A solid spherical ball of mass m is released from the topmost point of the shown semi-spherical shell. The track is sufficiently rough to enable pure rolling motion. The normal force between the ball and the shell at the lowest position is

Step-by-step solution

From the law of conservation of mechanical energy, Final energy = Translational kinetic energy + Rotational kinetic energy Hence, Final energy = 1 2 mv 2 + 1 2 Iω 2 = 1 2 mv 2 + 1 2 × 2 5 mR 2 × v R 2 = 7 10 mv 2 Hence, we finally obtain, mg R - r = 7 1 0 mv 2 .......(1) and N - mg = mv 2 R - r ..........(2) Solving Eqs. (1) and (2), we get N = 1 7 7 mg