Balance and Rotation - Lesson 4 - Angular Momentum

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Another Thing Conserved

A figure skater pulls her arms inward and spins faster.  An Olympic gymnast tucks in order to rotate quickly to stick his landing.   A comet travels fastest in its orbit when it is closest to the sun.  But why?  We can explain these phenomena in terms of another important conservation law—the conservation of angular momentum.     

In the previous section, we learned about the rotational counterpart to momentum—angular momentum.   The reason we are so interested in an object’s angular momentum is that...

In the absence of an external torque,
the angular momentum of a system stays constant.

This is the law of conservation of angular momentum.

In Lesson 3, we introduced open and closed systems in the context of energy.  A closed system is one in which the energy within a system stays constant.  That is, it is conserved.   An open system is one in which the energy within the system changes as work from an external force is done on the system.  The same is true for angular momentum.  In a closed system, the angular momentum stays constant.  In an open system, the angular momentum can change.  We saw in the previous section that the way we change the angular momentum of an object is by delivering an angular impulse to the system from a torque outside of the system.  There may be torques exerted on one object inside the system by another object inside the system, but since these are internal impulses, they cannot change the angular momentum of the system.

An illustration of a Closed vs. Open System in respect to Angular Momentum initial and final.  The closed system shows a box with L sub i (Initial Angular Momentum) = L sub f (Final Angular Momentum).  The Open System shows a perforated box with a green arrow on the left coming from outside and going into the box with the text + Angular Impulse (Delta L sub ext), an arrow on the right going from the box outwards with the text negative angular impulse (delta L sub ext), and in the box the equation L sub i + Delta L sub ext (Change from external Angular Momentum) = L sub f.

It is the closed system that we are most interested in here as we study the conservation of angular momentum.

We saw in the previous section that the angular momentum of a rotating object is equal to its moment of inertia times its angular velocity.  In other words, L = I ω.  In a closed system, since the angular momentum must be conserved, if the moment of inertia of the object changes, the angular velocity will adjust accordingly in order to conserve angular momentum.  Mathematically, we can write:

The Conservation of Angular Momentum equation I sub i (Initial Moment of Inertia) times omega sub i (initial angular velocity) = I sub f (Final Moment of Inertia) times omega sub f (Final angular velocity).

If you’re thinking, "What does this mean in real life?" then you’ll probably find the next three examples helpful in understanding how this conservation of angular momentum equation can be applied to real situations.

Example 1: Spinning Chair

Problem:  Whether or not you are a figure skater or a gymnast, you have probably sat in a chair that spins.  Imagine you’re sitting in this slowly spinning chair with weights in your outstretched hands.  You now bring your arms inward toward your body.  Like the figure skater or gymnast, your angular velocity increases significantly.  How come? 

Solution:  Let’s consider you, the chair, and the weights as our system. Provided there is no external torque acting on the system, angular momentum must be conserved.  When the weights are in your outstretched arms, you have a very large moment of inertia.  As you bring the weights in close to your body, however, your moment of inertia decreases significantly.  Since L = I ω, the only way to conserve angular momentum when the moment of inertia decreases is to have the angular velocity, ω, increase! 

Example 2: Conserved or Not Conserved

Problem:  A disk (ICM = 2 kg·m2) is rotating with an angular velocity of 6 rad/s about an axis through its center.  A ring (ICM = 1 kg·m2) is dropped on the disk so that their centers are aligned. 
(A) Is angular momentum conserved in this situation? 
(B) What is the final angular velocity of the disk-ring system?

A spinning disk with a blue angular velocity vector is shown, and a hollow ring is shown falling down onto the ring.

Solution: 
(A) If we define the system as both the disk and the ring, then the impulse that the ring exerts on the disk (which will slow it down) and the impulse that the disk exerts on the ring (which will speed it up) are both internal impulses.  Provided there are no external impulses, the angular momentum of this system must be conserved
(B) As a result, we can apply the conservation of angular momentum relationship to find that the final angular velocity of the disk-ring system together is 4 rad/s.

We start with I sub disk times omega sub disk + I sub ring times w sub ring = the combination of I sub disk + I sub ring times the omega sub f (final angular moment of both since they spin together).  Values are added, 2 kg meters squared times 6 rads per second + 0 (the ring has no angular momentum at first) = 2 + 1 kg meters squared times the omega sub f (final angular momentum), and solve for omega sub f to get 4 rads per second.

Example 3: The Spinning Wheel

Problem:  A person stands on a rotatable platform holding a bicycle wheel that is spinning counterclockwise with an angular momentum of Lwheel = +10 kg·m2/s.  He quickly inverts the spinning wheel so that it is now spinning clockwise.  Upon doing so, the person and platform start rotating.  Explain why this is the case?

