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University Physics Volume 1

11.3 Conservation of Angular Momentum

University Physics Volume 111.3 Conservation of Angular Momentum
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  1. Preface
  2. Unit 1. Mechanics
    1. 1 Units and Measurement
      1. Introduction
      2. 1.1 The Scope and Scale of Physics
      3. 1.2 Units and Standards
      4. 1.3 Unit Conversion
      5. 1.4 Dimensional Analysis
      6. 1.5 Estimates and Fermi Calculations
      7. 1.6 Significant Figures
      8. 1.7 Solving Problems in Physics
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    2. 2 Vectors
      1. Introduction
      2. 2.1 Scalars and Vectors
      3. 2.2 Coordinate Systems and Components of a Vector
      4. 2.3 Algebra of Vectors
      5. 2.4 Products of Vectors
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    3. 3 Motion Along a Straight Line
      1. Introduction
      2. 3.1 Position, Displacement, and Average Velocity
      3. 3.2 Instantaneous Velocity and Speed
      4. 3.3 Average and Instantaneous Acceleration
      5. 3.4 Motion with Constant Acceleration
      6. 3.5 Free Fall
      7. 3.6 Finding Velocity and Displacement from Acceleration
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    4. 4 Motion in Two and Three Dimensions
      1. Introduction
      2. 4.1 Displacement and Velocity Vectors
      3. 4.2 Acceleration Vector
      4. 4.3 Projectile Motion
      5. 4.4 Uniform Circular Motion
      6. 4.5 Relative Motion in One and Two Dimensions
      7. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    5. 5 Newton's Laws of Motion
      1. Introduction
      2. 5.1 Forces
      3. 5.2 Newton's First Law
      4. 5.3 Newton's Second Law
      5. 5.4 Mass and Weight
      6. 5.5 Newton’s Third Law
      7. 5.6 Common Forces
      8. 5.7 Drawing Free-Body Diagrams
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    6. 6 Applications of Newton's Laws
      1. Introduction
      2. 6.1 Solving Problems with Newton’s Laws
      3. 6.2 Friction
      4. 6.3 Centripetal Force
      5. 6.4 Drag Force and Terminal Speed
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    7. 7 Work and Kinetic Energy
      1. Introduction
      2. 7.1 Work
      3. 7.2 Kinetic Energy
      4. 7.3 Work-Energy Theorem
      5. 7.4 Power
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    8. 8 Potential Energy and Conservation of Energy
      1. Introduction
      2. 8.1 Potential Energy of a System
      3. 8.2 Conservative and Non-Conservative Forces
      4. 8.3 Conservation of Energy
      5. 8.4 Potential Energy Diagrams and Stability
      6. 8.5 Sources of Energy
      7. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
    9. 9 Linear Momentum and Collisions
      1. Introduction
      2. 9.1 Linear Momentum
      3. 9.2 Impulse and Collisions
      4. 9.3 Conservation of Linear Momentum
      5. 9.4 Types of Collisions
      6. 9.5 Collisions in Multiple Dimensions
      7. 9.6 Center of Mass
      8. 9.7 Rocket Propulsion
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    10. 10 Fixed-Axis Rotation
      1. Introduction
      2. 10.1 Rotational Variables
      3. 10.2 Rotation with Constant Angular Acceleration
      4. 10.3 Relating Angular and Translational Quantities
      5. 10.4 Moment of Inertia and Rotational Kinetic Energy
      6. 10.5 Calculating Moments of Inertia
      7. 10.6 Torque
      8. 10.7 Newton’s Second Law for Rotation
      9. 10.8 Work and Power for Rotational Motion
      10. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    11. 11 Angular Momentum
      1. Introduction
      2. 11.1 Rolling Motion
      3. 11.2 Angular Momentum
      4. 11.3 Conservation of Angular Momentum
      5. 11.4 Precession of a Gyroscope
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    12. 12 Static Equilibrium and Elasticity
      1. Introduction
      2. 12.1 Conditions for Static Equilibrium
      3. 12.2 Examples of Static Equilibrium
      4. 12.3 Stress, Strain, and Elastic Modulus
      5. 12.4 Elasticity and Plasticity
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    13. 13 Gravitation
      1. Introduction
      2. 13.1 Newton's Law of Universal Gravitation
      3. 13.2 Gravitation Near Earth's Surface
      4. 13.3 Gravitational Potential Energy and Total Energy
      5. 13.4 Satellite Orbits and Energy
      6. 13.5 Kepler's Laws of Planetary Motion
      7. 13.6 Tidal Forces
      8. 13.7 Einstein's Theory of Gravity
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    14. 14 Fluid Mechanics
      1. Introduction
      2. 14.1 Fluids, Density, and Pressure
      3. 14.2 Measuring Pressure
      4. 14.3 Pascal's Principle and Hydraulics
      5. 14.4 Archimedes’ Principle and Buoyancy
      6. 14.5 Fluid Dynamics
      7. 14.6 Bernoulli’s Equation
      8. 14.7 Viscosity and Turbulence
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
  3. Unit 2. Waves and Acoustics
    1. 15 Oscillations
      1. Introduction
      2. 15.1 Simple Harmonic Motion
      3. 15.2 Energy in Simple Harmonic Motion
      4. 15.3 Comparing Simple Harmonic Motion and Circular Motion
      5. 15.4 Pendulums
      6. 15.5 Damped Oscillations
      7. 15.6 Forced Oscillations
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    2. 16 Waves
      1. Introduction
      2. 16.1 Traveling Waves
      3. 16.2 Mathematics of Waves
      4. 16.3 Wave Speed on a Stretched String
      5. 16.4 Energy and Power of a Wave
      6. 16.5 Interference of Waves
      7. 16.6 Standing Waves and Resonance
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    3. 17 Sound
      1. Introduction
      2. 17.1 Sound Waves
      3. 17.2 Speed of Sound
      4. 17.3 Sound Intensity
      5. 17.4 Normal Modes of a Standing Sound Wave
      6. 17.5 Sources of Musical Sound
      7. 17.6 Beats
      8. 17.7 The Doppler Effect
      9. 17.8 Shock Waves
      10. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
  4. A | Units
  5. B | Conversion Factors
  6. C | Fundamental Constants
  7. D | Astronomical Data
  8. E | Mathematical Formulas
  9. F | Chemistry
  10. G | The Greek Alphabet
  11. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
    7. Chapter 7
    8. Chapter 8
    9. Chapter 9
    10. Chapter 10
    11. Chapter 11
    12. Chapter 12
    13. Chapter 13
    14. Chapter 14
    15. Chapter 15
    16. Chapter 16
    17. Chapter 17
  12. Index

