Calculating the Angular Momentum of a Spinning Skater

In the first part of the problem, we are looking for the skater’s angular velocity $\omega \prime$ after she has pulled in her arms. To find this quantity, we use the conservation of angular momentum and note that the moments of inertia and initial angular velocity are given. To find the initial and final kinetic energies, we use the definition of rotational kinetic energy given by

Because torque is negligible (as discussed above), the conservation of angular momentum given in $\mathrm{I\omega }=I\prime \omega \prime$ is applicable. Thus,

or

Solving for $\omega \prime$and substituting known values into the resulting equation gives

Rotational kinetic energy is given by

The initial value is found by substituting known values into the equation and converting the angular velocity to rad/s:

The final rotational kinetic energy is

Substituting known values into this equation gives

In both parts, there is an impressive increase. First, the final angular velocity is large, although most world-class skaters can achieve spin rates about this great. Second, the final kinetic energy is much greater than the initial kinetic energy. The increase in rotational kinetic energy comes from work done by the skater in pulling in her arms. This work is internal work that depletes some of the skater’s food energy.