A playground merry-go-round of radius R = 2.00 m has a moment of inertia I = 265 kg·m2 and is rotating at 12.0 rev/min about a frictionless vertical axle. Facing the axle, a 24.0 kg child hops onto the merry-go-round and manages to sit down on its edge. What is the new angular speed of the merry-go-round?
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The angular momentum is conserved in this case, this is
Io ωo = If ωf
where, in this case, the moment of inertia changes as the child hops on the merry-go-round. Solving for ωf
ωf = Io ωo / If = Io ωo / (Io+m R^2)
where the final moment of inertia is that if the merry-go-round plus the child's at the end of it (m R^2). Thus, the new angular speed is given by
ωf = (265*12) / (265 + 24*2^2) = 8.80 rev/min
Io ωo = If ωf
where, in this case, the moment of inertia changes as the child hops on the merry-go-round. Solving for ωf
ωf = Io ωo / If = Io ωo / (Io+m R^2)
where the final moment of inertia is that if the merry-go-round plus the child's at the end of it (m R^2). Thus, the new angular speed is given by
ωf = (265*12) / (265 + 24*2^2) = 8.80 rev/min