By Steve Marrero


2011-01-22 17:08:32 8 Comments

My friend thinks it's because she has less air resistance but I'm not sure.

5 comments

@MarkJB 2014-03-05 14:45:40

Yes the skater does increase the angular momentum by doing work; pulling her arms in. You do work on a swing (sitting up and down) to increase your angular momentum likewise.

@Mark Eichenlaub 2011-01-23 11:41:19

This is a fairly long answer, but I thought it would be fun to analyze the skater thinking solely about forces. Angular momentum comes in at the end, when it pops up inevitably. I'll give a qualitative answer describing the forces in the system and how they cause spinning, then a quantitative answer to calculate the spin rate.

Qualitative Answer

As an ice skater pulls in her arms a legs, her arms and legs exert a torque on her body, causing her to spin faster.

Set a plate in front of you. Try pushing on at in various places and in various directions. You can imagine an arrow pointing in the direction of the force, starting from the point where you're pushing. If this arrow points towards the center of the plate, the plate will not rotate. Otherwise, it will.

alt text

In this picture, the red arrows indicate forces on the gray plate. The force labeled "Spin" will induce some spinning in the plate (in addition to accelerating it as a whole) because the dotted line "j" of that force does not go through the center of the plate. The force labeled "NoSpin" will only accelerate the plate and not cause any spinning because it lies on a line that passes through the center of the plate.

There is nothing special about the circular plate in this example. Any other shape would work as well, but you would need to define the center point by the center of mass.

To see that the skater spins up as she pulls in her arms, we'll need to find the forces exerted on her body while she pulls her arms in. Then we can see whether these forces point straight towards the center of her body or not.

We'll model an ice skater as a circle of mass $M$ with a massless stick pointing through it, and two more circles of mass $m$ on either end of the stick. The skater can move the small circles inward or outward as she spins. Here's a picture of the setup, along with the path her "arms" (the small circles) trace out if she does not pull them in at all while spinning.

alt text

If you imagine holding something heavy in your arms while spinning, your arms would feel as if they're being pulled out of their sockets. In fact, they are. You arms exert a force on the weights straight inwards towards your body, and the weights exert equal and opposite force straight outward.

alt text

Each force is color-coded according to which body it acts on. $F_1$, for example, is the force exerted on the skater by the red arm. (This force is really exerted on the stick, which is rigidly attached to the skater.)

Both the blue forces lie on a line passing through the center of the skater, so the skater's spin rate doesn't change in this scenario. As long as she leaves her arms out, she'll remain spinning the same speed (neglecting friction or other losses of energy).

Now we imagine the skater pulling her arms in. If you watch the path the "arms" (small circles) trace out, you see a spiral shape.

alt text

Here, we show the two arms with both their past and future trajectories traced out. The arms are spinning around the skater while simultaneously being pulled in.

It will be more difficult to find the forces involved here. The arms are no longer moving in simple circles. However, at any given instant, there is a particular circle along which a given arm appears to move. This is the osculating circle. As before, there's a force on the arm pointing in towards the center of the osculating circle.

This is not the whole story, because we can no longer assume that the speed of the arms is constant. Hence there may also be a force on the arms in the direction of their motion. In the next picture, we'll draw only the forces on one arm, just to keep things from getting too cluttered.

alt text

This picture is zoomed in on one arm. The blue circle is still the skater. The green circle is the osculating circle. $F_r$ is the centripetal force towards the center of the osculating circle that curves the arm's path, and $F_t$ is a (small) tangential force in the direction of motion that speeds the arm up.

Without knowing just how fast the arms are moving, how much they're accelerating, and the equation describing their trajectory, it's hard to tell precisely what the sizes of these forces will be. However, we can see that the forces in general have no obligation to point towards the skater's center any more. She can change her spin rate because the forces may not necessarily lie on lines that pass through her center.

In order to see just what these forces are, we need to do an quantitative analysis.

Quantitative Answer

The plan of this answer is to find the forces on the arms as they spiral in, then use Newton's third law to find the forces the arms exert on the stick. Next, we'll relate these forces to the rate of change of the energy of the skater. The energy of the skater can also be calculated from her motion directly, so we'll do that, take a time derivative, and compare to our previous expression. This will reveal a conserved quantity, the angular momentum, which allows us to find the skater's rotation rate as a function of the initial conditions and the final distance of the arms from her center - the details of the shape of the spiral and how fast the arms are pulled in do not matter. Finally, we'll see that the skater spins faster and faster as she pulls her arms in.

