Question

Two men each of mass m stand on the rim of a horizontal circular disc diametrically opposite to each The disc has a mass M and is free to rotate about a vertical axis passing through its centre of mas Each mass 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. $\left(\frac{2m\pi }{4m+M}\right)$
• B. $\left(\frac{4m\pi }{4M+m}\right)$
• C. $\left(\frac{m\pi }{M+m}\right)$
• D. $\left(\frac{8m\pi }{4m+M}\right)$

Answer 1

The correct option is D $\left(\frac{8m\pi }{4m+M}\right)$

$\begin{array}{c}\text{Let disc rotates anti clockwise by an angle}\\ \theta \text{. Then angular displacement of each}\\ \text{men is}2\pi \theta \text{in clockwise. In absence}\\ \text{of external torque}\end{array}$

$\begin{array}{cc}\sum I_A{\theta }_A& =\sum I_C{\theta }_C\\ \frac{1}{2}m{R}^2\theta & =2\left(M{R}^2\right)(2\pi -\theta )\\ \theta & =\frac{8m\pi }{4m+M}\end{array}$