Q 11-06-168JEE MainJEE Main 2021 (27 Aug, Shift 2)Medium
Two discs have moments of intertia $I_1$ and $I_2$ about their respective axes perpendicular to the plane and passing through the centre. They are rotating with angular speeds, $\omega_1$ and $\omega_2$ respectively and are brought into contact face to face with their axes of rotation coaxial. The loss in kinetic energy of the system in the process is given by:
Answer: (C) $\frac{I_1I_2}{2(I_1+I_2)}(\omega_1 - \omega_2)^2$
Angular momentum is conserved: $\omega = \dfrac{I_1\omega_1 + I_2\omega_2}{I_1 + I_2}$.
$$\Delta K = \frac{1}{2}I_1\omega_1^2 + \frac{1}{2}I_2\omega_2^2 - \frac{(I_1\omega_1 + I_2\omega_2)^2}{2(I_1 + I_2)} = \frac{I_1I_2}{2(I_1 + I_2)}(\omega_1 - \omega_2)^2$$
Solution by Sreeraj P, M.Sc Physics