A rotating table completes one rotation in 2 sec. and its moment of inertia is 100 kg-m2. A person of 50 kg. mass is standing at the centre of the rotating table. If the person moves 2m. from the centre, the angular velocity of the rotating table will be:

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Q.

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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a

$2 π \frac{r a d}{s e c}$

b

$\frac{π}{3} \frac{r a d}{s e c}$

c

$\frac{2 π}{3} \frac{r a d}{s e c}$

d

$\frac{π}{2} \frac{r a d}{s e c}$

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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