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A circular platform is mounted on a frictionless vertical axle. Its radius $\mathrm{R=2 \mathrm{~m}}$ and its moment of inertia about the axle is $\mathrm{200 \mathrm{kgm}^2}$ . It's initally at rest. A $\mathrm{50 \mathrm{~kg}}$ man stands on the edge of the plate form and begins to walk along the edge at the speed of $\mathrm{1 \mathrm{~m} / \mathrm{s}}$ relative to the ground. Time taken by the man to complete one revolution is -

Option: 1
$\mathrm{\pi s}$

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

Option: 3
$\mathrm{2 \pi \: \mathrm{s}}$

Option: 4
$\mathrm{\frac{\pi}{2} s}$

Answers (1)

best_answer

According to the conservation of momentum -

$\mathrm{L_i=L_f }$

$\mathrm{m v R-I \omega=0 }$

$\mathrm{m vR=I \omega}$ ..............(1)

put the given all values-

$\mathrm{50 \times 1 \times 2=200 \times \omega }$

$\mathrm{\omega=1 / 2 }$

Now, $\mathrm{ (v+\omega R) t=2 \pi R }$

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

$\mathrm{ 2 t=2 \pi \times 2 }$

$\mathrm{ t=2 \pi \: \mathrm{sec} }$

Hence option 3 is correct.

Posted by

Devendra Khairwa

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