We'll use polar coordinates to describe the positions of the arms. The radial coordinate $r$ of arm 1 is some function of its angular coordinate $\theta$.

$$r = f(\theta)$$

With the definition $\omega = \dot{\theta}$, we have

$$\dot{r} = f'\omega$$ $$\ddot{r} = f''\omega^2 + f'\dot{\omega}$$

The acceleration in polar coordinates is

$$\vec{a} = \hat{r}(\ddot{r} - r \dot{\theta}^2) + \hat{\theta}(2\dot{r}\dot{\theta} ^2 + r\dot{\omega})$$

The force on this arm is given by Newton's second law, $\vec{F} = m\vec{a}$. The force the arm exerts on the stick is the negative of this, by Newton's third law. This force of the arm on the stick is what we're interested in.

As the stick spins, the force of the arm on the stick does work on the stick, which goes into the kinetic energy of the skater (the stick itself has no mass). The velocity of the point on the stick where the force is applied is $\vec{v} = f\omega\hat{\theta}$. The power delivered by this force, doubled to include the work done by the other arm, is

$$P = \vec{F} \cdot \vec{v} = -2m\omega f(2f'\omega^2 + f\dot{\omega})$$

This is the rate of change of the kinetic energy of the skater. That kinetic energy is

$$ T = \frac{1}{2}M (R \omega)^2$$

with $R$ the radius of the skater. If we equate the power to the time derivative of the kinetic energy, we get

$$MR^2\dot{\omega} = -2mf(2f'\omega^2 + f\dot{\omega})$$

If you stare at this and scratch your head a moment, it's mathematically equivalent to

$$\frac{\textrm{d}}{\textrm{d}t}\left(\omega(MR^2 + 2mf^2)\right) = 0$$

This means we've discovered something that doesn't change in time - a conserved quantity. It's called the angular momentum, and the part $MR^2 + 2mf^2$ is called the moment of inertia. We denote the angular momentum by $L$ and (switching back from $f$ to $r$) write

$$L = \omega(MR^2 + 2mr^2)$$

Because $L$ is a constant, we can find $\omega$.

$$\omega = \frac{L}{MR^2 + 2mr^2}$$

We've got the angular frequency as a function of $r$ only - the precise function form of $f$ didn't matter, and neither did how quickly we traversed the path. As long as we know $L$ from the initial conditions, we've solved the problem.

From

$$\frac{\textrm{d}\omega}{\textrm{d}r} = \frac{-4m L r}{(MR^2+2mr^2)^2}$$

we see that, assuming $L$ is positive, as $r$ goes down, $\omega$ goes up, so the skater goes faster and faster as she pulls in her arms.

@Carl Brannen 2011-01-24 05:33:34

Yes, I was considering mentioning this to my students as an explanation for why it is that W=Fd works to increase her KE when she pulls in those arms.

@Mark Eichenlaub 2011-01-24 05:40:15

@Carl What age range are your students?

@Carl Brannen 2011-01-24 05:52:11

Maybe 19 to 65. They're typically working on 2-year degrees in drafting or electronics technician. Teaching them physics is a humbling experience.

@Sklivvz 2011-01-22 21:06:02

A very simple explanation is the following: the arms of the ballerina are pulled outwards by the centripetal force she experiences by spinning. When she pulls her arms in, she is doing work by more than counteracting this force and this is what makes her spin faster.

This is due to the fact that the spin velocity is related to the effort that the ballerina makes in pulling in her arms. The closer the arms, the more force she has to use to pull them in future or keep them in position, and the faster she spins.

When I was in high school, we did a more hardcore version of this experiment by sitting on a chair that could spin, holding two heavy weights outwards. Then someone would spin the poor test subject and ask him to pull in the weights... The results were quite scary... :-)

@Joe Fitzsimons 2011-01-22 21:10:38

It has nothing to do with how hard she pulls in her arms (although it is related to the work done). If she pulls harder they come in faster, but her total change in angular velocity is the same.

@Sklivvz 2011-01-22 21:16:48

@Joe, what's your problem dude? This is a qualitative answer. The energy she uses to pull in the arms is exactly the energy she gains by spinning faster.

