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Fractal Moment

  1. May 1, 2003 #1
    Okay this problem sounded neat, but I'm stuck on it. How would one go about finding the moment of inertia of Sierpinski's triangle about an axis through its center and perpindicular to the triangle? Any thoughts?
  2. jcsd
  3. May 1, 2003 #2
    Here's a http://physics.harvard.edu/undergrad/prob9.pdf [Broken] to the problem so you can see what I'm talking about. This ones really throwing me for a loop. I thought maybe that you could use a symetry argument and say that it would be the same as for a regular triangle just with a smaller mass. But rethinking that I'm pretty sure its wrong. Then I thought that if you did what the problem statement described down to infinity you wouldn't have any mass left at the end which would make the moment zero. But again this seems wrong to me. This is really beginning to bother me. Can anyone help me out?
    Last edited by a moderator: May 1, 2017
  4. May 1, 2003 #3


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    Well, part of the trick is that the object has infinite mass density (because of the zero volume).

    I think you need to use a self-similarity argument to find the moment of the whole in terms of the moment of the individual parts, and coupled with the scale factor should yield a system you can solve.

  5. May 1, 2003 #4
    Okay, you lost me there....
  6. May 1, 2003 #5


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    Let I be the moment of inertia around the center of the fractal.

    Consider now just one section of the fractal. One section has one third of the mass and is scaled down by a factor of two, so the moment of inertia of one section about its center is (I / 12); multiply by one third because of the mass change, and by one half squared because of the scaling.

    By the parallel-axis theorem, to find the moment of inertia of one section about the center of the whole fractal, we add the mass of the section times the square of the distance between the axes. The mass of the section is (m / 3) because it's one third of the whole. The distance between the axes is (insert some geometrical reasoning) one third of the height which is (sqrt(3) / 2) * l, so the distance is l / (2 sqrt(3)), so the moment of inertia of one section is

    (I / 12) + (m / 3) * (l / (2 sqrt(3))^2
    = (I / 12) + (m / 3) * l^2 / 12
    = (I / 12) + m l^2 / 36

    The moment of inertia of the whole should be the sum of the moments of inertia of the three sections, so:

    I = 3 * ((I / 12) + m l^2 / 36)
    = I / 4 + m l^2 / 12

    3 I / 4 = m l^2 / 12
    I = m l^2 / 9
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