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 = 0.209 m and mass 4.89 kg, and two thin crossed rods of mass 7.37 kg each. You would like to replace the wheels with uniform disks that are 0.0651 m thick, made out of a material with a density of 8290 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? Number rd0.240 Im d-? 0.0651 m Incorrect

Explanation / Answer

Moment of inertia of current wheel

I = (M)(rh)^2 + 2[(1/12)(m)(2 rh)^2]

I = (M)(rh)^2 + (2/3)(m)(rh)^2

I = (rh)^2[M + (2/3)(m)]

I = (0.209 m)^2*(4.89 kg + 2/3*7.37 kg)

= 0.428 kg m^2

New disc wheel:

I = (1/2)(md)(rd)^2

where, md = pih(rd)^2 * pd [ mass = volume * density]

= pi * 0.0651 m * (rd)^2 * 8290 kg/m^3

= 1695.45*(rd)^2 kg

Therefore,

I = 1/2*md*(rd)^2

0.428 kg m^2 = 1/2*1695.45*(rd)^4

rd^4 = 0.428*2/1695.45

rd = (0.428*2/1695.45)^(1/4)

= 0.15 m

ANSWER

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