Question

Two men each of mass m stand on the rim of a horizontal circular disc, diametrically opposite to each other. The disc has a mass M and is free to rotate about a vertical axis passing through its centre of mass. Each man start simultaneously along the rim clockwise and reaches their original starting points on the disc. The angle turned through by the disc with respect to the ground ( in radian) is:

• A. $\frac{8m\pi }{4m+M}$
• B. $\frac{2m\pi }{4m+M}$
• C. $\frac{m\pi }{M+m}$
• D. $\frac{4m\pi }{2M+m}$

Answer 1

The correct option is A $\frac{8m\pi }{4m+M}$

Suppose the disc turned by an angle $\theta$ in anticlockwise direction.

$\therefore$ the angular distance moved my two men $=2\times \pi -\theta$

The angular momentum of the system (men + disc) must be conserved.

$\therefore L_{men}=L_{disc}$

$\Rightarrow 2\times \left(m{R}^2\right)\times (2\pi -\theta )=\frac{M{R}^2}{2}\times \theta$

$\Rightarrow 4\pi m-2m\theta =\frac{M}{2}\times \theta$

$\Rightarrow \theta =\frac{8m\pi }{4m+M}$