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 rads^(-1)`
B. `0.4rads^(-1)`
C. `2.0rads^(-1)`
D. `0.7rads^(-1)`

Answer

D

Explanation

Net external torque is zero. Therefore angular mometnumof system will remain conserved i.e, `L_(i)=L_(f)`
Initial angular momentum `L_(i)=L_(f)`
Initial angular momentum `L_(i)=0`
`:.` final angular mometum should be zero.
or angular momentum of man `=` angular momentum of platform is opposite direction
or `mv_(0)r=Iomega`
`omega=(mv_(0)r)/I=((70)(1.0)(2))/200`
`omega=0.7rad//s`