurement? (c) Why is the British system still used in a few countries? 3. (a) What apparatus is commonly used in the laboratory for the measurement of (i) length (ii) mass (ili) time? (b) Suggest other pieces of apparatus that could be used to measure these fundamental units more ac- curately. 4. 5. 6 . 7. 8 . decimetres (a) State the number of millimetres, centimetres and metre. How many metres are there In 1 kilometre? (b) Using the above units construct (i) square measure (ii) tables of in 1 cubic measure. (c) Define: litre, millilitre. (a) Distinguish between the mass and weight of an object. (b) State the number of milligrams, centigrams, decigrams, and grams in 1 kilogram. (a) Why was the apparent motion of the sun adopted as the basis for reckoning time? (b) Define: mean solar day, second. (a) Distinguish between exact and approximate numbers. Give an example of each. (b) What gives rise to approximate numbers? Why? (a) What do you mean by possible error? 16 9. (b) How do you calculate percentage error? (a) What is meant by significant digits? (b) State the number of significant digits in each of: 32060, 36.060, 0.32060, .032060, 3 X 10^, 3.56 X 105. 10. (a) What is meant by rounding off a number? (b) Round off the following to two significant digits: 36.7, 34.32, 37.495. 11. (a) State the rules for carrying out the following mathematical operations with approximate numbers: (i) addition and subtraction, (ii) multiplication and division. (b) Do the following: (i) 10.3575 + 9.75-8.65248. (ii) 7.935 X 2.4248 2.3. B 1. (a) Express in cm.: 15.2 m., 38 mm., 6 m. 5 cm. 4 mm. (b) Express in sq. cm.: 3 sq. m., 236 sq. mm., 6 sq. m. 5 sq. cm. 44 sq. mm. (c) Express in c.c.: 2.5 cu. m., 2300 cu. mm., 6 cu. m. 50 c.c. 465 cu. mm. 2. (a) Determine the number of inches In 950 mm., 40 cm., 10 dm. (b) Determine the number of (i) cm. in 1 foot, (ii) km. in 1 mile. (c) Which is the greater distance, 100 yd. or 100 metres? Express the difference In (i) (d) An object is at the rate of 40 miles per hour. Calculate the rate in (i) ft. per sec. (ii) metres per sec. (iii) kilometres per hour. 3. A tank is 50 cm. long, 3 dm. wide, ft. (ii) cm. travelling and 150 mm. high. (a) Calculate the area of a cover MEASUREMENT . Sec. 1:4 in sq. cm. (ii) in sq. for the tank (i) dm. (b) Calculate the volume of the tank, and express In (i) c.c. (ii) cu. dm. (iii) litres. 4 . (a) What mass of water will the above tank hold (1 c.c. of water weighs 1 gm.)? Express this mass in gm., mg., eg., dg., kg. (b) Calculate grams In an ounce (ii) kilograms in a number the of (i) ton? (c) Calculate your own weight in kilograms. 5. A beaker is 20 cm. high and has a diameter of 1 4 cm. Calculate its volume in (i) litres (ii) millilitres (iii) pints. 6 Using proper "rounding-off” tech- niques, make the following calculations: (a) Add 9.75+10.357+76.92 + 5.674. (b) Subtract (i) 10.357 (ii) 5.674 (iii) 9.75 from 76.92. (c) Multiply (i) 2.6X7.93X1.732. (ii) 77.5X1.4142X.0032. (iii) 46X23.55X0.25. (d) Divide (i) 154 by .1 1. (ii) 9.5 by 19.03. (iii) 134.5 by 15. 17 CHAPTER 3 DENSITY AND SPECIFIC GRAVITY 1 gm. per c.c. G.G.S, system of units is In the F.P.S. system, however, a cu. ft. of water is found to weigh 62.5 ib. (approx.), and hence the density of water in this system of units will be expressed as 62.5 lb. per cu. ft. Densiof various subties stances are given in the table on page 21. (in gm. per c.c.) Fig. 3:1 Relative Densities of A— Solids, B— Liquids. To find the corresponding densities in lb. per cu. ft. the numerical values must be multiplied by 62.5. I is e.g., iron “heavier” ; 5 MEANING OF DENSITY In ordinary conversation we often say that one substance is “heavi
er” than than another, aluminum. Obviously we cannot mean that any given piece of iron is heavier than every piece of aluminum, but rather that for pieces of equal size, the iron would be the heavier. In science we use the term density to express the physical difference implied in the above everyday statement, that is, we say that iron has a greater density than aluminum. Density is defined as the mass of a unit volume of a substance. The method of determining density is described in chapter 5, experiments 1, 2 and 3. numerical densities measure must always be accompanied by suitable units, e.g., gm. per c.c., lb. per cu. ft., etc., according to the units; in which the mass and volume of the substance have been measured. It should further be noted that the numerical value of the density of any given substance will depend on the system of Thus, since one gram of units used. water (at its maximum density) occupies a volume of one cubic centimetre (Sec. 1:2), the density of water in the stating the In 18 DENSITY AND SPECIFIC GRAVITY Sec. 1:6 Research Scientist Determining the Density of a Substance by Comparison with Standard Density Floats Suspended in a Solution of Known Density. Canadian Industries Ltd. We have previously defined mass as the quantity of matter in a body. According to modern theory, matter is comprised of molecules, the molecules of any given substance being identical Hence, to each other (Sec. differences in density between various substances (Fig. 3:1), are due to the relative masses of their molecules as spatial arrangement. well as Variations in the density of a given substance are due to changes which vary the closeness of packing of the molecules. Ill: 2). their to I of water is : 6 DENSITY OF WATER It is frequently said that the density 1 gm. per c.c. However, solids and gases, expand liquids, when heated and contract when cooled with no change whatever in mass. They therefore at different densities have like different temperatures. Careful experiments to show changes in density of water with changes in temperature can be carried out with the aid of the dilatometer shown in Fig. 3:2. If this instrument, filled with water. 19 Chap. 3 MECHANICS is placed in a water bath with a thermometer, and ice slowly added, the volume change can be observed on the Fig. 3:2 A Dilatometer. scale, for changes of temperature down to 0°C, The water contracts, i.e., density increases, until a temperature of its scale) to (not Volume is smallest which the volume is the temperature at which the density is the greatest, namely 4°C. We say that water has its maximum density at 4°C. It is at this temperature that 1 cubic centimetre of water has the mass of 1 gram. The changes mentioned above are shown graphically 3:3. (Most liquids show a gradual increase in goes down.) temperature density Fig. the in as at 0°G. This fact, together with the fact that there is a sudden expansion as water (evidenced by those freezes burst pipes in winter!) is of profound importance in nature. With wintry conditions the coldest layers of a pond or lake are those at the surface, and when these layers freeze the ice so formed remains on the surface because its density is less than that of the water beneath it. Without this unusual behaviour of water the pond would freeze solid from the bottom upwards, greatly to the detriment of all life and to aquatic life in particular. : 7 MEANING OF SPECIFIC GRAVITY I For many purposes, instead of density, it is found more convenient to use the density of a substance relative to that of water. The relationship so obtained is called the specific gravity of It may be calculated in the substance. the following way: g ^ _ density of the substance density of water Temperature Fig. 3:3 Graph to Show the Effect of Changes of Temperature on the Density of Water. o.o. — - „ mass of unit volume of the substance mass of unit volume of water g g _ mass of any volume of the substance mass of an equal volume of water reached. 4°G. is further cooling makes it expand, i.e., its density The temperature at decreases After that, again. Thus specific gravity is a ratio and no It is simply a numunits are required. 20 DENSITY AND SPECIFIC GRAVITY Sec. 1:8 7 ber stating how many times as heavy as water, bulk for bulk, the substance is. Further, it must be evident that the number giving the specific gravity of a substance will be the same whatever the units in which the masses are measured, (Chap, 5, Exp. 4). In the C.G.S. system the density of water is 1 gm. per mb, and thus in this system, density and specific gravity are numerically equal. The two tenns are not interchangeable, however, although frequently, but wrongly, so used. : 8 DENSITY OR SPECIFIC GRAVITY I OF VARIOUS SUBSTANCES or specific gravity The density of solids, liquids and gases is a property that helps us to identify them. Some of these you will have determined experimentally while others will be found in the accompanying table of specific gravi- ties. Table of Approximate Specific Gravities 10.5 11.4 13.6 19.3 21.5 8.4-8. 8. 7-8.9 2.6 0.6 0.8 1.74 2.70 7.15 7.30 7.85 8.90 Silver Lead Mercury Gold Platinum 7.0-7.7 7.1-7.7 Brass Bronze Sand Pine Oak 0.24 0.92 2.2 0.70 0.79 0.87 Metals Magnesium Aluminum Zinc Tin Iron (pure) Copper Alloys Steel Iron (cast)^ Miscellaneous Solids Cork Ice (0°C.) Salt Liquids Gasoline Alcohol Turpentine Gases (at S.T.P.) Hydrogen Helium Air Carbon tetrachloride Sea- water Cone, sulphuric acid 1.60 1.01-1.05 1.83 0.00009 0.00018 0.00129 Oxygen Carbon dioxide Chlorine 0.00143 0.00198 0.00322 Gases, being so light, generally have their densities expressed in grams Also, air or hydrogen is used as the standard, rather than water, for per litre, instead of grams per cubic centimetre. (preferably the latter) purposes of comparison. 