@Joe Fitzsimons 2011-01-22 21:19:26

@Sklivvz: This isn't a personal attack, it simply that the line "This is due to the fact that the spin velocity is a function of how hard you pull your arms in." is false, and I was pointing that out. It's an unfortunate coincidence that I have problems with your other question in parallel. I've been trying to raise the quality of the answers by flagging errors.

@Sklivvz 2011-01-22 21:22:49

@Joe, $F = m\omega^2r$ so F is a function of omega.

@Joe Fitzsimons 2011-01-22 21:27:17

@Sklivvz: She needs to apply a constant force to keep the arms in the new position, yes, but this doesn't mean that pulling in harder will make her spin faster, her arms reach a new position and to keep them there there is only one value of F she can apply. The harder she pulls in initially the faster her arms reach that new position, but in the end the force has to be the same.

@Sklivvz 2011-01-22 21:30:23

she pulls in her arms exactly by exceeding this force. So it is exactly by pulling harder than this that she spins faster.

@Carl Brannen 2011-01-22 21:30:38

It's a beautiful qualitative answer, an important one, and one that I give to my students regularly (of course I also talk about conservation of angular momentum). And thanks to the other answer providers, Steven is one of my students. We put the question onto Stack Exchange at the start of class at 9AM PST and the system gave an answer before the lab was over. Nice demonstration of a useful tool for students.

@Carl Brannen 2011-01-22 21:32:08

Oh, and the change in kinetic energy of rotation can be calculated by Work = Force x Distance. In a conceptual physics class this isn't done because it requires integration.

@Joe Fitzsimons 2011-01-22 21:42:14

@Carl: Exactly, it's the work done that counts, not the profile of force used.

@Chad Orzel 2011-01-22 21:50:38

The problem here is that second paragraph, which is phrased in a way that seems to reverse the causality. I can see how you might arrive at that wording from the correct understanding of the physics, but it's a really awkward formulation that makes it sound like the strength of the pull determines the speed, where later answers suggest the real intent was something closer to "the force required to pull her arms in is greater when she's spinning faster."

@Chad Orzel 2011-01-22 21:53:12

You can easily show that the spin rate doesn't depend on the force used by noting that the final spin rate doesn't depend on how quickly she brings her arms in, just the final position-- whether she brings them in slowly, or jerks them in quickly, she ends up spinning at the same rate. If you've got access to a good rotating platform, you can even do the experiment yourself. The quick pull uses a bigger force (at least at the start), but does the same work, leading to the same final speed.

@Sklivvz 2011-01-22 23:29:06

The force she exerts determines how fast she speeds up. I never said it changes the end rotational speed - nor I used the word force. You are reading in my words something I did not mean.

@Joe Fitzsimons 2011-01-23 05:48:59

@Sklivvz: you referred to how hard she pulled in her arms. That is clearly a reference to force. If you remove or correct that paragraph, I'll remove my downvote.

@Sklivvz 2011-01-23 09:30:10

@joe it's not meant to be as such, but then again English is not my first language.

@Joe Fitzsimons 2011-01-23 09:50:25

@Sklivvz: I have no objections to the new form, and have removed my downvote.

@TROLLHUNTER 2011-01-27 11:15:39

I like this answer, but is it possible to exlain why the work she does must go into kinetic energy ?

@Sklivvz 2011-01-27 13:41:33

@kake Mark has the complete answer.

@LuboŇ° Motl 2011-01-22 17:31:46

Joe's answer is of course right and I gave it +1. However, let me say some slightly complementary things.

Whenever the laws of physics don't depend on the orientation in space, a number known as the angular momentum is conserved. For a rotating body - including the body of a lady - the angular momentum $J$ may be written as the product of the moment of inertia $I$ and the angular frequency $\omega$ (the number of revolutions per second, multiplied by $2\pi=6.28$): $$ J = I \omega $$ The moment of inertia $I$ is approximately equal to $$ I = MR^2 $$ where $M$ is the mass and $R$ is equal to the weighted average distance of the atoms (weighted by the mass) from the axis. (More precisely, I need to compute the average $R^2$.)

alt text

It's up to you whether she is spinning clockwise or counter-clockwise.

So if the ballerina pulls in her arms, she becomes closer to the axis, and $R$ decreases. Her mass $M$ doesn't change but the moment of inertia $I$ decreases, too. Because $J=I\omega$ has to be conserved and $I$ decreased, $\omega$ inevitably increases.

