# Moment of inertia of a rod with varying density

1. Jan 21, 2009

### bigevil

1. The problem statement, all variables and given/known data

A thin rod is 10 ft long and has a density which varies uniformly between 4 and 24 lb/ft. Find:
a) the mass
b) the x-coordinate of centroid
c) Moment of inertia about an axis perpendicular to the rod
d) Moment of inertia about an axis perpendicular to the rod passing thru the heavy end.

3. The attempt at a solution

a) isn't difficult. I have got $$dm = (2x+4) dx$$, then integrate between 0 and 10 to get 140 lbs.

b) From $$\int \bar{x} dm = \int x dm$$, integrate and use earlier results to get 130/21.

c) I'm stuck at this one. I assume this means that the axis of MI passes through the centre of mass, so I set up the integration limits from 80/21 to -130/21 (rod is 10 foot). Then, taking $$I = \int x^2 dm = \int_{-130/21}^{80/21} 2x^3 + 4x^2 dx = 1.7(140) = 1.7m$$ (where m=140 lbs). But the answer (this question is from Mary L Boas' mathematical methods book) is 6.92m.

Am I missing something here? I am only calculating for the x-coordinate because the rod is "thin".

2. Jan 21, 2009

### darkSun

I haven't actually calculated anything, but the problem might be that in your integral you are making "x" refer to two different things.

In your expression for dm, x is referring to distance from the light end of the rod. But since you are finding moment of inertia and passing the axis through the center of mass, the "x^2" in the integral is referring to distance from the center of mass. The two reference points are not the same.

Maybe if you replaced x^2 by (130/21 - x)^2, it would end up working.

Oh, checked it and it does work. But you have to change the limits to 0 to 10, since I made x the distance from the light end.

3. Jan 22, 2009

### bigevil

Oh yeah... thanks Darksun... :)