Question

A ring of mass $m$ and radius $r$ has a particle of mass m attached to it at a point A. The ring can rotate about a smooth horizontal axis which is tangential to the ring at a point B diametrically opposite to A, The ring is released from rest when $AB$ is horizontal. When $AB$ has turned through an angle $\frac{\pi }{3}$, then

• A. angular velocity of the body is $\sqrt{\frac{6g\sqrt{3}}{11r}}$
• B. angular velocity of the body is $\sqrt{\frac{7g\sqrt{3}}{11r}}$
• C. angular acceleration of the body is $\frac{3g}{11r}$
• D. angular acceleration of the body is $\frac{7g}{11r}$

Answer 1

Answer: (A), (C)

The correct options are
A angular velocity of the body is $\sqrt{\frac{6g\sqrt{3}}{11r}}$
C angular acceleration of the body is $\frac{3g}{11r}$

Angular velocity From conservation of mechanical energy, decrease in gravitational $PE=$ increase in rotational $KE$

or

$mgr\sin {60}^{\circ }+mg\left(2r\sin {60}^{\circ }\right)$ $=\frac{1}{2}[\frac{3m{r}^2}{2}+m(2r{)}^2]{\omega }^2$
$\therefore \frac{3\sqrt{3}mgr}{2}=\frac{11}{4}m{r}^2{\omega }^2$
$\therefore \omega =\sqrt{\frac{6g\sqrt{3}}{11r}}$
Angular acceleration
$\alpha =\frac{\tau }{l}=\frac{mgr\cos {60}^{\circ }+mg\left(2r\cos {60}^{\circ }\right)}{[\frac{3m{r}^2}{2}+m(2r{)}^2]}$
$=\frac{\frac{3mgr}{2}}{\frac{11}{2}m{r}^2}=\frac{3g}{11r}$