What is the Moment of Inertia for a Rectangle?

Click For Summary

Homework Help Overview

The discussion revolves around calculating the moment of inertia for a rectangle, specifically using the area moment of inertia formulas and the parallel axis theorem. The original poster provides dimensions for the rectangle and seeks clarification on the application of these concepts.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster questions whether the thickness of the rectangle should be used as the base in the moment of inertia formula and seeks confirmation on calculating the area. There is uncertainty regarding how to determine the distance between the two axes involved in the calculations.

Discussion Status

Some participants have offered guidance regarding the use of the parallel axis theorem and the importance of clear diagrams. The conversation reflects a mix of attempts to clarify the problem and questions about the original poster's approach.

Contextual Notes

The original poster's drawing was noted as unclear, which may have contributed to the lack of immediate responses. There is also a mention of the need for better image quality in future submissions.

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 [tex]I_{cm}=I_0[/tex] about a certain axis, then [tex]I_x=I_0+mx^2[/tex], where x is the distance of the shifted axis from the axis of the center of mass (With moment of inertia [tex]I_0[/tex]).
 
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.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
Replies
3
Views
4K
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
15
Views
4K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K