Solution: Let’s consider the bicycle wheel, the person, and the platform as our system.  We’ll say that CCW is the positive direction of rotation.  While an angular impulse was exerted on the wheel to invert it (thus changing its angular momentum from +10 kg·m2/s to -10 kg·m2/s), this impulse was provided by another part of the system.  Thus, the angular momentum of the system must remain unchanged.  As a result, the wheel exerts a counter-torque on the person (thus changing his angular momentum from 0 kg·m2/s to +20 kg·m2/s), which causes the person and platform to rotate counterclockwise.

Two illustrations (before and after) are shown of a person holding a wheel on a rod.  The initial shows the bike wheel rotating Counterclockwise with L sub wheel = positive 10 kg meters squared per second, and the L sub person + platform = 0 kg meters squared per second.  The second illustration shows the wheel flipped, now L sub wheel is negative 10 kg meters squared per second and the rotation Clockwise.  A counterclockwise vector is now shown above the person, and L sub person + platform = positive 20 kg meters squared per second.

In the previous section, we introduced two equations for finding the angular momentum of an object with respect to an axis of rotation.  So far, our examples in this section have used L = I · ω rather than L = r p sin θ.  How might we demonstrate conservation with this second equation?  Satellite motion and a child jumping on a merry-go-round to get it rotating are two great examples to illustrate how our second equation is used in conservation of momentum scenarios.  Let’s go there next.

Satellite Motion

All satellites that orbit the Earth (or any central mass) follow an elliptical orbit.  As the satellite orbits, not only does its distance from the Earth vary, but its speed varies as well.  What is fascinating is that we can exactly predict the satellite’s speed at any point using conservation of angular momentum.  It is here, however, that we’ll use our second equation for angular momentum, L = r p sin θ.  Let’s consider two points in the satellite’s orbit (below).

Point A is the location where the satellite is closest to the Earth in its orbit.  Point B is where the satellite is furthest from Earth.  If we assume no external torque acts on the satellite, we can write...

We start with L sub A = L sub B (angular moment is conserved).  We replace each L with the Angular Momentum equation r times p times sine theta on both sides.  In our diagram the Theta is 90 degrees so Sine(90) is 1, and the p value is mass times velocity.  The mass is the same so those cancel out, leaving r sub a (radius at point A) times v sub a (velocity at point A) = r sub b (radius at point B) times v sub b (velocity at point B).  From this if the radius increases, the velocity must decrease, or vice versa.

If there is no external torque acting on the satellite, the angular momentum is constant throughout its orbit.  As a result, the speed decreases as the satellite moves from A to B and increases from B to A.

Example 4: An Orbiting Comet

Problem:  Halley’s comet orbits the sun in a very elliptical orbit.  It takes 76 years to complete one trip around the sun. At aphelion (furthest from the sun), the comet is nearly 60 times further from the sun than at perihelion (closest to sun).  At perihelion, it travels about 55 km/s relative to the sun.  How fast is Halley’s comet traveling at aphelion?

Solution:  Our first step is to recognize that the comet’s angular momentum is conserved throughout its orbit.  Setting the angular momentum at aphelion (position 1) equal to its angular momentum at perihelion (position 2), we are able to find that Halley’s comet is travelling at a leisurely 920 m/s when it is furthest from the sun.

We start with the angular momentum at aphelion (farthest from sun - labeled sub 1) = Angular Momentum at perihelion (closest to sun - labeled sub 2).  r sub 1 times p sub 1 times sine 90 = r sub 2 times p sub 2 times sine 90.  Sine 90 is 1, the p is replaced by mass times velocity (and masses cancel out), and we label r sub 2 as 1 and r sub 1 as 60 (60 times longer) so we get 60 * v sub 1 = 55 km per second.  V sub 1 = 0.92 km per second, or 920 meters per second.

Another Application of Conservation of Angular Momentum

Let’s consider another situation where we can use conservation of angular momentum to make sense of the motion of rotating objects.  In the previous section, we discussed that an object traveling in a straight line can have angular momentum with respect to an axis of rotation.  While finding its angular momentum might have made sense mathematically, it may have seemed like merely a plug-and-chug exercise to find the angular momentum of an object traveling in a straight line.  There is a very practical application, however. 

Consider a merry-go-round that is initially at rest, and a child comes running to jump on it.  If we consider the merry-go-round and the child as our system, is angular momentum conserved in this "collision"?   The answer is "yes."  Here is a case where we have an object (the child) moving in a straight line with angular momentum with respect to an axis of rotation (the center of the merry-go-round).  When the child jumps on the merry-go-round, the merry-go-round and the child rotate about the merry-go-round’s axis with the same angular momentum that the child had initially.  It is true that the child imparted an angular impulse to the merry-go-round when she grabbed hold; it is equally true that the merry-go-round imparted this same angular impulse to the child in the opposite direction.  Since these are internal impulses, they do not change the angular momentum of the system.  Thus, angular momentum is conserved, and we can use this to help us analyze the situation.  Let’s take a deeper look at this situation in the example below.  

