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Moment of Inertia - Elliptical beam

  1. Jul 8, 2008 #1
    Hi,

    I'm trying to work out the moment of inertia [itex]I[/itex] of an elliptical beam, about it's length-axis.

    The elliptical beam has a length [itex]L[/itex] along the z-axis, a semi-major axis of length [itex]a[/itex] along the y-axis, and semi-minor axis of length [itex]b[/itex] along the x-axis:
    [​IMG]


    Now, I have done some moment of inertia calculations in the past but all of them involved only the 'simple' single-integral formula:
    [tex]I = \int r^2 dm[/tex]

    Now however I am supposed to use the triple-integral formula:
    [tex]I = \int \int \int_V r^2 \, \rho \, dV[/tex]


    To satisfy myself that I understand the concept, I first tried to do a simple cylinder of height h and radius R. I know the moment of inertia to be [itex]1/2 m R^2[/itex] so I could check if it worked out:
    [tex]I = \int \int \int_V r^2 \, \rho \, dV[/tex]
    [tex]\rho = \frac{m}{\pi R^2 h}[/tex]

    Switching to cylinder coordinates:
    [tex]I = \int_0^h dz \int_0^{2 \pi} d \theta \int_0^R r^2 \frac{m}{\pi R^2 h} r \, dr[/tex]
    [tex]I = h 2 \pi \frac{m R^4}{4 \pi R^2 h} = \frac{1}{2}mR^2[/tex]

    I think this is all correct, no?


    Then I went on to the ellips. I figured I simply need to do a convenient coordinate transformation, just like I used the transformation to cylinder coordinates for the cylinder, right?

    So, I know the equation of the ellips is:
    [tex]\left( \frac{x}{b} \right)^2 + \left( \frac{y}{a} \right)^2 = 1[/tex]

    So I figured a good transformation would be:
    [tex]u = \frac{x}{b} \text{ , } v = \frac{y}{a}[/tex]
    Which would make it a circle:
    [tex]u^2 + v^2 = 1[/tex]

    Then, the conversion from dV = dx dy dz to dV = du dv dz is calculated with the Jacobian determinant, right? I don't know how to type this out in latex so I'll try to describe what I did.
    [tex]\frac{ \partial (u, v)}{\partial (x,y)} = \text{ that determinant } = \frac{1}{a} \times \frac{1}{b} - 0 = \frac{1}{ab}[/tex]

    So
    [tex]\frac{ \partial (x,y)}{\partial (u,v)} = \frac{1}{1/ab} = ab[/tex]

    So [tex]dV = dx \, dy \,dz = ab du \, dv \, dz[/tex]

    Then:
    [tex]I = \int_0^L dz \int_0^1 du \int_0^1 (u^2 + v^2) \rho \, ab \, dv[/tex]
    [tex]\rho = \frac{m}{\pi a b L}[/tex]
    [tex]I = \frac{m}{\pi a b L} \int_0^L dz \int_0^1 du \int_0^1 (u^2 + v^2)\, ab \, dv[/tex]
    [tex]I = \frac{m}{\pi a b L} a b L \int_0^1 (u^2 + \frac{1}{3}) du = \frac{2m}{3\pi}[/tex]

    And that does not make any sense to me... I am quite sure it is not independent of a and b...

    What did I do wrong? Thanks!
     
  2. jcsd
  3. Jul 8, 2008 #2

    dynamicsolo

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    Won't your density expression transform as well? It looks like the integral is set up correctly, with appropriate limits, but your density expression is absorbing the dimensions of the ellipse in the integral. So a and b actually are in your final result, but they've been "rescaled" to 1 each, making your moment look dimensionally incorrect -- when it isn't really -- because there are two "invisible" length terms.
     
  4. Jul 9, 2008 #3
    I'm not sure I understand what you are saying.

    Do you mean that
    [tex]\rho = \frac{m}{\pi a b L}[/tex]
    is wrong after changing to the (u,v) coordinates? How would I change it then?

    Because now, the abL is cancelled out by the ab resulting from the coordinate change and the L from the left-most integral...
     
  5. Jul 9, 2008 #4

    dynamicsolo

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    You rescaled two of the linear dimensions in the problem. Since density is mass divided by volume, wouldn't the volume element be transformed from abL to 1·1·L, so that density in your transformed system of coordinates is now just

    [tex]
    \rho = \frac{m}{\pi L}
    [/tex] ?
     
