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2 Pearson Education Limited Edinburgh Gate Harlow Essex CM20 2JE England and Associated Companies throughout the world Visit us on the World Wide Web at: Pearson Education Limited 2014 All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without either the prior written permission of the publisher or a licence permitting restricted copying in the United Kingdom issued by the Copyright Licensing Agency Ltd, Saffron House, 6 10 Kirby Street, London EC1N 8TS. All trademarks used herein are the property of their respective owners. The use of any trademark in this text does not vest in the author or publisher any trademark ownership rights in such trademarks, nor does the use of such trademarks imply any affiliation with or endorsement of this book by such owners. ISBN 10: ISBN 13: British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library Printed in the United States of America

NASA The basic dynamic principle for rotational motion (Equation 5, gt 5Ia) can be restated in terms of angular momentum. If the axis doesn t move within the rotating object, then I is constant.

3 NASA The basic dynamic principle for rotational motion (Equation 5, gt 5Ia) can be restated in terms of angular momentum. If the axis doesn t move within the rotating object, then I is constant. In that case, Dv Ia 5I Dt 5 D 1 Iv 2 Dt and Equation 5 can be rewritten as follows: 5 DL Dt, Torque and rate of change of angular momentum The rate of change of angular momentum of a rigid body with respect to any axis equals the sum of the torques of the forces acting on it with respect to that axis: Application Gearing down. Nanoengineering involves the construction of submicroscopic machines, often built through the manipulation of individual atoms. Engineers have recently made tiny gears from carbon nanotubes with interlocking teeth made of organic molecules precisely placed on the tube surfaces. The operation of these gears is based on the same physical principles of torque and angular momentum as are applied to their macroscopic counterparts. To turn the gears, a laser shines tangentially onto the atoms at the end of the cylinder, creating a torque. This torque can cause a gear to rotate up to several thousand times per nanosecond. O y mv sin f r S f S S p 5 mv m f l 5 r sin f FIGURE 16 A particle with mass m rotating in the x-y plane. The angular momentum of the particle has magnitude L 5 mvl. x gt 5 DL Dt. We ll use this result soon when we discuss a conservation principle for angular momentum. We can also define angular momentum for a single particle. In Figure 16, a particle moves in the x-y plane; its radial and tangential components of velocity are shown. The instantaneous velocity and linear momentum of the particle lie along a line at a perpendicular distance l from the axis through point O. We define the angular momentum of this particle as L 5 mvl. (11) S S In the figure, the directions of v and p correspond to a counterclockwise (positive) sense of rotation. The distance l serves as a moment arm for the velocity, analogous to the moment arm (or lever arm) of a force in the definition of torque. Equation 11 is consistent with our original definition of angular momentum of a rigid body. Suppose the object consists of a collection of particles with masses m A, m B, m C, c at distances r A, r B, r C, c from the axis through O. For each particle, the angle f in Figure 16 is 90, so the moment arm l for particle A is its distance r A from the axis of rotation. When the object rotates with angular velocity v, the magnitude v A of the velocity of particle A is v A 5vr A, and the angular momentum L A of this particle is L A 5 m A v A r A 5 m A 1 vr A 2 r A 5 m A r A 2 v. (10) If the object is rigid, all the particles must have the same angular velocity v; the total angular momentum of all the particles is then L 5 1 m A r A 2 1 m B r B 2 1 m C r C 2 1 c 2 v5iv, in agreement with our previous definition of the angular momentum of a rigid body. NOTE As with all angular quantities, angular momentum is defined with reference to a particular axis of rotation. It is essential to specify this axis and use it consistently when applying the principles we ve just discussed. Equation 10 will be particularly useful when we develop the principle of conservation of angular momentum in the next section. We ve discussed changes in angular momentum that occur when the angular velocity v of a body changes, but there are also cases in which v is constant and I changes because of a change of shape or rearrangement of component parts of a body. In cases where I isn t constant, Equation 5 1 gt 5Ia2 isn t valid. But the angular momentum is still given at each instant by L 5 Iv, and the rate of change of angular momentum is still given 354

