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 5.0cm. The disk and platform go from rest to an angular velocity of 8.5 rad/sin 2.5 seconds. Calculate the moment of inertia of the disk? What is its mass? Ignore the mass of the platform. Feel free to look up the moment of inertia of a disk. Show your work and/or explain your reasoning. How would the final angular velocity change if instead of a disk, the experiment was done using a ring with the same mass and radius as the disk? Show your work and/or explain your reasoning.

Explanation / Answer

Torque T =I*alpha

alpha is the angular accelaration = (w-wo)/t = 8.5/2.5 = 3.4 rad/s^2

But Torque is T = r*F = 0.05*m*g = 0.05*50*10^-3*9.8 = 0.0245 N-m

then

0.0245 = I*3.4

moment of inertia is I = 0.007205 Kg-m^2 = 7.205*20^-3 Kg-m^2

But I = 0.5*m*R^2 = 0.007205

0.5*m*0.05^2 = 0.007205

m = 5.764 kg is the mass of the disk

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I_ring = m*R^2 = 5.674*0.05^2 = 0.014185 kg-m^2

T = I_ring*alpha

0.0245 = 0.014185*alpha

alpha = 1.73 rad/s^2

alpha = (w-wo)/t = (w-0)/t = 1.73

w = 1.73*2.5

w = 4.325 rad/sec

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