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A child of mass $m$ runs tangentially at speed $v$ and jumps on the rim of a stationary playground merry-go-round (uniform disc, mass $M$, radius $R$). The angular speed just after is:
A$2mv/((M+2m)R)$
B$mv/(MR)$
C$v/R$
D$mv/((M+m)R)$
Answer & Solution
Correct answer: A. $2mv/((M+2m)R)$
Angular-momentum conservation about axle: $mvR = (I_{\text{disc}} + mR^2)\omega = (\tfrac{1}{2}MR^2 + mR^2)\omega$ ⇒ $\omega = mv / ((M/2 + m)R) = 2mv/((M + 2m)R)$.
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