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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. |
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Answer» Solution :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)` 
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