Angular Momentum and Its Conservation

Chapter contents: Chapter 6: Rotation

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

L

L = Iω .

(6.5.1)

This equation is an analog to the definition of linear momentum as

p = mv

. Units for linear momentum are

kg \cdot m /s

while units for angular momentum are

kg \cdot 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.

Example 6.5.1

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 6.3.3, the formula for the moment of inertia of a sphere is

I = \frac{2 {MR}^{2}}{5}

(6.5.2)

so that

L = Iω = \frac{2 {MR}^{2} ω}{5} .

(6.5.3)

Earth’s mass

M

5 . 979 \times 10^{24} kg

and its radius

R

6 . 376 \times 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

\begin{array}{l} L & = & 0.4 (5 . 979 \times 10^{24} kg) {(6 . 376 \times 10^{6} m)}^{2} (\frac{1 rev}{d}) \\ = & 9 . 72 \times 10^{37} kg \cdot m^{2} \cdot rev/d . \end{array}

(6.5.4)

Substituting

2π

rad for

1

rev and

8 . 64 \times 10^{4} s

for 1 day gives

\begin{array}{l} L & = & (9 . 72 \times 10^{37} kg \cdot m^{2}) (\frac{2π rad/rev}{8 . 64 \times 10^{4} s/d}) (1 rev/d) \\ = & 7 . 07 \times 10^{33} kg \cdot m^{2} /s . \end{array}

(6.5.5)

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

L

. The relationship between torque and angular momentum is

net τ = \frac{Δ L}{Δ t} .

(6.5.6)

This expression is exactly analogous to the relationship between force and linear momentum,

F = Δ p / Δ t

. The equation

net τ = \frac{Δ L}{Δ t}

is very fundamental and broadly applicable. It is, in fact, the rotational form of Newton’s second law.

Example 6.5.2

Calculating the Torque Putting Angular Momentum Into a Lazy Susan

Figure 6.5.1 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?

The given figure shows a lazy Susan on which various eatables like cake, salad grapes, and a drink are kept. A hand is shown that applies a force F, indicated by a leftward pointing horizontal arrow. This force is perpendicular to the radius r and thus tangential to the circular lazy Susan. — **Figure 6.5.1.** A partygoer exerts a torque on a lazy Susan to make it rotate. The equation $net τ = \frac{Δ L}{Δ t}$ gives the relationship between torque and the angular momentum produced.

Strategy

We can find the angular momentum by solving

net τ = \frac{Δ L}{Δ t}

for

Δ 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

L = Iω

Solution for (a)

Solving

net τ = \frac{Δ L}{Δ t}

for

Δ L

gives

Δ L = (net τ) Δt .

(6.5.7)

Because the force is perpendicular to

r

, we see that

net τ = rF

, so that

\begin{array}{l} L & = & rF Δ t = (0 . 260 m) (2.50 N) (0.150 s) \\ = & 9 . 75 \times 10^{- 2} kg \cdot m^{2} / s . \end{array}

(6.5.8)

Solution for (b)

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

L = Iω .

(6.5.9)

Solving for

ω

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

ω = \frac{L}{I} = \frac{L}{\frac{1}{2} {MR}^{2}} .

(6.5.10)

And substituting known values into the preceding equation yields

ω = \frac{9 . 75 \times 10^{- 2} kg \cdot m^{2} /s}{(0.500) (4.00 kg) (0.260 m)} = 0 . 721 rad/s .

(6.5.11)

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.

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 L = (net τ) Δ t

. 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 τ = \frac{Δ L}{Δ t}

for the situation in which the net torque is zero. In that case,

net τ = 0

(6.5.12)

implying that

\frac{Δ L}{Δ t} = 0 .

(6.5.13)

If the change in angular momentum

Δ L

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

L = constant (net τ = 0)

(6.5.14)

L = L' (net τ = 0) .

(6.5.15)

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 6.5.2, 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' .

(6.5.16)

Expressing this equation in terms of the moment of inertia,

Iω = I' ω',

(6.5.17)

where the primed quantities refer to conditions after she has pulled in her arms and reduced her moment of inertia. Because

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.

The image a shows an ice skater spinning on the tip of her skate with both her arms and one leg extended. The image b shows the ice skater spinning on the tip of one skate, with her arms crossed and one leg supported on another. — **Figure 6.5.2.** (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 6.5.3

Calculating the Angular Momentum of a Spinning Skater

Suppose an ice skater, such as the one in Figure 6.5.2, is spinning at 0.800 rev/ s with her arms extended. She has a moment of inertia of

2.34 kg \cdot m^{2}

with her arms extended and of

0.363 kg \cdot 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

{KE}_{rot} = \frac{1}{2} {Iω}^{2} .

(6.5.18)

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'

(6.5.19)

Iω = I' ω'

(6.5.20)

Solving for

ω'

and substituting known values into the resulting equation gives

\begin{array}{l} ω' & = & \frac{I}{I'} ω = (\frac{2.34 kg \cdot m^{2}}{0 .363 kg \cdot m^{2}}) (0.800 rev/s) \\ = & 5.16 rev/s. \end{array}

(6.5.21)

Solution for (b)

Rotational kinetic energy is given by

{KE}_{rot} = \frac{1}{2} {Iω}^{2} .

(6.5.22)

The initial value is found by substituting known values into the equation and converting the angular velocity to rad/s:

\begin{array}{l} {KE}_{rot} & = & (0 . 5) (2.34 kg \cdot m^{2}) {((0.800 rev/s) (2π rad/rev))}^{2} \\ = & 29.6 J. \end{array}

(6.5.23)

The final rotational kinetic energy is

{KE}_{rot}' = \frac{1}{2} I' {ω'}^{2} .

(6.5.24)

Substituting known values into this equation gives

\begin{array}{l} K E_{rot}' & = & (0.5) (0 .363 kg \cdot m^{2}) {[(5 . 16 rev/s) (2π rad/rev)]}^{2} \\ = & 191 J. \end{array}

(6.5.25)

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 6.5.3.)

The first figure shows a cloud of dust and gas,which is in the shape of a distorted circle, rotating in anti-clockwise direction with an angular velocity omega, indicated by a curved black arrow, and having an angular momentum L. The second figure shows an elliptical cloud of dust with the Sun in the middle of it, rotating in anti-clockwise direction with an angular velocity omega dash, indicated by a curved black arrow, and having an angular momentum L. The third figure depicts the Solar System, with the Sun in the middle of it and the various planets revolve around it in their respective elliptical orbits in anti-clockwise direction, which is indicated by arrows. The angular momentum remains L. — **Figure 6.5.3.** 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.

Exercise

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.

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 $net τ = \frac{Δ L}{Δ t} .$
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

Exercise 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)?

Exercise 14

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.

In figure A, there is a merry go round. A child is jumping radially outward. In figure B, a child is jumping backward to the direction of motion of merry go round. In figure C, a child is jumping from it to hang from the branch of the tree. In figure D, a child is jumping from the merry go round tangentially to its circumference. — **Figure 6.E.3.** A child may jump off a merry-go-round in a variety of directions.

Exercise 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 6.E.3).

Exercise 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.

Exercise 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.

Exercise 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?

Exercise 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.

Exercise 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?

Exercise 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.

Exercise 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.

Exercise 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.

The given figure shows a diver who curls her body while flipping and then fully extends her limbs to enter straight down into water. — **Figure 6.E.4.** The diver spins rapidly when curled up and slows when she extends her limbs before entering the water.

Exercise 24

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

Exercise 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?

**Figure 6.E.5.** 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)

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