$A$ man stands on a rotating platform with his arms stretched horizontally,holding a $5\; kg$ weight in each hand. The angular speed of the platform is $30$ revolutions per minute. The man then brings his arms close to his body,with the distance of each weight from the axis changing from $90\; cm$ to $20\; cm$. The moment of inertia of the man together with the platform may be taken to be constant and equal to $7.6\; kg\; m^2$.
$(a)$ What is his new angular speed? (Neglect friction.)
$(b)$ Is kinetic energy conserved in the process? If not,from where does the change come about?

(A) Initial moment of inertia of the man and platform system $I_{man+platform} = 7.6\; kg\; m^2$.
Initial moment of inertia of the weights $I_{w1} = 2 \times m r_1^2 = 2 \times 5 \times (0.9)^2 = 8.1\; kg\; m^2$.
Total initial moment of inertia $I_i = 7.6 + 8.1 = 15.7\; kg\; m^2$.
Initial angular speed $\omega_i = 30\; rev/min$.
Final moment of inertia of the weights $I_{w2} = 2 \times m r_2^2 = 2 \times 5 \times (0.2)^2 = 0.4\; kg\; m^2$.
Total final moment of inertia $I_f = 7.6 + 0.4 = 8.0\; kg\; m^2$.
By the law of conservation of angular momentum,$I_i \omega_i = I_f \omega_f$.
$\omega_f = \frac{I_i \omega_i}{I_f} = \frac{15.7 \times 30}{8.0} = 58.875\; rev/min \approx 58.88\; rev/min$.
$(b)$ Kinetic energy is not conserved. The final kinetic energy $K_f = \frac{1}{2} I_f \omega_f^2$ is greater than the initial kinetic energy $K_i = \frac{1}{2} I_i \omega_i^2$. The increase in kinetic energy is due to the work done by the man in pulling his arms inward against the centrifugal force.