Stuck on a Rolling Hoop

First of all make a diagram as I say Let $O$ be the centre of the hoop, $A$ be the initial position of the particle, $B$ be the instantaneous position of the particle $($ when $\angle AOB = \theta)$ and $P$ be the point on the ground where the hoop touches it $($ so, $\angle APB = \frac{\theta}{2})$ .

Assuming that the hoop rotates by $\theta$ (so that the particle also rotates by $\theta$ ) starting from its initial position, we find the angular speed of the hoop + particle. The instantaneous axis of rotation is located at $P$ . apply conservation of mechanical energy to get

$mgr(1 - cos(\theta)) = \frac{(2mr^{2})w^{2}}{2} + \frac{m(2rcos\frac{\theta}{2}w)^{2}}{2}$

(where $w =$ angular speed)

Differentiating and putting $\theta = 90$

Solving for $\alpha = w\frac{dw}{d\theta}$ and get

$\alpha = \frac{3g}{8r}$ Now, plugging the values we get answer $\alpha = \boxed{10}$