A solid hemisphere of weight P rests with its | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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

Similar Questions

$A$ body takes just twice the time to slide down a plane inclined at $30^\circ$ to the horizontal as it would if the plane were frictionless. The coefficient of friction between the body and the plane is:

$A$ body is placed on a rough inclined plane of inclination $\theta$. As the angle $\theta$ is increased from $0^o$ to $90^o$,the contact force between the block and the plane

$A$ particle is placed at rest inside a hollow hemisphere of radius $R$. The coefficient of friction between the particle and the hemisphere is $\mu = \frac{1}{\sqrt{3}}$. The maximum height up to which the particle can remain stationary is

In the figure,two blocks $M$ and $m$ are tied together with an inextensible and light string. The mass $M$ is placed on a rough horizontal surface with a coefficient of friction $\mu$,and the mass $m$ is hanging vertically. The pulley is frictionless. Choose the correct statement$(s)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module