  6. Jul 9, 2008 #5
  7. Jul 9, 2008 #6
    I'm starting to think my limits of integration are completely wrong.

    I am integrating both u and v from 0 to 1. That means I'm integrating over a rectangle, right? While the ellips is a circle in (u,v) coords, not a rectangle... Is that the problem?
     
  8. Jul 9, 2008 #7

    dynamicsolo

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    OK, I stand by my remark on the density, but there's another problem. Your triple integral is based on Cartesian coordinates. If you were working out the cylinder in such coordinates, what should the limits of integration be for the circular cross-section? (They won't be 0 to a and 0 to b...) There would be an analogous set of limits for the elliptical cross-section in your transformed variables.

    EDIT: Sorry, I typed this before I saw your new post. Yes, I started setting this up myself and I'd say that was the problem.
     
  9. Jul 9, 2008 #8
    Alright, I tried the following integral this time:
    [tex]I = \rho \int_0^L dz \int_0^1 du \int_{-\sqrt{1-u^2}}^{\sqrt{1-u^2}} (u^2 + v^2)\, ab \, dv[/tex]

    I had to use maple (math software) to calculate the integral (the first one over v I can do but the second, over u, became a bit more complicated). I haven't looked at it too long, but we were supposed to be able to do this at an exam, without even a calculator...

    However, I did calculate it using maple and now I got to:
    [tex]I = \rho \int_0^L \frac{1}{4}\pi \, ab \, dz[/tex]

    The 1/4 pi is promising when you look at the answer I got form that website (few posts up), but it's still not correct. The answer I get now is:
    [tex]I = \frac{m}{\pi L} L \frac{1}{4}\pi ab = \frac{1}{4}m ab[/tex]

    While it should apparently be
    [tex]I = \frac{1}{4}m (a^2 + b^2)[/tex]

    Hmm... Something else is wrong?
     
  10. Jul 9, 2008 #9
    Ok, I managed to get the correct answer now using Maple. I tried it once without using any substitutions, and it came out correct!

    The integral I had maple calculate was:

    [tex]I = \int_0^L dz \int_{-b}^b dx \int_{-a \sqrt{1-\frac{x^2}{b^2}}}^{a \sqrt{1-\frac{x^2}{b^2}}} (x^2 + y^2) \rho dy[/tex]
    [tex]I = \rho \, L \, \frac{1}{4} \pi (ab^3 + ba^3) = \frac{m}{\pi a b L}L \pi (ab^3 + ba^3) = \frac{1}{4}m(a^2 + b^2)[/tex]

    However, I am pretty sure I could not have done this by hand... So I suppose I am doing something wrong with the substitution... Can anyone see what?
     
  11. Jul 9, 2008 #10

    dynamicsolo

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    I just finished going back through this myself. Start from working out the moment about the longitudinal axis for a cylinder:

    [tex]I = 4\rho \int_0^L dz \int_0^a \int_{0}^{\sqrt{a^2-y^2}} (x^2 + y^2)dx dy

    = 4 \rho \cdot L \int_0^a \frac{1}{3}(a^2-y^2)^{3/2} + (a^2-y^2)^{1/2} \cdot y^2 dy
    [/tex] ,

    where the integral covers only one quadrant of the cylinder, which we then multiply by 4, and the density is [tex]\rho = \frac{m}{\pi a^2 L }[/tex].

    You have a couple trig substitution integrals to do, which yield for the two terms
    [tex]I = 4\rho \cdot L [ \frac{\pi}{16}a^4 + \frac{\pi}{16}a^4 ] = \frac{\pi}{2}\rho a^4 L = \frac{1}{2}ma^2[/tex] ,

    which is just what we expect, as it is the familiar result for a uniform "disk".

    To keep the change reasonably sane, now change just the x-coordinate to

    u = (a/b)·x or x = (b/a)·u ;

    this just stretches one of the axes. The integral from above will now just change to

    [tex]I = 4\rho L \int_0^a \int_{0}^{\sqrt{a^2-y^2}} (\frac{b^2}{a^2}u^2 + y^2) \, \frac{b}{a}du dy[/tex] .

