A uniform disc of mass m and radius R is pivo | vedclass

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Question

(A) The torque $	au$ acting on the disc about the pivot point $P$ is due to the gravitational force $mg$ acting at the centre of mass $C$.When the li

Vedclass · Accepted Answer

$\frac{2 g \cos 	heta}{3 R}$

Vedclass · Answer

(A) The torque $	au$ acting on the disc about the pivot point $P$ is due to the gravitational force $mg$ acting at the centre of mass $C$.When the line $PC$ makes an angle $	heta$ with the horizontal,the perpendicular distance from $P$ to the line of action of the gravitational force is $R \cos 	heta$.Therefore,the torque is $	au = mg(R \cos 	heta)$.The moment of inertia $I$ of the disc about an axis passing through the pivot point $P$ (which is on the rim) is given by the parallel axis theorem: $I = I_{cm} + mR^2 = \frac{1}{2} mR^2 + mR^2 = \frac{3}{2} mR^2$.Using the relation $	au = I \alpha$,we have:$mg(R \cos 	heta) = \left( \frac{3}{2} mR^2 ight) \alpha$Solving for the angular acceleration $\alpha$:$\alpha = \frac{mgR \cos 	heta}{\frac{3}{2} mR^2} = \frac{2g \cos 	heta}{3R}$.

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