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?

Options

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

Answer

C

Explanation

By conservation of anglar mometum on the man table system,
`L_(i)=L_(f)`
`0+0=I_(m)omega_(m)=I_(t)omega_(t)`
`impliesomega_(t)=(I_(m)omega_(m))/I_(t)` where `omega_(m)=v/r=1/2rad//s`
`implies omega_(t)=-(100(2)^(2)xx(1/2))/4000`
`implies omega_(m)=v/r=1/2rad//s`
`implies omega_(t)=-1/20rad//s`
Thus, the table rotates clockwise (opposite to the man) with angular velocity `0.05 rad//s.`