Question

A playground merry-go-round of radius R = 1.80 m has a moment of inertia I = 270 kg · m2 and is rotating at 8.0 rev/min about a frictionless vertical axle. Facing the axle, a 27.0-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

Answer:

$\dot n = 6.042\,rpm$

Explanation:

The final angle speed of the merry-go-round is determined with the help of the Principle of Angular Momentum Conservation:

$(270\,kg\cdot m^{2})\cdot \left(8\,rpm\right) = [270\,kg\cdot m^{2}+(27\,kg)\cdot (1.80\,m)^{2}]\cdot \dot n$

$\dot n = 6.042\,rpm$