What is the moment of inertia of a cone spinning about its symmetry axis?

[oliveyew1]

Homework Statement

Find the moment of inertia and center of mass of:
A uniform cone of mass M, height h, and
base radius R, spinning about its symmetry
(x) axis.

Homework Equations

I = ∫R^2dm

The Attempt at a Solution

I tried using I =∫R^2dm, solving for dm I got dm=(M/V)dV, with dV = piR^2*dx. Thus, ∫R^2*(M/V)piR^2*dx V = 1/3*piR^2*h, so ∫R^2*M*(1/(1/3*piR^2*h))piR^2dx. Pi and R^2 cancel, so ∫3(M/h^3)R^2x^2dx, which gets me MR^2, and the right answer is (3/10)MR^2

Homework Statement

 

[Simon Bridge]

The idea is that r is the perpendicular distance from the rotation axis of a small mass dm ... the moment of inertia of that mass is r²dm ... and you add up all the wee masses. It looks to me that you may have attempted to use a method intended for a point mass on a cylindrical one.

You appear to have adopted the strategy of slicing the cone into disks with x being the rotational axis. So the disks have thickness $dx$ and area $\pi r^2$ did you realize that the radius of the disk at x s a function of x?

You have a better shortcut though - you already know the moment of inertia of a disk ;) So why not just add them up? $$I=\int_0^h I_{disk}(x)dx$$

The best way to handle these is to write down you reasoning at each step - in words.
Avoids the need for this kind of guesswork on the part of people checking and/or marking your work ;)