Question

The wheels of a wagon can be approximated as the combination of a thin outer hoop, of radius ????h=0.209 m and mass 4.32 kg , and two thin crossed rods of mass 7.37 kg each. A farmer would like to replace his wheels with uniform disks ????d=0.0462 m thick, made out of a material with a density of 5990 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?

Answer 1

Answer:

r= 98.3 mm

Explanation:

For rim

R= 0.209 m

M= 4.32 kg

For rods

m= 7.37 kg

L= 2 R= 2 x 0.209 = 0.418 m

The Total moment of inertia of the  wagon

I=MR²+2 x 1/12 m L²

Now by putting the values

$I=4.32\times 0.209^2+2\times \dfrac{1}{12}\times 7.73\times 0.418^2\ kg.m^2$

I=0.413 kg.m²

For disk:

t= 0.0462 m

Density ρ = 5990 kg/m³

Lets take r is the radius of disk

So the mass of the disc

m'=ρ πr² t

The moment of inertia of disc

I'=1/2 m'r²

I'=1/2 x r² x ρ πr² t

Given that

I = I'

1/2 x r² x ρ πr² t = 0.413 kg.m²

1/2 x r³ x ρ π t = 0.413

r³ x ρ π t = 0.826

$r^3=\dfrac{0.826}{\pi \times 5990 \times 0.0462}$

r³=0.00095

r=0.0983 m

r= 98.3 mm