A playground merry-go-round of radius R = 2.00 m has a moment of inertia I = 250 kg.m2 and is rotating at 10.0 rev/min about a frictionless, vertical axle. Facing the axle, a 25.0-kg child hops onto the merry-go-round and manages to sit down on the edge. The new angular speed of the merry-go-round is

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answer is 3.

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Detailed Solution

From conservation of angular momentum, I1ω1 = Ifωf.(250 kg.m2)(10.0 rev/min) = [250 kg.m2+(25.0 kg)(2.0 m)2]ω2 ω2 = 7.14 rev /min

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