Finding moment of inertia for a right circular cone

[jforce93]

Homework Statement

Find the moment of inertia of a right circular cone of radius r and height h and mass m

Homework Equations

I = ∫r² dm
V = 1/3*π*r²*h

The Attempt at a Solution

Assume density is p

dm = p dv
divide both sides by dr
dm/dr = p dv/dr

dm/dr = p (d/dr * 1/3*π * r²*h)
so
dm/dr = p(2/3)*πrh
so:
dm = (2/3)pπrh dr

Sub that into the moment of inertia equation

∫(2/3)pπrh dr = I

I = (1/3) pπr²h
p = m/v
I = (1/3)(m/v)πr²h
I = v*(m/v)
I = m

What am I doing wrong?

 

[LawrenceC]

Hint: Set the problem up as a double integral problem. Lay the problem out with the cone's longitudinal axis being the x-axis. Use a ring for your dV.

If you need more hints, let me know.

 

[vela]

You forgot the factor of r² multiplying dm. What you found was $\int dm$, which unsurprisingly turns out to equal the mass of the cone.

About what axis are you supposed to be calculating the moment of inertia?