What is the Moment of Inertia for a Rectangle?

AI Thread Summary
The discussion focuses on calculating the moment of inertia for a rectangle using the area moment of inertia formula. The moment of inertia about the centroidal axis is given by Ix0 = 1/12bh^3, while the parallel axis theorem is applied to find the moment of inertia about a different axis. Participants clarify that the thickness t is not used for b in this context, and the area is calculated as width times height. There is also advice to improve the clarity of diagrams for better understanding. Ultimately, the original poster was able to resolve their confusion regarding the calculations.
rcmango
Messages
232
Reaction score
0

Homework Statement



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:


Homework Equations



Ix0 = 1/12bh^3

Ix = Ix0 + Ad^2x


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.

please help.
 
Last edited by a moderator:
Physics news on Phys.org
help please?
 
Re draw your diagram. The reason no one is helping is because instantly, when they see something that's too hard to read, they won't bother. I know that's what I thought when I saw the picture..
 
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).
 
how were you able to draw that. Is that found on image shack or somewhere on your computer. sorry for the off-topic comment
 
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.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
Back
Top