A disc of mass ' m ' and radius ' r ' rolls down an inclined plane of height ' h '. When it reaches the bottom of the plane, its rotational kinetic energy is ( g = acceleration due to gravity)

Step-by-step solution

The rotational kinetic energy of the disc rolling down the inclined plane without slipping is determined using conservation of mechanical energy. At height h , the potential energy mgh converts fully to kinetic energy at the bottom, comprising both translational and rotational components: mgh = 1 2 mv^2 + 1 2 I ^2 . For a disc, I = 1 2 mr^2 . With no slipping, = v/r , so the rotational term becomes 1 2 1 2 mr^2 ( v r )^2 = 1 4 mv^2 . Substituting, mgh = 1 2 mv^2 + 1 4 mv^2 = 3 4 mv^2 . Solving for mv^2 gives mv^2 =