21 Chap. 3 i:9 MECHANICS QUESTIONS A 1. (a) Define density and state what two measurements must be made in order to calculate the density of an 5. Find the mass of 20 cu. ft. of material whose density is 3.2 Ib./cu. ft. 6 . Find the volume of an object whose mass Is 42.7 lb. and whose density is 2.10 object. Ib./cu. ft. 2. (b) Calculate the density of a piece of aluminum whose volume is 150 c.c. and whose mass is 405 gm. (a) What effect does an increase in temperature have on the density of most substances? Why? (b) In what respect may water be said to be an unusual liquid? (c) Explain why this behaviour of water is important to life. (d) Why does it a much longer and more severe period of cold weather to cause a layer of ice to form on deep bodies of water than to form on shallow bodies? unusual require 3. (a) Define specific gravity and state clearly what is needed to find the specific gravity of an object. (b) Calculate the specific gravity of a substance whose volume is 20 c.c. and whose mass is 1 60 gm. 4. (a) Distinguish between density and specific gravity. (b) What relationship exists between the specific gravity of a substance and its density? (c) What is the specific gravity of the aluminum in 1. (b)? 7. Find the specific gravity of a substance whose mass is 148.5 gm. and whose volume is 30.5 c.c. 8 . The specific gravity of a substance is 1 .85. What volume of It weighs 1 00 gm.? Find the mass of 0.5 litres of a liquid 9. whose specific gravity is 1.6. 10. Find the specific gravity of a substance whose mass Is 3.2 lb. and whose volume is .75 cu. ft. (Density of water is 62.5 lb. per cu. ft.) 11. The specific gravity of a substance is 2.7. What volume of It will weigh 1 00 lb.? 12. Find the mass of 3.2 cu. ft. of material whose specific gravity is 8.9. 13. An irregular object has a mass of 72.6 gm. On placing it in water in a graduated cylinder the level rises from 12.0 ml. to 21.5 ml. Calculate its density. 14. A specific gravity bottle weighed 24.20 gm. when empty, 67.81 gm. when filled with turpentine and 74.20 gm. when with distilled water. What is the filled specific gravity of the turpentine? 15. A flask weighs 8.8 gm. when empty, 33.6 gm. when filled with water and 28.6 gm. when filled the with specific gravity of the alcohol. alcohol. Find B 1. Calculate the density of a rectangular solid 25 cm. long, 15 cm. wide, 3 cm. thick, whose mass is 5625 gm. 2. Find the mass of 75 ml. of a liquid whose density is 0.70 gm./ml. 3. Find the volume of an object whose mass is 750 gm. and whose density is 2.80 gm./c.c. 4. Find the density of water if 1 5 cu. ft. weigh 937.5 lb. 16. The density of a salt solution is 1.20 gm. per ml. If a flask weighs 1 2.6 gm. when empty and 62.8 gm. when filled with water, how much will it weigh when filled with the solution? 17. The composition of brass is 75% copper and 25% zinc by volume. Calculate its density. 1 8. An alloy of tin and lead has a specific gravity of 10.6. Calculate the proportion of tin present in the alloy (a) by volume (b) by weight. 22 2 DENSITY AND SPECIFIC GRAVITY Sec. 1:9 19 . A piece of wax whose real specific gravity is 0.96 has an apparent specific gravity of 0.92 owing to a bubble of air it. The volume of the being enclosed in whole is 10.0 c.c. Find the volume of the air enclosed, assuming the weight of the air to be negligible, 20 . A piece of metal 1 1.2 cm. long, 4.5 cm. wide and 1 mm. thick, has a mass of 30.2 gm. Find its density. 21 . Calculate the density of a cylinder 22 . Calculate the density of a sphere whose radius is 1.4 cm. and whose mass is 100 gm. 23 . A cube of ice whose side is 4 cm. is allowed to melt. The volume of the water is found to be 58.2 ml. Find the density of ice. 24 . Four solutions of salt, of densities 1.12, 1.17, 1.19 and 1.20 gm. per c.c., proportion are 1:2:3:4 by volume. Find the density of the together mixed the in whose length is 5.2 cm., diameter is mm., and whose mass fs 58 gm. 1 mixture. 23 CHAPTER 4 BUOYANCY force will be apparent after a consideration of the forces that water exerts on an object immersed in it (Fig. 4 : 1). The water exerts a downward pressure upon the top surface of the obje
ct, and an upward pressure upon the bottom. Because the bottom of the object is deeper in the liquid than is the top, and because pressure increases with depth, the upward pressure upon the bottom surface will exceed the downward pressure upon the top surface. There will, therefore, be a net upward pressure upon It is this upward force that the block. accounts for the apparent lightness of an object when immersed in the water. The same principle applies to all bodies immersed in any liquid or gas. Archimedes (287-212 b.g.), a Greek mathematician and inventor, was the first to study the buoyancy of liquids and to enunciate an important principle connected therewith. The story is told that Hiero, king of Syracuse, had sent his jeweller a known mass of gold to be made into a new crown. When the crown was delivered and tested it was found to have the right weight, but there was a suspicion that some silver had been substituted for gold in the interior of the Consequently Archimedes was crown. commissioned to determine whether or not the crown was pure gold, at the same time being instructed not to mar the crown in any way. Archimedes puzzled over this problem at great length. All that he had to work I : 10 INTRODUCTION TO BUOYANCY We are all familiar with the fact that things seem lighter under water. Those of us who have helped to build a dock know that a large stone which can only be raised with difficulty when out of the water may be raised quite easily when Similarly, we find that under water. Fig. 4:1 Explanation of Buoyancy. an anchor becomes heavier on emerging from the water. In general then, bodies immersed in water (or in any fluid) appear to lose some of their weight. This is, of course, due to the fluid exerting a buoyant force or lift on them. The reason why fluids exert a buoyant 24 BUOYANCY Sec. 1:11 with was the well-known fact that an object is easier to lift when immersed in water than when on land. medes’ Principle (Chap. 5, Exp. 5, 6) should be studied carefully in conjunction with the following statements: One day as he hopped into his bath, which had been filled brimful of water, he caused a considerable quantity to overflow onto the brick floor, and he was suddenly struck by the idea that the weight he lost on submerging was equal to that of the water which overflowed. He was so excited by this discovery, so the story goes, that he ran home “Eureka! Eureka!” (“I have found it! I have found it!”) As you that Archimedes enunciated, see if ypu can suggest the experiment that Archimedes probably performed to determine whether* or not the crown was pure gold. unclothed shouting, principle study the 1:11 ARCHIMEDES' PRINCIPLE The experiment to demonstrate Archi- 1. When a body is immersed in a liquid it displaces some of the liquid to make room for itself. Note that the volume of the liquid displaced is equal to the volume of the object. 2. The body is buoyed up by the liquid and therefore seems to weigh less than it does in air. 3. This loss in weight is exactly equal the weight of the liquid dis- to placed (Fig. 4:2). Archimedes’ Principle, therefore, may be stated as follows: An object, when immersed in a liquid, loses in apparent weight an amount equal to the weight of the liquid displaced. or an alternative statement is: The buoyant force of a fluid (liquid or gas) upon an object immersed in it, equal to the weight of the fluid displaced. is Examples 1. A rectangular piece of metal, 15 cm. long, 6 cm. wide and 3 cm. thick, weighs 2700 gm. in air. Find its weight when immersed in water. = 2700 gm. Weight of object in air = 15 X 6 X 3 = 270 c.c. Volume of object Volume of water displaced = 270 c.c. Weight of water displaced = 270 gm. (Density of water = 1 gm./cc.) Weight of object in water = 2700 — 270 (Archimedes’ Principle) = 2430 gm. 25 Chap. 4 MECHANICS . 