Learning Objectives

By the end of this section, you will be able to:
  • Apply conservation of angular momentum to determine the angular velocity of a rotating system in which the moment of inertia is changing
  • Explain how the rotational kinetic energy changes when a system undergoes changes in both moment of inertia and angular velocity

So far, we have looked at the angular momentum of systems consisting of point particles and rigid bodies. We have also analyzed the torques involved, using the expression that relates the external net torque to the change in angular momentum, Equation 11.8. Examples of systems that obey this equation include a freely spinning bicycle tire that slows over time due to torque arising from friction, or the slowing of Earth’s rotation over millions of years due to frictional forces exerted on tidal deformations.

However, suppose there is no net external torque on the system, τ=0. τ=0. In this case, Equation 11.8 becomes the law of conservation of angular momentum.

Law of Conservation of Angular Momentum

The angular momentum of a system of particles around a point in a fixed inertial reference frame is conserved if there is no net external torque around that point:

dLdt=0dLdt=0
(11.10)

or

L=l1+l2++lN=constant.L=l1+l2++lN=constant.
(11.11)

Note that the total angular momentum LL is conserved. Any of the individual angular momenta can change as long as their sum remains constant. This law is analogous to linear momentum being conserved when the external force on a system is zero.

As an example of conservation of angular momentum, Figure 11.14 shows an ice skater 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. Also, the friction is exerted very close to the pivot point. Both |F|and|r||F|and|r| are small, so |τ||τ| is negligible. Consequently, she can spin for quite some time. She can also 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

L=LL=L

or

Iω=Iω,Iω=Iω,

where the primed quantities refer to conditions after she has pulled in her arms and reduced her moment of inertia. Because II is smaller, the angular velocity ωω must increase to keep the angular momentum constant.

Two illustrations of a spinning ice skater. In figure a, on the left, the skater has her arms and one foot extended away from her body. She is spinning with angular velocity omega and L equals I times omega. In figure b, on the right, the skater has her arms and foot pulled close to her body. She is spinning faster, with angular velocity omega prime and L equals I prime times omega prime.
Figure 11.14 (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. (b) 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.

It is interesting to see how the rotational kinetic energy of the skater changes when she pulls her arms in. Her initial rotational energy is

KRot=12Iω2,KRot=12Iω2,

whereas her final rotational energy is

KRot=12I(ω)2.KRot=12I(ω)2.

