Question

A circular platform is mounted on a frictionless vertical axle. Its radius R = 2 m and its moment of inertia about the axle is 200 kg $m^2.$ It is initially at rest. A 50 kg man stands on the edge of the platform and begins to walk along the edge at the speed of 1 $m{s}^{-1}$ relative to the ground. Time taken by the man to complete one revolution is

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

Answer 1

The correct option is C $2\pi s$

Initial angular momentum, $L_i=0$
[$\because$ the system is at rest initially]
As per principle of conservation of angular momentum, initial momentum = final momentum i.e., $L_i=L_f=0$
$\therefore$ Angular momentum of man = Angular momentum of platform
or, mvR = I$\omega$
or, $\omega =\frac{mvR}{I}=\frac{50\times 1\times 2}{200}=0.5rad{s}^{-1}$
Now, angular velocity of man relative to the platform
${\omega }_{mp}=w+\frac{v}{R}=0.5+0.5=1rad{s}^{-1}$
But ${\omega }_{mp}=\frac{2\pi }{T}\Rightarrow =\frac{2\pi }{{\omega }_{mp}}=\frac{2\pi }{1}=2\pi sec.$
$[\because {\omega }_{mp}=1rad{s}^{-1}]$