Question

A playground merry-go-round of radius R = 1.60 m has a moment of inertia I = 255 kg·m² and is rotating at 11.0 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 its edge. What is the new angular speed of the merry-go-round?​

Answer 1

Radius of merry-go-round = R = 1.6 m

Moment of inertia of merry-go-round = I_m = 255 kg.m²

Initial rpm of the merry-go-round = N = 11 rev/min

Mass of child = m = 25 kg

Initial angular speed of the merry-go-round = _i

_i = 2N/60

_i = 2(11)/60

_i = 1.152 rad/s

Final angular speed of merry-go-round = _f

Moment of inertia of the system after the child hops onto the merry-go-round = I_f

I_f = I_m + mR²

I_f = 255 + 25(1.6)²

I_f = 319 kg.m²

By angular momentum conservation,

I_mi = I_ff

(255)(1.152) = (319)_f

_f = 0.921 rad/s

The new angular speed of the merry-go-round = 0.921 rad/s