There is a classic example that a spinning skater pulls his arms back. The angular momentum is conserved, the moment of inertia decreases. And therefore, it's angular velocity increases, so the rotational kinetic energy will increase. But what will happen if the skate pushes his arms outward? The rotational kinetic energy will decrease. But where will it go?

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    $\begingroup$ @Buraian I think that might be a little different. Pulling your arms in and relaxing your arms are two different things, and they are not the reverse of each other (your muscles don't get energy back when you relax them and your arms fly out). $\endgroup$ – BioPhysicist Sep 27 at 11:32
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    $\begingroup$ Absent the tiny amount of energy needed to move the arms, the total energy remains constant. $\endgroup$ – Hot Licks Sep 27 at 16:33

When the skater is pulling his arms towards his torso he is doing work.

Now to the converse of that:
First: To remain spinning fast the skater must sustain a large centripetal force: that's just to keep his arms close to his torso. By easing up that centripetal force he allows his arms to move away from his torso. The motion of his arms moving away from his torso is the converse of pulling in his arms. Due to this motion of his arms (away from his torso) the mass of his arms is doing negative work, decreasing his rotational kinetic energy.

During the stage that the skater is pulling his arms towards his torso his muscles are exerting a slight surplus of centripetal force, this surplus causes the contraction.

During the stage that the skater allows his arms to extend again he is still exerting a centripetal force, but now slightly less than the required centripetal force, such that his arms move away from this torso in a controlled manner.

Negative work
At first sight the concept of doing negative work may appear counter-intuitive, since it is always taught that energy is a scalar.

Textbook authors often rearrange the description of a setup to avoid the concept of negative work. Take for example the following setup: a cylinder with a piston and by applying force to compress a gas in the cylinder you are doing work upon the gas, compressing it. Now: when you are allowing that compressed gas to relax again you are doing negative work upon the gas. But textbook authors will describe that as follows: "The expanding gas is doing work". That is a change of perspective just for the purpose of avoiding the expression 'negative work'.

In the case of the skater allowing his arms to extend again to slow himself down: that change of perspective isn't available because the skater pulling in his arms is not done against some other force.

Where did the rotational kinetic energy go to?
In order to generate a lot of rotational kinetic energy the skater has to set up for that. He pushes off with his skates against the ice to give himself an initial angular velocity while holding his arms extended. Those two are both necessary. Without initial angular velocity contracting the arms doesn't do anything. And if his arms are already tight to his torse when giving himself initial angular velocity then he doesn't have any distance to do work. With both in place the skater has set up the ability to do work.

Of course, all of that setup would be useless without muscle power. The contracting force is provided by the skater's muscles. The energy source for the increase of rotational kinetic energy is muscle power.

When the skater extends his arms again he is decreasing his angular velocity. During that phase his muscles are absorbing energy. As we know, while muscles are good at extending in a controlled manner, muscles do not recoup energy. (In fact, extending a muscle in a controlled manner costs energy too.) So if a skater spins up and spins down repeatedly he will work up a sweat because each cycle his muscles absorb the rotational kinetic energy that was generated on spinning up.

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The other answers are stating some incorrect things here. Yes, when you pull your arms in your muscles are doing work on your arm, but if you

  1. relax your arms so they fly out, then you aren't doing work on your arms, or if you

  2. push your arms out so that you are doing work on your arms, then you are doing positive work on your arms. Furthermore, doing an opposite motion with other muscles does not give your muscles back energy.

So what is going on here? There has to be some radial force doing negative work here. It turns out it is still your arms, but you have to consider the entire motion of your hands and arms to see it.

When you either relax your arms$^*$ so that they start moving radially outwards, that isn't the end of the story. This can easily be seen by a simple model of an object that was moving around a horizontal circle on a string where the string is suddenly cut. Here the object will move tangentially away from the previous center of rotation, and the kinetic energy will remain constant since no forces are doing work.

Coming back to our scenario, obviously our hands and arms do not fly outwards forever, so there needs to be a radially inward force at the end to stop our arms (this can be applied willingly with muscles or unwillingly by whatever prevents our arms from popping out of place). This is where your negative work comes from.

$^*$ The following analysis works for pushing your arms out as well, but that just includes an additional step of adding energy to your hands before taking the energy away.

