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A man of mass 100 kg stands at the rim o...

A man of mass `100 kg` stands at the rim of a turtable of radius `2 m` and moment of inertia `4000 kgm^(2)` mounted on a vertical frictionless shaft at its centre. The whole system is initially at rest. The man now walks along the outer edge of the turntable with a velocity of `1m//s` relative to the earth
a. With what angular velocity and in what direction does the turntable rotate?
b. Through what angle will it have rotated when the man reaches his initial position on the turntable?
c. Through what angle will it have rotated when the man reaches his initial position relative to the earth?

A

`36^@` in clockwise direction

B

`36^@` in anticlockwise direction

C

`72^@` in clockwise direction

D

`72^@` in anticlockwise direction

Text Solution

Verified by Experts

The correct Answer is:
A

If the man completes one revolution relative to the earth, then
`theta_(m)=2pi`
`Time=(2pi)/theta_(m)=(2pi)/0.5`
During this time the angular displacement of the table
`theta_(t)=omega_(t)` (time)`=-0.05xx((2pi)/0.5)`
`=-pi/5`radians
`theta_(t)=36^@` in clockwise direction.
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Knowledge Check

  • A circular platform is mounted on a vertical frictionless axle. Its radius is r = 2 m and its moment of inertia I = 200 kg m^2 . It is initially at rest. A 70 kg man stands on the edge of the platform and begins to walk along the edge at speed v_0 = 1 m s^-1 relative to the ground. The angular velocity of the platform is: a) 1.2 rad/s b) 0.4 rad/s c) 0.7 rad/s d) 2 rad/s

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