Moment of Inertia of a Right Triangle

[fajoler]

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 = \intr^{2}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 = \rho dA.

then from there, the integral can be written as I = \ointr^{2}\rho(h/b)rdr


A little bit of simplifying will give us I = \intr^{3}\rho(h/b)dr

bounded from 0 to b.

If we solve the integral, we get I = (1/4)b^{4}\rho(h/b) - 0


Which we can simplify further to I = (1/4)b^{3}\rhoh



The only problem is, the question asks for I to be in terms of the given values, so I have to somehow change \rho 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!

 

[Delphi51]

It has mass and area, so it has density = mass/area = m/(.5bh).

I fear you may have a mistake in the early stage where you have not shown enough detail for me to follow. I made a diagram showing a "slice" of the triangle at x, part way between x=0 and b along the base. The slice is trapezoidal but can be safely approximated as a rectangle. It's area dA = y*dx where y is the height of the slice. I think your r is my x. Tricky little relationship between y and x; I guess it is the formula for the line forming the hypotenuse. If I were you I would write out those steps carefully to make sure you have the right expression before integrating.

 

[fajoler]

so y = mx + b and we can calculate that y = (-h/b) x assuming the point of rotation is aligned at the origin right?which means dA = (-h/b) x dxNow since r is the distance from the origin, or the point of rotation, can we say that r is equal to x and dx is equal to dr?then now that we know that density = mass/area, density equals mass/(0.5*b*h) right?
This is how I derived my integral for Inertia, but it gives me the wrong answer. What did I do wrong?
By the way the answer I got was
I = 1/2 M b^2

 

[Delphi51]

so y = mx + b and we can calculate that y = (-h/b) x assuming the point of rotation is aligned at the origin right?

It says "rotation about the side of length h" so it seems to me that at x = 0, we must have y = h. At x = b, y = 0. If you plug those two points into y = mx + b, you'll get a different and more complex formula for y.

 

[fajoler]

ah youre right.so would the equation for y then be y = -(h/b)x + h?If I use this as my equation for y, and plug it into the integral for inertia, I come up with an answer of 1/6 Mb^2.this is the correct answer. thanks for walking me through this problem. I greatly appreciate it!

 

[Delphi51]

Most welcome!