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.

Options

A. `1.2 rad s^(-1)`
B. `0.4 rad s^(-1)`
C. `0.7 rad s^(-1)`
D. `2 rad s^(-1)`

Answer

C

Explanation

As the system is initially at ret, therefore, initial angualr momentum, `L_(i)=0`.
According to the law of conservation of angular momentum, final angular momentum, `L_(f)=0`.
According to the law of conservation of angular momentum, final angular momentum , `L_(f)=0`
`therefore` Angular momentum of man = angular momentum of platform in opposite direction
i.e., `mv_(0)r= Iomega`
or `omega= (mv_(0)r)/(I) = (70(1.0)(2))/(200) = 0.7 rad s^(-1)`