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?

Explanation

As for the man and table system, initial angular momentum is zero and it will remain zero as no external torque about rotation axis. When the man starts moving in the anticlockwise direction to make angular momentum zero, the table will start rotating in the clockwise direction.

a. By conserving of angular momentum on the man -table system
`L_(i)+L_(f)`
`implies0+0=I_(m)omega_(m)+I_(t)omega_(t)`
`impliesomega_(t)=(-I_(m)omega_(m))/I_(r)`, where `omega_(m)=v/r=1/2rad//s`
`=(100(2)^(2)xx(1/2))/4000`
(negative sign represents clockwise direction)
`impliesomega_(m)=v/r=1/2rad//simpliesomega_(t)=-1/20rad//s`
Thus, the table rotates clockwise (opposite to the man) with angulr velocity `0.05 rad//s`.
b. If the man completes one revolution relative to the table. Angular displacement of the man w.r.t table is
`theta_(mt)=2piimplies2pi=theta_(m)-theta_(t)`
`2pi=omega_(m)t-momega_(r)t` (where `t` is the time taken)
`t=(2pi)/(omega_(m)-omega_(t))=(2pi)/(0.5-(-0.05))=(2pi)/0.05`
`implies` Angular displacement of table is
`theta_(t)=omega_(t)t=0.05xx((2pi)/0.55)=-((2pi)/11)rad`
The table rotates through `2pi//11` rad clockwise.
c.If the man completes one revolution relative to the earth, then
`theta_(m)=2pi`
Time`=2pi/omega_(m)=((2pi)/0.5)`
During this time, angular displacement of the table
`theta_t=omega_t(time)=-0.05xx((2pi)/0.5)`
`=-pi/5radians =36^@ ` clockwise direction