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Question

A man of mass 100 kg standsat 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. Through what angle will it have rotated when the man reaches his initial position relative to earth?

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Solution

The correct option is **B** −π5

The angle covered by man w.r.t ground is 2π (as he comes back to the original position)

θM=2π

ωM=vR=12

Let the time taken by the man to come back to the original position w.r.t ground be t

t=θMωM=4π s

By conservation of angular mometum on the man table system,

Intially when the man is stationary, Li=0

Lf=Imωm+ITωT

Lf=100×22×12+4000×ωT

Using conservation of angular momentum

ωT=−120=−0.05 rad/s

The angular velocity of turn table is given by

θT=ωT×t=−π5

For detailed solution watch next video.

The angle covered by man w.r.t ground is 2π (as he comes back to the original position)

θM=2π

ωM=vR=12

Let the time taken by the man to come back to the original position w.r.t ground be t

t=θMωM=4π s

By conservation of angular mometum on the man table system,

Intially when the man is stationary, Li=0

Lf=Imωm+ITωT

Lf=100×22×12+4000×ωT

Using conservation of angular momentum

ωT=−120=−0.05 rad/s

The angular velocity of turn table is given by

θT=ωT×t=−π5

For detailed solution watch next video.

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