Two rings of masses 'm' (smaller one) and '4m' (bigger one) are placed in the set up as shown in the diagram. The smaller ring is hung through a horizontal frictionless thread. The bigger ring is joined to the smaller ring in such a way that it can rotate freely about the point of suspension. The planes of both the rings are perpendicular to the plane of the paper. The bigger ring has a radius 'R'. After the bigger ring is released from the initial position as shown in the figure, find the angular velocity of the bigger ring when its plane becomes vertical i.e the planes of both the rings become vertical.
The Attempt at a Solution
First of all, when the rings reach the final position, we can calculate the relation between the velocities of the center of masses of both the rings (their velocity vectors will be horizontal and since there is no ext. force in the horizontal direction we can conserve the linear momentum). Now when the bigger ring reaches the final position the decrease in the potential energy of the system is, (4m)gR, which will be converted into kinetic energies of the bodies. But i have a doubt; the K.Es of the smaller ring will be only its linear K.E, but for the bigger ring, what will be the K.E? Will it just the linear K.E of the COM of the ring at that instant? or it will be the rotational energy about the point of suspension? or both?
The answer to the question is (20g/17R)^(1/2), but in no ways i am getting to it.