Question

A uniform circular ring of mass $M$ and radius $R$ with a particle of mass $m=\frac{M}{2}$ rigidly attached to its topmost point is placed on a rough horizontal surface. When a negligible push is given to the ring, it starts rolling without slipping relative to the rough horizontal surface as shown. Now, when the particle is getting at the right end of the horizontal diameter of the ring, choose the correct option(s).

• A. The angular velocity of the ring is $\sqrt{\frac{g}{3R}}$.
• B. The angular acceleration of the ring is $\frac{g}{6R}$.
• C. The angular acceleration of the ring is $\frac{2g}{3R}$.
• D. The acceleration of the centre of the ring is $\frac{g}{6}$.

Answer 1

Answer: (A), (B), (D)

The correct options are
A The angular velocity of the ring is $\sqrt{\frac{g}{3R}}$.
B The angular acceleration of the ring is $\frac{g}{6R}$.
D The acceleration of the centre of the ring is $\frac{g}{6}$.


Moment of inertia of the system about instantaneous axis of rotation is
$I_p=2M{R}^2+2m{R}^2=6m{R}^2$
Distance of centre of mass of the system from centre of the ring is $r=\frac{mR+0}{m+M}=\frac{R}{3}$
Using conservation of energy of the system
$mgR=\frac{1}{2}{I}_p{\omega }^2$
$mgR=\frac{1}{2}6m{R}^2{\omega }^2$
$\therefore \omega =\sqrt{\frac{g}{3R}}$

Now, torque at point $P$(stationary point on ground) is
${\tau }_p=I_p\alpha$
$3mg\frac{R}{3}=6m{R}^2\alpha$
$\therefore \alpha =\frac{g}{6R}$
For pure rolling motion $a=\alpha R=\frac{g}{6R}R$
$\therefore$ acceleration of the centre of the ring, $a=\frac{g}{6}$