Question

A circular platform is mounted on a vertical frictionless axle. Its radius is $r=2m$ and its moment inertia is $I=200kg$ $m^2$. It is initially at rest. A $70kg$ man stands on the edge of the platform and begins to walk along the edge at speed $v_0=10{ms}^{-1}$ relative to the ground. When the man has walked once around the platform so that he is at his original position on it, what is his angular displacement relative to ground

• A. $\frac{6}{5}\pi$
• B. $\frac{5}{6}\pi$
• C. $\frac{4}{5}\pi$
• D. $\frac{5}{4}\pi$

Answer 1

Answer: (B)

$\frac{5}{6}\pi$

Let after time $t$ , man comes to its initial position.

About O, $L_i=L_f$

$0=Iw-m{v}_{o}r$

$200\times w=70\times 10\left(2\right)⟹w=7rad/s$

Also, $\theta =wt⟹\theta =7t$ .............(1)

and $2\pi -\theta =w_{m}t$

$w_m=\frac{v_o}{r}=\frac{10}{2}=5rad/s$

$⟹2\pi -\theta =5t$ ...................(2)

Solving (1) and (2), we get $\theta =\frac{7\pi }{6}$

Angular displacement $=2\pi -\theta =2\pi -\frac{7\pi }{6}=\frac{5\pi }{6}$