Moment of inertia of a car tire problem

[bootfeet]

Homework Statement

A side view of a car tire. Model it as having two sidewall of uniform thickness 0.635 cm and a tread wall of uniform thickness 2.50 cm and width 20.0 cm. Assume the rubber has a uniform density equal to 1.10 X 10^3 kg/(m^3). Radius from the center of the tire to the rubber is 16.5 cm. Radius from the center to the tread of the tire is 33.0 cm.

Homework Equations

None given

The Attempt at a Solution

p = density = m/V
dm = p dV
I = integral of p(r^2) dV
surface mass density: o = pt where t = uniform thickness

 

[jbsteele]

I = integral from r1 to r2 of (p*r^2) drI = p * integral from r1 to r2 of r^2 drI = p * [(r^3)/3] from r1 to r2For the sidewall:I = 1.10 X 10^3 kg/(m^3) * integral from 16.5 cm to 33.0 cm of (r^3/3) dr = 1.10 X 10^3 kg/(m^3) * [33.0^3/3 - 16.5^3/3] = 3.96 X 10^4 kg m^2For the tread wall:I = 1.10 X 10^3 kg/(m^3) * integral from 33.0 cm to 49.5 cm of (r^3/3) dr = 1.10 X 10^3 kg/(m^3) * [49.5^3/3 - 33.0^3/3] = 4.67 X 10^4 kg m^2Surface mass density of sidewall: o = pt = (1.10 X 10^3 kg/(m^3)) * (0.635 cm) = 6.99 X 10^1 kg/m^2Surface mass density of tread wall: o = pt = (1.10 X 10^3 kg/(m^3)) * (2.50 cm) = 2.75 X 10^2 kg/m^2