2. Find the weight of the object in example 1, if it were immersed in carbon tetrachloride (S.G. = 1.60). Weight of object in air = 2700 gm. Volume of object = 15 X 6 X 3 = 270 c.c. Volume of carbon tetrachloride displaced = 270 c.c. 1 c.c. of carbon tetrachloride weighs 1.60 gm. (S.G. = 1.60) Weight of carbon tetrachloride displaced = 270 X 1.60 = 432 gm. Weight of object in carbon tetrachloride = 2700 — 432 Archimedes’ Principle is used for the accurate determination of the specific gravity of solids and liquids (Chap. 5, Exp. 7-9) (Archimedes’ Principle) = 2268 gm. : 12 PRINCIPLE OF FLOTATION I Archimedes’ Principle applies to floating bodies as well as to those which are submerged in a fluid. When a body is floating in a liquid it appears to lose all its weight, as can be shown by lowering into water a block of wood, suspended by a thread from the hook of a spring balance (Fig. 4:3). As the wood settles more deeply into the water the reading of the spring balance decreases until, when the wood is floating, it records zero weight. The entire weight of the wood is now being supported by the upthrust of the water, this being equal to the weight of water displaced. We thus have, for floating bodies, the Principle of Flotation, which states that the weight of a floating object is equal to the weight of the fluid (liquid or gas) it displaces when floating. A fairly heavy object will of course sink lower in the liquid in which it is floating than will a lighter In either case the submerged object. displaced an amount of portion has liquid equal to the weight of the floating object. This statement is embraced in Archimedes’ Principle, i.e., a body which floats has lost its whole weight. (Remember loss is an apparent one, not that this real. The pull of the earth on the floating body is still the same.) Experimental proof for this Principle is supplied in Chap. 5, Exp. 10. Fig. 4:3 Principle of Flotation. 26 BUOYANCY Examples Sec. 1:13 1. An object loaded onto a flat barge 18 feet long and 10 feet wide, causes it to settle 2 inches deeper into the water. Calculate the weight of the object. Volume of water displaced =z 18 X 10 X 2/12 = 30 cu. ft. Weight of water displaced = 30 X 62.5 = 1875 lb. (Density of water = 62.5 Ib./cu. ft.) .'. weight of object = 1875 lb. (Principle of Flotation) 2. A plastic tray 25 cm. long, 15 cm. wide is floating on water. A lead weight whose mass is 300 gm. when placed in it causes it to sink deeper into the water. Calculate the depth to which it sinks. Let the depth to which it sinks be x cm. Volume of water displaced = 25 X 15 X x = 375x c.c. Density of water = 1 gm./c.c. Weight of water displaced = 375 x gm. Weight of water displaced = weight of floating object. (Principle of Flotation) Weight of water displaced = 300 gm. 375 x = 300 - = ^“ = .8 375 Depth to which the tray sinks is .8 cm. : 13 HYDROMETERS AND THEIR USES I (a) Structure displaced its own weight of the liquid, the lighter the liquid, the deeper the 5, for the Exp. (Chap. instrument The Principle of Flotation finds application in the hydrometer, which is a convenient rapid determination of the specific gravity of 12). A liquids common type of hydrometer consists of a cylindrical stem, graduated or containing a paper scale, an expansion of the stem called the float, and a bulb weighted with mercury or lead shot, the ballast, to make it float upright (Fig. 4:4). The float increases the buoyancy of the hydrometer. 11, 0.80 0.85 * 0.90 Stem 0.95 Scale 1.00N / Float The liquid whose specific gravity is to be determined is poured into a tall jar. The hydrometer is gently lowered into the liquid until it floats freely. The specific gravity of the liquid is indicated by the number on the scale which is even with the surface liquid. Since the hydrometer sinks until it has of the Weighted Bulb Fig. 4:4 Structure of a Hydrometer. hydrometer will sink in it. the largest specific gravity readings are Therefore, 27 Chap. 4 MECHANICS long and having a cross-sectional area of 1 sq. cm. It is weighted at one end with a plug of lead. On one face is marked at the bottom of the scale, and the smallest at the top. Hydrometers used lighter than water have a for liquids large float, and scale gradations starting with a specific gravity of 1.000 at the bottom. Those used for liquids heavier than water have a small float, and scale gradations starting with 1 .000 at the top. A universal hydrometer has 1.000 in the centre so that it may be used for Fig. 4:6 Principle of the Hydrometer. a centimetre scale, and the rod has been rendered impervious to water by dipping in hot paraffin. Float the rod in water and note the depth to which it sinks, for example, 15 cm. Hence it displaces 15 c.c. or 15 gm. of water, and thus the weight of the hydrometer is 15 gm. Now float it in some liquid whose specific gravity is to be determined, and note the level to which it sinks, for example, 20 cm. Hence the volume of the liquid displaced is 20 c.c. and the mass of an equal volume of water is 20 gm. But note, the weight of liquid displaced is equal to the weight of the hydrometer, namely, 15 gm. That is, the mass of liquid displaced is 15 gm. and the mass of an equal volume of water is 20 gm. therefore, that the specific It follows, gravity of the liquid is 15/20 or .75, and its density is .75 gm. per c.c. (c) Uses Fig, 4:5 Hydrometers of Different Types. A— for liquids less dense than water B— universal C— for liquids denser than water both kinds of liquids (Fig. 4:5). Since few scale readings can be accommodated above and below, such an instrument has a narrow range of usefulness. (b) Principle The principle of the hydrometer can best be illustrated by means of the simple hydrometer (Fig, 4:6). This consists of a straight wooden rod about 25 cm. Hydrometers, of course, are widely used beyond the walls of the physics laboratory for testing liquids. Industries using or preparing syrups, salt solutions, acids and heavy petroleum products, 28 BUOYANCY Sec. 1:14 is used for chemicals make extensive use of them. The storage-battery hydrometer (Fig. testing the specific 4:7) gravity of the acid
in a battery, and hence for estimating the degree to which the battery is charged. Antifreeze hydrometers are used to measure the specific gravity of antifreeze and thereby indicate how low a temperature it can stand without freezing. Similarly lactometers are used in of milk for checking possible dilution, and alcoholometers the strength of beers and wines. estimating testing used are the for Courtesy of Exide Automotive Division Fig. 4:7 Storage Battery Hydrometer. : 14 OTHER APPLICATIONS OF I BUOYANCY Archimedes’ Principle and the Principle of Flotation apply wherever fluids displaced, and the following exare The Plimsoll mark is painted amidships on the ship's hull to indicate safe-loading levels under different conditions (scale: 1 14 in. = 1 ft.). 29 Chap. 4 MECHANICS amples are important these principles are: a few which show how its being hollow, because, (a) Ships Even when its hull is made of iron (density 7.9 gm. per c.c.) a ship floats on water average density is less than that of water. When it is launched it will sink to the level at which the weight of the water it displaces is equal to the entire weight of the ship, and hence the meaning of the term “displacement”. A ship like the Queen Elizabeth, of 85,000 tons displacement, displaces 85,000 tons of sea-water when afloat. To ensure that it does not sink more deeply than is safe in the water, each ship has a safe-loading line, known as the Plimsoll line (Fig. 4:8) painted on its hull. Actually there are several such lines, to allow for regional These marks and seasonal variations. indicate the depth to which the ship may be safely loaded under the different conditions. (b) Submarines A submarine is a vessel with a cylindrically shaped enclosed hull fitted with ballast tanks at bow and stern and with smaller tanks on either side amidships. In order to submerge, the submarine allows water to enter these tanks until the total weight of the boat and ballast is nearly as great as that of the water it can displace. The submarine is now in “diving trim” and, by a proper use of horizontal or diving rudders, it can submerge completely and maintain its depth below the surface. In order to resurface, it forces the water out of its ballast tanks with compressed air. This action reduces its total weight until, when the weight is less than the upthrust, the ship rises. (c) Floating Docks When the hull of a great liner is to be serviced and reconditioned, a floating dock is used to lift it out of the water. 30 This dock consists, essentially, of a large flat tank divided into several compartments which can be filled with water to Fig. 4:9 Floating Dry-dock. sink the dock to a sufficient depth for the ship to be drawn into it by tugs. The water is then blown out of the -tanks by compressed air, and the dock rises, lifting the ship with it (Fig. 4:9). (d) Balloons When we apply Archimedes’ Principle to air, we see that a body in air will experience a force of buoyancy equal to the weight of air it displaces. A balloon will rise if the weight of the air displaced is greater than the weight of the balloon envelope and its attachments. It will continue to rise until it reaches a level of more rarified air where the weight of the air displaced is equal to that of the balloon. The first balloons, built in 1783, employed hot air for upthrust, having open bottoms with burning braziers slung underneath them, but by August of the same year the French scientist Charles had sent up the first balloon to be filled with the newly discovered gas, hydrogen. Two years later a trip was made from France to England in such a balloon, motive power being provided by oars. Modern balloons are made of a gastight silk fabric fitted with valves. Increased height is obtained by releasing water or sand ballast, while the descent BUOYANCY Sec. 1:14 is brought about by slowly releasing the gas from the envelope. Then, since atmospheric pressure decreases with height, the