The wheels of a wagon can be approximated as the combination of a thin outer hoo

Question

The wheels of a wagon can be approximated as the combination of a thin outer hoop, of radius rh = 0.262 m and mass 4.51 kg, and two thin crossed rods of mass 9.95 kg each. You would like to replace the wheels with uniform disks that are 0.0462 m thick, made out of a material with a density of 7830 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?

I tried r=0.2278 meters but was marked wrong.

Explanation / Answer

moment of inertia remain same

I = Ihoop + 2 * Irod

Ihoop= mr^2 = 4.51 x (0.262 ^2) = 0.3096 kg .m^2

Irod = ML^2/12

L = 2 * Rhoop

Irod = 9.95 x ( 2 * 0.262)^2/12

I = Ihoop + 2*Irod = 0.3096 + 0.455 = 0.7646 kg.m^2

the inertia for disk

I = mr^2/2

m = rho * V

V = pi * r^2 * 0.0462

m = (pi *r^2 * 0.0462 ) * 7830

I = pi * r^4* 0.0462 * 7830 /2

r^4 = 2I / 7830 * pi * 0.0462

r = 0.1915 m

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