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Question

A man of mass 100 kg stands at the rim of a turntable of radius 2 m and moment of inertia 4000 kgm2 mounted on a vertical frictionless shaft at its center. The whole system is initially at rest. The man now walks along the outer edge if the turntable(anticlockwise) with a velocity of 1 m/s relative to earth. With what angular velocity and in what direction does the turntable rotate?

A
The turntable rotates anticlockwise(in the direction of the man motion) with angular velocity 0.05 rad/s.
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B
The turntable rotates clockwise(opposite to the man motion) with angular velocity 0.1 rad/s.
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C
The turntable rotates clockwise(opposite to the man motion) with angular velocity 0.5 rad/s.
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D
The turntable rotates anticlockwise(in the direction of the man motion) with angular velocity 0.5 rad/s.
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Solution

The correct option is A The turntable rotates anticlockwise(in the direction of the man motion) with angular velocity 0.05 rad/s.

By conservation of angular mometum of the man table system,
Intially when the man is stationary, Li=0
The angular velocity of man ωm=vR=12
Im=mr2=100×22
Lf=Imωm+Itωt
Using conservation of angular momentum
ωT=ImωmIt
Where, Im is the moment of inertia of man, ωm angular velocity of man and It moment of inertia of the table.
By putting all the values we get
ωT=120=0.05 rad/s
negatice sign indicates anticlockwise direction.

For detailed soultion watch next video.

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