Since Iω=Iω,Iω=Iω, we can substitute for ωω and find

KRot=12I(ω)2=12I(IIω)2=12Iω2(II)=KRot(II).KRot=12I(ω)2=12I(IIω)2=12Iω2(II)=KRot(II).

Because her moment of inertia has decreased, I<I,I<I, her final rotational kinetic energy has increased. The source of this additional rotational kinetic energy is the work required to pull her arms inward. Note that the skater’s arms do not move in a perfect circle—they spiral inward. This work causes an increase in the rotational kinetic energy, while her angular momentum remains constant. Since she is in a frictionless environment, no energy escapes the system. Thus, if she were to extend her arms to their original positions, she would rotate at her original angular velocity and her kinetic energy would return to its original value.

The solar system is another example of how conservation of angular momentum works in our universe. Our solar system was born from a huge cloud of gas and dust that initially had rotational energy. Gravitational forces caused the cloud to contract, and the rotation rate increased as a result of conservation of angular momentum (Figure 11.15).

An illustration of the formation of the solar system from a cloud of gas and dust. At first, the gas cloud is rotating with angular velocity omega and has angular momentum L. It forms a fairly continuous disc in the plane of rotation. Later, the disc is rotating with angular velocity omega prime but still has angular momentum L. The disc starts to break up into concentric rings. The gaps between the rings grow. Eventually, the gas in the rings forms a star in the center and planets whose orbits trace the rings from which they came. In all cases, the angular velocity is in the same direction as that of the original gas cloud and the angular momentum is L.
Figure 11.15 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. (credit: modification of work by NASA)

We continue our discussion with an example that has applications to engineering.

Example 11.7

Coupled Flywheels A flywheel rotates without friction at an angular velocity ω0=600rev/minω0=600rev/min on a frictionless, vertical shaft of negligible rotational inertia. A second flywheel, which is at rest and has a moment of inertia three times that of the rotating flywheel, is dropped onto it (Figure 11.16). Because friction exists between the surfaces, the flywheels very quickly reach the same rotational velocity, after which they spin together. (a) Use the law of conservation of angular momentum to determine the angular velocity ωω of the combination. (b) What fraction of the initial kinetic energy is lost in the coupling of the flywheels?

In the drawing on the left, two flywheels are shown. Their axes are vertical and aligned, and the wheels face each other, but the wheels are separate from one another. The lower wheel has moment to inertia I sub 0 and is spinning counterclockwise as viewed from above. The upper wheel has moment to inertia 3 I sub 0 and is at rest. In the drawing on the right, the wheels are coupled together and spin counterclockwise as viewed from above.
Figure 11.16 Two flywheels are coupled and rotate together.

Strategy Part (a) is straightforward to solve for the angular velocity of the coupled system. We use the result of (a) to compare the initial and final kinetic energies of the system in part (b).

Solution a. No external torques act on the system. The force due to friction produces an internal torque, which does not affect the angular momentum of the system. Therefore conservation of angular momentum gives

I0ω0=(I0+3I0)ω,ω=14ω0=150rev/min=15.7rad/s.I0ω0=(I0+3I0)ω,ω=14ω0=150rev/min=15.7rad/s.

b. Before contact, only one flywheel is rotating. The rotational kinetic energy of this flywheel is the initial rotational kinetic energy of the system, 12I0ω0212I0ω02. The final kinetic energy is12(4I0)ω2=12(4I0)(ω04)2=18I0ω02.12(4I0)ω2=12(4I0)(ω04)2=18I0ω02.

Therefore, the ratio of the final kinetic energy to the initial kinetic energy is

18I0ω0212I0ω02=14.18I0ω0212I0ω02=14.

Thus, 3/4 of the initial kinetic energy is lost to the coupling of the two flywheels.

Significance Since the rotational inertia of the system increased, the angular velocity decreased, as expected from the law of conservation of angular momentum. In this example, we see that the final kinetic energy of the system has decreased, as energy is lost to the coupling of the flywheels. Compare this to the example of the skater in Figure 11.14 doing work to bring her arms inward and adding rotational kinetic energy.

Check Your Understanding 11.4

A merry-go-round at a playground is rotating at 4.0 rev/min. Three children jump on and increase the moment of inertia of the merry-go-round/children rotating system by 25%25%. What is the new rotation rate?

Example 11.8

Dismount from a High Bar An 80.0-kg gymnast dismounts from a high bar. He starts the dismount at full extension, then tucks to complete a number of revolutions before landing. His moment of inertia when fully extended can be approximated as a rod of length 1.8 m and when in the tuck a rod of half that length. If his rotation rate at full extension is 1.0 rev/s and he enters the tuck when his center of mass is at 3.0 m height moving horizontally to the floor, how many revolutions can he execute if he comes out of the tuck at 1.8 m height? See Figure 11.17.

