A cylinder of mass 250 kg and radius 2.60 m is rotating at 4.00 rad/s on a frict

Question

A cylinder of mass 250 kg and radius 2.60 m is rotating at 4.00 rad/s on a frictionless surface when two more identical non-rotating cylinders fall on top of the first. Because of friction between the cylinders they will eventually all come to rotate at the same rate. What is this final angular velocity

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A cylinder of mass 250 kg and radius 2.60 m is rotating at 4.00 rad/s on a frictionless surface when two more identical non-rotating cylinders fall on top of the first. Because of friction between the cylinders they will eventually all come to rotate at the same rate. What is this final angular velocity

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Explanation / Answer

I = mr^2/2 = 250*2.6^2/2 = 845 kg m^2

after falling 2 more cylinder, I' = 845 + 2*845 = 2535 kg m^2

using conservation of angular momentum

Iw = I'w'

w' = Iw/I' = 845*4/2535 = 1.33 rad/s

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