**Question:
**

A man stands on a rotating platform, with his arms stretched horizontally holding a $5 \mathrm{~kg}$ 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 $90 \mathrm{~cm}$ to $20 \mathrm{~cm}$. The moment of inertia of the man together with the platform may be taken to be constant and equal to $7.6 \mathrm{~kg} \mathrm{~m}^{2}$.

(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?

**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.

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