A small particle of mass `m` is attached at `B` to a hoop of mass `m` and radius `r`, whole system is placed on the rough horizontal ground. The system is released from rest when `B` is dierctly above `A` and rolls without slipping. Find the angular acceleration of the system at the instant when `AB` becomes horizontal as shown in the figure.

Explanation

Assuming that the hoop rotates by `theta` (so that the attached 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`. we apply conservation of mechanical energy to get
`mgR(1-costheta)=(1)/(2)(2mR^(2))omega^(2)+(1)/(2)m(2Rcos"(theta)/(2)omega)^(2)`
Differentiating and putting `theta=90^(@)`
We solve for `alpha = omega(domega)/(d theta)` and get `4alpha=(3g)/(2R)`
or `alpha=(3g)/(8R)`