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Home , Angular momentum, Circular motion, Torque

AND ANGULAR MISN-0-34 IN CIRCULAR by Kirby Morgan, Charlotte, Michigan

TORQUE AND 1. Introduction ...... 1 2. Torque and Angular Momentum IN a. Definitions ...... 1 b. Relationship: ~τ = dL/dt~ ...... 1 c. Motion Confined to a ...... 2 d. Circular Motion of a ...... 3 3. Systems of a. Total Angular Momentum ...... 4 shaft b. Total Torque ...... 4 c. Motion About a Fixed Axis ...... 5 d. Example: ...... 5 e. Kinetic of ...... 6 f. Linear vs. Rotational Motion ...... 6 4. Conservation of Angular Momentum a. Statement of the Law ...... 7 b. If the External Torque is not Zero ...... 7 c. Example: Two ...... 7 d. of the Two Flywheels ...... 8

5. Nonplanar Rigid Bodies ...... 9

Acknowledgments...... 9

Glossary ...... 9

ProjectPHYSNET·Bldg.·MichiganStateUniversity·EastLansing,MI

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2 ID Sheet: MISN-0-34

THIS IS A DEVELOPMENTAL-STAGE PUBLICATION Title: Torque and Angular Momentum in Circular Motion OF PROJECT PHYSNET Author: Kirby Morgan, HandiComputing, Charlotte, MI The goal of our project is to assist a network of educators and scientists in Version: 4/16/2002 Evaluation: Stage 0 transferring physics from one person to another. We support manuscript processing and distribution, along with communication and information Length: 1 hr; 24 pages systems. We also with employers to identify basic scientific skills Input Skills: as well as physics topics that are needed in science and technology. A number of our publications are aimed at assisting users in acquiring such 1. Vocabulary: kinetic energy (MISN-0-20), torque, angular acceler- skills. ation (MISN-0-33), angular momentum (MISN-0-41). 2. Solve constant angular problems involving torque, Our publications are designed: (i) to be updated quickly in response to of , angular , rotational and field tests and new scientific developments; (ii) to be used in both class- (MISN-0-33). room and professional settings; (iii) to show the prerequisite dependen- 3. Justify the use of conservation of angular momentum to solve prob- cies existing among the various chunks of physics knowledge and skill, lems involving torqueless change from one state of uniform circular as a guide both to mental organization and to use of the materials; and motion to another (MISN-0-41). (iv) to be adapted quickly to specific user needs ranging from single-skill instruction to complete custom textbooks. Output Skills (Knowledge): New authors, reviewers and field testers are welcome. K1. Define the torque and angular momentum vectors for (a) a single (b) a system of particles. PROJECT STAFF K2. Starting from ’s 2nd law, derive its rotational analog and state when it can be written as a equation. Andrew Schnepp Webmaster K3. Start from the equation for linear kinetic energy and derive the Eugene Kales Graphics corresponding one for rotational kinetic energy. Peter Signell Project Director K4. Explain why conservation of angular momentum may not hold in one system but may if the system is expanded. ADVISORY COMMITTEE Output Skills (Problem Solving): D. Alan Bromley Yale University S1. For in circular motion at fixed radii, solve problems relating E. Leonard Jossem The Ohio State University torque, , , rotational kinetic A. A. Strassenburg S. U. N. Y., Stony Brook energy, work, and angular momentum. S2. Given a system in which angular momentum is changing with time Views expressed in a module are those of the module author(s) and are due to a specified applied torque, reconstruct the minimum ex- not necessarily those of other project participants. panded system in which total angular momentum is conserved. Describe the reaction torque which produces the compensating °c 2002, Peter Signell for Project PHYSNET, Physics-Astronomy Bldg., change in angular momentum. Mich. State Univ., E. Lansing, MI 48824; (517) 355-3784. For our liberal use policies see: http://www.physnet.org/home/modules/license.html.

