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 through `2pi//11` radians anticlockwise
B. The table rotates through `4pi//11` radians clockwise
C. The table rotates through `4pi//11` radians anticlock-wise
D. The table rotates through `27pi//11 radians clockwise

Answer

D

Explanation

If the man completes one revolution relative to the table, then
`theta_(mt)=2pi`
`implies 2pi=theta_(m)-theta_(t)`
`2pi=theta_(m)t-omega_(t)t` (where `t` is the time taken)
`t=(2pi)/(omega_(m)-omega_(t))=(2pi)/(0.5-(-0.05))=(2pi)/0.55s`
Angular displacement of the table is
` theta_(t)=omega_(r)t=-0.05xx((2pi)/0055)`
`=-((2pi)/11)` radians
The table rotates through `2pi//11` radians clockwise.