Finding the moment of inertia for a specific figure (illustrated)

AI Thread Summary
The discussion focuses on calculating the moment of inertia for a specific figure, with users referencing a provided image for dimensions. The solution involves understanding the symmetry of the figure, specifically using the vertical length h/2 to account for both halves. There is confusion regarding the derivation of a particular fraction used in the calculations, with hints that it relates to similar triangles within the figure. Users emphasize the importance of identifying the base of the smaller triangle in relation to the larger triangle's dimensions. Overall, the conversation aims to clarify the steps needed to arrive at the correct moment of inertia.
PhyIsOhSoHard
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Homework Statement


http://img577.imageshack.us/img577/2069/inertia.gif

The following figure with these known lengths are given. I need to find the moment of inertia.

Homework Equations


According to my book's facit, this is the solution:
http://img845.imageshack.us/img845/6725/inertiaformula.png

The Attempt at a Solution


This is what I understand so far...
The fraction h/2 refers to half the figure (when cut from horizontal). So you start from the middle of the figure, and then move up to the end in which case you have traveled the vertical length h/2. Then by multiplying the entire integral with 2, you take the bottom half of the figure into account as well (since it's symmetrical).

I have no idea about the fraction. I've tried, but I have no idea how they came up with that. All I know is that "y" is the vertical axis for the figure...
 
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Welcome to PF!

Hi PhyIsOhSoHard! Welcome to PF! :smile:
PhyIsOhSoHard said:
I have no idea about the fraction. I've tried, but I have no idea how they came up with that. All I know is that "y" is the vertical axis for the figure...

it's similar triangles …

you want the base of the smaller triangle, and you know that the base of the larger triangle is b :wink:
 

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