Question

A small body of mass m is attached at B to a
hoop of mass 3m and radius r. The system is
released from rest with 0 = 90° and rolls without
sliding. Just after release find​

Answer 1

Answer:

The angular acceleration just after release is $\frac{g}{8r}$

Explanation:

We have to find the angular acceleration just after release.

Taking the centre of the hoop as (0, 0)

At $\theta=90^\circ$, the centre of mass of the system will be

$X_{cm}=\frac{3m\times 0+m\times r}{3m+m}$

$\implies X_{cm}=\frac{r}{4}$

We can assume that whole mass of 4m is acting at point r/4 from the centre of the hoop

If the friction force is $f$ and moment of inertia $I$ then

Net torque = $I\alpha$

or, $4mg\times \frac{r}{4}-f\times r=I\alpha$

$\implies mgr-fr=I\alpha$

If linear acceleration is a then

$a=r\alpha$

And $f=4ma$

$\implies f=4mr\alpha$

Therefore,

$mgr-4mr\alpha\times r=(3mr^2+mr^2)\alpha$

$mgr-4mr^2\alpha=4mr^2\alpha$

$\implies mgr=8mr^2\alpha$

$\implies \alpha=\frac{g}{8r}$

Hope this helps.