A circular platform is mounted on a frictionless vertical axle. Its radius $R=2 \mathrm{~m}$ and its

Question

A circular platform is mounted on a frictionless vertical axle. Its radius $R=2 \mathrm{~m}$ and its moment of inertia about the axle is $200 \mathrm{~kg} \mathrm{~m}^{2}$. 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 $1 \mathrm{~m} \mathrm{~s}^{-1}$ relative to the ground. Time taken by the man to complete one revolution is\

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(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, $L_{f}=0$.\ $	herefore$ Angular momentum of platform $=$ Angular momentum of man in opposite direction of platform.\ i.e., $m v R=I \omega$\ or $\omega=\frac{m v R}{I}=\frac{50 	imes 1 	imes 2}{200}=\frac{1}{2} \mathrm{rad} \mathrm{s}^{-1}$\ Angular velocity of man relative to platform is $$ \omega_{r}=\omega+\frac{v}{R}=\frac{1}{2}+\frac{1}{2}=1 \mathrm{rad} \mathrm{~s}^{-1} $$ Time taken by the man to complete one revolution is $$ T=\frac{2 \pi}{\omega_{r}}=\frac{2 \pi}{1}=2 \pi \mathrm{~s} $$