A circular platform is mounted on a frictionless vertical axle. Its radius $R=2 ext{ m}$ and moment of inerti

Question

A circular platform is mounted on a frictionless vertical axle. Its radius $R=2	ext{ m}$ and moment of inertia about the axle is $200	ext{ kg m}^2$. It is initially at rest. A $50	ext{ kg}$ man stands on the edge of the platform and begins to walk along the edge at the speed of $1	ext{ ms}^{-1}$ relative to the ground. Time taken by the man to complete one revolution is:

Accepted Answer

Since no external torque acts on the system (man + platform), the total angular momentum is conserved. The initial angular momentum is zero.
Let the angular velocity of the platform be $\omega_p$. The velocity of the man relative to the ground is $v = 1	ext{ m/s}$. His angular velocity relative to the ground is $\omega_m = \frac{v}{R} = \frac{1}{2} = 0.5	ext{ rad/s}$.
According to the law of conservation of angular momentum:
$L_{	ext{man}} + L_{	ext{platform}} = 0$
$m v R + I \omega_p = 0$
$(50)(1)(2) + 200 \omega_p = 0 \implies 100 + 200 \omega_p = 0 \implies \omega_p = -0.5	ext{ rad/s}$.
The negative sign indicates that the platform rotates in the opposite direction to the man's motion.
The angular velocity of the man relative to the platform is:
$\omega_{	ext{rel}} = \omega_m - \omega_p = 0.5 - (-0.5) = 1.0	ext{ rad/s}$.
The time taken by the man to complete one revolution on the platform is:
$T = \frac{2\pi}{\omega_{	ext{rel}}} = \frac{2\pi}{1.0} = 2\pi	ext{ s}$.

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