The wheels of a wagon can be approximated as the combination of a thin outer hoop, of radius rh = 0.156 m and mass 5.65 kg, and two thin crossed rods of mass 9.95 kg each. You would like to replace the wheels with uniform disks that are 0.0588 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? in meters

To realize this problem we must calculate the moment of inertia of the wheel formed by a thin circular ring plus the two bars with an axis that passes through its center.

The moments of inertia of the bodies are additive quantities whereby we can add the mounts inertia of the ring and the two bars.

Moment of inertia ring I1 = MR²

Moment of inertia bar I2 = 1/12 ML²

Moment of inertia disk I3 = ½ mR²

Let's calculate the moment of inertia of the wheel

I = I1 + 2 I2

I = MR² + 2 1/12 ML²

The length of the bar is ring diameter

L = 2R

I = 5.65 0.156² + 1/6 9.95 (2 0.156)²

I = 0.1375 + 0.1614

I = 0.2989 kg m²

This is the same moment of inertia of the solid disk,

Disk

I3 = I

I3 = ½ MR²

They give us disk density

ρ = M / V

M = ρ V

M = ρ (pi R² e)

Done is the thickness of the disc, in general it is e= 1 cm = 0.01 m

Let's replace

I3 = ½ ( ρ π R²) R²

I3 = ½ ρ π e R⁴

R⁴ = 2 I3 / ( ρ π e)

R = ( 2 I3 / ( ρ π e)

[tex])^{1/4}[/tex]

R⁴ = 2 0.2989 / (5990 π 0.01)

R = 0.237 m