4 by Equation 10. The next section includes some examples in which we have to take this more general view. We ve discussed angular momentum in the context of macroscopic bodies, but it is vitally important in physics at all scales, from nuclear to galactic. Angular momentum plays an essential role in classifying the quantum states of nuclei, atoms, and molecules and in describing their interactions. Angular momentum considerations are also an important tool in analyzing the motions of stars, galaxies, and nebulae. Along with mass, energy, and momentum, angular momentum is one of the unifying threads woven through the entire fabric of physical science. EXAMPLE 7 A kinetic sculpture A part of a mobile suspended from the ceiling of an airport terminal building consists of two metal spheres, each with mass 2.0 kg, connected by a uniform metal rod with mass 3.0 kg and length s m. The assembly is suspended at its midpoint by a wire and rotates in a horizontal plane, making 3.0 revolutions per minute. Find the angular momentum and kinetic energy of the assembly. SOLUTION SET UP Figure 17 shows our sketch. We draw the assembly as viewed from above (or below), in its plane of rotation. SOLVE First, we need to find the moment of inertia of the assembly about an axis through its midpoint; we ll assume that each sphere can be treated as a point. Then the angular momentum L is given by L 5 Iv, and the kinetic energy is K Iv2. The moment of inertia of each sphere (treated as a point) is I sphere 5 m1 s kg m kg # m 2. From Table 9.2, the moment of inertia of a rod about an axis through its midpoint and perpendicular to its length is I rod Ms kg m kg # m 2. So the total moment of inertia of the assembly is I total kg # m kg # m kg # m 2. The angular velocity is given in revolutions per minute; we must convert it to radians per second: 3.0 rev v53.0 rev/min min 21 1 min 2p rad 60 s 21 1 rev rad /s. FIGURE 17 Now we can calculate the angular momentum of the assembly: L 5 Iv kg # m rad/s kg # m 2 /s. The kinetic energy is K Iv kg # m rad/s kg # m 2 /s J. REFLECT The mass of the rod is comparable to that of the spheres, but it contributes less to the total moment of inertia because most of its mass is closer to the axis of rotation than are the masses of the spheres. Practice Problem: How long would an equal-mass rod have to be in order for its contribution to the total moment of inertia to equal that of the two spheres together? Answer: 8.0 m. 5 Conservation of Angular Momentum We ve seen that angular momentum can be used in an alternative statement of the basic dynamic principle for rotational motion. Angular momentum also forms the basis for a very important conservation principle: the principle of conservation of angular momentum. Like conservation of energy and conservation of linear momentum, this principle appears to be a universal conservation law, valid at all scales, from atomic and nuclear systems to the motions of galaxies. The principle is based on the concept of an isolated system. We generalize this term to mean a system in which the total external force is zero and the total external torque is zero. Then our new conservation principle states simply that the total angular momentum of an isolated system is constant. PhET: Torque ActivPhysics 7.14: Ball Hits Bat 355

Application The day the day got shorter.

plate, allowing the India plate to move downward by about 15 m. This downward shift of mass reduced our planet s moment of inertia slightly.

Scientists estimated that the quake shortened the length of the day by 2.7 microseconds. The proof of this principle follows directly from Equation 10 1 gt 5 DL/Dt 2.

