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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:

[tex]\rho =8,000[/tex] kg/m[tex]{}^2[/tex],

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, [tex]\inline{I_G}[/tex] 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, [tex]\inline{I_O}[/tex].

[tex]

\begin{align}

dI_G = r^2\,dm\\

I_O = I_G + m\,d^2 \mathrm{(parallel \,axis \,thrm)}\\

dm = \rho\,dV

\end{align}

[/tex]

The way I tried to solve it:

[tex]

\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}

[/tex]

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 [tex]\inline{dm}[/tex] on it's own, solve for [tex]\inline{\rho}[/tex] and plug it into my result above, things cancel and I get the sought after [tex]\frac{m\,r^2}{2}[/tex]

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!

Anyway, here's my problem:

## Homework Statement

**Given:**A washer-like ring with:

[tex]\rho =8,000[/tex] kg/m[tex]{}^2[/tex],

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, [tex]\inline{I_G}[/tex] at (0, .25, 0). (z-axis is taken to be orthogonal to your monitor.)

I think I made that clear...

## Homework Equations

[tex]

\begin{align}

dI_G = r^2\,dm\\

I_O = I_G + m\,d^2 \mathrm{(parallel \,axis \,thrm)}\\

dm = \rho\,dV

\end{align}

[/tex]

## The Attempt at a Solution

The way I tried to solve it:

[tex]

\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}

[/tex]

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 [tex]\inline{dm}[/tex] on it's own, solve for [tex]\inline{\rho}[/tex] and plug it into my result above, things cancel and I get the sought after [tex]\frac{m\,r^2}{2}[/tex]

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!

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