Moment of Inertia of Washer-like Ring

[InSaNiUm]

First of all hi to everyone. This is my first post, though I've been reading (read: lurking) for a while. A lot of good, smart people willing to help each other. I've learned so much by just browsing...
Anyway, here's my problem:

Homework Statement

Given: A washer-like ring with:
$$\rho =8,000$$ kg/m$${}^2$$,
thickness .01 m,
.250 m outer radius,
and .125 m radius hole cut out of the center.
It sits on top of the x-axis, symmetric about the y-axis, with it's center, $$\inline{I_G}$$ at (0, .25, 0). (z-axis is taken to be orthogonal to your monitor.)
I think I made that clear...
​

We are to find the moment of inertia about the origin, $$\inline{I_O}$$.

Homework Equations

$$\begin{align} dI_G = r^2\,dm\\ I_O = I_G + m\,d^2 \mathrm{(parallel \,axis \,thrm)}\\ dm = \rho\,dV \end{align}$$

The Attempt at a Solution

The way I tried to solve it:
$$\begin{align} dI_G = r^2\,dm\\ dm = \rho\,dV\\ = \rho\,r\,dr\,dz\,d\theta\\ \hookrightarrow dI_G = r^2\,\rho\,r\,dr\,dz\,d\theta\\ I_G = \int_{0}^{2\pi}\!\!\!\int_{0}^{.01}\!\!\!\int_{.125}^{.25}\,\rho{}\,r^3\,dr\,dz\,d\theta\\ ...\hookrightarrow I_G = \rho\,2\pi\,z\,\frac{r^4}{4} \end{align}$$
Before even plugging in the limits this is clearly wrong, but I'm not finding any calculation errors. My logic was that if I integrated over r from the inner to outer radius I wouldn't have to do the problem in two separate parts (disk minus smaller disk).
If I integrate $$\inline{dm}$$ on it's own, solve for $$\inline{\rho}$$ and plug it into my result above, things cancel and I get the sought after $$\frac{m\,r^2}{2}$$
I'm thoroughly confused... This seems recursive. Didn't I include the density when I set up my integrals?
I'm sure there's a concept here that's just eluding me. Anyone feel like explaining this one?
Thanks in advance!

 

[Doc Al]

Before even plugging in the limits this is clearly wrong, but I'm not finding any calculation errors.

What makes you think it is wrong? (Overkill, surely, but not wrong.) Your answer is in terms of density; to relate it to the more familiar equation in terms of mass, plug in $\rho = m/V = m/(\pi r^2 z)$.

My logic was that if I integrated over r from the inner to outer radius I wouldn't have to do the problem in two separate parts (disk minus smaller disk).

That's fine. But unless this is an exercise in calculus, wouldn't it be easier to do it in two parts using the known formula for the rotational inertia of a disk?

 

[InSaNiUm]

I realize that my approach isn't necessarily the simplest. I'm just baffled as to why the answer is numerically different if I use eqn. (6) and plug in the density that was given in the problem; versus simplifying it into terms of mass like you suggested and then solving for said mass seperately and plugging that in. It seems like the equations should be yielding identical results if they are to describe the same quantity of the same system. Again - there's got to be something I'm just not seeing/understanding.

As per your second question: It probably would be easier but I'm supposed to be ``proving'' the formula as part of the process.

And thank you for replying! I appreciate you taking the time to help.

 

[Doc Al]

I realize that my approach isn't necessarily the simplest. I'm just baffled as to why the answer is numerically different if I use eqn. (6) and plug in the density that was given in the problem; versus simplifying it into terms of mass like you suggested and then solving for said mass seperately and plugging that in. It seems like the equations should be yielding identical results if they are to describe the same quantity of the same system. Again - there's got to be something I'm just not seeing/understanding.

Are you sure you are plugging numbers into your equation (6) correctly? I get the same answer either way.

Remember that each variable in the triple integral is evaluated independently. That means you'd get a value of:
$$I_G = \rho\,2\pi\,z\,\frac{r^4}{4} = \rho 2 \pi \frac{(0.01)}{4} (r_2^4 - r_1^4)$$