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. Through what angle will it have rotated when the man reaches his initial position on the turntable?

• A. The table rotates through $\frac{2\pi }{11}$ radians clockwise
• B. The table rotates through $\frac{4\pi }{11}$ radians clockwise
• C. The table rotates through $\frac{4\pi }{11}$ radians anticlockwise
• D. The table rotates through $\frac{2\pi }{11}$ radians anticlockwise

Answer 1

The correct option is D The table rotates through $\frac{2\pi }{11}$ radians anticlockwise

By conservation of angular mometum on the man table system,
Intially when the man is stationary, $L_i=0$
The angular velocity of the man w.r.t ground is ${\omega }_M=\frac{v}{R}=\frac{1}{2}$
$L_f=I_m{\omega }_m+I_T{\omega }_T$
$L_f=100\times 2^2\times \frac{1}{2}+4000\times {\omega }_T$
Using conservation of angular momentum
${\omega }_T=-\frac{1}{20}=-0.05rad/s$
Since the man is coming back to his original position w.r.t turntable
If ${\theta }_{MT}=2\pi$
${\theta }_{MT}={\theta }_M-{\theta }_T$
$={\omega }_{M}t-{\omega }_{T}t=2\pi$
$\therefore t=\frac{40\pi }{11}s$
The angle turned by the turntable when the man reaches the same point on table is
${\theta }_T=t\times {\omega }_T$
$=-\frac{2\pi }{11}$

For detailed solution watch next video.