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+2m))omega^(2)R^(2)`
C. `(Mn)/((M-2m))omega^(2)R^(2)`
D. `((M+m)M)/((M+2m))omega^(2)R^(2)`

Answer

B

Explanation

According to law of conservation of angular momentum
`l_(1)MR^(2)," "l_(1)omega_(1)=l_(2)omega_(2)`
`l_(2)=MR^(2)+2mR^(2)`
`therefore" "omega_(2)=(l_(1)omega_(1))/(l_(2))=(MR^(2)omega)/((MR^(2)+2mR^(2)))=(momega)/(M+2m)`
`therefore" "DeltaKE=K_(1)-K_(2)" "(because" K.E decreases")`
`=(1)/(2)l_(1)omega_(1)^(2)-(1)/(2)l_(2)omega_(2)^(2)`
`=(1)/(2)MR^(2)omega^(2)-(1)/(2)(MR^(2)+2mR^(2)).((Momega)/(M+2m))^(2)`
`=(1)/(2)MR^(2)omega^(2)[1(M(M+2m))/((M+2m)^(2))]`
`=(1)/(2)MR^(2)omega^(2)[(M+2m-M)/((M+2m))]`
`=(MR^(2)omega^(2)m)/((M+2m))`