Two identical rings A and B of same mass and radius are revolving, ring A arounds its own diameter and ring B about tangential axis in its own plane. Both the rings A and B have same rotational kinetic energy. The ratio of the angular velocity of ring B to that of ring A is

Step-by-step solution

Moment of inertia of ring A about its diameter is I_A = 1 2 MR^2 . Moment of inertia of ring B about a tangential axis in its own plane is obtained using the parallel axis theorem: I_B = I_A + MR^2 = 1 2 MR^2 + MR^2 = 3 2 MR^2 . Rotational kinetic energy is given by K = 1 2 I ^2 . Given that both rings have the same rotational kinetic energy, K_A = K_B : 1 2 I_A _A^2 = 1 2 I_B _B^2 _B^2 _A^2 = I_A I_B _B _A = 1 2 MR^2 3 2 MR^2 = 1 3 = 1 3 The ratio of the angular velocity of ring B to that of ring A is 1 : 3 . Answ