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

1. Apr 9, 2012

### PhyIsOhSoHard

1. The problem statement, all variables and given/known data
http://img577.imageshack.us/img577/2069/inertia.gif [Broken]

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

2. Relevant equations
According to my book's facit, this is the solution:
http://img845.imageshack.us/img845/6725/inertiaformula.png [Broken]

3. 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 travelled 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...

Last edited by a moderator: May 5, 2017
2. Apr 10, 2012

### tiny-tim

Welcome to PF!

Hi PhyIsOhSoHard! Welcome to PF!
it's similar triangles

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