An illustration of a Merry go round (top view) that is initially at rest being impacted by a child running onto the merry-go-round. The radius of the merry-go-round is labeled r, and the child running has m mass and v velocity.  It runs and jumps onto the merry-go-round at the bottom (at a 90 degree angle to the radius).  The Merry-go-round will start rotating once the child jumps on.

Example 5: Merry Go Round

Problem: A 40 kg child running at 3 m/s jumps on a 100 kg merry-go-round that has a radius of 2 m.  The merry-go-round can be approximated as a uniform disk.  What is the magnitude of the angular velocity of the child and the merry-go-round once the child has jumped on?

Solution: Let’s consider our system as both the child and the merry-go-round. The child has angular momentum with respect to the axis of the merry-go-round.  By conservation of momentum, this value must equal the angular momentum of the child and disk as they rotate together after the child jumps on the merry-go-round.  We will find it convenient to use our second equation for angular momentum (L = r p sin θ) to represent the initial value.  In the end, since both rotate about a common axis, we’ll find it most convenient to use the first equation for angular momentum (L = I ω).  The child will act like as ring in its Center of Momentum calculation.  Setting these equal, we find the magnitude of the angular velocity to be 0.67 rad/s.

We start with Conservation of Angular Momentum equation again.  For the initial state, we will use r p sine theta for the child (the merry-go-round has no angular momentum initially) which will equal the sum of the Moment of inertia of the child and the Moment of inertia of the disk multiplied by the same Angular velocity omega sub f.  We will use the ring equation for the child's Moment of Inertia (M R squared) and the disk equation for the Merry go round's Moment of Inertia and plug in our values.  2m times 3 meters per second times 40 kg time 1 (initial Angular Momentum) = the sum of 40 kg times 2 meters squared plus 1/2 times 100 kg times 2 meters squared.  Solving for omega sub f we get 0.67 rads per second.

We’ve seen that conservation of angular momentum is a powerful tool in explaining so many situations that involve rotation.  Now it’s your turn to see if you can apply these ideas to the questions below.  

Check Your Understanding

Use the following questions to assess your understanding. Tap the Check Answer buttons when ready.

1: A high diver is rotating once per second when he tucks (pulls his arms and legs inward) to increase his rotation to three times per second. 
(A) Did his moment of inertia increase or decrease in this process?
(B) By how much did his moment of inertia change?

Check Part A Answer

Check Part B Answer

2: A figure skater spins with an angular velocity of 4.0 rad/s and has a moment of inertia of 2.0 kg·m² when her arms and legs are extended. When she brings her arms and legs inward, she can reduce her momentum of inertia to 0.4 kg·m².  What angular velocity does she spin with now?

Check Answer

3: As Halley’s comet orbits the sun, its speed relative to the sun changes. Where in its orbit is it slowing down?
a. when it is closest to the sun
b. when it is furthest from the sun
c. when it is approaching the sun
d. when it is moving away from the sun

Check Answer

4: When studying Kepler's Laws, we learned how he discovered that satellites sweep out equal areas in equal amounts of time.  Two students offer an explanation as to how the conservation of momentum explains this observation.  With which student do you agree?
Student 1: The speed of a satellite increases as its distance from the sun increases since the moment of inertia varies directly with angular velocity.
Student 2: When a satellite is closer to the sun, it has a smaller moment of inertia and thus it must have a larger angular velocity, so it is traveling faster.

Check Answer

5: A toy dart (mass = 0.2 kg) is shot at 20 m/s perpendicular to the very end of a meterstick that has been allowed to pivot in a horizontal plane about its center without friction.  The meterstick has a moment of inertia is 0.1 kg·m2.  If the dart sticks to the end of the meterstick, with what angular velocity does the meterstick-dart system rotate? (Hint:  Once it sticks, the moment of inertia of the dart itself will be 0.05 kg·m2 with respect to the pivot).

The top view of a Meter stick (pivot at center) with a dart that hits the edge of the meter stick and 'sticks', showing the meter stick rotating after being struck and the dart sticking to the end of the stick.

Check Answer

Looking for additional practice? Check out the CalcPad (RD8) for additional practice problems on both this section and the previous section.


 
Figure 1 Borrowed from Wikimedia Commons https://commons.wikimedia.org/wiki/File:Change-in-angluar-speed-due-to-change-in-moment-of-inertia.svg
Figure 2 Image generated using some MS Word Iconography.
Figure 3 Borrowed from Wikimedia Commons https://commons.wikimedia.org/wiki/File:Halley%27s_Comet_animation.gif


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