In the apparatus below a 50 g mass is hung from the pulley and released. A disk

Question

In the apparatus below a 50 g mass is hung from the pulley and released. A disk is placed on the platform. The platform has a radius of 5.0 cm. The disk and platform go from rest to an angular velocity of 8.5 rad/s in 2.5 seconds. What is the moment of inertia of the disk? What is its mass? Ignore the mass of the platform. Show your work and/or explain your reasoning. How would your answer change if instead of a disk, the experiment was done using a ring with the same mass as the disk? Show your work and/or explain your reasoning.

Explanation / Answer

angular accelration of disk alpha = (wf-wi)/ t = (8.5-0)/2.5 = 3.4 rad/s^2

tangetial acceleration a = r*alpha = 0.05*3.4 = 0.17 m/s^2

for the hanging mass

mg - T = m*a

T = mg - ma

for the disk

T*r = Ialpha

(mg - ma)*r = Ialpha

(0.05*9.8 - 0.05*0.17)*0.05 = I*3.4

I = 0.0071 kg m^2 <<<----------answer

Idisk = (1/2)*mdisk*r^2

0.0071 = (1/2)*mdisk*0.05^2

mdisk = 5.68 kg   <<<----------answer

====================

Iring = mring*r^2

T*r = I*alpha

m*(g-a) = I*wf/t

but a = r*alpha = r*wf/t

m*(g-(r*wf/t)*r = mring*r^2*wf/t

0.05*(9.8-(0.05*wf/2.5)) = 5.68*0.05*wf/2.5

wf = 4.27 rad/s <<<---------answer

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