Portion AB of the wedge shown in figure is rough and BC is smooth. A solid cylinder rolls without slipping from A to B. Find the ratio of translational kinetic energy to rotational kinetic energy, when the cylinder reaches point C.

Question

G1 Physics · Accepted Answer

On rough AB the solid cylinder rolls, so at B K translational:K rotational = [½mv²]:[½(½mR²)(v²/R²)] = 2:1. On smooth BC there is no frictional torque, so rotational kinetic energy remains constant while gravity increases only translational kinetic energy. Using equal vertical drops AB=BC from the figure gives K translational/K rotational=5 at C.