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TORQUE AND ANGULAR MOMENTUM q IN CIRCULAR MOTION q p F by

Kirby Morgan, Charlotte, Michigan r r 1. Introduction 0 (a) 0 (b) Just as for translational motion (motion in a straight line), circular or rotational motion can be separated into and . Since Figure 1. Vector relationships for: (a) torque (b) angular rotational kinematics is covered elsewhere,1 the discussion here will center momentum (both directed out of the page). on rotational dynamics. Our goal is to derive the rotational analog of Newton’s law and then apply it to the circular motion of a single Starting from Newton’s second law, written in the form particle and to systems of particles. In particular we wish to develop dp~ the relationship between torque and angular momentum and discuss the F~ = , (3) circumstances under which angular momentum is conserved. dt the torque is: dp~ 2. Torque and Angular Momentum ~τ = ~r × F~ = ~r × . (4) dt 2a. Definitions. The torque and angular momentum are defined as This can be rewritten using the expression for the of a cross vector products of , and momentum. Suppose a force F~ product: Help: [S-1] acts on a particle whose position with respect to the origin O is the d d~r displacement vector ~r. Then the torque “about the point 0 and acting on ~τ = (~r × p~) − × p~ 2 dt dt the particle,” is defined as: (5) dL~ ~ = − ~v × p.~ ~τ = ~r × F . (1) dt Now p~ = m~v so ~v × p~ = 0 (because the vector product of parallel vectors Now suppose the particle has a linear momentum P relative to the is zero), so the torque is:3 origin. Then the angular momentum of the particle is defined as: dL~ ~τ = . (6) L~ = ~r × p.~ (2) dt Thus the time rate of change of the angular momentum of a particle is equal to the torque acting on it. The directions of ~τ and L~ are given by the right-hand rule for cross products (see Fig. 1). 2c. Motion Confined to a Plane. The expression ~τ = dL~ /dt for a particle takes on a scalar appearance when the motion of the particle is ~ 2b. Relationship: ~τ = dL/dt. Using the definitions of torque and confined to a plane. angular momentum, we can derive a useful relationship between them. Consider a particle constrained to move only in the x-y plane, as 1 See “Kinematics: Circular Motion” (MISN-0-9) and “Torque and Angular Accel- shown in Fig. 2. The torque on the particle is always to eration for Rigid Planar Objects: Flywheels” (MISN-0-33). 2 See “Force and Torque” (MISN-0-5). 3This equation is valid only if ~τ and L~ are measured with respect to the same origin.

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v Similarly, the torque is:

dL 2 dω 2 F m τ = = mr = mr α, (11) dt dt p 6 r where α is the particle’s angular acceleration. r 3. Systems of Particles x 3a. Total Angular Momentum. The total angular momentum of Figure 2. ~r, F~ , and p~ are Figure 3. For a a system of particles is simply the sum of the angular momenta of the ~ ~ ~ ~ all coplanar for motion in a particle in circular individual particles, added vectorially. Let L1, L2, L3, . . ., LN , be the plane. motion, ~r and ~v are respective angular momenta, about a given point, of the particles in the perpendicular. system. The total angular momentum about the point is:7

N ~ ~ ~ ~ this plane as is the angular momentum [work this out using Eqs. (1) and L = L1 + L2 + . . . = Li. (12) i=1 (2)]. Equivalently, we say that ~τ and L~ have only z-components. Since X their directions remain constant, only their magnitudes change. Then: As time passes, the total angular momentum may change. Its rate of change, dL~ /dt, will be the sum of the rates dL~ i/dt for the particles in dL τ = (motion in a plane). (7) the system. Thus dL/dt~ will equal the sum of the acting on the dt particles. This equation holds only if F~ and p~ are in the same plane; if not (and 3b. Total Torque. The total torque on a system of particles is just they won’t be for non-planar motion), the full Eq. (6) must be used.4 the sum of the external torques acting on the system. The torque due to 2d. Circular Motion of a Mass. The torque and angular momentum internal is zero because by Newton’s third law the forces between for the special case of a single particle in circular motion can be easily any two particles are equal and opposite and directed along the line con- related to the particle’s angular variables. Suppose a particle of mass m necting them. The net torque due to each such -reaction force pair moves about a of r with v (not necessarily constant) is zero so the total internal torque must also be zero. Then the total as shown in Fig. 3. The particle’s angular momentum is: torque on the system is just equal to the sum of the external torques: N ~ L = ~r × m~v, (8) ~τ = ~τi,ext (system of particles). (13) i=1 but since ~r and ~v are perpendicular,5 the magnitude of L~ is: X For the system, then: dL~ L = mvr (9) ~τ = . (14) dt and the direction is out of the page. Equation (9) may be rewritten in In words, the time rate of change of the total angular momentum about a terms of the (since v = ωr) as: given point, for a system of particles, is equal to the sum of the external torques about that point and acting on the system. L = mr2ω. (10) 6See “Torque and Angular Acceleration for Rigid Plane Objects: Flywheels” 4 The component equations are τx = dLx/dt, τy = dLy/dt, τz = dLz/dt. (MISN-0-33). 5See “Kinematics: Circular Motion” (MISN-0-9). 7For continuous mass distributions the summation becomes an integration.

