Dear Mayank,

Since the cylinder is tipping without slipping, it is rotating around the line where it meets the edge of the block. Because the force (normal + friction) exerted by the block does no work on the cylinder (the point where the force is applied is motionless), therefore, the only work on the cylinder is by gravity.

The center of mass of the cylinder is a height R above the edge of the block initially. When the cylinder has turned through an angle theta, the height of its center is now Rcos(theta). Therefore, it has lost a gravitational potential energy mgR(1-cos(theta)). Since it is pivoting around a line on its surface (instead of rotating around its own center axis) the cylinder''s rotational inertia is 1/2?mR^2 + mR^2 by the parallel axis theorem. Therefore its kinetic energy is 1/2(3/2?mR^2)(omega)^2. By conservation of energy, mgR(1-cos(theta)) = 3/4?mR^2(omega)^2, so omega^2 = (3g)/(4R)?(1-cos(theta).

Now think about when the cylinder will lose contact with the block. The normal force exerted by the block on the cylinder can only point toward the center of the cylinder, because the block is not sticky. This is what tells you when the cylinder will leave contact with the block. The normal force will depend on the acceleration of the center of the cylinder by F=ma.

The acceleration of the cylinder center will have a centripetal component R?omega^2 = (3g/4)?(1-cos(theta)) and also a tangential component. The tangential component can be found from a_t = R?alpha, where alpha = torque/I. The torque around the contact point comes from gravity, and it is mgR?sin(theta), so a_t = R?mgR?sin(theta)/(3/2?mR^2) = 2g/3?sin(theta).

Since the normal force acts along the line joining the point of contact with the center of the cylinder, it is aligned with the centripetal acceleration. Therefore, the tangential acceleration is not needed for solving the problem. Summing forces in the centripetal direction gives mg?cos(theta) - N = m(3g/4)?(1-cos(theta)), so that N = mg(cos(theta) - 3/4?(1-cos(theta))) = mg/4?(7?cos(theta) - 3). Therefore N>0 when cos(theta) > 3/7 and N<0 when cos(theta)<3/7. Therefore, the cylinder loses contact with the block when theta = arccos(3/7).

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