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The wheels of a wagon can be approximated as the combination of a thin outer hoop, of radius rh = 0.315 m and mass 5.65 kg, and two thin crossed rods of mass 9.09 kg each. You would like to replace the wheels with uniform disks that are 0.0525 m thick, made out of a material with a density of 5990 kilograms per cubic meter. If the new wheel is to have the same moment of inertia about its center as the old wheel about its center, what should the radius of the disk be?
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Added Sun, 26 Jul '15
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let the radius of the disk be R.

Mass of new disk wheel, M2 = 5990 *(pi*R^2 *0.0525) = 987.45 *R^2

Moment of inertia of old wheel = I(hoop) +I(rods)
=> I1 = [5.65*(0.315)^2] + [2*9.09*(2*0.315)^2]/12
=> I1 = 0.56 + 0.6 = 1.16 kg-m^2

Moment of inertia of new wheel = M2*R^2/2
=> I2 = [(987.45*R^2)*R^2]/2

Since, I1 = I2
=> [(987.45*R^2)*R^2]/2 = 1.16
=> R = 0.22 m
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Added Sun, 26 Jul '15
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