Question

A point mass m is rigidly attached at the circumference of uniform Ring of same mass. It is at same horizontal position as centre of ring. The minimum coefficient of friction between ring and ground so that there is pure rolling of ring on surface just after system is released is:

• A. 3/5
• B. 2/3
• C. 1/3
• D. 2/7

Answer 1

The correct option is D 2/7

Acceleration of the center of mass in the vertical direction
$=\alpha dsin\theta =\alpha \frac{R}{2}$(here $\alpha$ is the angular acceleration w.r.t the instantaneous center of rotation)
Thus force balance in the vertical direction
$2mg-N=2m\cdot \alpha \frac{R}{2}...............\left(i\right)$
Force balance in the horizontal direction
$f=2ma......................\left(ii\right)$
Now torque balance about the instantaneous center we get
$2mg\frac{R}{2}=4m{R}^2\alpha$($2m{R}^2$ for the ring about the instantaneous center and $m(R\sqrt{2}{)}^2=2m{R}^2$ for the point mass)
Thus
$\frac{g}{4R}=\alpha ....................\left(iii\right)$
Solving i, ii and iii we get
$\frac{7mg}{4}=N$ and $f=\frac{mg}{2}$
$f\le \mu N$
$\frac{mg}{2}\le \mu \frac{7mg}{4}$ $\frac{2}{7}\le \mu$
Ring and ground so that there is pure rolling of ring on surface just after system is released
$\mu =\frac{2}{7}$