5 Application The day the day got shorter. The earthquake that caused the devastating tsunami of December 26, 2004, represented a rupture along a 1200 km stretch of the fault where the India plate of the earth s crust dives under the Burma plate, allowing the India plate to move downward by about 15 m. This downward shift of mass reduced our planet s moment of inertia slightly. Its angular momentum about its axis of rotation is very nearly constant; to conserve this angular momentum, the speed of rotation increased, shortening the length of the day. Scientists estimated that the quake shortened the length of the day by 2.7 microseconds. The proof of this principle follows directly from Equation 10 1 gt 5 DL/Dt 2. When a system has several parts, the internal forces that the parts exert on each other cause changes in the angular momenta of the parts, but the total angular momentum of the system doesn t change. Here s an example: When two bodies A and B interact with each other but not with anything else, A and B together form an isolated system. Suppose body A exerts a force F S A on B on body B; the corresponding torque (with respect to whatever axis we choose) is t A on B. According to Equation 10, this torque is equal to the rate of change of angular momentum of B: Action reaction pair S F B on A Particle A Shared moment arm l S F A on B Line of action of both forces Particle B The torques cancel: t A on B = 1Fl; t B on A = 2Fl FIGURE 18 Why internal forces cannot change the total angular momentum of a system. Chris Trotman/Corbis l O NASA Duomo/Corbis t A on B 5 DL B Dt. At the same time, body B exerts a force torque t B on A, and t B on A 5 DL A Dt. on body A, with a corresponding From Newton s third law, F S B on A 52F S A on B. Furthermore, if the forces act along the same line, their moment arms with respect to the chosen axis are equal. Figure 7 demonstrated this point; we repeat it here as Figure 18. Thus, the torques of these two forces are equal and opposite, and t B on A 52t A on B. So when we add the preceding two equations, we find that DL A Dt 1 DL B Dt F S B on A 5 0, or, because L A 1 L B is the total angular momentum L of the system, DL (12) Dt 5 0. That is, the total angular momentum of the system is constant; we also say that it is conserved. The torques of the internal forces can transfer angular momentum from one interacting object to another, but they can t change the total angular momentum of the system. Conservation of angular momentum When the sum of the torques of all the external forces acting on a system is zero, the total angular momentum of the system is constant (conserved). FIGURE 19 As this skater reduces her moment of inertia by pulling in her arms, she spins faster, conserving angular momentum. An ice skater performing a pirouette on the toe of one skate takes advantage of this principle. Figure 19 shows the skater at the start of her pirouette, with her arms extended and with a counterclockwise angular momentum. As she pulls in her arms, she reduces her moment of inertia I. Because her angular momentum L 5 Iv must remain constant as I decreases, her angular velocity v increases; she spins faster. That is, I i v i 5 I f v f. (13) 356

6 A diver going into a tuck uses the same principle. Only by reducing his moment of inertia is he able to rotate multiple times before hitting the water. Notice that the skater and the diver can both be treated as isolated systems. The net external forces and torques on the skater are slight. Gravity has a major effect on the diver s translational motion, but once he is in the air it exerts no net torque on him. Thus, in terms of angular momentum, he is isolated. EXAMPLE 8 Two rotating disks interacting Figure 20 shows two disks, one an engine flywheel, the other a clutch plate attached to a transmission shaft. Their moments of inertia are I A and I B ; initially, they are rotating with constant angular velocities v A and v B, respectively. We then push the disks together with forces acting along the axis, so as not to apply any torque on either disk. The disks rub against each other and eventually reach a common final angular velocity v f. Derive an expression for v f. BEFORE F S I A v A v B I B 2F S v f AFTER F S 2F S FIGURE 20 I A 1 I B SOLUTION SET UP The two disks are an isolated system. The only torque acting on either disk is the torque applied by the other disk; there are no external torques. Thus the total angular momentum of the system is the same before and after they are pushed together. At the end, they rotate together as one body with total moment of inertia I f 5 I A 1 I B and angular velocity v f. SOLVE Conservation of angular momentum gives I A v A 1 I B v B 5 1 I A 1 I B 2 v f, v f 5 I A v A 1 I B v B I A 1 I B. REFLECT This is the analog of a completely inelastic collision of two particles. The final kinetic energy is always less than the initial value. This can be proved in general, using the above expression for v f. Practice Problem: Suppose I B 5 2I A. How must v A and v B be related if the entire system comes to rest after the interaction? Answer: v A 522v B. Quantitative Mass on a string: I Analysis 3 In Figure 21, a block slides in a circular path on a horizontal frictionless plane under the action of a string that passes through a hole in the plane and is held vertically underneath it. By pulling on the lower end of the string, we reduce the radius to half of its original value. If the initial speed of the block is v i, then the speed of the block after the string is shortened is 1 A. 2 v i. B. 2v i. C. 4v i. SOLUTION From Equation 11 the angular momentum of the block about a vertical axis through the hole is mvr. The force from the string pulls directly toward the axis. Therefore, it exerts no torque with respect to this axis, and it cannot change the block s angular momentum. Thus the angular momentum mvr of FIGURE 21 the block must be constant. If r is reduced by half, then v must double to keep the angular momentum constant (answer B). 357