gas inside the balloon expands as it rises (in accordance with Boyle’s law) and thus there is danger that the balloon For this will burst at great heights. reason, in balloon ascents to high altitudes, the envelope is not filled to capacity at ground level. The greatest height reached by a manned balloon, 72,395 feet (over I 3/2 miles) was made by Stevens and Anderson, two United States Army officers, at Rapid City, South Dakota, in 1935. Their trip was made in a tightly-sealed hollow metal sphere attached to a huge helium-filled balloon. (e) Weather or Sounding Balloons The main use of hydrogen balloons today is to collect information about the upper atmosphere for meteorological pur- poses. These sounding balloons, as they are called, expand as they rise, until they spring a leak or burst, when the meteorological instruments which they carry are parachuted to earth. Temperature, pressure and humidity reports at various altitudes are automatically sent to weather stations by radio transmitters. Analysis of the readings have shown that sounding balloons have reached heights of 25 miles or more. (f) Airships An airship is a balloon built on a light rigid framework, propelled by air-screws and steered by rudders. The design of this type of craft developed rapidly in the early decades of this century. Although long flights were successfully made in these ships, a series of disasters resulting from the ease with which they caught fire, has caused their further development to be abandoned. 31 Chap. 4 : 15 I A MECHANICS QUESTIONS 1. (a) State Archimedes’ Principle. 7. air a piece of iron whose (b) In 1 00 c.c. has a mass of volume Is 890 gm. When this is immersed in water, calculate: (i) the buoyant force of the water on it. (ii) the weight of the iron in water. 2. Describe briefly how Archimedes’ Principle is used to determine the specific gravity of (a) a solid denser than water (b) a liquid. 3. (a) State the Principle of Flotation. (b) Show that it is a modification of Archimedes’ Principle. (c) When a piece of wood is floated in water in a graduated cylinder the level rises from 1 5.7 ml. to 1 8.3 ml. (i) Calculate the mass of the wood. (ii) If its specific gravity is 0.60, what is its volume? 4. Explain how Archimedes could have determined whether King Hiero’s crown was pure gold or a mixture of gold and silver. 5. Explain how Archimedes’ Principle or the Principle of Flotation applies to each of the following statements: (a) In landing a fish, you find that it seems to weigh more when it is pulled out of the water than it does beneath the water surface. (b) As a ship in harbour is being unloaded, it slowly rises higher in the water. In order to make a submarine (c) submerge, large tanks aboard it are filled with water. (d) The same ship with the same cargo will ride higher on the Atlantic Ocean than on the Great Lakes. (e) Plimsoll lines are used on ships. (a) What is hydrometer? (b) Why are the smaller numbers of purpose the of a 6 . 32 the hydrometer scale near the top? (c) Compare the size of float required for denser and less dense liquids. Explain the difference. for hydrometers used State three methods used for finding the specific gravity of a liquid. Which of the three do you think Is the most accurate? Why? B 1. 15 c.c. material weigh of (a) If 45 gm. in air, find the weight when immersed in water. 2. (b) If 3 cu. ft. of a substance weigh 350 lb. in air, find the weight when immersed in water. (a) Iron has a density of 7.8 gm. per c.c. Find the weight of 10 c.c. of it when immersed in water. (b) A substance has a density of 1 87.5 lb. per cu. ft. Find the weight when 5 cu. ft. of it are immersed in water. 3. (a) If an object weighs 140 gm. in air and 1 1 5 gm. in water, what is the volume of the water displaced? (b) If an object weighs 140 lb. in air and 115 lb. in water, what is the volume of the water displaced? 4. A piece of silver weighs 65.1 gm. in air and 58.9 gm. in water. Find its specific gravity. 5. A piece of metal weighs 500 gm. in air and 430 gm. in water, (i) What is specific gravity? (Ii) What is its volume? Its 6 . A 15 lb. weight weighs only 9 lb. in (ii) Find water, (i) Find its specific gravity, Its volume, (iii) Find its density. 7. A piece of metal weighing 1 20.4 gm. has a volume of 14.5 c.c. (i) What will it weigh In water? (Ii) Find also the density of the metal. 8. An object weighs 42.2 gm. in air, 29.4 gm. in water and 25.6 gm. in a liquid. . BUOYANCY Sec. 1 : 15 9 Calculate the specific gravity of the liquid. A metal weighs 56.3 gm. in air, 45.8 gm. when immersed in water and 48.6 gm. when immersed in a liquid. Calculate the specific gravity of (i) the metal (ii) the liquid. 10. A piece of metal weighs 138.8 gm. in air, 123.2 gm. in water and 125.7 gm. in a liquid. Find the specific gravity of the metal and of the liquid. 11. An object weighs 42.2 gm. in air and 29.4 gm. in water. How much will it weigh in a liquid of density 0.80 gm. per c.c.? 12 . Was King Hiero’s crown made of pure gold if in air it weighed 1500 gm., and when immersed in water it weighed 1 400 gm.? 13 . A rectangular block of soap 8 cm. long and 6 cm. wide floats in water with 2.5 cm. of its thickness submerged. Calculate the mass of the soap. 14. A cube of wood, side 50 cm., floats in water with its base horizontal and 6 cm. of its height above the surface. Find its density. 15. A wooden raft 5 ft. long and 4 ft. wide floats in water. When a person steps on the raft it sinks 1.5 inches deeper into the water. Calculate the person’s weight. 16 . A cork of volume 60 c.c. and density 0.24 gm. per c.c. floats in a liquid of density 0.85 gm. per c.c. Find the least weight required to sink it. 17 . What volume of lead of density 11.2 gm. per c.c. will be required to sink a piece of wood in water, the weight of the wood being 425 gm. and its volume 556 C.C.? 18 . A block of wood of volume 100 c.c. floats in a liquid of specific gravity 1.2 with 75 c.c. immersed. Calculate the density of the wood. 19 . A wooden hydrometer sin
ks in water to a depth of 1 8 cm. and in a liquid to a depth of 1 4 cm. What is the specific gravity of the liquid? 20 . A hydrometer sinks in water to a depth of 1 5 cm. How far would it sink in a liquid whose specific gravity Is 0.80? 21 . A hydrometer sinks to a depth of 1 2 cm. in a liquid whose specific gravity is 1.7. To what depth would it sink in water? 22 . A piece of wood whose volume is 1 50 c.c. floats with of its volume submerged in water. Find its mass. 23 . A piece of wood whose mass is 75.0 gm. floats in water with ^ of its volume above the surface. Find its volume. 24 . A piece of cork of density 0.25 gm. per c.c. floats in a liquid of density 1.2 gm. per c.c. What proportion of the volume of the cork will be immersed? 25 . An object floats in water with half its volume submerged. How much will be submerged when it floats in a liquid of specific gravity 1.5? 26 . To what depth will a block of wood 20 cm. high and of density 0.63 gm. per c.c. sink in a liquid of density 0.90 gm. per c.c.? • 33 CHAPTER 5 EXPERIMENTS ON MECHANICS INTRODUCTION Before proceeding with the following experiments the student should review the following techniques used in measurement. A. — Use of the ruler 1 . Avoid using the ends of the ruler. 2. Place your eye directly above the point where the reading is to be taken to avoid the error due to parallax. B / Measuring Length with a Fig. 5:1 Ruler—How to Avoid the Error Due to Parallax. Fig. 5:2 Measuring Volume with a Graduated Cylinder, B, — Use of the graduated cylinder Place your eye directly opposite the centre of the meniscus curve. Take the reading at this level. C. — Use of the balance 1. Clean and level the balance. 2. Support the beam of the balance on the knife-edge as demonstrated by the instructor. 3. Adjust for zero reading with all weights removed. Note that in all readings the pointer should swing an equal number of divisions on either side of the zero mark on the scale. Do not wait for the pointer to come to rest. 4. Place object to be weighed on the centre of the left pan of the balance. 5. Commencing with a weight that is definitely too heavy on the right- 34 EXPERIMENTS ON MECHANICS A B Fisher Scientific Co. Canadian laboratory Supplies Ltd. Fig. 5:3 The Balances A—Triple-beam balance. B— Equal-arm balance and box of weights. hand pan of the equal-arm balance (or on the arm of the triple-beam balance), systematically reduce the weight until balance is attained. 6. Total the weights used. 7. Disengage the knife-edge and return all weights to their box or to their zero position. EXPERIMENT 1 To determine the density of a regular solid. (Ref. Sec. 1:5) Apparatus Rectangular solid, ruler (graduated in mm.) , balance. 1. Carefully measure to the nearest millimetre the length, width and 35 Chap. 5 MECHANICS thickness of a rectangular solid.* Record your measurements, and calculate the volume of the object in cubic centimetres. 2. Determine the mass of the object and record it in grams. Observations = Length = Width Thickness = Mass =: cm. cm. cm. gm. Calculations 1. What is the volume of the object? 2. Determine the mass of one cubic centimetre. 