# Find the moment of inertia

1. Mar 7, 2007

### rcmango

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

heres a very messy drawing for an idea of what were looking at: http://img89.imageshack.us/img89/1764/untitledyi0.png

area moment of inertia Ix0 of a rectangle about the axis x0 that passes through the centroid is: Ix = 1/12bh^3

the moment of inertia about an axis x that is parallel to the x0 is given by: Ix = Ix0 + Ad^2x

where A is the area of the rectangle, and dx is the distance between the two axes.

the inputs are width w, height h, and thickness t.

w = 200mm h = 300 mm, and t = 22mm

find the moment of inertia:

2. Relevant equations

Ix0 = 1/12bh^3

Ix = Ix0 + Ad^2x

3. The attempt at a solution

my questions are..

is t the what plugs into b in this problem?

also, is area of the rectangle just Base * height, so 200 * 300?

not sure how to find the distance between the two axis.

2. Mar 8, 2007

3. Mar 9, 2007

### Gib Z

Re draw your diagram. The reason no one is helping is because instantly, when they see something thats too hard to read, they wont bother. I know thats what I thought when I saw the picture..

4. Mar 13, 2007

### chaoseverlasting

I didnt get your drawing, but from the wording it seems you need to use the parallel axis theorem. If $$I_{cm}=I_0$$ about a certain axis, then $$I_x=I_0+mx^2$$, where x is the distance of the shifted axis from the axis of the center of mass (With moment of inertia $$I_0$$).

5. Mar 14, 2007

### Corkery

how were you able to draw that. Is that found on image shack or somewhere on your computer. sorry for the off-topic comment

6. Apr 18, 2007

### rcmango

Thanks anyways, I was able to figure this out eventually.

Sorry for the awkward drawing. I used paintbrush to draw it, then hosted it through imageshack. However, i recommend hosting a bmp file instead of a low quality jpeg.