Question

A circular platform is mounted on a frictionless vertical axle. Its radius `R = 2m` 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 at the platform and begins to walk along the edge at the speed of `1ms^(-1)` relative to the ground. Time taken by the man to complete one revolution is :
A. `pi s`
B. `(3pi)/(2) s`
C. `2 pi s`
D. `(pi)/(2) s`

Answer 1

Correct Answer - C
Here, `R = 2m, I = 200 kg m^(2), u = 0,M = 50 kg`
`v = 1 m//s, t = ?`
Using principle of conservation of angular
momentum, `L_(t) = L_(0)`
`M v R - I omega = 0, omega = (MvR)/(I) = (50 xx 1 xx 2)/(200) = (1)/(2)`
If `r` is time taken by man to complate one revolution, then
`(v + R omega) t = 2piR`
`(1+2 xx (1)/(2))t = 2pi xx 2`
`t = 2pi sec`