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)

$2 \pi s$

Answer 2

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
$T =\frac{2\pi}{\omega_r} = \frac{2\pi}{1} = 2\pi \,s$