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A circular platform is mounted on a vertical frictionless axle. Its radius is r= $2 m$ and its moment of inertia $I = 200 kg m^{2}$ . It is initially at rest. A $70 kg$ man stands on the edge of the platform and begins to walk along the edge at speed $v_{0} = 1 m s^{- 1}$ relative to the ground. The angular velocity of the platform is

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$0.4 {rads}^{- 1}$

$0.7 {rads}^{- 1}$

$1.2 {rads}^{- 1}$

$2 {rads}^{- 1}$

answer is C.

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Detailed Solution

As the system is initially at rest, therefore, initial angular momentum, $L_{i} = 0 .$

According to the law of conservation of angular momentum, final angular momentum, $L_{f} = 0$

$∴$ Angular momentum of man = angular momentum of platform in opposite direction

i.e., $m v_{0} r = I ω$

or $ω = \frac{m v_{0} r}{I} = \frac{70 (1.0) (2)}{200} = 0.7 {rads}^{- 1}$

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