A uniform disc of mass $M$ and radius $R$ is rotating with angular speed $\omega_0$ about its axis. It is gently placed on a ring of the same mass $M$ and radius $R$, initially at rest, so that their axes coincide. Friction between them soon makes them rotate together. The fraction of the initial kinetic energy lost is
Answer: (A) $\dfrac23$
Concept. Friction between the two is internal, so angular momentum about the common axis is conserved; kinetic energy is not.
Step 1. Moments of inertia. $I_{disc} = \dfrac12MR^2$, $I_{ring} = MR^2$.
Step 2. Common angular speed.
$$\frac12MR^2\omega_0 = \left(\frac12 + 1\right)MR^2\omega \;\Rightarrow\; \omega = \frac{\omega_0}{3}$$
Step 3. Kinetic energies.
$$K_i = \frac12\cdot\frac12MR^2\omega_0^2 = \frac{MR^2\omega_0^2}{4},\qquad K_f = \frac12\cdot\frac32MR^2\cdot\frac{\omega_0^2}{9} = \frac{MR^2\omega_0^2}{12}$$
Step 4. Fraction lost.
$$\frac{K_i - K_f}{K_i} = 1 - \frac13 = \frac23$$
Answer. $\dfrac23$ (Option A).
Solution by Sreeraj P, M.Sc Physics