In the diagram shown, a rod of mass M has been fixed on a ring of the same mass. The whole system has been placed on a perfectly rough surface. The system is gently displaced so that the ring starts rolling. The velocity of the centre of the ring when the rod becomes horizontal is (the length of the rod is equal to the radius of the ring)

Step-by-step solution

The moment of inertia of the system about instantaneous axis of rotation is given by, I = I ring + I rod ⇒ I = ( MR 2 + MR 2 ) + 1 12 MR 2 + MR 1 2 (From parallel axis theorem) ⇒ I = 2 MR 2 + 1 12 MR 2 + M 5 R 2 2 ∵ R 1 = R 2 + R 2 2 = 5 R 2 ⇒ I = 2 MR 2 + 1 12 MR 2 + 5 4 MR 2 = 2 MR 2 + 4 3 MR 2 = 10 3 MR 2 Hence, I system = 10 3 MR 2 . Now from work - energy theorem, we get, Mg R 2 = Rotational kinetic energy of the system about instantaneous axis of rotation (here R 2 is the distance thro