Question

A playground merry-go-round of radius R = 1.40 m has a moment of inertia I = 245 kg · m2 and is rotating at 9.0 rev/min about a frictionless vertical axle. Facing the axle, a 24.0-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: 11.65rpm

Step-by-step explanation:

Moment of inertia of the child = mr²

Inertia = 24 * 1.4 * 1.4 = 47.04kgm²

Converting rpm to rad/s we have

12rpm = 12 * π/30

12rpm = 37.7/30

12rpm = 1.257 rad/s

Angular momentum = Iω

(245 * 1.257) + (47.04 * 0) = 355kgm²

355 = (245 + 47.04) * ω

355 = 292.04ω

ω = 355/292.04

ω = 1.22 rad/s

1.22 rad/s = 1.22 * 9.55 = 11.65 rpm

Thus, the new angular speed of the merry go round is 11.65rpm