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Question

(A) Let $R$ be the radius of the hemisphere. The center of gravity of the solid hemisphere is at a distance $3R/8$ from the center of the plane surfac

Vedclass · Accepted Answer

$\mu = \frac{Q}{\sqrt{P(P + 2Q)}}$

Vedclass · Answer

(A) Let $R$ be the radius of the hemisphere. The center of gravity of the solid hemisphere is at a distance $3R/8$ from the center of the plane surface. However,for the given setup,let the contact point be $O$. Taking moments about the contact point $O$ on the inclined plane,for the plane surface to be horizontal,the net torque must be zero. Let $\phi$ be the angle of inclination of the plane. The weight $P$ acts at the center of gravity of the hemisphere,and $Q$ acts at the rim. From the geometry of the system,the condition for equilibrium is $P(R \sin \phi) = Q(R - R \sin \phi)$,where $R$ is the radius. Simplifying,$P \sin \phi = Q(1 - \sin \phi) \implies \sin \phi (P + Q) = Q \implies \sin \phi = \frac{Q}{P+Q}$. For the hemisphere to remain in equilibrium on a rough inclined plane,the coefficient of friction $\mu$ must satisfy $\mu \ge 	an \phi$. Since $\sin \phi = \frac{Q}{P+Q}$,we have $\cos \phi = \sqrt{1 - \sin^2 \phi} = \sqrt{1 - \left(\frac{Q}{P+Q}ight)^2} = \sqrt{\frac{(P+Q)^2 - Q^2}{(P+Q)^2}} = \frac{\sqrt{P^2 + 2PQ}}{(P+Q)}$. Thus,$	an \phi = \frac{\sin \phi}{\cos \phi} = \frac{Q/(P+Q)}{\sqrt{P^2 + 2PQ}/(P+Q)} = \frac{Q}{\sqrt{P(P+2Q)}}$. Therefore,the minimum coefficient of friction is $\mu = \frac{Q}{\sqrt{P(P+2Q)}}$.

$A$ solid hemisphere of weight $P$ rests with its curved surface in contact with a rough inclined plane. $A$ weight $Q$ is placed at some point on the rim of the hemisphere to keep its plane surface horizontal. Then its minimum coefficient of friction is

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