A body of radius R and mass m is rolling smoothly with speed v on a horizontal surface. It then rolls up a hill to a maximum height h. If h = 3v24g, what might the body be?

Interpreting h as the height increase for all the centre of mass of the body, then mechanical energy conservation, Ki = Uf, leads to12mvcom2+12Iω2 = mgh ⇒ 12mv2+12I(vR)2                                    = mg(3v24g)from which v cancels and we obtain I = 12mR2This moment of inertia is the moment of inertia of solid cylinder.