I'm trying to compile a Feynman technique-esque definition for pertinent physical concepts. Here is mine for what I understand of angular momentum, in particular, conservation of angular momentum

Conservation of Angular Momentum

Basically, it means the energy from changes of angular momentum is conserved. If an ice skater moves her arms to her chest, she gains energy from the system of her rotation. That sentence sounds bizarre but bare with me. When she's spinning with her arms out, her arms feel an inward force that keep her arms from falling off her body tangent to the path of motion. When she moves along that direction when pulling in her arms, that inward force is now doing work on her. The work done is giving her kinetic energy, which amounts to an increase in the speed in her arms, and thus a higher angular velocity. When she brings her arms from close to her body to away, she is now doing negative work from the same basic concept from before, but this work is now going to be due to her, so she transfers energy into the system of her rotation. As long as no external force from her body, say chemical energy, arises, and her moving is a constant velocity, the energy between these states are conserved.

Okay, so I don't like the sentences "If an ice skater her arms to her chest, she gains energy from the system of her rotation." and "So she transfers energy into the system of her rotation." If what I'm saying is true, what exactly is this "system of rotation" I'm trying to describe? When her arms go outward again, and her speed decreases, she clearly transferred energy, but to where, exactly?


1 Answer 1


So the skater's angular momentum never changes, because there is no external torque on her. When the arms come in, the moment of inertia goes down, hence the angular velocity goes up, per:

$I(t)\omega(t) = L(t) = L $ (=constant).

That's pretty much all you can say about angular momentum with this example.

Of course the energy of the system:

$E(t) = \frac{1}{2}I(t)\omega(t)^2 = \frac{L^2}{2I(t)}$

can not also be constant--it increases as the moment of inertia decreases--that is, as the arms come in.

What's going on with energy? The skater does work by pulling her arms in--the energy it takes to change her $I$ goes into a higher $\omega$.

Reformulating the problem with 2 masses on a spring would show that the spring energy matches the change in rotational energy (that's always murky when muscles are involved).

  • $\begingroup$ Where does the work end up going though when the skater pulls her arms away and in? $\endgroup$
    – sangstar
    Nov 18, 2017 at 22:12
  • $\begingroup$ @sangstar pulling is force times distance, that's work. Like I said, rethink the problem with masses and springs and it should be clearer than with muscles. $\endgroup$
    – JEB
    Nov 20, 2017 at 4:33

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