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# Angular Momentum and Its Conservation

Module by: OpenStax College. E-mail the author

Summary:

• 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.

Note: You are viewing an old version of this document. The latest version is available here.

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 LL size 12{L} {} as

L=.L=. size 12{L=Iω} {}
(1)

This equation is an analog to the definition of linear momentum as p=mvp=mv size 12{p= ital "mv"} {}. Units for linear momentum are kgm/skgm/s size 12{"kg" cdot m rSup { size 8{2} } "/s"} {} while units for angular momentum are kgm2/skgm2/s size 12{"kg" cdot m rSup { size 8{2} } "/s"} {}. As we would expect, an object that has a large moment of inertia II size 12{I} {}, such as Earth, has a very large angular momentum. An object that has a large angular velocity ωω size 12{ω} {}, such as a centrifuge, also has a rather large angular momentum.

## Making Connections:

Angular momentum is completely analogous to linear momentum, first presented in Uniform Circular Motion and Gravitation. It has the same implications in terms of carrying rotation forward, and it is conserved when the net external torque is zero. Angular momentum, like linear momentum, is also a property of the atoms and subatomic particles.

## 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=L= size 12{L=Iω} {}. First, according to (Reference), the formula for the moment of inertia of a sphere is

I = 2 MR 2 5 I = 2 MR 2 5 size 12{I= { {2 ital "MR" rSup { size 8{2} } } over {5} } } {}
(2)

so that

L = = 2 MR 2 ω 5 . L = = 2 MR 2 ω 5 . size 12{L=Iω= { {2 ital "MR" rSup { size 8{2} } ω} over {5} } } {}
(3)

Earth’s mass MM size 12{M} {} is 5.979×1024 kg5.979×1024 kg size 12{5 "." "979" times "10" rSup { size 8{"24"} } "kg"} {} and its radius RR size 12{R} {} is 6.376×106m6.376×106m size 12{6 "." "376" times "10" rSup { size 8{6} } m} {}. The Earth’s angular velocity ωω size 12{ω} {} is, of course, exactly one revolution per day, but we must covert ωω size 12{ω} {} to radians per second to do the calculation in SI units.

Solution

Substituting known information into the expression for LL size 12{L} {} and converting ωω size 12{ω} {} to radians per second gives

L=0.45.979×1024kg6.376×106 m21revd=9.72×1037kgm2rev/d.L=0.45.979×1024kg6.376×106 m21revd=9.72×1037kgm2rev/d.alignl { stack { size 12{L=0 "." 4 left (5 "." "979" times "10" rSup { size 8{"24"} } " kg" right ) left (6 "." "376" times "10" rSup { size 8{6} } " m" right ) rSup { size 8{2} } left ( { {1" rev"} over {d} } right )} {} # " "=9 "." "72" times "10" rSup { size 8{"37"} } " kg" cdot m rSup { size 8{2} } "rev/d" {} } } {}
(4)

Substituting size 12{2π} {} rad for 11 size 12{1} {} rev and 8.64×104s8.64×104s size 12{8 "." "64" times "10" rSup { size 8{4} } s} {} for 1 day gives

L=9.72×1037kgm2 rad/rev8.64×104 s/d1 rev/d=7.07×1033 kgm2/s.L=9.72×1037kgm2 rad/rev8.64×104 s/d1 rev/d=7.07×1033 kgm2/s.alignl { stack { size 12{L= left (9 "." "72" times "10" rSup { size 8{"37"} } " kg" cdot m rSup { size 8{2} } right ) left ( { {2π" rad/rev"} over {8 "." "64" times "10" rSup { size 8{4} } " s/d"} } right ) left (1" rev/d" right )} {} # " "=7 "." "07" times "10" rSup { size 8{"33"} } " kg" cdot m rSup { size 8{2} } "/s" {} } } {}
(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 LL size 12{L} {}. The relationship between torque and angular momentum is

net τ=ΔLΔt.net τ=ΔLΔt. size 12{"net "τ= { {ΔL} over {Δt} } } {}
(6)

This expression is exactly analogous to the relationship between force and linear momentum, F=Δp/ΔtF=Δp/Δt size 12{F=Δp/Δt} {}. The equation net τ=ΔLΔtnet τ=ΔLΔt size 12{"net "τ= { {ΔL} over {Δt} } } {} is very fundamental and broadly applicable. It is, in fact, the rotational form of Newton’s second law.

