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Homework Help: Moment Inertia w/ density F(x)

  1. Aug 17, 2012 #1
    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
    [itex]\int ρr^2 dm[/itex]

    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.
    [itex]\int ρr^2 dm[/itex] dm=ρdx=3xdx
    =[itex]\int 3x(x)^2 dx[/itex]

    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. jcsd
  3. Aug 17, 2012 #2


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    hi fisselt! :smile:
    hardly surprising …

    half of your rod has negative density! :biggrin:
  4. Aug 17, 2012 #3
    Should I be evaluating from 0 to L then?

    Meaning it would be (3L^4)/4
  5. Aug 17, 2012 #4


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

    hmm … i don't really understand the question …

    where is x being measured from?

    and how can it be uniform if ρ = 3x ? :confused:
  6. Aug 17, 2012 #5
    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.
  7. Aug 18, 2012 #6
    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.
  8. Aug 18, 2012 #7
    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.
  9. Aug 18, 2012 #8
    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.
  10. Aug 18, 2012 #9
    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.
  11. Aug 18, 2012 #10
    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.
  12. Aug 18, 2012 #11
    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.
  13. Aug 18, 2012 #12
    While integrating with the 2/3 and 1/3, however I'm getting a negative answer. Should I switch the signs in this case?
  14. Aug 18, 2012 #13
    What are your bounds?
  15. Aug 18, 2012 #14
    I had -(2L)/3 to L/3.
  16. Aug 18, 2012 #15
    I got - (5L^4)/36
  17. Aug 18, 2012 #16
    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

  18. Aug 18, 2012 #17
    Yeah if you switch the bounds you get +5L^4/36
  19. Aug 18, 2012 #18
    Yeah, thats what I was figuring. Thanks a lot for the help. This discussion helped me out a lot!
  20. Aug 18, 2012 #19
    No problem! Good luck with everything else.
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