A horizontal disc of mass $M$, radius $R$ spins freely at $\omega$. A second identical disc is dropped vertically onto it and they stick. The fraction of original kinetic energy lost is:

$L$ conserved: $I\omega = 2I\omega'$ ⇒ $\omega' = \omega/2$. KE$_i = \tfrac{1}{2}I\omega^2$. KE$_f = \tfrac{1}{2}(2I)(\omega/2)^2 = \tfrac{1}{4}I\omega^2$. Fraction lost $= 1 - 1/2 = 1/2$. (Half the rotational KE goes to heat/sound when the discs lock up.)