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

Figure 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?

Strategy

We can find the angular momentum by solving net τ= Δ L Δ t net τ= Δ L Δ t size 12{"net "τ= { {ΔL} over {Δt} } } {} for ΔLΔL size 12{ΔL} {}, 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, ΔL=LΔL=L size 12{ΔL=L} {}. To find the final velocity, we must calculate ωω size 12{ω} {} from the definition of LL size 12{L} {} in L=L= size 12{L=Iω} {}.

Solution for (a)

Solving net τ=ΔLΔtnet τ=ΔLΔt size 12{"net "τ= { {ΔL} over {Δt} } } {} for ΔLΔL size 12{ΔL} {} gives

ΔL=net τΔt.ΔL=net τΔt. size 12{ΔL= left ("net "τ right ) cdot Δt} {}
(7)

Because the force is perpendicular to rr size 12{r} {}, we see that net τ=rFnet τ=rF size 12{"net "τ= ital "rF"} {}, so that

L=rFΔt=(0.260 m)(2.50 N)(0.150 s)=9.75×102kgm2/s.L=rFΔt=(0.260 m)(2.50 N)(0.150 s)=9.75×102kgm2/s.alignl { stack { size 12{L= ital "rF" cdot Δt= $$0 "." "260"m$$ $$2 "." "50"N$$ $$0 "." "150"s$$ } {} # size 12{ {}=9 "." "75" times "10" rSup { size 8{ - 2} } "kg" cdot m rSup { size 8{2} } /s} {} } } {}
(8)

Solution for (b)

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

L=.L=. size 12{L=Iω} {}
(9)

Solving for ωω size 12{ω} {} and substituting the formula for the moment of inertia of a disk into the resulting equation gives

ω=LI=L12MR2.ω=LI=L12MR2. size 12{ω= { {L} over {I} } = { {L} over { { size 8{1} } wideslash { size 8{2} } ital "MR" rSup { size 8{2} } } } } {}
(10)

And substituting known values into the preceding equation yields

ω=9.75×102 kgm2/s0.5004.00 kg0.260 m=0.721 rad/s.ω=9.75×102 kgm2/s0.5004.00 kg0.260 m=0.721 rad/s. size 12{ω= { {9 "." "75" times "10" rSup { size 8{ - 2} } " kg" cdot m rSup { size 8{2} } "/s"} over { left (0 "." "500" right ) left (4 "." "00"" kg" right ) left (0 "." "260"" m" right )} } =0 "." "721"" rad/s"} {}
(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.

## Example 3: Calculating the Torque in a Kick

The person whose leg is shown in Figure 2 kicks his leg by exerting a 2000-N force with his upper leg muscle. The effective perpendicular lever arm is 2.20 cm. Given the moment of inertia of the lower leg is 1.25 kgm21.25 kgm2 size 12{1 "." "25""kg" cdot m rSup { size 8{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 57.57. size 12{"57" "." 3°} {} (1.00 rad)?

Strategy

The angular acceleration can be found using the rotational analog to Newton’s second law, or α=net τ/Iα=net τ/I size 12{α="net "τ/I} {}. The moment of inertia II size 12{I} {} is given and the torque can be found easily from the given force and perpendicular lever arm. Once the angular acceleration αα size 12{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 αα size 12{α} {} is

α=net τI.α=net τI. size 12{α= { {"net "τ} over {I} } } {}
(12)

Because the force and the perpendicular lever arm are given and the leg is vertical so that its weight does not create a torque, the net torque is thus

net τ=rF=0.0220 m2000 N=44.0 Nm.net τ=rF=0.0220 m2000 N=44.0 Nm.
(13)

