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This may be a very basic question but I am not seeing how it works. Consider the standard example of an ice skate rotating about his/her center of mass and pulling in his/her arms. The torque is zero so we have conservation of angular momentum. This implies that $\omega$ increases to keep $I\omega$ constant, but then $K_{rot}=\frac{1}{2}I\omega^2$ doesn't stay constant, it increases. This implies that there is work done, but what force is doing this work?

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The energy comes from the ice-skater's muscles; they have to work to pull their arms in.

There is no external work done on the skater - the energy is converted from the chemical potential energy stored in the skater's body to kinetic energy.

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  • $\begingroup$ This is best seen if the system is defined to be just the skater. Then we see a change in internal energy. $\endgroup$ – user11266 Dec 3 '12 at 2:02
  • $\begingroup$ @JoeH "Best" is subjective, but I agree that this is a valuable point of view. I wrote about it extensively on this question: physics.stackexchange.com/a/3675/74 $\endgroup$ – Mark Eichenlaub Dec 3 '12 at 3:13
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Work is being done by the force that rearranges distribution of matter. If you want to pull in your arms, you have to fight centrifugal force.You can see it as being in some kind of a force field, much like gravitational, for example.Work done by your muscles while contracting your arms is just integral of a force over some path.

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Here's a few intuitive thoughts of my own regarding speeding up when a skater or ballerina pulls their arms in. The same logic should apply to slowing down when an arm is put out.

  1. In any lossless moving system, surely we can assume conservation of both kinetic energy and momentum. That’s how we analyse lossless collisions between objects.
  2. I’m not yet convinced that any energy is required to pull the skaters arms in. It seems logical looking at the system in a rotating frame of reference, but consider a fixed frame of reference. As the skater pulls their arms in, the centre of momentum of their body is altered - it moves closer to their axis of rotation. So, to conserve momentum, the rotation speeds up. And the kinetic energy is just redistributed throughout the body. The kinetic energy of the hands is just transferred to the rest of the body. Momentum and kinetic energy should be conserved.

Late addition: OK, I’ve finally found an analysis of what I thought, posted by Mark Eichenlaub. Why does a ballerina speed up when she pulls in her arms? Pulling your arms in does require a force, but not all of it is directly towards yourself. Because you are spinning, your hands are moving more at right angles to you. You think you are pulling your hands totally towards you, because you are seeing them in a rotating frame of reference. They are moving towards you slowly, of course, but mostly at an angle. It depends how fast you are spinning. Anyway, the movement at right angles is the torque that is speeding you up if you pull your arms in.

You need to use a fixed frame of reference, not a rotating frame of reference, to appreciate this. That’s why some people have trouble believing it. But I admit to having a problem with my current thoughts: If angular momentum is to stay constant, then shouldn't the net torque should be zero. Maybe, because every force has an equal and opposite reaction, they cancel out. I need to study Mark’s analysis more carefully. Whoops, I’ve just noticed he’s posted above, and stated that the arms muscles input energy to increase rotation. I need to delete my post - damn.

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