Question

What is the new angular speed of the merry-go-round (in rev/min)?

A playground merry-go-round of radius R = 1.3 m has a moment of inertia of I = 290 kg*m^2. and is rotating at a rate of ω = 11 rev/min around a frictionless vertical axis. Facing the axle, a 30 kg child hops onto the merry-go-round and manages to sit down on the edge.

Answer 1

The merry-go-round has a moment of inertia `I`. When the child hops it on and sits on the edge the moment of inertia changes to

`I^'=I+mr^2`
where `m` is mass of child and `r` radius of merry-go-round.

As the child hops onto the edge there is no external force acting on the system. As such angular momentum `L` is conserved. Therefore we have

`L^′=L`

Let `omega^'` be changed angular velocity. Using the definition of angular momentum we have

`I^′omega^'=Iomega`
`=>omega^'=(Iomega)/I^′`

Inserting given values we get

`omega^'=(290xx11)/(290+30xx(1.3)^2)`
`omega^'=9.4\ "rpm"`