A small spherical ball of mass m rolls down a loop track. The ball is released on the linear portion at a vertical height H from the lowest point. The circular part shown has a radius R . Evaluate the normal force and the frictional force acting on the ball if H = 60 cm , R = 10 cm , = 0 and m = 70 g .

Step-by-step solution

Let the horizontal position on the loop track correspond to = 0 . At this point, the ball is at a height R from the lowest point. By the principle of conservation of mechanical energy from the release point to the horizontal position: mgH = mgR + 1 2 mv^2 + 1 2 I ^2 For a solid spherical ball, the moment of inertia is I = 2 5 mr^2 . For rolling without slipping, v = r . Substituting these into the energy equation: mg(H - R) = 1 2 mv^2 + 1 2 ( 2 5 mr^2 ) ( v r )^2 mg(H - R) = 7 10 mv^2 v^2 = 10 7 g(H - R) The normal