Question

A playground merry-go-round of radius 1.9 m has a moment of inertia 139 kg.m2 and is rotating at 9 rev/min about a frictionless vertical axle. Facing the axle, a 27 kg child hops onto the merry-go-round, and manages to sit down on the edge. What is the new angular speed of the merry-go-round?

Answer 1

Using angular momentum conservation

L1 = L2

I1w1 = I2w2

I1 = Initial Moment of inertia of ride = 139 kg-m^2

I2 = final Moment of inertia of ride + child = 139 kg-m^2 + m*r^2

m = mass of child = 27 kg

r = radius of ride = 1.9 m

I2 = 139 + 27*1.9^2 = 236.47

w1 = Initial Angular Speed = 9 rev/min

w2 = final angular speed = ?

So,

w2 = w1*(I1/I2)

Using above values:

w2 = 9 rev/min*(139 kg-m^2/236.47 kg-m^2)

w2 = 5.3 rev/min = new angular speed of merry-go-round

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