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 m2 . 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 ms-1 relative to the ground. Time taken by the man to complete one revolution is

Question

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 m2 . 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 ms-1 relative to the ground. Time taken by the man to complete one revolution is (A)  π s  (B)  (3 π/2) s  (C)  2 π s  (D)  (π/2) s

Tardigrade Admin · Accepted Answer

As the system is initially at rest, therefore, initial angular momentum

L_{i} = 0

According to the principle of conservation of angular momentum, final angular momentum,

L_{f} = 0

∴

Angular momentum = Angular momentum of man is in opposite direction of platform.
i.e

m v R = I ω

or

ω = \frac{m v R}{I} = \frac{50 \times 1 \times 2}{200} = \frac{1}{2} r a d s^{- 1}

Angular velocity of man relative to platform is

ω_{r} = ω + \frac{v}{R} = \frac{1}{2} = 1 r a d s^{- 1}

Time taken by the man to complete one revolution is

T = \frac{2 π}{ω _{r}} = \frac{2 π}{1} = 2 π s

$π s$

$\frac{3 π}{2} s$

$2 π s$

$\frac{π}{2} s$

πs

23π​s

2πs

2π​s

$π s$

$\frac{3 π}{2} s$

$2 π s$

$\frac{π}{2} s$