The wheels of a wagon can be approximated as the combination of a thin outer hoop, of radius r h = 0.262 m and mass 5.46 kg , and two thin crossed rods of mass 8.66 kg each. A farmer would like to replace his wheels with uniform disks t d = 0.0525 m thick, made out of a material with a density of 5530 kg 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?

0.136m

Explanation:

Given

rh = 0.262m

m1 = 5.46 kg

m2 = 8.66 kg

td = 0.0525m

density, ρ= 5530 kg/m³

Calculating the moment of inertia of the new disk;

M1 = m1r²

M1 = 5.46 * 0.262²

M1 = 0.375 kg·m²

Calculating the moment of inertia of the rod

M2 = m2r²

M2 = 8.66 * (2 * 0.262)²/12

M2 = 0.198 kg·m²

Moment of inertia of the old wheel;

M3 = M1 * M2

M3 = 0.375 + 2 * 0.198

M3 = 0.771 kg·m²

Calculating Moment of Inertia of the disk

M4 = mr²/2

To get the mass of the disk, W

We'll need to calculate the mass from its volume V and density ρ

m = V * ρ

Where V = Area * Thickness

m = Area * Thickness * ρ

m = πr² * 0.0525 * ρ

m = 0.525ρπr²

So, M4 = 0.525ρπr² * r²/2

M4 = 0.2625ρπr⁴/2

Solving for r

r⁴ = M4/0.2625ρπ

M4 must equal M3 = 0.771

So, r⁴ = 0.771/0.2625ρπ/2

r⁴ = 0.771/(0.2625 * 5530 * 3.14/2)

r⁴ = 0.000338298667044

r = (0.000338298667044)^¼

r = 0.135620415830572

r = 0.136m --- Approximately

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