A child stands at the centre of a frictionless turntable with his two arms outstretched. The turntab

Question

A child stands at the centre of a frictionless turntable with his two arms outstretched. The turntable is rotating with an initial angular speed. If he folds his hands back and thereby reduces his moment of inertia to $2/5$ times the initial value, what is the ratio of his new kinetic energy of rotation to the initial kinetic energy?

Professor PC Thomas Classes · Answer

Since no external torque acts on the system, the angular momentum $L$ is conserved. The kinetic energy of rotation is given by $K = \frac{L^2}{2I}$. With the new moment of inertia $I' = \frac{2}{5}I$, the new kinetic energy $K' = \frac{L^2}{2(\frac{2}{5}I)} = \frac{5}{2} \left(\frac{L^2}{2I}\right) = 2.5K$. Therefore, the ratio is $2.5$.