A playground merry-go-round of radius 2.3 m has a moment of inertia 174 kg.m2 and is rotating at 13 rev/min about a frictionless vertical axle. Facing the axle, a 35 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 in units of rev/min.
Using conservation of angular momentum
I1W1 = I2W2
New Moment of inertia will be = I1 +mr^2
174 13 = (174 + 352.3*2.3) W2
W2 = 6.3 rev/min