# Moment Inertia w/ density F(x)

fisselt

## Homework Statement

Calculate the moment of inertia of a uniform rigid rod of length L and mass M lying along the x-axis which rotates about an axis perpendicular to the rod (the y axis) and passing through it’s center of mass. The rod has a line density that is a function of location such that =3x.

## Homework Equations

$\int ρr^2 dm$

## The Attempt at a Solution

I thought this would be done simply by replacing ρ with 3x and r with x. However, while integrating I just keep getting 0 as my answer.
$\int ρr^2 dm$ dm=ρdx=3xdx
=$\int 3x(x)^2 dx$
=$\frac{3x^4}{4}$

Now when I evaluate from L/2 and -L/2 I always get zero.

I assume I'm setting up my integral incorrectly as I should probably have an odd exponent that I totally forgot how to do calculate an integral over the summer. Appreciate the help.

Homework Helper
hi fisselt!
… Now when I evaluate from L/2 and -L/2 I always get zero.

hardly surprising …

half of your rod has negative density!

fisselt
Should I be evaluating from 0 to L then?

Meaning it would be (3L^4)/4

Homework Helper
Calculate the moment of inertia of a uniform rigid rod of length L and mass M lying along the x-axis which rotates about an axis perpendicular to the rod (the y axis) and passing through it’s center of mass. The rod has a line density that is a function of location such that =3x.
Should I be evaluating from 0 to L then?

Meaning it would be (3L^4)/4

(try using the X2 button just above the Reply box )

hmm … i don't really understand the question …

where is x being measured from?

and how can it be uniform if ρ = 3x ?

fisselt
I think this may be part of my problem as well. I know typically with rods spinning about the perp. axis we evaluate from the center point (hence the L/2 AND -L/2). For all the other problems we have done this has worked but I have never done one with density as a function. I expected an answer that would be at least similar to the general form but I can't find it. I'm stuck doing it the same way over and over.

Home internet is down so no TEX on this phone, too time consuming.

Arghzoo
Because the line density is ρ=3x, wouldn't the center of mass not be in the center? And so when evaluating the integral of inertia wouldn't the bound not be from -L/2 to L/2? You'd first have to find the location of the actual center of mass, and then evaluate from the endpoints when the center of mass is zero.

Arghzoo
When making the substitution from the original form of the inertia integral you substitute M*R^2/L dr. But M/L is simply the linear density (for any uniform object). So when substituting in dm=ρ*r^2 dr you can use the 3x even though you have a negative lower bound because definite integrals allow for that (just keep in mind that you have to evaluate where the center of mass is placed at x=0).

Sorry I don't know how to use LaTeX lol.

fisselt
Isn't this effectively what I've done originally? Perhaps the range is incorrect and I should figure center of mass of a 1 meter rod of this varying density? Something like -(L2)/3 to L/3.

Arghzoo
Something like that.

You know how to perform a center of mass integral, right?

integrate 3*x^2 dx from 0 to L and divide by M, giving L^3/M.

So that's the center of mass. It seems like it should be from -2L/3 to L/3, but it's from -L^3/M to (L-L^3/M) because the actual calculated center of mass is at L^3/M.

Arghzoo
The length could still be L, but you're right on with the -L/3 to 2L/3 if the mass of the rod is 1kg.

Arghzoo
Also, the M in the center of mass is not M given, but mass obtained when integrating the linear density across the rod. Working through the calculations, we find the center of mass at 2L/3, which verifies your guess.

fisselt
While integrating with the 2/3 and 1/3, however I'm getting a negative answer. Should I switch the signs in this case?

Arghzoo

fisselt

fisselt
I got - (5L^4)/36

Arghzoo
If you switch the bounds, you'd get the same answer, just positive. That should be right...

integral(3*x^3 dx) from -L/3 to 2L/3

?

Arghzoo
Yeah if you switch the bounds you get +5L^4/36

fisselt
Yeah, that's what I was figuring. Thanks a lot for the help. This discussion helped me out a lot!

Arghzoo
No problem! Good luck with everything else.