JEE Main2013PhysicsRotational MotionActual
A ring of mass M and radius R is rotating about its axis with angular velocity . Two identical bodies each of mass m are now gently attached at the two ends of a diameter of the ring. Because of this, the kinetic energy loss will be :
Options
- Am(M+2 m) M ^2 R^2
- BM m (M+m) ^2 R^2
- CM m (M+2 m) ^2 R^2
- D(M+m) M (M+2 m) ^2 R^2
Correct answer
C. M m (M+2 m) ^2 R^2
Step-by-step solution
Kinetic energy _ (rotational) K _ R = 1 2 I ^2 Kinetic energy _ (translational) K _ T = 1 2 Mv ^2 aligned & ( v = R ) & M.I. _ (initial) I _ ring = MR ^2 ; _ initial = & M.I. _ (new) I ^ _ (system) = MR ^2+2 mR ^2 & _ ( system) ^ = M M+2 m & aligned Solving we get loss in K.E. = Mm ( M +2 ~m ) ^2 R ^2