# Moment of inertia equations

1. Nov 25, 2012

### aaronfue

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

I was given a formula sheet that shows the moment of inertia equations for three shapes: rectangle, circle, and triangle.

2. Relevant equations

There seems to be two sets of MOI equations.

Here are the rectangular equations:

Rectanglular:
Ix=$\frac{bh^3}{3}$, Iy=$\frac{hb^3}{3}$

$\bar{I}$x=$\frac{bh^3}{12}$, $\bar{I}$y=$\frac{hb^3}{12}$

What is the difference between the two, besides the denominators and order of variables? How can I remember which ones to use and when to use them?

Last edited: Nov 25, 2012
2. Nov 25, 2012

### SteamKing

Staff Emeritus
The equation is selected depending about which axis the inertia is calculated.
Check your sheet with the diagrams of the figures.

3. Nov 25, 2012

### aaronfue

I've attached the diagram that I'm using. I'm going to be finding the MOI of both axes so I know I have to use both equations, but there are the equations with the "bar" and ones without? That's what is confusing.

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4. Nov 25, 2012

### SteamKing

Staff Emeritus
The equations with the bar denote the inertia about axes through the centroid of the figure. The equations with out the bar denote the inertia about some other axes, as shown on your diagram.
In your figure, the centroidal axes are labeled x0 and y0.

5. Nov 25, 2012

### aaronfue

Using the parallel axis theorem, I would create a " x' " axis through the centroid for all of the figures, which would be 0.5 in from the original x axis. And then I would use the x-bar equation? Then for the y axis I would use the equations without the bar? I just want to make sure that I understood your response.

6. Nov 25, 2012

### SteamKing

Staff Emeritus
Take the rectangle for instance.
Using the centroidal axes x0-y0, Ix with the bar is (bh^3)/12
Applying the parallel axis theorem to find Ix, then Ix = Ix-bar + Ad^2
Ix = (bh^3)/12 + bh * (h/2)^2 = (bh^3)/12 + (bh^3)/4
Ix = (bh^3)*(1+3)/12 = 4(bh^3)/12 = (bh^3)/3