A drawing of a gymnast dismounting from a 3 m tall high bar. He starts the dismount at full extension above the bar, then tucks when he is moving horizontally to the floor, level with the bar. The gymnast is 1.8 meters tall.
Figure 11.17 A gymnast dismounts from a high bar and executes a number of revolutions in the tucked position before landing upright.

Strategy Using conservation of angular momentum, we can find his rotation rate when in the tuck. Using the equations of kinematics, we can find the time interval from a height of 3.0 m to 1.8 m. Since he is moving horizontally with respect to the ground, the equations of free fall simplify. This will allow the number of revolutions that can be executed to be calculated. Since we are using a ratio, we can keep the units as rev/s and don’t need to convert to radians/s.

Solution The moment of inertia at full extension is I0=112mL2=11280.0kg(1.8m)2=21.6kg·m2I0=112mL2=11280.0kg(1.8m)2=21.6kg·m2.

The moment of inertia in the tuck is If=112mLf2=11280.0kg(0.9m)2=5.4kg·m2If=112mLf2=11280.0kg(0.9m)2=5.4kg·m2.

Conservation of angular momentum: Ifωf=I0ω0ωf=I0ω0If=21.6kg·m2(1.0rev/s)5.4kg·m2=4.0rev/sIfωf=I0ω0ωf=I0ω0If=21.6kg·m2(1.0rev/s)5.4kg·m2=4.0rev/s.

Time interval in the tuck: t=2hg=2(3.01.8)m9.8m/s=0.5st=2hg=2(3.01.8)m9.8m/s=0.5s.

In 0.5 s, he will be able to execute two revolutions at 4.0 rev/s.

Significance Note that the number of revolutions he can complete will depend on how long he is in the air. In the problem, he is exiting the high bar horizontally to the ground. He could also exit at an angle with respect to the ground, giving him more or less time in the air depending on the angle, positive or negative, with respect to the ground. Gymnasts must take this into account when they are executing their dismounts.

Example 11.9

Conservation of Angular Momentum of a Collision A bullet of mass m=2.0gm=2.0g is moving horizontally with a speed of 500.0m/s.500.0m/s. The bullet strikes and becomes embedded in the edge of a solid disk of mass M=3.2kgM=3.2kg and radius R=0.5m.R=0.5m. The cylinder is free to rotate around its axis and is initially at rest (Figure 11.18). What is the angular velocity of the disk immediately after the bullet is embedded?

Illustrations of a bullet before and after striking a disk. On the left is the before illustration. The bullet is travelling to the left at 500 meters per second, toward the front edge of a horizontal disk of radius R. The vertical axis through the center of the disc is shown as a vertical line connecting points A above and A prime below the center. On the right is the after illustration. The bullet is embedded in the edge of the disk, which is rotating about the vertical axis through the center. The rotation is counterclockwise as viewed from above.
Figure 11.18 A bullet is fired horizontally and becomes embedded in the edge of a disk that is free to rotate about its vertical axis.

Strategy For the system of the bullet and the cylinder, no external torque acts along the vertical axis through the center of the disk. Thus, the angular momentum along this axis is conserved. The initial angular momentum of the bullet is mvRmvR, which is taken about the rotational axis of the disk the moment before the collision. The initial angular momentum of the cylinder is zero. Thus, the net angular momentum of the system is mvRmvR. Since angular momentum is conserved, the initial angular momentum of the system is equal to the angular momentum of the bullet embedded in the disk immediately after impact.

Solution The initial angular momentum of the system is

Li=mvR.Li=mvR.

The moment of inertia of the system with the bullet embedded in the disk is

I=mR2+12MR2=(m+M2)R2.I=mR2+12MR2=(m+M2)R2.

The final angular momentum of the system is

Lf=Iωf.Lf=Iωf.

Thus, by conservation of angular momentum, Li=LfLi=Lf and

mvR=(m+M2)R2ωf.mvR=(m+M2)R2ωf.

Solving for ωf,ωf,

ωf=mvR(m+M/2)R2=(2.0×10−3kg)(500.0m/s)(2.0×10−3kg+1.6kg)(0.50m)=1.2rad/s.ωf=mvR(m+M/2)R2=(2.0×10−3kg)(500.0m/s)(2.0×10−3kg+1.6kg)(0.50m)=1.2rad/s.

Significance The system is composed of both a point particle and a rigid body. Care must be taken when formulating the angular momentum before and after the collision. Just before impact the angular momentum of the bullet is taken about the rotational axis of the disk.

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