System of Particles and Rotational Motion - Result Question 33
33. A circular platform is mounted on a frictionless vertical axle. Its radius $R=2 m$ and its moment of inertia about the axle is $200 kgm^{2}$. 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) $\frac{3 p}{2} s$
(c) $2 \pi s$
(d) $\frac{\pi}{2} s$
[2012M]
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Answer:
Correct Answer: 33. (c)
Solution:
- (c) According to the conservation of angular momentum,
When no external torque acts on system, the angular momentum does not change.
$L _{\text{system }}=L _{\text{man }}+L _{\text{platform }}=$ constant
$L _{\text{man }}=-L _{\text{platform }}$
$L _{\text{man }}=mvR$
where $m$ is mass of man, $v$ is speed of man relative to ground, $R$ is the radius of platform,
$L _{\text{platform }}=I \omega$
where I is moment of inertia, $\omega$ is angular velocity.
$\Rightarrow \omega=\frac{L _{\text{platform }}}{I}=-\frac{L _{\text{man }}}{I}=\frac{-mvR}{I}$
Speed of man relative to platform is
$v _{\text{platform }}=v-\omega R=v+\frac{m v R}{I} R$
$ =v[1+\frac{mR^{2}}{I}] $
Time taken by man to complete one revolution is,
$ \begin{aligned} & t=\frac{2 \pi R}{v _{\text{platform }}}=\frac{2 \pi r}{v[1+\frac{mR^{2}}{I}]}=2 \times \frac{2}{1(1+1)} \pi \\ & \Rightarrow t=2 \pi \text{ second } \end{aligned} $
