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

 

Answers (1)
V Vakul

As discussed in

Law of conservation of angular moment -

vec{	au }= frac{vec{dL}}

- wherein

If net torque is zero

i.e. frac{vec{dL}}= 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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