A playground merry-go-round of radius R = 2.00 m has a moment of inertia I = 250 $k g . m^{2}$ 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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5.25 rev/min

3.14 rev/min

8.45 rev/min

7.14 rev/min

answer is C.

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Detailed Solution

From conservation of angular momentum,

$I_{1} ω_{1} =$ $I_{f} ω_{f}$ .

(250 $k g . m^{2}$ )(10.0 rev/min) = [250 $k g . m^{2}$ +(25.0 kg)(2.0 m)²] $ω_{2}$

$ω_{2} = 7.14 r e v / m i n$

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