A uniform disk has a mass of 4.9 kg and a radius of 0.25 m. The disk is mounted

Question

A uniform disk has a mass of 4.9 kg and a radius of 0.25 m. The disk is mounted on frictionless bearings and is used as a turntable. The turntable is initially rotating at 60 rpm. A thin-walled hollow cylinder has the same mass and radius as the disk. It is released from rest, just above the turntable, and on the same vertical axis. The hollow cylinder slips on the turntable for 0.20 s until it acquires the same final angular velocity as the turntable.
(I_disk=0.5MR2, I_ring=MR2)
What is the final angular velocity of the disk in rpm?
What is the kinetic coefficient of friction between the disk and hollow cylinder?

Explanation / Answer

Use law of conservation of angular momentum

(I_disk*w1) = (I_disk+I_ring)*w2

I-disk =0.5*M*R^2 = 0.5*4.9*0.25*0.25 = 0.153125 kg*m^2

I_ring = M*R^2 = 4.9*0.25*0.25 = 0.30625 kg*m^2

then (0.153125*60*2*3.142/60) = (.153125+0.30625)*w2

w2 = 2.09 rad/s = 2.09*60/(2*3.142) = 19.95 rpm

alpha = w2/t = 2.09/0.2 = 10.45 rad/s^2

accealration a = r*alpha = 0.25*10.45 = 2.6125 m/s^2

frictional force f = m*a

mu_k*m*g*2 = m*a

mu_k = a/(2*g) = 2.6125/(2*9.81) = 0.133

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