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mash [69]
3 years ago
13

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?
Physics
1 answer:
Arte-miy333 [17]3 years ago
3 0

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

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