Question

The wheels of a wagon can be approximated as the combination of a thin outer hoop, of radius rh = 0.156 m and mass 4.70 kg, and two thin crossed rods of mass 9.09 kg each. You would like to replace the wheels with uniform disks that are 0.0525 m thick, made out of a material with a density of 7370 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

here,

mass of hoop , m1 = 4.7 kg

rh = 0.156 m

mass of each rod , m2 = 9.09 kg

the moment of inertia of old wheel ,I1 = m1 * rh^2 + 2 * m2 * (2 * rh)^2 /12

I1 = 4.7 * 0.156^2 + 2 * 9.09 * ( 2 * 0.156)^2 /12 kg.m^2 = 0.2619 kg.m^2

thickness of disk , t = 0.0525 m

let the radius of new disk be r2

the volume of new wheel , V = pi * r2^2 * t

density , p = 7370 kg/m^3

as they have same moment of inertia

0.5 * ( mass of new disk) * r2^2 = I1

0.5 * ( volume of new disk * p) * r2^2 = I1

0.5 * ( pi * r2^2 * 0.0525 * 7370) * r2^2 = 0.2619

solving for r2

r2 = 0.14 m

the radius of new disk is 0.14 m