A solid cylinder of radius R rolls without slipping with a center of mass speed v₀ = gR 3 on a horizontal surface with a vertical edge, as shown in the figure. Here, g is the acceleration due to the gravity. At the moment when the cylinder loses contact with the surface due to rotation around the corner, the speed of its center of mass is:

Step-by-step solution

Let the mass of the cylinder be m and its radius be R . When the cylinder is rolling on the horizontal surface, its kinetic energy is: K_i = 1 2 mv₀^2 + 1 2 I_ cm ₀^2 Since it rolls without slipping, ₀ = v₀ R and I_ cm = 1 2 mR^2 . K_i = 1 2 mv₀^2 + 1 2 ( 1 2 mR^2 ) ( v₀ R )^2 = 3 4 mv₀^2 Given v₀ = gR 3 , we have v₀^2 = gR 3 . K_i = 3 4 m ( gR 3 ) = 1 4 mgR Let the corner be the reference level for potential energy. The initial total mechanical energy of the cylinder just as it reaches the corner is: E_i = K_i + U