Question

A man of mass 100 kg stands at the rim of a turntable of radius 2m and moment of inertia 4000$kg-m^2$ mounted on a vertical frictionless shaft at its center. The whole system is initially at rest. The man now walks along the outer edge of a turntable with a velocity of 1 m/s relative to the earth. With what angular velocity and in what direction does the turntable rotates?

• A. $\frac{1}{40}rad/s$
• B. $\frac{1}{80}rad/s$
• C. $\frac{1}{20}rad/s$
• D. $\frac{1}{10}rad/s$

Answer 1

The correct option is C

$\frac{1}{20}rad/s$

Let the man be moving anti-clockwise

(a)By conservation of angular momentum on the man-table system

$\Rightarrow L_1=L_f\Rightarrow 0+0=I_m{\omega }_m+I_t{\omega }_t\Rightarrow {\omega }_t=-\frac{I_m{\omega }_m}{I_t}where{\omega }_m=\frac{V}{r}=\frac{1}{2}rad/s\Rightarrow {\omega }_t=-\frac{1002^2\times \left(\frac{1}{2}\right)}{4000}$

$\Rightarrow {\omega }_m=\frac{V}{r}=\frac{1}{2}rad/s{\omega }_t=-\frac{1}{20}rad/s$ Thus the table rotates clock-wise (opposite to man) with angular velocity 0.05 rad/s.