# 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.

*angular momentum*

*L*as

*L*=

*Iω*.

*p = mv*. Units for linear momentum are kg ⋅ m/s while units for angular momentum are kg ⋅ m

^{2}/s. As we would expect, an object that has a large moment of inertia

*I*, such as Earth, has a very large angular momentum. An object that has a large angular velocity

*ω*, such as a centrifuge, also has a rather large angular momentum.

**Making Connections**

### Example 1. 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 *L*=

*Iω*. First, according to Figure 1, the formula for the moment of inertia of a sphere is

*M*is 5.979 × 10

^{24}kg and its radius

*R*is 6.376 × 10

^{6}m. The Earth’s angular velocity

*ω*is, of course, exactly one revolution per day, but we must covert

*ω*to radians per second to do the calculation in SI units.

**Solution**

Substituting known information into the expression for *L*and converting

*ω*to radians per second gives

^{4}s for 1 day gives

**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.*L*. The relationship between torque and angular momentum is

*F*= Δ

*p*/Δ

*t*. The equation

### Example 2. Calculating the Torque Putting Angular Momentum Into a Lazy Susan

**Strategy**

We can find the angular momentum by solving*L*, 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

*L*=

*L*. To find the final velocity, we must calculate

*ω*from the definition of

*L*in

*L*=

*Iω*.

**Solution for (a)**

Solving *L*gives

*r*, we see that

**Solution for (b)**

The final angular velocity can be calculated from the definition of angular momentum,*L*=

*Iω*.

*ω*and substituting the formula for the moment of inertia of a disk into the resulting equation gives

**Discussion**

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.### Example 3. Calculating the Torque in a Kick

^{2}, (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)?

**Strategy**

The angular acceleration can be found using the rotational analog to Newton’s second law, or *I*is given and the torque can be found easily from the given force and perpendicular lever arm. Once the angular acceleration

*α*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 *α*is

*α*gives

**Solution to (b)**

The final angular velocity can be calculated from the kinematic expression*ω*

^{2}just found and the given value for the moment of inertia. The kinetic energy is then

**Discussion**

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**

## Conservation of Angular Momentum

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 *τ *= 0

*L*is zero, then the angular momentum is constant; thus,

*L* = constant (net *τ* = 0)

*L* = *L*′ (net *τ* = 0).

*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 4, 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

*F*and

*r*are small, and so

*τ*is negligibly small.) Consequently, she can 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

*L *= *L*′.

*Iω *= *I*′*ω*′,

*I*′ is smaller, the angular velocity

*ω*′ must increase to keep the angular momentum constant. The change can be dramatic, as the following example shows.

### Example 4. Calculating the Angular Momentum of a Spinning Skater

Suppose an ice skater, such as the one in Figure 4, is spinning at 0.800 rev/s with her arms extended. She has a moment of inertia of 2.34 kg ⋅ m^{2}with her arms extended and of 0.363 kg ⋅ m

^{2 }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 *ω*′ 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

**Solution for (a)**

Because torque is negligible (as discussed above), the conservation of angular momentum given in *Iω*=

*I*′

*ω*′ is applicable. Thus,

*L*=

*L*′

*Iω*=

*I*′

*ω*′

*ω*′ and substituting known values into the resulting equation gives

**Solution for (b) **

Rotational kinetic energy is given by**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.### Check Your Undestanding

**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.## Section Summary

- Every rotational phenomenon has a direct translational analog , likewise angular momentum
*L*can be defined as*L = Iω*. - This equation is an analog to the definition of linear momentum as
*p = mv*. The relationship between torque and angular momentum is$\text{net }\tau =\frac{\Delta L}{\Delta t}\\$. - 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.

### Conceptual Questions

1. 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)?

2. Suppose a child walks from the outer edge of a rotating merry-go round to the inside. Does the angular velocity of the merry-go-round increase, decrease, or remain the same? Explain your answer.

3. 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 6).

4. 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.

5. 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.

6. 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?

7. 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.

8. 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?

9. 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.

10. 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.

11. 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.

12. Draw a free body diagram to show how a diver gains angular momentum when leaving the diving board.

13. In terms of angular momentum, what is the advantage of giving a football or a rifle bullet a spin when throwing or releasing it?

### Problems & Exercises

1. (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.

2. (a) What is the angular momentum of the Moon in its orbit around Earth?

(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.

3. 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?

4. 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.

5. 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?

6. (a) Calculate the angular momentum of an ice skater spinning at 6.00 rev/s given his moment of inertia is 0.400 kg. (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?

7.

**Construct Your Own Problem**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.

## Glossary

- angular momentum:
- the product of moment of inertia and angular velocity

- 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

### Selected Solutions to Problems & Answers

1. (a) 2.66 × 10^{40}kg ⋅ m

^{2}/s (b) (a) 7.07 x 10

^{33}kg ⋅ m

^{2}/s

The angular momentum of the Earth in its orbit around the Sun 3.77 × 10

^{6}is times larger than the angular momentum of the Earth around its axis.

3. 22.5 kg ⋅ m

^{2}/s

5. 25.3 rpm