A 3.00-m-diameter merry-go-round is initially rotating with an angular speed of 1.65 rad/s. The merry-go-round is a solid, uniform disk with mass 120 kg. A child, 35.0 kg, and a parent, 65.0 kg, are standing at the edge. a) Because there are no rails on this merry-go-round, static friction provides the centripetal force. What minimum value for the coefficient of static friction between the child’s shoes and the merry-go-round surface is required to keep the child on the merry-go-round? (Apply Newton’s 2nd law in the radial direction.) b) What is the initial angular momentum of the system (merry-go-round+child+parent) about the central rotation axis? c) What is the initial rotational kinetic energy of the system

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A 3.00-m-diameter merry-go-round is initially rotating with an angular speed of 1.65 rad/s. The merry-go-round is a solid, uniform disk with mass 120 kg. A child, 35.0 kg, and a parent, 65.0 kg, are standing at the edge.


a) Because there are no rails on this merry-go-round, static friction provides the centripetal force. What minimum value for the coefficient of static friction between the child’s shoes and the merry-go-round surface is required to keep the child on the merry-go-round? (Apply Newton’s 2nd law in the radial direction.)

b) What is the initial angular momentum of the system (merry-go-round+child+parent) about the central rotation axis?

c) What is the initial rotational kinetic energy of the system?

The parent and child each move inward from the edge of the merry-go-round, stopping at a point halfway between the centre and the edge.

d) What is the new angular speed of the merry-go-round? (Consider conservation of angular momentum.)

e) What is the new rotational kinetic energy of the system? (Was mechanical energy gained or lost from the system during the move?

m, = 35 kg'
R=1.50 m
m, = 65 kg
M = 120 kg
o, = +1.65 rad/s
Transcribed Image Text:m, = 35 kg' R=1.50 m m, = 65 kg M = 120 kg o, = +1.65 rad/s
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