# A circular platform is mounted on a frictionless vertical axle. Its radius R=2m 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 adge at the speed of 1 $ms^{-1}$ relative to the ground. Time taken by the man to complete one revolution is Option 1) $\pi\:s$ Option 2) $\frac{3\pi}{2}\: s$ Option 3) $2\pi\: s$ Option 4) $\frac{\pi}{2}\: s$

V Vakul

As discussed in

Law of conservation of angular moment -

$\vec{\tau }= \frac{\vec{dL}}{{dt}}$

- wherein

If net torque is zero

i.e. $\frac{\vec{dL}}{{dt}}= 0$

$\vec{L}= constant$

angular momentum is conserved only when external torque is zero.

$L_{i}=L_{f}$

$L_{i}=0$    (initial moment)

$L_{f}=mvR-I \omega$ (final moment)

$mvR-I \omega=0$

$mvR=I \omega\:\:\:\:\:\:\Rightarrow\omega=\frac{1}{2}$

$(v+ \omega R)t=2 \pi R$

$t\left( 1+ \frac{1}{2}\times 2 \right )=2 \pi \times 2$

$\Rightarrow t=2 \pi s$

Option 1)

$\pi\:s$

This option is incorrect.

Option 2)

$\frac{3\pi}{2}\: s$

This option is incorrect.

Option 3)

$2\pi\: s$

This option is correct.

Option 4)

$\frac{\pi}{2}\: s$

This option is incorrect.

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