3. What is the average of the results obtained by the class? Conclusion What is the density of this material? Questions 1. What is the correct value for the density of this material? (Table P-21). 2. Express the difference between the class average and the true value as a percentage of the true value. This is a measure of the experimental error. 3. Suggest sources of experimental error. * Note The density of other regular solids, such as a cylinder (Volume = 77 h), or a sphere (Volume = - tt 4 3 may be determined in a similar way (Fig. 5:4). EXPERIMENT 2 To determine the density of an irregular solid (rubber stopper), (Ref. Sec, 1:5) 36 EXPERIMENTS ON MECHANICS Apparatus Rubber stopper, thread, graduated cylinder, water, balance. Method 1. Determine the mass of the object and record it in grams. 2. Half fill a graduated cylinder with water. Note and record the Tie a thread to the object and carefully volume of the water. is completely immersed. Note and lower it into the water until it record the final volume of the water and object. Determine the volume of the object and record it in cubic centimetres. Observations Mass of object Initial volume of water Final volume of water and object .'. Volume of object ' = = = rr gm. c.c. c.c. c.c. Conclusion What is the density of this material? Question Why should the object be weighed before its volume is determined? EXPERIMENT 3 To determine the density of a liquid by measurement, (Ref. Sec. 1:5) Apparatus Beaker, graduated cylinder, balance, liquid (water, alcohol, etc.). Method 1. Determine the mass of a clean dry beaker. 2. Add about 50 c.c.^of the liquid to the beaker, weigh, and hence determine the mass of the liquid used. 3. Pour the liquid into a graduated cylinder and determine its volume. Observations = Mass of beaker Mass of beaker plus liquid = = .'. Mass of liquid Volume of liquid gm. gm. gm. c.c. Calculations Determine the mass of one cubic centimetre. What is the average of the results obtained by the class? Conclusion What is the density of this liquid? 37 Chap. 5 MECHANICS Questions 1. What is the correct value for the density of this liquid? (Table p. 21) 2. Calculate the percentage error. 3. Suggest sources of experimental error. 4. If the balance at your disposal is suitable, a graduated cylinder, instead of a beaker, could be used in steps 1 and 2 in the above experiment. Why should this tend to reduce the experimental error? EXPERIMENT 4 To determine the specific gravity of a liquid by means of the specific gravity bottle. (Ref. Sec. 1:7) Apparatus Specific-gravity bottle, water, balance, liquid (carbon tetrachloride, alcohol, etc.) X Fig. 5:6 Method 1. Carefully clean and dry the bottle. Determine its mass. 2. Fill completely with the liquid whose specific gravity is to be Insert the stopper, wipe off any excess liquid that determined. exudes through the opening. Determine the mass of the bottle plus the liquid. 3. Pour out the liquid. Rinse out the bottle with water. Fill the bottle completely with water. Determine the mass of the bottle plus the water. 38 EXPERIMENTS ON MECHANICS Observations = Mass of specific-gravity bottle empty Mass of specific-gravity bottle full of liquid = = Mass of specific-gravity bottle full of water = = .’. Mass of water Mass of liquid gm. gm. gm. gm. gm. Conclusion What is the specific gravity of the liquid? Questions 1. Define specific gravity. 2. Calculate the percentage error. 3. Suggest sources of experimental error. 4. What is the purpose of the hole through the centre of the stopper of the specific-gravity bottle? EXPERIMENT 5 To demonstrate Archimedes' Principle. (Ref. Sec. 1:11) Apparatus Balance, bucket and cylinder apparatus, beaker, water. Method 1. Hook the cylinder A on the bottom of the bucket B. Suspend them from the hook on the balance. Adjust the weights until the balance is in equilibrium. 2. Completely immerse the cylinder in a beaker of water. Be sure that the cylinder does not touch the bottom or sides of the beaker. Note the effect on the equilibrium. 3. Carefully add water to the bucket until it is completely full. Again note the effect on the equilibrium. 39 Chap. 5 MECHANICS Observations 1. What was observed when the cylinder was completely immersed in the water? 2. What was observed when the bucket was filled with water? Conclusions 1 . Why was the equilibrium disturbed in step 2 ? 2. Why was the equilibrium restored in step 3? 3. State Archimedes’ Principle. Questions 1. Why do objects apparently weigh less when immersed in a liquid? 2. What would be the effect of immersing the object in a denser liquid ? EXPERIMENT 6 Alternative method to demonstrate Archimedes' Principle. (Ref. Sec. 1:11) Apparatus Balance, object (glass stopper), beaker, overflow can, catch bucket, water, several other liquids (alcohol, carbon tetrachloride, brine, etc.) Method 1. Suspend the object from the hook on the balance. Determine its mass in air. 40 EXPERIMENTS ON MECHANICS 2. Completely immerse the object in a beaker of water. Be sure that it does not touch the beaker. Weigh the object while immersed in water. 3. Weigh a dry empty catch bucket. 4. Completely fill an overflow can with water. Let any excess water flow freely from the spout and discard it. Do not disturb the overflow can. Place the catch bucket under the spout. Carefully lower the object into the water and catch all of the overflow in the bucket. Weigh the catch bucket and overflow water. 5. Repeat the above using other liquids and fill in the table below. Observations Liquid Used Water Alcohol Carbon TETRAC H LORIDE Weight of object in air Weight of object in liquid .'. Apparent loss of weight in liquid Weight of empty catch bucket Weight of bucket plus displaced liquid .*. Weight of displaced liquid Conclusions 1. How does the apparent loss of weight compare with the weight of liquid displaced? 2. State Archimedes’ Principle. Questions 1. Why is a glass stopper an ideal solid to use in the above experiment? 2. What type of solid must be avoided? EXPERIMENT 7 To determine the specific gravity of a solid which is more dense than water using Archimedes' Principle. (Ref. Sec. 1:11) Apparatus Balance, beaker, water, thread, several solid objects more dense than water. Fig. 5:9 41 ) . Chap. 5 MECHANICS Method 1. Suspend a solid object by a thread from the hook on the balance. Determine its weight in air. 2. Completely immerse the object in a beaker of water. Be sure that it Determine the weight of the object does not touch the beaker. when immersed in water. 3. Repeat the above weighings for other objects supplied and fill in the table below. Observations Object Weight in Air Weight in Water Apparent Loss IN Weight 1. 2. 3. Calculations Weight of object in air = Apparent loss in weight = Weight of water displaced = gm. gm. gm. (Archimedes’ Principle) Spec
ific gravity = .’. Specific gravity = Weight of object in air Weight of equal volume of water Conclusion What is the specific gravity of the object? Questions 1. Calculate the percentage error. 2. Suggest sources of experimental error. 3. Explain how the weight of an equal volume of water was deter- mined. 4. State the density of the above objects in the metric system of units. 5. Discuss advantages and disadvantages of this method for finding density compared with the method used formerly (Exp. 2) EXPERIMENT 8 To determine the specific gravity of a liquid using Archimedes' Principle. (Ref. Sec. 1 : 1 1 Apparatus Balance, glass stopper, thread, several beakers, water, liquid (alcohol, carbon tetrachloride, etc.). Method 1. Suspend the object by a thread from the hook on the balance. Determine its weight in air. 42 EXPERIMENTS ON MECHANICS 2. Immerse the object in a beaker of water, being careful not to let it touch the beaker. Determine the weight of the object when immersed in water. 3. Rinse the object in a reserve supply of the liquid whose specific gravity is to be determined. Immerse the object in a beaker of this liquid and again weigh. 4. Repeat for other liquids supplied. Weight of Object in Air Weight of Object in Water Weight of Object in Liquid Observations Liquid Used Alcohol Carbon tetrachloride Calculations Mass of water displaced = Mass of liquid displaced = Calculate the specific gravity of this liquid. gm. gm. Conclusion What is the specific gravity of the liquid used? Questions 1. Calculate the percentage error. 2. Suggest sources of error. 3. Explain how the mass of water displaced or of liquid displaced was obtained in the above experiment. 4. Why was it correct to say that the volumes of water displaced and of liquid displaced were equal? EXPERIMENT 9 To demonstrate the Principle of Flotation. (Ref. Sec. 1:12) Apparatus Balance, paraffined wooden block, overflow can and catch bucket, water, other liquids. 43 Chap. 5 MECHANICS Method 1. Place the overflow can on the pan of the balance. Fill the overflow can to the spout with water (Exp. 6) . Balance it. 2. Without adjusting the weights, and after placing the catch bucket under the spout of the overflow can, carefully lower the wooden block into the water. Let it float freely being careful not to let it touch the sides of the can. Note all changes that occur until the water ceases to flow. 3. Repeat this experiment using the other liquids provided. Observations Describe the changes in equilibrium that occurred. Conclusions 1. How does the mass of the floating block compare with the mass of liquid it displaces? 