Substituting this value for the torque and the given value for the moment of inertia into the expression for αα size 12{α} {} gives

α=44.0 Nm1.25 kgm2=35.2 rad/s2.α=44.0 Nm1.25 kgm2=35.2 rad/s2. size 12{α= { {"44" "." 0" N" cdot m} over {1 "." "25"" kg" cdot m rSup { size 8{2} } } } ="35" "." 2" rad/s" rSup { size 8{2} } } {}
(14)

Solution to (b)

The final angular velocity can be calculated from the kinematic expression

ω 2 = ω 0 2 + 2 αθ ω 2 = ω 0 2 + 2 αθ size 12{ω rSup { size 8{2} } =ω rSub { size 8{0} } rSup { size 8{2} } +2 ital "αθ"} {}
(15)

or

ω 2 = 2 αθ ω 2 = 2 αθ size 12{ω rSup { size 8{2} } =2 ital "αθ"} {}
(16)

because the initial angular velocity is zero. The kinetic energy of rotation is

KE rot = 1 2 2 KE rot = 1 2 2 size 12{"KE" rSub { size 8{"rot"} } = { {1} over {2} } Iω rSup { size 8{2} } } {}
(17)

so it is most convenient to use the value of ω2ω2 size 12{ω rSup { size 8{2} } } {} just found and the given value for the moment of inertia. The kinetic energy is then

KErot=0.51.25 kgm270.4 rad2/s2=44.0 J.KErot=0.51.25 kgm270.4 rad2/s2=44.0 J.alignl { stack { size 12{"KE" rSub { size 8{"rot"} } =0 "." 5 left (1 "." "25"" kg" cdot m rSup { size 8{2} } right ) left ("70" "." 4" rad" rSup { size 8{2} } /s rSup { size 8{2} } right )} {} # " "="44" "." 0" J" {} } } {}
(18)

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:

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, ΔL=(netτ)ΔtΔL=(netτ)Δt size 12{ΔL= $$ital "net"τ$$ cdot Δt} {}. 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= Δ L Δ t net τ= Δ L Δ t size 12{"net "τ= { {ΔL} over {Δt} } } {} for the situation in which the net torque is zero. In that case,

net τ=0net τ=0 size 12{"net "τ=0} {}
(19)

implying that

Δ L Δ t =0. Δ L Δ t =0. size 12{ { {ΔL} over {Δt} } =0} {}
(20)

If the change in angular momentum ΔLΔL size 12{ΔL} {} is zero, then the angular momentum is constant; thus,

L = constant net τ = 0 L = constant net τ = 0 size 12{L="constant " left ("net "τ=0 right )} {}
(21)

or

L = L net τ = 0 . L = L net τ = 0 . size 12{L=L'" " left ("net "τ=0 right )} {}
(22)

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 3, 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 FF size 12{F} {} and rr size 12{r} {} are small, and so ττ size 12{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.L=L. size 12{L=L'} {}
(23)

Expressing this equation in terms of the moment of inertia,

=Iω,=Iω, size 12{Iω=I'ω'} {}
(24)

where the primed quantities refer to conditions after she has pulled in her arms and reduced her moment of inertia. Because II size 12{I'} {} is smaller, the angular velocity ωω size 12{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 3, is spinning at 0.800 rev/ s with her arms extended. She has a moment of inertia of 2.34kgm22.34kgm2 size 12{2 "." "34""kg" cdot m rSup { size 8{2} } } {} with her arms extended and of 0.363kgm20.363kgm2 size 12{0 "." "363""kg" cdot m rSup { size 8{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 ωω size 12{ { {ω}} sup { ' }} {} 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

KErot=122.KErot=122. size 12{"KE" rSub { size 8{"rot"} } = { {1} over {2} } Iω rSup { size 8{2} } } {}
(25)

Solution for (a)

Because torque is negligible (as discussed above), the conservation of angular momentum given in =Iω=Iω size 12{Iω= { {I}} sup { ' } { {ω}} sup { ' }} {} is applicable. Thus,

