Two discs of moment of inertia `I_(1)` and `I_(2)` and angular speeds `omega_(1)` and `omega_(2)` are rotating along the collinear axes passing through their center of mass and perpendicular to their plane. If the two are made to rotate combindly along the same axis the rotational `K.E.` of system will be

Options

A. `(I_(1)omega_(1)+I_(2)omega_(2))/(2(I_(1)+I_(2)))`
B. `((I_(1)+I_(2))(omega_(1)+omega_(2))^(2))/(2)`
C. `((I_(1)omega_(1)+I_(2)omega_(2))^(2))/(2(I_(1)+I_(2)))`
D. None of these

Answer

C

Explanation

`L_(1)+L_(2)=L`
implies `I_(1)omega_(1)+I_(2)omega_(2)=(I_(1)+I_(2))omegaimplies omega=(I_(1)omega_(1)+I_(1)omega_(2))/(I_(1)+I_(2))`
`K_(R)=(1)/(2)(I_(1)+I_(2))omega^(2)=(1)/(2)(I_(1)+I_(2))((I_(1)omega_(1)+I_(2)omega_(2))/(I_(1)+I_(2)))^(2)`
= `((I_(1)omega_(1)+I_(2)omega_(2))^(2))/(2(I_(1)+I_(2)))`