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A circular platform is mounted on a frictionless vertical axle. Its radius $R = 2 m$ and its moment of inertia about the axle is $200 k g m^{2}$ . It is initially at rest. A $50 k g$ man stands on the edge of the platform and begins to walk along the edge at the speed of $1 m s^{- 1}$ relative to the ground. Time taken by the man to complete one revolution w.r.t to the disc is :-

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$2 π s$

$\frac{π}{2} s$

$π s$

$\frac{3 π}{2} s$

answer is A.

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Detailed Solution

$M I$ of platform $= 200 k g - m^{2},$

$M I$ of man $= m R^{2} = 200 k g - m^{2}$

For system (platform + man) by using $C O A M$

$I ω_{0} = m v R \Rightarrow ω_{0} = \frac{50 \times 1 \times 2}{200} = \frac{1}{2} \frac{r a d}{s}$

Angular velocity of man w.r.t. platform (rotation of platform is in opposite direction to the moment of the man) $= \frac{v}{R} + ω_{0} = \frac{1}{2} + \frac{1}{2} = 1 \frac{r a d}{s}$

Time taken by the man to complete one revolution

$= \frac{2 π rad}{1 \frac{r a d}{s}} = 2 π s$

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