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Moment of inertia of a square lamina through a diagonal

  1. Jul 11, 2013 #1
    1. The problem statement, all variables and given/known data
    Square lamina (of side a) of uniform density. Find I about a diagonal.


    2. Relevant equations
    I = ∫ dm*l^2


    3. The attempt at a solution

    So I drew a square and its diagonal and I imagine a differential mass drawn somewhere on the lamina. The distance squared to that differential mass from the diagonal axis is x^2 + y^2.

    So I did the following integral:
    [tex]
    \rho \int_ \frac{-a}{2}^\frac{a}{2} \int_\frac{-a}{2}^\frac{a}{2} (x^2 + y^2) dx dy
    [/tex]

    thinking this would be the correct answer, however I get
    [tex]
    \frac{1}{6}Ma^{2}
    [/tex]

    which is wrong by a factor of [tex] \frac{1}{2} [/tex]

    I don't know where I'm going wrong. Any help would be greatly appreciated. Thanks so much.
     
  2. jcsd
  3. Jul 11, 2013 #2

    TSny

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    Hello. From the limits on your integral, it appears that you are choosing the plate to lie in the xy plane with edges parallel to the x and y axes and with the center of the plate at the origin of the coordinate system.

    If (x,y) are the coordinates of a mass element ρdxdy, then x2 + y2 would be the square of the distance of the mass element form the origin of the coordinate system rather than from a diagonal of the plate.
     
    Last edited: Jul 11, 2013
  4. Jul 11, 2013 #3
    Hey, thanks so much for the response. So I guess I'm lost then as to find the distance from the axis to the differential mass piece.

    I don't know if I'm going about this inefficiently by not using the perpendicular axis theorem or the parallel axis theorem somehow, but I wanted to try to do this just through the definition of the moment of inertia.

    Is there an easy way to find that distance from the axis to the differential mass piece in question? (ie, that l^2 term)

    Thanks again
     
  5. Jul 11, 2013 #4

    haruspex

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    Not sure it's the easiest, but:
    - what is the slope of a line perpendicular to the diagonal axis?
    - what therefore is the equation of such a line through the point (x0, y0)?
    - where does that line intersect the diagonal?
    - what is the distance from (x0, y0) to that intersection?
     
  6. Jul 11, 2013 #5

    TSny

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    If you are familiar with the cross product, think about what you would get if you took the cross product of the position vector ##\vec{r}## locating the mass element at (x,y) and a unit vector ##\hat{u}## along the diagonal.
     
  7. Jul 12, 2013 #6

    Filip Larsen

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    Instead of a direct integral along the diagonal you can also use the fact that for a lamina Iz = Ix + Iy and that Ix + Iy = 2Ix for any rotation around z of the lamina where the lamina is identical distributed around both x and y. Since there are two geometrically different rotations of a square lamina where Ix and Iy are identical (with x and y along the diagonals being one such rotation), you can convert your integral along diagonals into an integral along the axis of the "other rotation".
     
  8. Jul 12, 2013 #7

    haruspex

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    Very neat.
     
  9. May 24, 2014 #8
    How would you solve this problem using direct integration across the diagonal?

    I can't seem to figure out how to get the distance from the diagonal axis.

    Thanks
     
  10. May 24, 2014 #9

    tms

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    I think it would be easier to put the origin at the center of the square and the diagonals along the ##x## and ##y## axes. This results in a single integral, most simply along the ##x## axis. Due to symmetry you only have to do the integration in the first quadrant, and multiply that answer by 4.
     
  11. Jun 15, 2017 #10

    kuruman

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    Put the y-axis along the diagonal. Consider a right triangle that is 1/4 of the square with one right side along y and the other along x. A strip on the lamina parallel to the x-axis of length x and width dy located at distance y from the x-axis has moment of inertia (about the diagonal) given by ##dI = \frac{1}{3} ~ dm ~x^2## where ##dm = \frac{m}{a^2} ~x ~dy##.
    Note that because we have a square, the length of the strip, x, equals its distance from the y-axis, y.

    Therefore ##y=x## and ##dy = dx##. Integrate from ##x=0## to ##x=a/\sqrt{2}## and don't forget to multiply the result by 4 to get the moment of inertia of the full square.

    On edit: This is an elaboration of the post by @tms.
     
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