# Moment Inertia w/ density F(x)

1. Aug 17, 2012

### fisselt

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

2. Relevant equations
$\int ρr^2 dm$

3. 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.

2. Aug 17, 2012

### tiny-tim

hi fisselt!
hardly surprising …

half of your rod has negative density!

3. Aug 17, 2012

### fisselt

Should I be evaluating from 0 to L then?

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

4. Aug 17, 2012

### tiny-tim

(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 ?

5. Aug 17, 2012

### 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.

6. Aug 18, 2012

### 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.

7. Aug 18, 2012

### 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.

8. Aug 18, 2012

### 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.

9. Aug 18, 2012

### 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.

10. Aug 18, 2012

### 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.

11. Aug 18, 2012

### 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.

12. Aug 18, 2012

### 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?

13. Aug 18, 2012

### Arghzoo

14. Aug 18, 2012

### fisselt

15. Aug 18, 2012

### fisselt

I got - (5L^4)/36

16. Aug 18, 2012

### 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

?

17. Aug 18, 2012

### Arghzoo

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

18. Aug 18, 2012

### fisselt

Yeah, thats what I was figuring. Thanks a lot for the help. This discussion helped me out a lot!

19. Aug 18, 2012

### Arghzoo

No problem! Good luck with everything else.