A circular platform is mounted on a frictionless vertical axle. Its radius $R = 2 m$ and its moment 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 is-

\frac{3 π}{2} s

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2 π s

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\frac{π}{2} s

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π s

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Solution

The correct option is B $2 π s$
By applying the law of conservation of angular momentum,
$I_{1} ω_{1} + I_{2} ω_{2} = 0$
$200 (\frac{1}{2}) + 200 ω_{2} = 0$
$ω_{2} = - \frac{1}{2}$
$ω_{R} = ω_{1} - ω_{2} = 1 r a d / s e c$
$T = \frac{2 π}{ω} = 2 π s e c$

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Similar questions

Q. 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 kg

m^{2} .

It is initially at rest. A 50 kg 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 is

A circular platform is mounted on a frictionless vertical axle. Its radius R=2m and its moment of inertia about the axle is $200 k g m^{2}$ . It is initially at rest. A 50 kg 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 compete one revolution with respect to disc is: