Question

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

• A. $ \pi s $
• B. $ \frac{3 \pi}{2} s $
• C. $ 2 \pi s $
• D. $ \frac{\pi}{2} s $

Answer 1

Answer: (C)

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 = Angular momentum of man is in opposite direction of platform.
i.e $ mvR = I\omega$
or $ \omega = \frac{mvR}{I} = \frac{50 \times 1 \times 2}{200} = \frac{1}{2}\, rad\, s^{-1}$
Angular velocity of man relative to platform is
$ \omega_r = \omega + \frac{v}{R} = \frac{1}{2} = 1 \,rad \,s^{-1} $
Time taken by the man to complete one revolution is
$ T =\frac{2\pi}{\omega_r} = \frac{2\pi}{1} = 2\pi \,s $