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A 36.5 kg child stands at the center of a 125 kg playground merry-go-round which rotates...

A 36.5 kg child stands at the center of a 125 kg playground merry-go-round which rotates at 3.10 rad/s. If the child moves to the edge of the merry-go-round, what is the new angular velocity of the system? Model the merry-go-round as a solid disk.
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Answer #2

Mass of child = m = 36.5 kg

Mass of merry-go-round = M = 125 kg

Radius of the merry-go-round = R

Initial angular velocity of the merry-go-round = 1 = 3.1 rad/s

Final angular velocity of the merry-go-round = 2

Moment of inertia of merry-go-round = I = MR2/2

Speed of child at the edge of the merry-go-round = V = 2R

By angular momentum conservation,

I1 = I2 + mVR

(MR2/2)1 = (MR2/2)2 + m(2R)R

M1/2 = M2/2 + m2

(125)(3.1)/2 = 2(125/2 + 36.5)

2 = 1.957 rad/s

The new angular velocity of the system = 1.957 rad/s

answered by: anonymous
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