Question

The wheels of a wagon can be approximated as the combination of a thin outer hoop, of radius rh = 0.209 m and mass 4.51 kg, and two thin crossed rods of mass 7.80 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?

Answer 1

Expressing what we know in equation form:

(1) Id must = Io,

where:
Id is the moment of inertia (MoI) of the new disk; and
Io = MoI of the old wheel

(2) Io = Ihoop + 2 * Irod

(3) Ihoop = Mhoop * Rhoop^2

= 4.51 * 0.209^2

= 0.197 kg·m^2

(4) Irod = Mrod * Lrod^2 / 12

= 7.80 * (2 * Rhoop)^2 / 12

= 7.80 * (2 * 0.209)^2 / 12

= 0.1136 kg·m^2

Substituting the values into (2) we get:

(5) Io = 0.197 + 2 * 0.1136

= 0.4242 kg·m^2

The MoI for a disk is:

(6) Id = m * r^2 / 2

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

(7) m = V * ρ

The volume is area X thickness:

(8) V = π * r^2 * 0.0588

Substituting into (7):

(9) m = (π * r^2 * 0.0588) * ρ

and substituting for m into (6):

(10) Id = ((π * r^2 * 0.0588) * ρ) * r^2 / 2

= π * r^4 * 0.0588 * ρ / 2


Solving for r the radius:

(11) r^4 = Id / (π * 0.0588 * ρ / 2)

Since Id must equal Io, which we have from (5),

(12) r = ( Io / (π * 0.0588 * ρ / 2) )^(1/4)

= ( 0.4242 / (3.14 * 0.0525 * 5990 / 2) )^(1/4)

= 0.1712 m = radius of the disk