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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
$\frac{\pi}{2}$ s
B
$\pi$ s
C
$\frac{3\pi}{2}$ s
D
$2\pi$ s
Detailed Solution
Angular velocity of the man relative to the ground: $\omega_m = \frac{v}{R} = \frac{1}{2}$ rad/s
Moment of inertia of the man about the axle: $I_m = mR^2 = 50\times(2)^2 = 200$ kg $m^2$
The system starts from rest and there is no external torque, so the platform turns the opposite way: $I_p\omega_p = I_m\omega_m$
$200\times\omega_p = 200\times\frac{1}{2} \Rightarrow \omega_p = \frac{1}{2}$ rad/s (opposite direction)
Angular velocity of the man relative to the platform: $\omega = \omega_m - (-\omega_p) = \frac{1}{2} + \frac{1}{2} = 1$ rad/s
Time for one revolution: $T = \frac{2\pi}{\omega} = 2\pi$ s
Moment of inertia of the man about the axle: $I_m = mR^2 = 50\times(2)^2 = 200$ kg $m^2$
The system starts from rest and there is no external torque, so the platform turns the opposite way: $I_p\omega_p = I_m\omega_m$
$200\times\omega_p = 200\times\frac{1}{2} \Rightarrow \omega_p = \frac{1}{2}$ rad/s (opposite direction)
Angular velocity of the man relative to the platform: $\omega = \omega_m - (-\omega_p) = \frac{1}{2} + \frac{1}{2} = 1$ rad/s
Time for one revolution: $T = \frac{2\pi}{\omega} = 2\pi$ s