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3c. Rigid Body Motion About a Fixed Axis. A rigid body is a dm m2 system of particles whose positions are all fixed relative to each other. m3 r Since L = mvr = mr2ω for each particle in the body, we may write for 2 r r the total angular momentum: 3 m1 r1 r4 2 L = miri ω, (15) Ã i ! X m4 where we have assumed that the body is rotating about a fixed axis with Figure 4. Mass Figure 5. Rota- angular velocity ω. The quantity in parentheses, on the rim of a fly- tional Ek equals the 2 (heavy line). sum of particles’ E . I ≡ miri , (16) k i X is called the “moment of inertia” of the body.8 Thus we write 3e. Kinetic Energy of Rotation. The total kinetic energy of a sys- tem can be written in terms of the system’s moment of inertia and angular L = Iω (17) velocity. We start with the statement that the total kinetic energy of the and, since the axis of rotation is fixed, equations with a scalar appearance system of particles, each of which is in circular motion about a fixed axis hold for the torque: of rotation, is equal to the sum of the kinetic of the individual dL dω particles. τ = = I = Iα. (18) dt dt Each individual particle of mass mi moves in a circle of radius ri 3d. Example: Flywheel. As a visual example, we calculate the about the axis of rotation. If the positions of the particles are all fixed torque and angular momentum for the special case of a flywheel where relative to each other (as in a rigid body), then the angular velocity ω is the entire mass is uniformly distributed around the rim. Let “dm” be the same for all particles. The kinetic energy of each particle is thus: the mass of an infinitesimal segment of the rim as shown in Fig. 4. The 1 1 E = m v2 = m r2ω2 angular momentum dL of the mass dm is: ik 2 i i 2 i i dL = dm r2ω. (19) The total kinetic energy of the rotating body is therefore Since r and ω are the same for all points on the rim, the total angular N 1 2 2 1 2 2 1 2 2 E = m1r ω + m2r ω + . . . = m r ω momentum for the flywheel is: k 2 1 2 2 2 i i "i=1 # X By Eq. (16) the sum, m r2, is just the moment of inertia I of the body, L = dL = r2ω dm = Mr2ω = Iω, (20) i i so the rotational kinetic energy for a rigid body can finally be written: Z Z P where M is the total mass of the flywheel. This is identical to Eq. (17) 1 2 E = Iω . (22) for a single mass M in circular motion. The total torque is also the same, k 2 i.e., dL £ See if you can determine the moment of inertia and kinetic energy of τ = = Mr2α = Iα. (21) dt our flywheel in Sect. 3d. Help: [S-2]

8See “Uniform Circular Motion: Moment of Inertia, Conservation of Angular Mo- 3f. Linear vs. Rotational Motion. Here is a comparison of the mentum, Kinetic Energy, ” (MISN-0-41). equations of dynamics in “normal” and “rotational” form:

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Table 1. General forms. Normal Rotational F~ = dP~ /dt ~τ = dL~ /dt 2 2 Ek = MV /2 Ek = Iω /2

