$(a)$ $A$ child stands at the centre of a turntable with his two arms outstretched. The turntable is set rotating with an angular speed of $40\; rev/min$. How much is the angular speed of the child if he folds his hands back and thereby reduces his moment of inertia to $2/5$ times the initial value? Assume that the turntable rotates without friction.
$(b)$ Show that the child's new kinetic energy of rotation is more than the initial kinetic energy of rotation. How do you account for this increase in kinetic energy?

(A) Initial angular velocity,$\omega_{1} = 40\; rev/min$.
Final angular velocity $= \omega_{2}$.
The moment of inertia of the boy with stretched hands $= I_{1}$.
The moment of inertia of the boy with folded hands $= I_{2}$.
The two moments of inertia are related as: $I_{2} = \frac{2}{5} I_{1}$.
Since no external torque acts on the system,the angular momentum $L$ is conserved.
Hence,$I_{2} \omega_{2} = I_{1} \omega_{1}$.
$\omega_{2} = \frac{I_{1}}{I_{2}} \omega_{1} = \frac{I_{1}}{\frac{2}{5} I_{1}} \times 40 = \frac{5}{2} \times 40 = 100\; rev/min$.
$(b)$ Final kinetic energy $E_{F} = \frac{1}{2} I_{2} \omega_{2}^{2}$.
Initial kinetic energy $E_{I} = \frac{1}{2} I_{1} \omega_{1}^{2}$.
$\frac{E_{F}}{E_{I}} = \frac{\frac{1}{2} I_{2} \omega_{2}^{2}}{\frac{1}{2} I_{1} \omega_{1}^{2}} = \frac{2}{5} \times \left(\frac{100}{40}\right)^{2} = \frac{2}{5} \times \left(\frac{5}{2}\right)^{2} = \frac{2}{5} \times \frac{25}{4} = 2.5$.
Therefore,$E_{F} = 2.5 E_{I}$.
The increase in rotational kinetic energy is due to the work done by the child in folding his arms,which is provided by his internal muscular energy.