A child stands at the centre of a turntable with his two arms outstretched. The turntable is set rot

Question

A child stands at the centre of a turntable with his two arms outstretched. The turntable is set rotating with an angular speed of 40 rev/min. If he folds his hands back, reducing his moment of inertia to 2/5 times the initial value, what is the new angular speed of the child? (Assume the turntable rotates without friction)

Professor PC Thomas Classes · Answer

Since there is no external torque, angular momentum is conserved: $I_1\omega_1 = I_2\omega_2$. Given $I_2 = \frac{2}{5}I_1$ and $\omega_1 = 40	ext{ rev/min}$. Substituting these values gives $I_1 	imes 40 = (\frac{2}{5}I_1) 	imes \omega_2 \implies \omega_2 = \frac{40 	imes 5}{2} = 100	ext{ rev/min}$.