Spinning Skater

Combining the equations for angular momentum, $L = I\omega$ , and rotational kinetic energy, $K = \tfrac12I \omega^2$ , we get $K = \frac{L^2}{2I}.$ The angular moment is conserved, because pulling in the arms involves only internal forces. Thus the kinetic energy $K$ is inversely proportional to the moment of inertia $I$ .

Since $I' = \tfrac35I$ , it follows that $K'=\tfrac53K$ . This means that the kinetic energy increases by $\tfrac23 K$ ; this increase is equal to the amount of work done. Thus the ratio is $\boxed{2:3}$ .

Let initial angular velocity be $\omega_i$ . Then, by conservation of angular momentum, $\omega_f = \frac {5}{3}\omega_i$ (because $I_f = \frac {3}{5}I_i$ ). Also, we know that $W = \Delta K.E_{rot}$ . So we have -

$\frac {W}{K} = \frac {\Delta K.E_{rot}}{K_i} = \frac {\frac {3}{10}I_i \cdot \frac {25}{9}\omega_i^{2} - \frac {1}{2}I_i\omega_i^{2}}{\frac {1}{2}I_i\omega_i^{2}} = \frac {2}{3}$