2. State the Principle of Flotation. Questions 1. Why was the block of wood used in the above experiment coated with a thin film of paraffin? 2. Why does a steel ship float? EXPERIMENT 10 To show the principle of the hydrometer, (Ref. Sec. 1:13) Apparatus Simple hydrometer (Fig. 4:6), two tall cylindrical vessels, water, other liquids. Method 1. Float the hydrometer in a cylinder of water, being careful that it does not touch the sides. Note the depth to which it sinks in the water. Calculate the mass of the hydrometer. 2. Rinse the hydrometer in a reserve supply of the liquid to be used. Then float the hydrometer in a cylinder of the liquid, again being careful not to let it touch the sides. Record the depth to which it sinks in the liquid. 3. Repeat part 2 for other liquids. Observations = 1 . Depth to which hydrometer sinks in water 2. Depth to which hydrometer sinks in the liquid = Calculations Calculate the specific gravity of the liquid. 44 . EXPERIMENTS ON MECHANICS Conclusions 1. What is the specific gravity of the liquid? 2. On what principle does the use of the hydrometer depend? Questions 1. How do you find the mass of the hydrometer? 2. How do you find the mass of the liquid displaced? 3. How is the depth that the hydrometer sinks related to the specific gravity of the liquid? 4. What is the use of the hydrometer? EXPERIMENT 11 To determine the specific gravity of a liquid using a hydrometer, (Ref. Sec. 1:13) Apparatus Several tall cylindrical vessels, several liquids (brine, alcohol, etc.), three hydrometers (one for heavy liquids, one for light liquids, one universal) Method 1. Float an appropriate hydrometer in a cylinder containing the liquid whose specific gravity is to be determined. Be careful not to let it touch the sides of the vessel. Determine the specific gravity by reading the hydrometer scale at the liquid surface level. 2. Repeat for other liquids. Note This hydrometer method can be used to check the specific gravities of liquids obtained in previous experiments. Observations Li^^uid Specific Gravity Conclusion State the specific gravity of each liquid used. Questions 1 . Describe the construction of a hydrometer. 2. Why is the lower end of the hydrometer weighted (with mercury or lead shot) ? 3. Is the flotation bulb on a hydrometer for low-density liquids larger or smaller than the bulb on a hydrometer for high-density liquids? Explain. 4. Why is the hydrometer scale graduated with the smallest readings at the top and the largest at the bottom? Explain. 45 UNIT II SOUND Describe the different ways in which the sound of this depth charge would be heard below and above the water, and why. star Newspaper Service CHAPTER 6 PRODUCTION AND TRANSMISSION OF SOUND 11:1 INTRODUCTION From earliest childhood our ears grow . accustomed to sounds about us : first the sound of our mother’s voice, then the : sounds of home, of nature, and the busy world. Our consideration of their nature rarely goes further than calling the sounds we dislike noises, and some of the more pleasant ones music. These, however, are often subjective definitions, 1 as can be seen from the fact that a “hot-rodder” may drive for miles in his unmuffled car and think the sound it is making is “music”, while he will hurriedly turn off the radio because of the “noise” Beethoven is making. I I I [ I I Man, in fact, has been interested in inventing devices for making music and noises for much longer than in investigating the nature of sound. References to musical instruments in the Old Testament date back to 4000 b.c. Remember I the story of Joshua and the walls of ! Jericho. Yet, although Aristotle and the early philosophers knew something of the physical nature of sound, it is only in the last four or five hundred years that a fuller understanding of it has been gained. [ j i The word sound has been used frequently already, but no effort has so far been made to define it. One definiit the sensation that results tion calls when the auditory nerve is stimulated, while another refers physical the to causes of this sensation, in terms of the three necessary agencies for any sound: a source, a medium and a receiver. It is this second definition of sound which will be our concern in this unit. 11:2 THE ORIGIN OF SOUND When we ring a bell, bow a string, or strike a tuning-fork, and bring each into contact with a light object such as a pith ball, the object moves away as if being struck regularly (Fig. 6:1). Some demonstrators may prefer to touch the sounding body to some water and note Fig. 6:1 Sounding Bodies Vibrate. the splash and waves set up. This is ample proof that the sounding body is vibrating, i.e., moving to and fro. It may be concluded that sound always 49 Chap. 6 SOUND originates in a rapidly vibrating body. of Only sufficiently rapid vibrations cause sound, but some vibrations, whether or not they emit sound, have other effects, such as the destruction buildings, bridges, or parts of moving machinery. To take one instance, the vibration set up by a body of troops marching would be sufficient to destroy some bridges and so troops marching across them have to break of vibratory motion is one of great importance. Obviously the study step. 11:3 A STUDY OF VIBRATORY MOTION (a) The Pendulum and Transverse Vibrations A simple device for demonstrating vibrations is the pendulum (Chap. 10, Exp. 1 ) . This consists of a weight, called the bob, attached to the free end of a vertical cord, the other end of which is securely attached to a support (Fig. / \ \ \ \—V- \ \ / / / / / / / / Simple Pendulum in the Mean Position Amplitude of Vibration y- Movement of the Bob during one Vibration (Cycle) Fig. 6:2 Transverse Vibrations. 6:2). When the weight is drawn aside and allowed to swing, it will be -“seen the pendulum swings back “and that forth in a regular manner and that it 50 have moves about the same distance on either side of the point of rest or mean position. The advantage of using a pendulum to study vibrations is that the action is slow enough to be noted in every detail. For example, if we draw the bob aside and let it move to and fro once pendulum will performed the Most objects one vibration or cycle. would vibrate too fast to allow this to be seen. The horizontal displacement that the bob undergoes on either side of its natural rest position is the amplitude of the vibration. This is the destructive element of vibratory motion that affects rigid and brittle components of buildings, bridges, machinery and the like. Moreover, we shall see later (Sec. 11:11) that it affects the loudness of sound. To the movement of the pendulum, which is obviously at right angles to the mean position, is ascribed the term transverse vibration. Although there are other kinds of vibrations, you will recognize this kind in some of the objects used to demonstrate the origin of sound. Counting the number of vibrations in a given time, say 30 seconds, enables us to determine two other important propIf we calculate the erties of vibrations. number of vibrations per second, we know the frequency of vibration. Differences in the frequencies of musical notes account for differences in pitch of the sounds (Sec. 11:12). If we determine the time in seconds required for one vibration, we know the period of vibration. You will discover that all pendulums of the same length will hav
e the same period. It is for this reason that the pendulum is used in For that physics as a timing device. reason also, it is the primary component of large clocks. Were you to experiment with pendulums of different lengths, the period of a short one would be less than It is for this reason that of a long one. that you shorten the pendulum of a frequently ; V PRODUCTION AND TRANSMISSION OF SOUND Sec. II:4 clock which loses time and lengthen it for one that gains. Another demonstration may be arranged using a long rnetal rod, clamped (b) LoyigitiuUnal Vibrations A coil spring with a weight attached is supported vertically from a strong support as in Fig. 6:3. When it is vibrated its pith centre, having a at in contact with one end as in Fig. 6:4. When the half of the rod farthest from the pith ball is stroked with a chamois shrill sound is coated with resin, ball a emitted. Meanwhile the pith ball is displaced in the direction of the axis of the rod, thus showing longitudinal vibrations, although these are too rapid to be seen in detail. These examples indiare cate characterized by motion back and forth along the length of the vibrating body. longitudinal vibrations that 11:4 MEDIA FOR THE TRANSMISSION OF SOUND The need for a material medium for the transmission of sound is studied in Fig. 6:3 Longitudinal Vibrations. so that the weight moves vertically by alternately stretching and compressing the spring, longitudinal vibrations result. Fortunately vibrations slow the are Fig. 6:4 Longitudinal Vibrations. enough to enable us to observe all the details as in the pendulum (Chap. 10, Exp. 