A merry-go-round of radius R = 3.00 m has a moment of inertia I = 250...
A merry-go-round of radius R = 3.00 m has a moment of inertia I = 250 kg m2 and is rotating at 10.0 rev/min about a vertical axis with friction (coefficient of 0.3). Facing the axle, a 30.0 kg child hops onto the merry go-round and manages to sit down on the edge.
(a) What is the new angular speed of the merry-go-round?
(b) What would the angular speed the merry-go-round be if the child
would sit at a distance of 1.0 m from the axle?
Homework Answers
using conservation of angular momentum
I w1 = ( I + Mr^2) w2
250 * 10 = ( 250 + 30* 3^2) w2
w2 = 4.808 rev/min
==========
b)
w2 = 25* 10 / ( 250 + 30* 1)
w2 = 8.93 rev/min
========
Comment in case any doubt, will reply for sure. Goodluck
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