Two discs of moments of inertia `l_(1)` and `l_(2)` and angular speeds `omega_(1)` and `omega_(2)` are rotating along collinear axes pass through their centre of mass and perpendicular to their plane. If the two are made to rotate jointly along the same axis, the rotational K.E. of system will be

Options

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

Answer

A

Explanation

By conservation of angular momentum,
`I_(1)omega_(1)+I_(2)omega_(2)=(I_(1)+I_(2))omega`
Angular velocity of the system `omega=(I_(1)omega_(1)+I_(2)omega_(2))/(I_(1)+I_(2))`
Rotational kinetic energy `=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)))`