Problem — The skater who pulls herself in

Problem

A skater spins on the spot with her arms outstretched, then pulls them in to her body. (a) What happens to her angular velocity, and why? (b) What happens to her rotational kinetic energy? (c) Where does the change in energy, if any, “come from”?

Solution

(a) The angular velocity increases. No external torque acts about the vertical axis, so angular momentum $L = I\omega$ is conserved; by pulling in her arms the moment of inertia $I$ decreases, and to keep $L$ constant $\omega$ must grow.

(b) The rotational kinetic energy increases. Writing it as $E_\text{cin,rot} = \dfrac{L^2}{2I}$, with $L$ constant and $I$ decreasing, the energy grows.

(c) The extra energy comes from the muscular work the skater does to bring her arms in, overcoming the apparent centrifugal force that tends to push them outward.

$\ev{\omega \uparrow \;(L=I\omega \text{ conserved});\quad E_\text{cin,rot}\uparrow \;(=L^2/2I);\quad \text{energy from muscular work}}$

Connections

Topics: Dinamica rotazionale Concepts: Conservazione del momento angolare · Energia cinetica rotazionale