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