Question

A man of mass $100kg$ standsat the rim of a turntable of radius 2 $m$ 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 relative to earth?

• A. $\frac{\pi }{5}$
• B. $-\frac{\pi }{5}$
• C. $\frac{2\pi }{3}$
• D. $-\frac{2\pi }{3}$

Answer 1

The correct option is B $-\frac{\pi }{5}$

The angle covered by man w.r.t ground is $2\pi$ (as he comes back to the original position)
${\theta }_M=2\pi$
${\omega }_M=\frac{v}{R}=\frac{1}{2}$
Let the time taken by the man to come back to the original position w.r.t ground be $t$

$t=\frac{{\theta }_M}{{\omega }_M}=4\pi s$
By conservation of angular mometum on the man table system,
Intially when the man is stationary, $L_i=0$
$L_f=I_m{\omega }_m+I_T{\omega }_T$
$L_f=100\times 2^2\times \frac{1}{2}+4000\times {\omega }_T$
Using conservation of angular momentum
${\omega }_T=-\frac{1}{20}=-0.05rad/s$

The angular velocity of turn table is given by
${\theta }_T={\omega }_T\times t=-\frac{\pi }{5}$

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