- #1
fajoler
- 11
- 1
Homework Statement
A metal sign for a car dealership is a thin, uniform right triangle with base length b and height h. The sign has mass m. What is the moment of inertia of the sign for rotation about the side of length h?
Homework Equations
I = [tex]\int[/tex]r[tex]^{2}[/tex]dm
The Attempt at a Solution
Alright so I started off by trying to convert the integral so that it is in terms of dr rather than dm. So, I calculated that dm = [tex]\rho[/tex] dA.
then from there, the integral can be written as I = [tex]\oint[/tex]r[tex]^{2}[/tex][tex]\rho[/tex](h/b)rdr
A little bit of simplifying will give us I = [tex]\int[/tex]r[tex]^{3}[/tex][tex]\rho[/tex](h/b)dr
bounded from 0 to b.
If we solve the integral, we get I = (1/4)b[tex]^{4}[/tex][tex]\rho[/tex](h/b) - 0
Which we can simplify further to I = (1/4)b[tex]^{3}[/tex][tex]\rho[/tex]h
The only problem is, the question asks for I to be in terms of the given values, so I have to somehow change [tex]\rho[/tex] into something in terms of b, h, and m. I'm not sure how to do this though because a 2D object shouldn't even have density, should it?
Thanks for the help!