Question

A man of mass $100kg$ stands at the rim of a turntable of radius $2m$ and moment of inertia $4000kg{m}^2$ 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 $1m/s$ relative to earth. With what angular velocity and in what direction does the turntable rotate?

• A. The turntable rotates anticlockwise(in the direction of the man motion) with angular velocity $0.05rad/s$.
• B. The turntable rotates clockwise(opposite to the man motion) with angular velocity $0.1rad/s$.
• C. The turntable rotates clockwise(opposite to the man motion) with angular velocity $0.5rad/s$.
• D. The turntable rotates anticlockwise(in the direction of the man motion) with angular velocity $0.5rad/s$.

Answer 1

The correct option is A The turntable rotates anticlockwise(in the direction of the man motion) with angular velocity $0.05rad/s$.


By conservation of angular mometum of the man table system,
Intially when the man is stationary, $L_i=0$
The angular velocity of man ${\omega }_m=\frac{v}{R}=\frac{1}{2}$
$I_m=m{r}^2=100\times 2^2$
$L_f=I_m{\omega }_m+I_t{\omega }_t$
Using conservation of angular momentum
${\omega }_T=-\frac{I_m{\omega }_m}{I_t}$
Where, $I_m$ is the moment of inertia of man, ${\omega }_m$ angular velocity of man and $I_t$ moment of inertia of the table.
By putting all the values we get
${\omega }_T=-\frac{1}{20}=-0.05rad/s$
negatice sign indicates anticlockwise direction.

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