# Spinning Ice Skater and Coriolis Force

among the explanations for the Coriolis force I've seen is a mention of the ice skater's rotational acceleration by pulling arms in radially.

I've always understood the spinning ice skater as conservation of angular momentum. However, I do see that the Coriolis force cross product formulation of -2m omega x v vector (cross product of rotational velocity and the velocity of pulling the arm in ) yields a vector that is tangential to the arm's spiral sweep and should cause rotational acceleration.

Can someone confirm that the rotating ice skater system is a satisfactory example of this. I've already read through an analysis of forces in the inertial frame that is posted at Why does a ballerina speed up when she pulls in her arms?

which seems pretty thorough but doesn't mention Coriolis so I'm not sure that my perspective is correct.

Thanks,

• My view of this is that the author of the excellent answer you linked concentrated on a very full account of the conservation of angular momentum, that he simply didn't think of the viewpoint from a reference frame that incorporated the Coriolis Force. I would think, obviously enough I suppose, if you could "map" every aspect of the ice skater to a standard CF example, you could confirm/ refute it. My point is that just because the author didn't mention it probably has no significance. – user108787 Nov 3 '16 at 0:09
• Right. Angular momentum is conserved. The term "coriolis force" is just the name for the force she feels making her speed up.. – Mike Dunlavey Nov 3 '16 at 0:14

When the rotating skater pulls his arm in close to the body, his moment of inertia $I$ decreases. As the angular momentum $L=I\omega$, where $\omega$ is the angular velocity, is conserved, this means that the angular velocity $\omega$ has to increase so that $L$ stays the same. This has nothing to do with the Coriolis force.