7 Quantitative Mass on a string: II Analysis 4 Let s consider again the whirling block shown in Figure 21. By letting additional string pass through the hole in the plane, we increase the radius to twice its original value. If the initial tension in the string is T, then the tension in the string after the radius is increased is 1 A. 2 T. 1 B. 4 T. C. 1 8 T. SOLUTION From Quantitative Analysis 3, we see that in this situation the angular momentum of the block doesn t change: if r is doubled, then v must decrease by half. The tension T provides the needed centripetal force, according to the relationship T 5 mv 2 /r. In the situation presented here, r doubles and v is reduced to half its original value. Thus, in the expression for T, the denominator doubles and the numerator is reduced to onefourth its initial value, for an overall reduction of T to one-eighth of its initial value (answer C). EXAMPLE 9 Anyone can be a ballerina An acrobatic physics professor stands at the center of a turntable, holding his arms extended horizontally, with a 5.0 kg dumbbell in each hand (Figure 22). He is set rotating about a vertical axis, making one revolution in 2.0 s. His moment of inertia (without the dumbbells) is 3.0 kg # m 2 when his arms are outstretched, and drops to 2.2 kg # m 2 when his arms are pulled in close to his chest. The dumbbells are 1.0 m from the axis initially and 0.20 m from it at the end. Find the professor s new angular velocity if he pulls the dumbbells close to his chest, and compare the final total kinetic energy with the initial value. FIGURE 22 BEFORE AFTER SOLUTION SET UP If we neglect friction in the turntable, there are no external torques with respect to the vertical axis, and the angular momentum about this axis is constant. That is, I i v i 5 I f v f, where I i and v i are the initial total moment of inertia and angular velocity, respectively, and and are the final values. SOLVE In each case, I 5 I prof 1 I dumb ; thus, I i kg # m kg m kg # m 2 I f kg # m kg m kg # m 2 1 rev v i 5 2p 2.0 s 5p rad /s. v f From conservation of angular momentum, I f 1 13 kg # m 2 21p rad/s kg # m 2 2 v f, v f 5 5.0p rad/s rev/s. That is, the angular velocity increases by a factor of five. The initial kinetic energy is K i I i v 2 i kg # m 2 21p rad/s J. The final kinetic energy is K f I f v 2 f kg # m p rad/s J. REFLECT Where did the extra 256 J of kinetic energy come from? Practice Problem: Suppose the professor drops the dumbbells and then pulls his arms close to his chest. What is his final angular velocity? Answer: 0.68 rev/s. EXAMPLE 10 Angular momentum in a crime bust A uniform door 1.0 m wide with a mass of 15 kg is hinged at one side so that it can rotate without friction about a vertical axis. The door is unlatched. A police detective fires a bullet with a mass of 10 g and a speed of 400 m/s into the exact center of the door in a direction perpendicular to the plane of the door. Find the angular velocity of the door just after the bullet embeds itself in it. Is kinetic energy conserved? Continued 358

Student Workbook for Physics for Scientists and Engineers: A Strategic Approach with Modern Physics Randall D. Knight Third Edition

Student Workbook for Physics for Scientists and Engineers: A Strategic Approach with Modern Physics Randall D. Knight Third Edition Pearson Education Limited Edinburgh Gate Harlow Essex CM20 2JE England