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

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 $50 \mathrm{~kg}$ man stands on the edge of the platform and begins to walk along the edge at the speed of $1 \mathrm{~ms}^{-1}$ relative to the ground. Time taken by the man to complete one revolution is
  1. $\pi \sec$
  2. $\frac{3 \pi}{2} \sec$
  3. $2 \pi \mathrm{sec}$
  4. $\frac{\pi}{2} \sec$

Solution

From conservation of angular momentum $\begin{aligned} I \omega & =m v r \\ 200 \times \omega & =50 \times 2 \times 1 \\ \omega & =\frac{1}{2} \mathrm{rad} / \mathrm{s} \end{aligned}$ $\begin{aligned} \quad v & =r \omega=1 \mathrm{~m} / \mathrm{s} \\ \therefore \quad T & =\frac{2 \pi r}{1-(-1)}=\frac{2 \pi r}{2}=\pi r=2 \pi \mathrm{s} \end{aligned}$

Asked in: NEET 2012 (Mains)

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