# Why does a ballerina speed up when she pulls in her arms?

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

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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.

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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 $R^2$.)

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

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You're definitely better at expository answers +1. –  Joe Fitzsimons Jan 22 '11 at 17:36
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. –  Joe Fitzsimons Jan 22 '11 at 18:27
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? –  Noldorin Jan 22 '11 at 18:31
Just one correction: \omega is actually the angular velocity in this context. –  Nick Pascucci Jan 22 '11 at 18:34
@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. –  Marek Jan 24 '11 at 8:43

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.

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.

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.

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.

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.

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.

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.

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

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

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

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.

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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. –  Carl Brannen Jan 24 '11 at 5:33
@Carl What age range are your students? –  Mark Eichenlaub Jan 24 '11 at 5:40
Maybe 19 to 65. They're typically working on 2-year degrees in drafting or electronics technician. Teaching them physics is a humbling experience. –  Carl Brannen Jan 24 '11 at 5:52

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... :-)

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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. –  Joe Fitzsimons Jan 22 '11 at 21:10
@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. –  Sklivvz Jan 22 '11 at 21:16
@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. –  Joe Fitzsimons Jan 22 '11 at 21:19
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 Jan 22 '11 at 21:30
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 Jan 22 '11 at 21:50

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.

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