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What is the new angular velocity when a man moves on a rotating merry-go-round?
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[QUOTE="BrainMan, post: 4795895, member: 508011"] [h2]Homework Statement [/h2] A merry-go-round rotates at an angular velocity of 0.2 rev/s with an 80 kg man standing at a point 2 m form the axis of rotation. what is the new angular velocity when the man walks to a point 1 m from the center? Assume the merry-go-round is a solid cylinder of mass 25 kg and radius 2 m. [h2]Homework Equations[/h2] I= MR^2 L = Iω [h2]The Attempt at a Solution[/h2] First I found the moment of inertia of the platform 25 x 4 = 100 Then I found the original moment of inertia of the man 80(4) = 320 Then I found the total angular momentum L = (100 + 320)0.2/2pi L = 42/pi Then I found the moment of inertia after the man moved I = 80(1) I = 80 Then I compared the old momentum to the new momentum 42/pi = 180ω ω = .074 rad/sec .074 x 2pi = .467 rev/sec The correct answer is .569 rev/sec [/QUOTE]
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What is the new angular velocity when a man moves on a rotating merry-go-round?
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