How Do You Calculate the Moment of Inertia for a Cone?

[christensent]

Homework Statement

Use integration to determine the moment of inertia of a right circular homogeneous solid cone of height H, base radius R, and mass density \rho about its symmetry axis

Homework Equations

Volume of cone = 1/3*pi*r^2*h
I = \int r^2 dm
\rho = m/v

The Attempt at a Solution

m=\rho v
m=1/3 \rho \pi r^2 h
dm=2/3 \rho \pi r h dr

\rho = m/v
\rho = \frac{m}{(1/3 \pi r^2 h)}

I=\int r^2 dm
I=\int \rho \frac{2}{3} \pi r^3 h dr
I = \frac{1}{6} \rho \pi r^4 h dr
(now substituting rho out)
I = \frac{m \pi r^4 h}{2 \pi r^2 h}
MY ANSWER: I = \frac{mr^2}{2}
CORRECT ANSWER: \frac{3mr^2}{10}

I can't figure out where my error is

 

[mg0stisha]

Not completely sure, and i didn't look too intensely, but shouldn't the \frac{2}{3} in I=\int\rho\frac{2}{3}\pi r^{3}hdr be \frac{1}{3}? Once again, I could be wrong.

 

[rl.bhat]

Consider a disc ofmass m, radius r,thickness dh at a distance h from the vertex of the cone.
Its moment of inertia about the axis of symmetry is 1/2*m*r^2.----(1)
Volume of the disc = π*r^2*dh
Density of the cone ρ= M/(1/3*πR^2*H) = 3M/π*R^2*H
There mass of disc m = (3M/π*R^2*H)* π*r^2*dh -------(2)
From the simple geometry it can be shown that
h/H = r/R or dh/H = dr/R or dh = (H/R)*dr
Substitute the value of dh in eq(2) and substitute the value of m in eq.(1).
Now find the integration between the limits r = o to r = R.

 

[christensent]

Thanks! That explains it