For a rigid body rotating about a fixed axis, the following comparisons hold. Figure 6. Two flywheels, one spinning, one at rest, are brought together. Table 2. Rigid-body forms. Normal Rotational to be exerted on the spinning one and, since this torque comes from an external source, the angular momentum of the spinning flywheel is not P = Mv L = Iω conserved. If, however, our system consists of both flywheels, the torques F = dp/dt = Ma τ = dL/dt = Iα each exert on the other are internal to the system and so there is no longer an external torque. Thus for the expanded system, containing both fly- Some of the equations shown in the Table 2 can be rewritten in vector/ten- , angular momentum is conserved. This tells us that: sor form so they have validity beyond linear and circular motion. 2 2 2 2 Mr ω0 = Mr ωf + Mr ωf = 2Mr ωf , (24)

4. Conservation of Angular Momentum where ωf is the final angular velocity at which the two flywheels rotate. Once they are together—the spinning one slowed down (ω0 → ωf ), the 4a. Statement of the Law. The total angular momentum of a system other having gone from zero to ωf —the two act as a single flywheel, with of particles is conserved if there are no external torques acting on the mass 2 M, rotating with angular velocity ωf = ω0/2. Help: [S-3]. system. Thus: 4d. Kinetic Energy of the Two Flywheels. When flywheels ro- dL~ ~τ = 0 = (23) tating at different are brought into contact, it can be expected dt that kinetic energy is not conserved. This is because the nonconservative so L~ is a constant. For a rigid body rotating about a fixed axis, L = Iω, frictional force acts on the two flywheels. The kinetic energy is initially so that if I changes, there must be a compensating change in ω in order 1 2 2 E 0 = Mr ω , (25) for L to remain constant. k 2 0 4b. If the External Torque is not Zero. If the external torque on and afterward it is: Help: [S-4] a system is not zero then angular momentum is not conserved for the 2 2 1 2 2 system. However, all is not lost, for if the system is expanded to include Ekf = Mr ωf = Mr ω0. (26) whatever is causing the external torque on it (therefore changing it into an 4 internal torque), angular momentum will be conserved for the expanded Therefore the ratio of the final kinetic energy to the initial kinetic energy system. is: E 1 kf = . (27) 4c. Example: Two Flywheels. Consider what happens when two Ek0 2 identical flywheels, one spinning and one at rest, are suddenly brought to- This means that half of the initial kinetic energy has been dissipated due 2 gether. The spinning flywheel has total angular momentum L0 = Mr ω0 to the frictional forces between the surfaces of the two flywheels. Thus before it is allowed to come in contact with the second flywheel. When the the total angular momentum of the system is conserved even though the two flywheels come together, the between them causes a torque total kinetic energy is not.

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5. Nonplanar Rigid Bodies Although the flywheel has been used as a specific example of rota- PROBLEM SUPPLEMENT tional motion, the concepts described can be applied to all rigid bodies whether they are planar or not. The rotational motion of nonplanar rigid Problem 5 also occurs on this module’s Model Exam. bodies is discussed elsewhere.9

1. A flywheel of radius 0.50 m and mass 500.0 kg has a torque of 30.0 N m Acknowledgments acting on it. This module is based on one by J. Borysowicz and P. Signell. Prepa- a. Calculate its moment of inertia. ration of this module was supported in part by the National Science Foun- b. Find its angular acceleration. dation, Division of Science Education Development and Research, through Grant #SED 74-20088 to Michigan State University. c. Assuming the flywheel is initially at rest, calculate the time required to complete the first two revolutions. Glossary d. What is its angular velocity and tangential velocity after two revo- lutions? • angular momentum (vector): L~ = ~r × p~, where p~ is the linear e. Find its kinetic energy after two revolutions. momentum of a particle at a position ~r with respect to the origin. 2. A very lightweight circular platform has a of 300.0 N placed on it • conservation of angular momentum: the total angular momentum at a 25 cm from its center. The platform is placed horizontally of a system is conserved if the external torque on the system is zero. on a pedestal such that a frictional drag force acts on the platform at the point where the weight is. The coefficient of friction is µ = 0.050. • moment of inertia (of a system of particles): the sum I ≡ 2 miri . a. Sketch the forces acting on the platform (assume it has zero mass) and find their numerical values. • Prigid body: a system of particles whose positions are fixed relative to each other. b. Calculate and sketch all torques acting on the system. c. If the platform initially rotates at 8π /s, find how long it • rotational dynamics: the application of dynamics to rotating bod- takes for it to slow down and stop. ies. d. What else must be included in the system in order for angular mo- • rotational kinetic energy: the kinetic energy of a particle or system mentum to be conserved? of particles rotating about a fixed point. 3. Suppose the stabilizing of a ship has a rotor of mass 5.0 × • torque (vector): ~τ = ~r × F~ , where F~ is the force acting on a particle 104 kg, all located on the rim at a radius of 0.20 m. The rotor is started located at ~r. from rest by a constant force of 1.00 × 103 N applied through a on the rim by a motor.

