A man stands on a rotating platform,

Solution:

(a) 58.88 rev/min (b) No

(a)Moment of inertia of the man-platform system $=7.6 \mathrm{~kg} \mathrm{~m}^{2}$

Moment of inertia when the man stretches his hands to a distance of $90 \mathrm{~cm}$ :

$2 \times m r^{2}$

$=2 \times 5 \times(0.9)^{2}$

$=8.1 \mathrm{~kg} \mathrm{~m}^{2}$

Initial moment of inertia of the system, $I_{i}=7.6+8.1=15.7 \mathrm{~kg} \mathrm{~m}^{2}$

Angular speed, $\omega_{1}=300 \mathrm{rev} / \mathrm{min}$

Angular momentum, $L_{\mathrm{i}}=I_{\mathrm{i}} \omega_{\mathrm{i}}=15.7 \times 30$ $\ldots(i)$

Moment of inertia when the man folds his hands to a distance of $20 \mathrm{~cm}$ :

$2 \times m r^{2}$

$=2 \times 5(0.2)^{2}=0.4 \mathrm{~kg} \mathrm{~m}^{2}$

Final moment of inertia, $I_{\mathrm{f}}=7.6+0.4=8 \mathrm{~kg} \mathrm{~m}^{2}$

Final angular speed $=\omega_{r}$

Final angular momentum, $L_{\mathrm{i}}=I_{\mathrm{f}} \omega_{\mathrm{f}}=0.79 \omega_{\mathrm{f}} \ldots$ (ii)

From the conservation of angular momentum, we have:

$I_{\mathrm{i}} \omega_{\mathrm{i}}=I_{\mathrm{f}} \omega_{\mathrm{f}}$

$\therefore \omega_{\mathrm{f}}=\frac{15.7 \times 30}{8}=58.88 \mathrm{rev} / \mathrm{min}$

(b)Kinetic energy is not conserved in the given process. In fact, with the decrease in the moment of inertia, kinetic energy increases. The additional kinetic energy comes from the work done by the man to fold his hands toward himself.