2). Chap. 10, Exp. 3. There it will be found that as the air is removed from the belljar in which there is a vibrating bell (Fig. 6:5), the sound gradually becomes fainter. The bell is heard again when air is reintroduced into the bell- jar. Since Of THE UNIVERSITY Of ALBERTA 51 Chap. 6 SOUND from or towards the point of origin of the waves. Again, if part of our rope were chalk-marked, this part would be seen as a white line perpendicular to the length of the rope when the latter was vibrated at one end to set up a train of Electromagnetic waves waves along it. (Sec. IV : 38) , which include light waves, are also transverse in character. Fig. 6:6 shows the displacements of the particles of a medium transmitting a transverse wave. Particles at the crests (B, F, etc.) of the wave are undergoing a maximum displacement upwards, those at the bottom of the troughs (D, H, etc.) a maximum displacement downwards. Some terms used in wave motion follow: Amplitude is the maximum displacement from the mean position (BBi or DDi, etc.). Phase. Particles at the same distance from their mean positions and which are moving in the same direction are said to be in the same phase. Thus particles P and Q are in the same phase, as also are particles B and F. On the other hand, particles B and D are completely out of phase. Wave-Length is the distance, (/) usually expressed in centimetres or inches, between two consecutive particles in the same phase. Thus the distance BF between two adjacent crests, or the distance DH from one trough to the next, gives the wave-length of the disturbance. This we could see the vibrating bell throughout the entire experiment but could not hear the sound when there was no air present, we conclude that sound, unlike light, cannot travel through a vacuum. Sounds can also be conveyed through liquids and most solids. Thus the noise of the engines of a submarine can be picked up by underwater microphones, and the sound of vibrating telegraph wires can be clearly heard by an ear pressed against a telegraph post. 11:5 WAVE MOTION (a) Transverse Waves The disturbances set up by a vibrating body are propagated in the form of waves in the medium (Chap. 10, Exp. 4). A wave may be defined as a disturbance of any kind which travels without change of form and without the medium moving bodily with it. Simple examples of waves are to be seen when one end of a taut rope is jerked or when a stone is thrown into a pond and makes ripples on the surface. Waves produced in this way are known as transverse waves, since the disturbances in the medium are perpendicular or transverse to the direction of propagation of the waves. The ripples on water can be seen to travel outwards from the centre of the disturbance, but a cork floating on the surface will execute an up-anddown motion without moving away 52 PRODUCTION AND TRANSMISSION OF SOUND Sec. II: 5 represents the distance that the motion has travelled during the execution of one com]3lete vibration. Period of a \ ibratioii is the time of one complete vibration, or is the time taken by a particle in travelling from its mean position through the maximum displacement first in one direction and then in the other, finally returning to its mean position. Frequency (n) is the number of vibra- tions in one second. Wave-Train is a succession of waves caused by continuous vibration of the source. DISPLACEMENT OF END A . % Vibration Vi Vibration Ff V* Vibration n 1 Vibration DISTANCE TRAVELLED BY DISTURBANCE IN CORD AB g '/4 Wave-length Vi Wave-length Ti Wave-length 1 Wave-length Fig. 6:7 Proving that V = nl. Velocity (F) is the distance covered in a unit of time (a second). Since during one vibration the disturbance travels I cm. (Fig. 6:7), then during n vibrations, the disturbance travels nl cm. (n wave-lengths). Now, (one wave-length) if n the frequency is vibrations per second, the disturbance travels nl cm. in one second, i.e., the velocity is nl cm. per second. This gives us the wave formula found to be equally useful in all branches of physics. Velocity = frequency X wave-length or V = nl. R I -C .1 R 53 Chap. 6 SOUND audible in all directions, each wave-front, i.e., the leading edge of each wave, must be spherical. How the air responds to the vibrating bell may be illustrated with the aid of the apparatus shown in Fig. 6:10. The Fig. 6:10 Illustrating Longitudinal Wave Motion. steel balls are suspended so that they just touch. When the first one is drawn aside and allowed to hit the ball next in line, none moves except the one at the opIt flies out about as far as posite end. the first was drawn aside. Because steel is elastic, the impact is passed through Such waves (b) Longitudinal Waves Sound waves differ from those described above in that the particles of the medium are displaced from their mean positions backwards and forwards along the line of travel of the wave are known as motion. longitudinal waves, and may be illustrated by reference to Fig. 6:8. There we have a coil spring stretched between two supports (S, Si). A piece of cloth is tied near its centre. When several coils of the spring are squeezed together a compression, formed. When the coils are released, their elasticity causes them to return to their normal position. The momentum so produced causes them to move past this position, thereby forming a stretched region or rarefaction (R). or condensation (C) is A study of the jerking of the cloth to and fro along the length of the spring will give ample proof of what is happening. It will now be good practice to draw a longitudinal wave-train in a coil spring and label amplitude, particles in the same phase, and one wave-length. Longitudinal waves are always characterized by condensations and rarefactions. Rarefoction Sound Waves from a Vibrating Tuning-fork. the line as a compression wave or condensation followed by an expansion or rarefaction. This is similar to the way in which sound waves are transmitted through air. Ii:6 THE SUPERPOSITION OF WAVES (a) Interference Interesting effects are observed when Fig. Sound waves from a vibrating bell are 6:9. The transverse depicted in vibrations of the gong give rise to condensations and rarefactions alternately, i.e., longitudinal waves. As the sound is 54 PRODUCTION AND TRANSMISSION OF SOUND Sec. II: 6 two waves are simultaneously propagated through the same medium. The resulting displacement at any point of the medium is the algebraic sum of the displacements produced by the two separate waves. When these are in the same direction the effects are thus reinforced, and when In the opposed they are diminished. special case where the two waves are of the same frequency and amplitude, as in Fig. 6:11 (a), each wave will assist the other at all points when die two waves crests and phase, are ! troughs of the two waves exactly coinciding throughout the medium. If, however, they are completely out of phase * as in Fig. 6: 11(b) the two waves are i in exact opposition at all points, the crests of one coinciding with the troughs of the other, and accordingly the result- exactly the in I ; j I ing effect in the medium is nil. These phenomena are referred to under the general heading of interference. Special cases of particular interest to us in our study of sound will be presented in Sec. 11:24. (b) Standing Waves A very important case of interference is seen when two trains of waves of the same frequency and amplitude travel in opposite directions through a medium, for example, original and reflected waves (Chap. 10, Exp. 5). To demonstrate this, attach a light flexible silk cord to one prong of a large tuning-fork. Pass the other end of the cord over a pulley and attach a weight to it. The tuningfork should be activated by an electromagnet to give continuous vibration. In (b) Two Waves In Opposite Phase They Nullify Each Other Fig. 6:11 Production of Standing Waves. 55 Chap. 6 SOUND Vi Period After (a) '/2 Period After (a) Vi Period After (o) ^ Period After (o) A combination of (a) (b) (c) (d) (e) Fig. 6:12 Superposition of Waves. 56 PRODUCTION AND TRANSMISSION OF SOUND Sec. II to place of the tuning-fork an electric bell with gong removed may be used, the cord being tied the clapper, or a special vibrator as shown in Fig. 6:11 (c) may be employed. For reasons to be discussed later, the length of the cord is adjusted until it takes the form