    This keeps almost everything the same as before (including the expression for density, and the integrations do not need to be re-worked), with the first term being altered to

    [tex]I = 4\rho \cdot L \frac{b}{a} \int_0^a \frac{1}{3}\cdot \frac{b^2}{a^2}(a^2-y^2)^{3/2} + (a^2-y^2)^{1/2} \cdot y^2 dy[/tex] ,

    with the numerical terms becoming

    [tex]I = 4\rho \cdot L \frac{b}{a}[ \frac{b^2}{a^2} \cdot \frac{\pi}{16}a^4 + \frac{\pi}{16}a^4 ] = \frac{\pi}{4}\rho L\frac{b}{a}(a^2b^2 + a^4) [/tex] .

    The mass is [tex]m = \pi \cdot a^2\frac{b}{a}L\rho [/tex], giving

    [tex]I = \frac{1}{4}m(a^2 + b^2)[/tex] .

    The transformation of the one coordinate alters the result from just one of the integrals for the circular cylinder's moment, explaining why the change in the expression for the elliptical cylinder includes a sum of squared terms.

    (Well, I learned something from this...)
     
    Last edited: Jul 9, 2008
  12. Jul 9, 2008 #11
    Interesting!

    Is this 'one axis stretching' similar to an actual coordinate transformation (the type I kept trying to do)? Since it doesn't really look like it much, at least not how I learned to do it...

    Well thanks for your help anyway!
     
  13. Jul 9, 2008 #12

    dynamicsolo

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    Yes, "stretching" the x-axis is just another way of saying that we're transforming the x-coordinate. I did it that way because it minimizes the amount of re-working of the cylindrical calculations; it turns out to only introduce a constant factor into one of the integral terms. (And I think I'm finally done editing all that LaTeX...)
     
  14. Jul 9, 2008 #13
    Alright that makes sense. So instead of stretching the x-axis with a factor b, and the y-axis with a factor a, you stretch only the x-axis with a factor (b/a), which should be equivalent?

    And we don't need to use the jacobian determinant to calculate the conversion from dxdydz to dudydz?
     
  15. Jul 9, 2008 #14

    dynamicsolo

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    The way I look at it, you are really only stretching one dimension relative to the other, so you could just multiply one of the variables by the relative factor.

    Well, I don't see that you need to use the Jacobian, since we are just doing a linear transformation of one of the Cartesian coordinates. But if you did calculate it, you should find the same result as I used.
     
  16. Jul 9, 2008 #15
    Alright!

    I still wanted to know how to do it using the substitution I have learned, and I finally saw what I did wrong.

    I wrote the r^2 as u^2 + v^2, but that's wrong ofcourse...
    [tex]r^2 = x^2 + y^2 = (bu)^2 + (av)^2[/tex]

    Using that in the following integral yields the correct result!
    [tex]I = \rho \int_0^L dz \int_{-1}^1 du \int_{- \sqrt{1-u^2}}^{\sqrt{1-u^2}} \left( (bu)^2 + (av)^2 \right) ab \, dv = \frac{1}{4}m(a^2 + b^2)[/tex]

    This is still done with maple however, I am going to try to do it by hand now, see how far I get...
     
  17. Jul 9, 2008 #16
    Got it :D

    Using the substitution x = bu, y = av, I can set up the integral until I arrive at:
    [tex]I = \rho \int_0^L dz \int_{-1}^1 du \int_{- \sqrt{1-u^2}}^{\sqrt{1-u^2}} \left( (bu)^2 + (av)^2 \right) ab \, dv = \frac{m}{\pi} \int_{-1}^1 du \int_{- \sqrt{1-u^2}}^{\sqrt{1-u^2}} \left( (bu)^2 + (av)^2 \right) dv[/tex]

    Then, I use another substitution to polar coordinates:
    [tex]v = r \sin \theta[/tex]
    [tex]u = r \cos \theta[/tex]

    Which changes the integral to:
    [tex]I = \frac{m}{\pi} \int_0^{2\pi} d\theta \int_0^1 \left( b^2 \cos^2 \theta + a^2 \sin^2 \theta \right) r^3 \, dr[/tex]
    [tex]I = \frac{m}{4\pi} \int_0^{2\pi} \left( b^2 \cos^2 \theta + a^2 \sin^2 \theta \right)[/tex]

    Both the cos^2 and sin^2 integrals evaluate to pi, so:
    [tex]I = \frac{m}{4 \pi}(b^2 \pi + a^2 \pi) = \frac{1}{4}m(a^2 + b^2)[/tex]

    Yay!

    Thanks for all the help, now at least I know how to do this for the next exam hehe...
     
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