Q. Two discs of moment of inertia $I_{1}$ and $I_{2}$ and angular speeds $\omega_{1}$ and $\omega_{2}$ are rotating along collinear axes passing through their centre of mass and perpendicular to their plane. If the two are made to rotate combindly along the same axis the rotational $KE$ of system will be
A) $\frac{I_{1}\omega_{1} + I_{2}\omega_{2}}{2(I_{1} + I_{2})}$
B) $\frac{\left( I_{1} + I_{2} \right)\left( \omega_{1} + \omega_{2} \right)^{2}}{2}$
C) $\frac{\left( I_{1}\omega_{1} + I_{2}\omega_{2} \right)^{2}}{2(I_{1} + I_{2})}$
D) None of these
Explanation
(c) Conservation of angular momentum
$I_{1}\omega_{1} + I_{2}\omega_{2} = (I_{1} + I_{2})\omega$
Angular velocity of system $\omega = \frac{I_{1}\omega_{1} + I_{2}\omega_{2}}{I_{1} + I_{2}}$
$\therefore$ Rotational kinetic energy $= \frac{1}{2}\left( I_{1} + I_{2} \right)\omega^{2}$
$= \frac{1}{2}\left( I_{1} + I_{2} \right)\left( \frac{I_{1}\omega_{1} + I_{2}\omega_{2}}{I_{1} + I_{2}} \right) = \frac{\left( I_{1}\omega_{1} + I_{2}\omega_{2} \right)^{2}}{2(I_{1} + I_{2})}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.