A particle of mass m is placed on a fixed, smooth sphere of radius R at a position where the radius passing through the particle makes an angle of 30^ with the vertical. The particle is released from this position. Determine the force exerted by the sphere on the particle just after the release.

Step-by-step solution

Let the normal reaction exerted by the sphere on the particle be N . The forces acting on the particle are its weight mg acting vertically downwards and the normal reaction N acting radially outwards. Resolving the weight along the radial direction, the component is mg 30^ . The equation of motion in the radial direction is given by mg 30^ - N = mv^2 R . Just after the release, the velocity of the particle is zero, so v = 0 . Substituting v = 0 into the equation, we get mg 30^ - N = 0 . Therefore, N = mg 30^ = 3 mg