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Question

A circular disc of mass 300 gm and radius 20 cm can rotate freely about a vertical axis passing through is center O. A small insect of mass 100 gm is initially at a point A on the rim. The insect initially stationary starts walking from rest along the rim of disc with such a time varying relative velocity that the disc rotates in the opposite direction with a constant angular acceleration = 2πrad/s2. After some time T, the insect is back at the point A

. What is the time taken by insect to reach the original position?


A

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B

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C

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D

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Solution

The correct option is A


Let the force applied by insect on the disc be F.

It is clear from the contact,

R×F=Idαd

R×F=MR22×2π

F=πMdR ..........(1)

The same amount of force will be applied to the insect by the disc,

Considering its torque about the centre of disc,

πMdR×R=Iiαii=insect

MiR2αi=MdπR2

αi=MdMi×π{MdMi=3}

αi=3π

now with respect to disc's frame of reference, αid=5π {Both α are opposite in direction}

now relative angular displacement = 2π

2π=12αidt2t=54=52s


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