A circular disc of mass $M$ and radius $R$ is rotating about its axis with angular speed $\omega_1$. If another stationary disc having radius $\dfrac R2$ and same mass $M$ is dropped co-axially on to the rotating disc, gradually both discs attain constant angular speed $\omega_2$. The energy lost in the process is $p\%$ of the initial energy. Value of $p$ is ______.
Numerical value type. Enter your answer.
Answer: 20
$I_1 = \dfrac{MR^{2}}{2}$, $I_2 = \dfrac{M(R/2)^{2}}{2} = \dfrac{I_1}{4}$.
Angular momentum: $I_1\omega_1 = \dfrac54I_1\omega_2 \Rightarrow \omega_2 = \dfrac45\omega_1$.
$$\frac{K_f}{K_i} = \frac{\tfrac54I_1\omega_2^{2}}{I_1\omega_1^{2}} = \frac54\cdot\frac{16}{25} = \frac45$$
Energy lost $= 20\%$, so $p = 20$.
Solution by Sreeraj P, M.Sc Physics