Question

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

• A. $8mx/4m+M$
• B. $2mx/4m+M$
• C. $mx/M+m$
• D. $4mx/2M+Mm$

Answer 1

The correct option is A $8mx/4m+M$

$Given:$ Mass=$m$(horizontal circular disc) ; Mass=$M$ ( Vertical circular disc)

$Solution:$ As we know that,

$\text{Angular momentum}=Conserve$

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

$4xm-2m\theta =\frac{m}{2}\theta$

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

$So,the$ $correct$ $option:A$