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A rotating table completes one rotation in $2 s e c$ . and its moment of inertia is $100 k g - m^{2}$ . A person of $50 k g$ . mass is standing at the centre of the rotating table. If the person moves $2 m$ . from the centre, the angular velocity of the rotating table will be:

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$\frac{π}{3} \frac{r a d}{s e c}$

$\frac{π}{2} \frac{r a d}{s e c}$

$\frac{2 π}{3} \frac{r a d}{s e c}$

$2 π \frac{r a d}{s e c}$

answer is A.

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Detailed Solution

$ω_{1} = \frac{2 π}{2} = π \frac{rad}{s} .$

By conservation of Angular momentum

$I_{1} ω_{1} = I_{2} ω_{2} \Rightarrow 100 π = [100 + 50 {(2)}^{2}] ω \Rightarrow ω = \frac{π}{3}$

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