Moment of inertia for a cone

1. Oct 19, 2011

jforce93

1. The problem statement, all variables and given/known data
Find the moment of inertia of a right circular cone of radius r and height h and mass m

2. Relevant equations

I = ∫r2 dm
V = 1/3*π*r2*h

3. 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?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 19, 2011

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.

3. Oct 19, 2011

vela

Staff Emeritus
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?