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 :

Options

A. `(m(M+2m))/(M)omega^(2)R^(2)`
B. `(Mm)/((M+m))omega^(2)R^(2)`
C. `(Mm)/((M+2m))omega^(2)R^(2)`
D. `((M+m)M)/((M+2m))omega^(2)R^(2)`

Answer

C

Explanation

Kinetic `"energy"_(("rotational"))K_(R)=(1)/(2)Iomega^(2)`
Kinetic `"energy"_(("rotational"))K_(T)=(1)/(2)Mv^(2)(v=Romega)`
`M.I._(("initial"))I_("ring")=MR^(2),omega_("initial")=omega`
`M.I._(("new"))I._("system")=MR^(2)+2mR^(2)`
`omega._(("system"))=(Momega)/(M+2m)`
Solving we get loss in K.E. = `(Mm)/((M+2m))omega^(2)R^(2)`