question_answer

  A ring of mass M and radius R is rotating about its axis with angular velocity \[\omega \]. 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:                                                                                                 [JEE ONLINE 25-04-2013]

A)  \[\frac{\operatorname{m}(\operatorname{M}+2m)}{\operatorname{M}}{{\omega }^{2}}{{R}^{2}}\]

B)         \[\frac{\operatorname{Mm}}{(\operatorname{M}+m)}{{\omega }^{2}}{{R}^{2}}\]

C)   \[\frac{\operatorname{Mm}}{(\operatorname{M}+2m)}{{\omega }^{2}}{{R}^{2}}\]

D)        \[\frac{\left(\operatorname{M}+m \right)\operatorname{M}}{(\operatorname{M}+2m)}{{\omega }^{2}}{{R}^{2}}\]

Correct Answer: C

Solution :

[c] Kinetic energy

Kinetic energy

Solving we get loss in K.E.