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We can write angular momentum as:

$$ \frac{L}{I} = \omega$$

and, hence we can write energy as:

$$ K = \frac{1}{2} I \omega^2 = \frac{L^2}{2I}$$

So, since the angular momentum of the system is conserved in absence of external torques (when you move your arms you generate internal torques which sum to zero). Hence, the Kinetic energy is purely a function of rotational inertia:

$$ K(I) = \frac{C}{I}$$

So, suppose we reduce inertia by pulling our arms in closer then our kinetic energy increases and when we increase inertia by pulling outwards kinetic energy increases.

The increase and decrease of energy can be associated with the changes in configuration of a system. When you move your arm, you have to do a work with that.

Here is an idea for understanding this better:

Think of your whole arms as a spring , so when you put your arms more outwards, it's like extending a spring. The energy at a given extension is given as:

$$ E= \frac{k}{2} x^2 + \frac{C}{I}$$

So, when I increase my $x$ my kinetic energy has to adjust in such a way that my energy is a constant by the conservation of energy.

Note: this is a really gross approximation.


Simple university article

Muscles as locomoter springs (pubmed)

Spring mass model for jumping (pubmed)

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  • $\begingroup$ Pulling your arms is not the same thing as compressing a spring. You don't store energy in your arms when you bring them in and then release that energy when you extend them (or vice versa, as you talk about here). $\endgroup$ – BioPhysicist Sep 27 at 11:51
  • $\begingroup$ I'm pretty sure you can compress/ extend your arm beyond it's natural length. For example consider how we jump. We push compress our leg muscle and then when we jump, we quickly let them extend outwards just like a spring. So, quickly thrusting our arms to be more outwards could be thought of as compression and extensions. $\endgroup$ – Buraian Sep 27 at 11:56
  • $\begingroup$ Finally I am not at all talking about an exact model. I did not say they are same thing, I said they can be grossly approximated as so. $\endgroup$ – Buraian Sep 27 at 11:57
  • $\begingroup$ downloads.lww.com/wolterskluwer_vitalstream_com/sample-content/… see page -55 , pic-4.14 $\endgroup$ – Buraian Sep 27 at 12:03
  • $\begingroup$ You can make analogies with springs, but I don't think you can take the analogy to a place where movement of your muscles stores energy in them. In your jumping scenario, it's not like bending down stores energy, and then to jump you just release that stored energy. $\endgroup$ – BioPhysicist Sep 27 at 12:10

These answers seem to be missing the point. Ignoring the relatively small amount of energy required to change the position of the arms, and ignoring friction, energy is conserved.

If we model the skater as a circle with circumference of one meter, with two 1kg weight at the extremes, rotating at 1 revolution per second, the kinetic energy will be 6 joules.


If the circumference is reduced to 1/2 meter, the kinetic energy will be preserved, causing the rotational rate to increase to 2 revs per sec, preserving the 1 meter/second velocity.

"Angular momentum" is just a way to describe overall momentum for a rotating mass that does not have point masses but instead has the mass distributed across it's diameter somehow. But it can be approximated by "dicing" the mass into smaller pieces and calculating the velocity of each, then applying the rules for linear momentum/energy. (Or, if you're into self-abuse, calculate the integral.)

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  • $\begingroup$ In your approximation it seems right but this contradicts the other answers but is still a reasonably good explanation of what is going on... $\endgroup$ – Buraian Sep 28 at 11:37

The skater does work with her arms when bringing them closer to her body or pushing them further out, and that work done manifests as a decrease or increase in rotational kinetic energy. So the TOTAL energy does not actually go anywhere since total energy (and momentum) is conserved. Work done results in a change of kinetic (or potential in other cases) energy. If she brings here arms back, the rotational kinetic energy returns to its original value, so overall the total energy is conserved (assuming no energy losses to friction, air resistance etc).

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  • $\begingroup$ If you push your arms out then you're doing positive work on your arms, so why would the energy decrease during that step? $\endgroup$ – BioPhysicist Sep 27 at 11:21

The rotational kinetic energy will decrease. But where will it go?

It goes into the linear kinetic energy of the arms as they accelerate from the body outward.

When the arms are slowed (either by muscles or by pulling on tendons), that energy is transformed into heat.

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It doesnt go anywhere.
All this assumes you ignore friction and energy used to move the arms.
In the long run all the skaters energy becomes heat which is where it went.

In the simple example it is the opposite of pulling the arms in, the same way you analysed that case, but in reverse.

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