Moment of inertia for a cone

Homework Statement

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

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?

The Attempt at a Solution

Related Introductory Physics Homework Help News on Phys.org
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
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.