A cylinder of mass $M$ is placed on an inclined plane of angle $\alpha$ with its axis horizontal. A

Question

A cylinder of mass $M$ is placed on an inclined plane of angle $\alpha$ with its axis horizontal. A small block of mass $m$ is placed inside the cylinder. The coefficient of static friction between the block and the inner surface of the cylinder is $\mu$. Assume that the friction between the cylinder and the inclined plane is sufficiently high to prevent the cylinder from slipping. What is the maximum angle of inclination $\alpha$ for which the cylinder can remain at rest?

Professor PC Thomas Classes · Answer

"An extended object supported at a point may be static if its center of mass lies directly above or below that point."
For equilibrium, the center of mass (CM) of the cylinder-block system must lie vertically above the contact point $A$ on the incline. Let $O$ be the cylinder center and $R$ its radius. The CM lies at a distance $OB = \frac{mR}{M+m}$ from $O$ along the line joining $O$ and the block. In $	riangle OAB$, the angle $OAB = \alpha$. For the block to be on the verge of slipping, the radius to the block must make an angle $\phi$ with the vertical such that $	an\phi = \mu$, meaning $\sin\phi = \frac{\mu}{\sqrt{1+\mu^2}}$. Applying the sine rule in $	riangle OAB$, we get $\frac{OB}{\sin\alpha} = \frac{OA}{\sin(180^\circ - \phi)} = \frac{R}{\sin\phi}$. Substituting $OB$, we get $\sin\phi = \frac{R}{\frac{mR}{M+m}}\sin\alpha = \left(1 + \frac{M}{m}ight)\sin\alpha$. Equating the two expressions for $\sin\phi$ gives the maximum angle $\alpha$.