You may also explain the increased angular frequency of the rotation in terms of forces and torques. If the arms move closer to the axis, they exert a torque on the ballerina that speeds her up. I would need some cross products here but I am afraid that wouldn't be fully appreciated.

These issues were also discussed here:

Why do galaxies and water going down a plug hole spin?

Why do galaxies and bathtub whirlpools spin?

Cheers LM

@Joe Fitzsimons 2011-01-22 17:36:08

You're definitely better at expository answers +1.

@Joe Fitzsimons 2011-01-22 18:27:59

As a side note regarding the new ballerina picture, it seems there is a right and a wrong answer, since the picture appears to use perspective rather than a parallel projection.

@Noldorin 2011-01-22 18:31:31

Hah, I've seen this image so many times... every time I'm sure she's spinning clockwise. Is it even possible to argue she's spinning the other way?

@Nick Pascucci 2011-01-22 18:34:17

Just one correction: \omega is actually the angular velocity in this context.

@Noldorin 2011-01-22 18:37:04

@codeMonk: Angular velocity is the vector; he's just using the magnitude of it (angular frequency) here.

@Joe Fitzsimons 2011-01-22 18:38:54

@Noldorin: You could settle it once and for all by pausing it when she has the foot on the left hand side of the image elevated and pointing directly towards you, and measuring the minimum thickness of the ankle, and then doing the same thing once she has rotated 180 degrees. That would give you the definitive answer.

@Noldorin 2011-01-22 21:21:15

Hmm. But I don't see how it's even possible for the animation to represent her spinning in the counter-clockwise direction.

@Nick Pascucci 2011-01-23 00:58:36

@Noldorin: Angular momentum is a vector quantity; both magnitude and direction are conserved, thus the need to use angular velocity vs. angular frequency.

@Marek 2011-01-23 10:42:54

@Noldorin: I can make the image go whichever way I want. Even few times cw. then few times ccw. or even just oscillate with half-turns. It's all in the brain because the animation doesn't contain enough information. But @Joe is right that it might be possible to measure it microscopically. However, brain ignores these microscopic details completely. And by the way, you are not the first person who could only see it spin one way, few of my friends also had the same problem. But then with practice they were also able to see it go the other way too :)

@Noldorin 2011-01-23 16:19:00

@Marek: Yeah, this is meant to signify that a certain half of your brain is dominant, I believe. In my case, I believe it's the right half!

@Noldorin 2011-01-23 16:19:59

Also, surely one can trace the position of her left hand, for example, and it will represent the locus of a circle in 3D space. From this, it is clear that the point is orbiting clockwise. surely?

@Marek 2011-01-24 00:34:55

@Noldorin: think about how would that circle look if it were lying in a plane intersecting your eyes. It would be a line! Then you'd surely wouldn't be able to tell rotation from pure oscillation. If you tilt that circle then your brain will interpret it as a true rotation because you'll think of one direction as "to the back" and the second "to the front" and you'd expect that these two motions periodically switch for a rotating object. But with training you can force the brain to attach any interpretation to the movement leaving the side, not just the natural one.

@Noldorin 2011-01-24 00:57:32

@Marek: Ah, I see exactly what you mean now. Thanks for explaining. At a glance, it appeared that her foot was travelling in an ellipse (when projected onto the 2D plane), but in fact it's simply a line. This indeed does not distinguish between direction of rotation. However, does the head not resolve the matter? i.e. Does the fact that one sees her nose/eyes/hair gradually appear from the right side not suggest clockwise rotation?

@Marek 2011-01-24 08:43:39

@Noldorin: only thing by which you could distinguish the rotation besides colors and texture (which are both absent here) is depth. But the relative changes in distance are too small here and you don't see the biggest changes anyway because they are hidden by the body. As for the nose/eyes that's again a matter of interpretation. For ccw. rotation you see them appear from the left side.

@Joe Fitzsimons 2011-01-22 17:16:33

No, it's caused by conservation of angular momentum. Reducing air resistance won't cause her (or anything else) to speed up without an external force.

Like linear momentum ($m v$), angular momentum ($r \times mv$) is a conserved quantity, where $r$ is the vector from the center of rotation. For a skater holding a static pose, for each particle making up her body, the contribution in magnitude to the total angular momentum is given by $m_i r_i v_i$. Thus bringing in her arms reduces $r_i$ for those particles. In order to conserve angular momentum, there is then an increase in the angular velocity.

Hope this answers your question.

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