A small spherical ball is released from a height h on the rough track shown in the figure. Assuming no slipping occurs anywhere, determine its linear speed as it rolls along the horizontal section of the track.

Step-by-step solution

By the principle of conservation of mechanical energy, the loss in potential energy is equal to the gain in total kinetic energy. mgh = 1 2 mv^2 + 1 2 I ^2 For a solid spherical ball, the moment of inertia about its center of mass is I = 2 5 mr^2 . Since the ball rolls without slipping, the linear speed v and angular speed are related by v = r . Substituting I and into the energy equation: mgh = 1 2 mv^2 + 1 2 ( 2 5 mr^2 ) ( v r )^2 mgh = 1 2 mv^2 + 1 5 mv^2 mgh = 7 10 mv^2 v^2 = 10gh 7 v = 10gh 7