a. Draw a diagram showing all forces acting on the rotor. b. Show all torques acting on the rotor. c. Compute the length of time needed to bring the rotor up to its normal speed of 9.00 × 102 rev/min. 9See “Rotational Motion of a Rigid Body” (MISN-0-36).

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d. State what else you must include in the system so that the total an- 5. A vertical flywheel of radius R contains (virtually) all of its mass M gular momentum will be conserved. That is, name the other object on its rim and has a handle on one spoke at a radius r which is less whose about the rotor shaft changes oppositely than the radius of the rim. You stand next to the flywheel, grasp the as the rotor picks up speed. Describe the reaction torque which handle, and apply a constant tangential force F to the handle. produces the angular acceleration of the other object. a. Sketch all forces acting on the flywheel including that due to . 4. A system consists of two massless struts, F b. Write down and justify the value of the net (total) force on the rigidly connected to a sleeve as shown in flywheel. the diagram. The force F~ is always at right r to its strut and the is vertical, c. Write down or sketch all torques acting on the system. enabling the system to rotate freely in a hor- axle R At the end of one-half of a revolution, find the rim’s: izontal plane (the sketch is a view “looking m sleeve down”). Neglect gravity. d. moment of inertia about the axle a. State why gravity can be neglected in this problem. e. angular acceleration b. Sketch all forces acting on the system. f. time c. Write down and justify the value of the (net) total force acting on g. angular velocity the system. h. tangential velocity d. Write down or sketch all torques acting on the system. i. kinetic energy At the end of the system’s first complete revolution, derive the mass j. work done on it m’s: k. total energy e. moment of inertia about the axle l. angular momentum f. angular acceleration m. Describe the mechanism by which angular momentum is conserved in this case. g. time h. angular velocity

i. tangential velocity Brief Answers: j. kinetic energy 2 2 2 2 k. work done on it ( F~ · d~x) 1. a. I = miri = Mr = (500 kg)(0.50 m) = 125 kg m b. τ = Iα l. total energy R P τ 30 N m m. change in angular momentum about the axle. α = = = 0.24 rad/s2 I 125 kg m2 Describe a plausible expanded system within which angular momentum 1 c. θ = αt2: θ = 4π is conserved in the above case. Specifically: 2 1 2 1 2 2θ / (2)(4π) / n. Describe the reaction torque which produces the compensating an- t = = = 10.2 s α 0.24 s gular momentum. µ ¶ · ¸ 2 d. ω = ω0 + αt = 0 + (0.24/ s )(10.2 s) = 2.4 rad/s

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v = ωr = (2.4 rad/s)(0.50 m) = 1.2 m/s 4. a. The force and torque produced by gravity are exactly cancelled by 1 1 the force and torque exerted on the system to keep it in a horizontal e. E = Iω2 = (125 kg m2)(2.4 s)2 = 360 J k 2 2 plane. 2. a. F ( gravity) = 300 N gravityforce b. F F ( reaction) = 300 N F ( friction) = µN(normal) r = (0.05)(300 N) frictionalforce axleforce = 15 N reactionforceofpedestal c. is zero because the system goes nowhere (does not accel- erate away from its present location). b. τ( friction) = rF ( friction) d. ~τ = rF , into the paper = (0.25 m)(15 N) 0.25m e. I = mR2 = 3.75 N m 15N f. α = τ/I = F r/(mR2)

