A playground merry-go-round of radius R=2.00 m has a moment of inertia I=250 kg ⋅ 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

Question

A playground merry-go-round of radius R=2.00 m has a moment of inertia I=250 kg ⋅ 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 (A) 5.25 rev/min (B) 8.45 rev/min (C) 7.14 rev/min (D) 3.14 rev/min

Tardigrade Admin · Accepted Answer

From conservation of angular momentum, Ii ωi =If ωf(250 kg ⋅ m 2)(10.0 rev / min ) =[250 kg ⋅ m 2+(25.0 kg )(2.0 m )2] ω2 ω2 =7.14 rev/min

5.25 rev/min

8.45 rev/min

7.14 rev/min

3.14 rev/min

From conservation of angular momentum, Ii​ωi​=If​ωf​(250kg⋅m2)(10.0rev/min) =[250kg⋅m2+(25.0kg)(2.0m)2]ω2​ ω2​=7.14rev/min

From conservation of angular momentum,
$I_{i} ω_{i} = I_{f} ω_{f} (250 k g \cdot m^{2}) (10.0 re v / min)$
$= [250 k g \cdot m^{2} + (25.0 k g) (2.0 m)^{2}] ω_{2}$
$ω_{2} = 7.14 re v / min$