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 ms^(-1)` relative to the ground . Time taken by the man to complete one revolution is

Options

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

Answer

C

Explanation

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`.
`therefore` Angular momentum of man on platform = Angular momentum of system i.e., `mvR = I omega`
or `omega = (mv R)/(I) = (50 xx 1 xx 2)/(200) = (1)/(2)` rad `s^(-1)`
Angular velocity of man relative to platform is
`omega_(r) = omega + (v)/(R) = (1)/(2) + (1)/(2) = 1 rad s^(-1)`
Time taken by the man to complete one revolution is `T = (2pi)/(omega_(r)) = (2pi)(1) = 2 pi s`