Q. A circular disk of moment of inertia $I_{t}$ is rotating in a horizontal plane, about its symmetry axis, with a constant angular speed $\omega_{1}$. Another disk of moment of inertia $I_{b}$ is dropped coaxially onto the rotating disk. Initially the second disk has zero angular speed. Eventually both the disks rotate with a constant angular speed $\omega_{1}$. The energy lost by the initially rotating disc to friction is
A) $\frac{1}{2}\frac{I_{b}I_{t}}{(I_{t} + I_{b})}\omega_{i}^{\ \ 2}$
B) $\frac{1}{2}\frac{I_{b}^{\ \ 2}}{(I_{t} + I_{b})}\omega_{i}^{\ \ 2}$
C) $\frac{1}{2}\frac{I_{t}^{\ \ 2}}{(I_{t} + I_{b})}\omega_{i}^{\ \ 2}$
D) $\frac{I_{b} - I_{t}}{(I_{t} + I_{b})}\omega_{i}^{\ \ 2}$
Explanation
(a) By conservation of angular momentum
$I_{t}\omega_{i} = \left( I_{t} + I_{b} \right)\omega_{f} \Rightarrow \omega_{f} = \left( \frac{I_{t}}{I_{t} + I_{b}} \right)\omega_{i}$
Loss in kinetic energy $= \frac{1}{2}I_{t}\omega_{i}^{2} - \frac{1}{2}\left( I_{t} + I_{b} \right)\left( \omega_{f}^{2} \right)$
$= \frac{1}{2}\left( \frac{I_{b}I_{t}}{I_{b} + I_{t}} \right)\omega_{i}^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.