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 with respect to disc is

Options

A. `pi s`
B. `(3pi)/2 s`
C. `2pi s`
D. `pi/2 s`

Answer

C

Explanation

Using angular momentum conservation,
`L_i=0 , L_f= mvR-I omega`, so , `mvR = I omega`
`omega=(mvR)/I=(50xx1xx2)/200=1/2`
For one complete revolution,
`(v+omegaR)t=2piR`
`(1+1/2xx2)t=2pixx2 rArr t=2pi s`.