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. `0.5 rad//sec`
B. `0.1 rad //sec`
C. `0.05 rad//sec`
D. `0.2 rad//sec`

Answer

C

Explanation

By conservation of angular momentum on the mantable system,
`vec(L)_(i)=vec(L)_(f)` or `0+0=I_(m)omega_(m)+I_(t)omega_(t)`
`omega_(t)=-(I_(m)omega_(m))/(I_(1))` where `omega_(m)=v/r=1/2 rad//s`
`omega_(t)=-100(2)^(2)xx(1//2)/4000=-1/20 rad//s`
Thus the table rotats clockwise (opposite to man) with angular velocity `0.05 rad//s`.