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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:
A$0$ (KE conserved)
B$1/2$
C$1/4$
D$1/3$
Answer & Solution
Correct answer: B. $1/2$
$L$ 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.)
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