shown in the figure. The hazy oval regions where displacement of the cord is greatest are called loops. The points of quiet tlie ends where reflection as well It might be occurs are called nodes. imagined that a loop would occur where the cord meets the vibrator but, in reality, the amplitude of the vibrator is so small compared to that of the cord that it must be considered to be a node. In any case some reflection occurs there, as depends The reason for adjusting the length of to be the cord is rather too difficult explained here. Flowever, this much can be said, that the wave-length of the disturbance tension (caused by the weight) and the frequency of the source. Since each loop is half a wave-length and there must be a whole number of loops in the cord, the length must be adjusted to accommodate a whole number of loops. the on An examination of Fig. 6:12 will help us to understand the phenomenon. The wave composed of dashes is proceeding to the right, the dotted one to the left. They have the same amplitude and wave-length. The solid line is the resultant of the former two. At the start let us assume that the waves are completely out of phase as in (a). We know that the result will be Onea line of undisturbed particles. quarter of a period later (b), each will have shifted one-quarter wave-length but in opposite directions. -The waves will be in phase now and will reinforce each the resultant other. has wider amplitude than either of the original waves. Diagrams (c), (d), and (e), may be explained as above except For this reason, that the phase is different after each quarter vibration. The combined effect is shown in (f). 4, 8, Examination of the diagrams reveals 10 which are that points 2, 6, one-half a wave-length apart are always at rest and hence constitute the nodes. Points 1, 3, 5, 7, 9 move from rest to a point of maximum displacement on one side then back through the point of rest to a point of maximum displacement on the other and return. These are the loops. Thus standing waves consist of nodes and loops. The distance between successive nodes or is one-half a wave-length. These waves will be useful in understanding vibrations in strings and air columns which will be presented in succeeding chapters. loops 11:7 REFLECTION OF SOUND WAVES Sound waves travelling through the air and striking a smooth hard surface undergo reflection, obeying the same laws of reflection as light waves. That is so can be demonstrated using this the apparatus shown in Fig. 6:13. Sound A B Fig. 6:13 Reflection of Sound. waves from a source S (a watch) are directed by a tube to a hard surface, AB (a drawing-board is suitable for the purpose). A receiver, R (the ear), is 57 Chap. 6 SOUND Canadian National Exhibition Band Shell. Canadian National Exhibition placed at the end of a second tube which, to detect the signals, must be inclined to AB at the same angle as the first In short, the angle of reflection tube. equals the angle of incidence (i). (r) Also, the incident sound (SO), the perpendicular (CO) and the reflected sound The (RO) are in screen CD acts as a shield to protect R from the sound waves transmitted directly from S. the same plane. A very interesting demonstration of the reflection of sound is found in the Museum of Science and Industry at Chicago. Two concave mirrors are arranged a long distance apart. A person standing at the focus of one may whisper softly and the words will be clearly audible to a person standing at the focus of the other. Echoes are due to reflection. They are produced as the result of a sound 58 PRODUCTION AND TRANSMISSION OF SOUND Sec. II: 8 all etc. will sides, signal directed towards a distant ob(e.g., a wall) being returned to stacle the listener, who thus hears a repetition of the sound a short period after it has been produced. Forests, clifTs, hillreflect sound and cause the formation of echoes. If there are a number of reflecting surfaces at different distances from the source, a series of repetitions of the sound signal, each following the other, will be received. These are known as multiple echoes, and they are heard, for example, when an Alpine horn is blown amidst a number of mountain peaks. The reflection of sounds may have unpleasant results in auditoria that are not properly constructed. The sounds echo and re-echo from the walls so that the effects of one sound have not died The away before the next is made. resulting jumble of sounds is known as reverberation. Further reference of sound will be made in the next section and in section 11:32. reflection to 11:8 THE VELOCITY OF SOUND (a) Methods of Measuring 1. By Direct Measurement The early experiments to determine the velocity of sound in air, made in the late seventeenth and early eighteenth centuries, were based on estimating the difference in time for the light and the of sound to travel to an observer from a cannon fired some distance away (Fig. 6:15). Since the speed of light is very great (186,000 miles per second), the the explosion is seen almost flash instantaneously. Hence the time elapsing between an observer seeing the flash and hearing the report of the explosion may be taken as the time needed for the sound to travel the measured distance between the cannon and observer, and this enables the speed of sound to be calculated. There are two main objections to these simple “flash-bang” experiments. If there is a wind the result will be greater or less than the true value according to whether the wind is blowing from the cannon to the observer or in the opposite direction. The other chief source of error is the “reflex time” of the observer. Thus, if a stop-watch is used in such an experiment there is a difference between the response time in seeing the flash and the starting report and stopping it. This error varies with different observers, and with the same individual from time to time, and also even with the loudness of the sound. If the cannon is one mile from the obtake about 5 server seconds to travel the distance, and hence a personal error of 1/5 second will introduce an error of four per cent in the the watch, and hearing the sound will final result. 59 Chap. 6 SOUND The accepted values for the velocity of sound in air are tabulated below. It will be noted that for a change of 1 centigrade is a corresponding change in the velocity of sound in air of 2 feet per second or 0.6 metres per second. degree, there Temperature 0°C. 10°C. 20°C. Velocity Ft. per Sec. Velocity Metres PER Sec. 1089 1109 1129 332 338 344 The first accurate determination of the speed of sound was carried out by 1738. Two the French Academy in cannons were used, separated by a distance of about 18 miles to reduce the error due to the “reflex time”. To eliminate the effect of wind, timings were taken (by means of pendulums) in both directions and the average used to calculate the speed of sound. The result obtained was 337 metres per sec. at 6°G., or 332 metres per sec. at 0°C. Later experiments, carried out in the same way, but using chronometers, accurate to one-tenth of a second, gave a mean result of 331 metres per sec. at 0°C. 2. By Echoes If a sound signal is directed to a distant wall or obstacle an echo will be received some time later, the sound wave having travelled twice the distance between the wall and the observer during This suggests a possible the period. method of measuring the velocity of sound. For clear echoes, however, the wall (or reflecting surface) should not be too far distant, and this involves the difficulty of measuring very practical short time-intervals. With a wall 100 yards distant, for example, the echo of a sound signal will be received only about one-half second later. This difficulty has been overcome by using a metronome or 60 an electrically controlled tapper to send out a sequence of sound signals at regular intervals. The distance of the metronome from the wall is adjusted until the echo of one click is heard simultaneously with the next click. The sound signal has clearly travelled to and from the wall during the interval between the clicks. This time-interval is accurately known, and so the velocity of sound can be It should be noted that since found. the sound signal has to travel back along its track this method automatically eliminates the effect of the wind. Example A metronome was 280 feet from the interval between wall and the time clicks was 1/2 second. Distance covered in 1 /2 sec. was 560 feet In 1/2 sec. sound has travelled 560 feet In 1 sec. sound has travelled 1/2 ‘ . . the velocity of sound ^ = 1120 feet per sec. 3. By Resonance In section 11:22, and experiment 10, chapter 10, we shall study resonance experimentally, and shall use the results in the formula V — nl to determine the velocity of sound. (h) Breaking the Sound Barrier In recent years breaking the sound barrier, i.e., flying at the speed of sound (742.5 miles per hour at 0°C.) and more, has become a great test both of the skill of operation and the construction of aircraft. When an advancing craft catches up with its own sound waves a giant barrier of compressed molecules of air must be penetrated, and this requires the power of jet-type engines and an extremely strong construction of the aircraft. This situation PRODUCTION AND TRANSMISSION OF SOUND Sec. II: 9 presents just one of the many practical problems which can only be dealt with because of our modern grasp of the nature of sound. The Velocity of Sound in Various Media Medium Velocity Gases feet per sec. metres per sec. Carbon Dioxide (0°C.) Oxygen (0°C.) Air (0°C.) Hydrogen (0°C.) Liquids Water (9°C.) Sea-Water (9°G.) Solids Brass Oak Glass Iron Aluminum 846 1041 1089 4165 4708 4756 11480 12620 16410 16410 16740 i , 11:9 QUESTIONS 258 317 332 1269.5 1435 1450 3500 3850 5000 5000 5104 A 1. Which of the following statements is correct? (a) All vibrating objects produce sound, (b) All sound is produced by vibrating objects. Explain your answer fully. 2. 3. betwee

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