L = L L = L size 12{L=L'} {}
(26)

or

= I ω = I ω size 12{Iω=I'ω'} {}
(27)

Solving for ωωand substituting known values into the resulting equation gives

ω = I I ω = 2.34 kg m 2 0 .363 kg m 2 0.800 rev/s = 5.16 rev/s. ω = I I ω = 2.34 kg m 2 0 .363 kg m 2 0.800 rev/s = 5.16 rev/s.
(28)

Solution for (b)

Rotational kinetic energy is given by

KErot=122.KErot=122. size 12{"KE" rSub { size 8{"rot"} } = { {1} over {2} } Iω rSup { size 8{2} } } {}
(29)

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

KE rot =( 0 . 5) 2 . 34 kg m 2 0 . 800 rev/s rad/rev 2 = 29 . 6 J KE rot =( 0 . 5) 2 . 34 kg m 2 0 . 800 rev/s rad/rev 2 = 29 . 6 J alignl { stack { size 12{"KE" rSub { size 8{"rot"} } =0 "." 5 left (2 "." "34"" kg" cdot m rSup { size 8{2} } right ) left [ left (0 "." "800"" rev/s" right ) left (2π" rad/rev" right ) right ] rSup { size 8{2} } } {} # " "="29" "." 6" J" {} } } {}
(30)

The final rotational kinetic energy is

KErot=12Iω2.KErot=12Iω2.
(31)

Substituting known values into this equation gives

K E rot = 0 . 5 0 .363 kg m 2 5 . 16 rev/s 2π rad/rev 2 = 191 J. K E rot = 0 . 5 0 .363 kg m 2 5 . 16 rev/s 2π rad/rev 2 = 191 J. alignl { stack { size 12{K { {E}} sup { ' } rSub { size 8{"rot"} } = left (0 "." 5 right ) left (0 "." "363"" kg" cdot m rSup { size 8{2} } right ) left [ left (5 "." "16"" rev/s" right ) left (2π" rad/rev" right ) right ] rSup { size 8{2} } } {} # " "="191"" J" {} } } {}
(32)

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 4.)

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.

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 LL size 12{L} {} can be defined as L=.L=. size 12{L=Iω} {}
• This equation is an analog to the definition of linear momentum as p=mvp=mv size 12{p= ital "mv"} {}. The relationship between torque and angular momentum is net τ= Δ L Δ t .net τ= Δ L Δ t . size 12{"net "τ= { {ΔL} over {Δt} } } {}
• 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 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)?

### Exercise 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.

### Exercise 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 5).

### Exercise 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.

### Exercise 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.

### Exercise 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?

### Exercise 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.

### Exercise 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?

### Exercise 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.

### Exercise 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.

### Exercise 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.

### Exercise 12

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

### Exercise 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

### Exercise 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.

#### Solution

(a) 2.66×1040kgm2/s2.66×1040kgm2/s size 12{2 "." "66" times "10" rSup { size 8{"40"} } "kg" cdot m rSup { size 8{2} } "/s"} {}

(b) 7.07×1033kgm2/s7.07×1033kgm2/s size 12{7 "." "07" times "10" rSup { size 8{"33"} } "kg" cdot m rSup { size 8{2} } "/s"} {}

The angular momentum of the Earth in its orbit around the Sun is 3.77×1063.77×106 size 12{3 "." "76" times "10" rSup { size 8{6} } } {} times larger than the angular momentum of the Earth around its axis.

### Exercise 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.

### Exercise 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?

#### Solution

22 . 5 kg m 2 /s 22 . 5 kg m 2 /s size 12{"22" "." "5 kg" cdot m rSup { size 8{2} } "/s"} {}

### Exercise 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.

### Exercise 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?

25.3 rpm

### Exercise 6

(a) Calculate the angular momentum of an ice skater spinning at 6.00 rev/s given his moment of inertia is 0.400kgm20.400kgm2 size 12{0 "." "400"`"kg" cdot m rSup { size 8{2} } } {}. (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?

### Exercise 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

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