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 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:

Question

G1 Physics · Accepted Answer

Immediately after contact with the corner, angular momentum about the corner is conserved. During the subsequent motion about the corner, mechanical energy is conserved until contact is lost. Combining the angular-momentum relation with the loss-of-contact condition gives v² = 5gR/7. Therefore v = √(5gR/7).