A solid sphere of mass m is released from rest at the rim of a hemispherical cup such that it rolls down the surface. Given that the rim of the hemisphere is kept horizontal, determine the normal force exerted by the cup on the ball as it reaches the bottom.

Step-by-step solution

Let R be the radius of the hemispherical cup and r be the radius of the solid sphere. As the sphere rolls down from the rim to the bottom, its center of mass descends by a height h = R - r . By conservation of mechanical energy, the loss in potential energy is equal to the gain in translational and rotational kinetic energy: mg(R - r) = 1 2 mv^2 + 1 2 I ^2 For a solid sphere, the moment of inertia about its center of mass is I = 2 5 mr^2 . For pure rolling, the velocity of the center of mass is v = r . Substituting