1 2 frictionaltorque g. θ = θ0 + ω0t + αt 2 2 c. τ = Iα; I = Mr 1 F r 2 2π = 0 + 0 + 2 t τ τ 3.75 N m 2 2 mR α = = 2 = 2 2 = 1.96 /s µ ¶ µ ¶ I Mr (300 N/9.8 m/s )(0.25 m) 1 2 4πmR2 / ω = αt t = F r ω 8π/ s µ ¶ t = = 2 = 12.8 s 1 2 1 2 α 1.96/ s F r 4πmR2 / F r4π / h. ω = ω0 + αt = 0 + = d. If the pedestal and are included then the earth will acquire mR2 F r mR2 µ ¶ µ ¶ the compensating angular momentum so angular momentum is con- 1 2 F r4π / served for the combined system. i. v = ωR = m 3. a. µ ¶ belt axle 1 2 j. E = mv = F r2π k 2 mounting ~ · gravity k. W = F d~x = F r dθ = F r2π bolt l. Etot =R Ek = F r2Rπ 1 2 motor rotor rotor m. ∆L = mvR = (mF r4π) / R (into paper) n. If the axle is attached to the earth and if F ’s reaction force is against b. only one torque (from belt): out of paper. the earth, then it is the earth which acquires the compensating c. 1.57 × 101 min. angular momentum about the axle. d. The motor exerts a torque on the ship through the motor’s mounting bolts and mounting bed, while the ship transmits it to the water and the water transmits some of it to the earth.

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5. a. F SPECIAL ASSISTANCE SUPPLEMENT axle axleforce S-1 (from TX-2b) gravityforce d(~r × p~) d~r dp~ = × p~ + ~r × dt dt dt b. zero; no acceleration of the (center of the) flywheel. so: dp~ d d~r ~r × = (~r × p~) − × p~ c. F r, into page dt dt dt d. MR2 2 e. F r/(MR ) S-2 (from TX-3e) 1 2 2πMR2 / f. F r I = r2 dm = Mr2 µ ¶ 1 2 F r2π / Z g. 1 2 1 2 2 MR2 Ek = Iω = Mr ω µ ¶ 2 2 1 2 F r2π / h. M µ ¶ S-3 (from TX-4c) i. F rπ 2 2 2 2 j. F rπ Mr ω0 = Mr ωf + Mr ωf = 2Mr ωf

k. F rπ =⇒ ω0 = 2ωf l. (MF r2π)1/2 R 1 ωf = ω0 m. The expanded system consists of the flywheel, you, and the earth. 2 The (frictional) force of your feet and the reaction force of the mounting bolts against the earth causes the earth’s angular mo- S-4 (from TX-4d) mentum about the axle to change in a compensating fashion.

1 2 2 1 2 2 2 2 2 1 2 1 2 2 E = Mr ω + Mr ω = Mr ω = Mr ( ω0) = Mr ω kf 2 f 2 f f 2 4 0

1 2 2 E Mr ω0 1 =⇒ kf = 4 = E 0 1 2 k Mr2ω2 2 0

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MODEL EXAM

1. See Output Skills K1-K4 in this module’s ID Sheet. The actual exam may contain one or more, or none, of these skills. 2. A vertical flywheel of radius R contains (virtually) all of its mass M on its rim and has a handle on one spoke at a radius r which is less than the radius of the rim. You stand next to the flywheel, grasp the handle, and apply a constant tangential force F to the handle. a. Sketch all forces acting on the flywheel including that due to gravity. b. Write down and justify the value of the net (total) force on the flywheel. c. Write down or sketch all torques acting on the system. At the end of one-half of a revolution, find the rim’s: d. moment of inertia about the axle e. angular acceleration f. time g. angular velocity h. tangential velocity i. kinetic energy j. work done on it k. total energy l. angular momentum m. Describe the mechanism by which angular momentum is conserved in this case.

Brief Answers:

1. See this module’s text. 2. See this module’s Problem Supplement, problem 5.

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