Question

The radius R of collapsing spinning star drops to 1/3 its initial value wheres the mass M remains unchanged. The star spins as a rigid body and its rotational inertia I =2/5 MR^2. The ratio of the new rotational kinetic energy to the initial rotational kinetic energy is

Answer 1

From the conservation of angular momentum:

$$I\omega=I'\omega'$$

$$\frac{2}{5}MR^2\omega=\frac{1}{3^2}\frac{2}{5}MR^2\omega'$$

$$\omega'=9\omega$$

$$I'=\frac{1}{9}I$$

The ratio of the new rotational kinetic energy to the initial rotational kinetic energy is

$$\frac{0.5I'\omega'^2}{0.5I\omega^2}=9^2\frac{1}{9}=9$$