Q 11-06-097JEE MainJEE Main 2024 (30 Jan, Shift 1)Medium
Consider a disc of mass $5\ \text{kg}$ and radius $2\ \text{m}$, rotating with angular velocity of $10\ \text{rad s}^{-1}$ about an axis through its centre perpendicular to its plane. An identical disc is kept gently over the rotating disc along the same axis. The energy dissipated so that both the discs continue to rotate together without slipping is ______ J.
Numerical value type. Enter your answer.
Answer: 250
$I = \tfrac12 MR^2 = \tfrac12(5)(2)^2 = 10\ \text{kg m}^2$.
Angular momentum is conserved: $10\times10 = 20\,\omega \Rightarrow \omega = 5\ \text{rad s}^{-1}$.
$$K_i = \tfrac12(10)(10)^2 = 500\ \text{J},\qquad K_f = \tfrac12(20)(5)^2 = 250\ \text{J}$$
Energy dissipated $= 250\ \text{J}$.
Solution by Sreeraj P, M.Sc Physics