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Centroids of various triangles.

  1. Apr 10, 2008 #1
    I've been having some confusions regarding the centroid coordinates of triangles.

    I've been taught that the centroid of a triangle lies at 1/3rd of the perpendicular distance from any selected base to the corresponding top point of the triangle.

    I tried to use this shortcut to find the centroid of a triangular shape formed by three uniform-thickness rods, and my answer came out wrong. I consulted my teacher, and he told me that this shortcut applies only to planar triangles of a uniform lamina. The explanation doesn't make sense to me. Why can it not apply? Doesn't every solid object share the same centroid with its hollow counterpart?

    What about other variations of the triangle? What if its isoceles, or equilateral? Does the 1/3rd distance rule apply to them or not, then? What about triangles formed with 3 rods, consisting of a circular disc in the middle which touches all three rods ? What if this triangle is equilateral? Does the shortcut apply?
  2. jcsd
  3. Apr 16, 2008 #2
    The centroid is 1/3 the perpendicular distance from any selected base to the opposite vertex marked out on the median from that base to that vertex. It works for all planar triangles. For the center of mass to be coincident with the centroid the lamina comprising the triangle must be of uniform density and symmetric relative to the centroid. The only case where the centroid and the center of mass would align using three uniform rods would be in the case of the equilateral triangle. In all other cases the alignment would fail. In the picture below you can see how the width of each leg is different, hence the failure using uniform rods.


    Attached Files:

  4. Apr 19, 2008 #3

    First of all, thanks for replying.

    Secondly, you seem to be treating the centroid and center of mass separately. Are they not the same? Please clarify me on this.

    "Symmetric relative to the centroid". What do you mean by this? I don't see any form of symmetry, like in the centroid.png file, unless the triangle is equilateral.

    And I still didn't understand your bit about the alignment fail in a non-equilateral triangle made of uniform rods. The pictures only show the centroids. Where/what is the centroid and the center of mass of such a triangle then in a non-equilateral triangle?

    With my assumption that a centroid and center of mass are the same, I tried to calculate the centroid of a right-angled triangle and an equilateral triangle. Both triangles were formed by three rods each, each rod of the same mass. The centroids came out 1/3rd of the perpendicular distance from each base to the opposite vertex. So I thought that it will probably apply to all such triangles.

    I have just realized something. Still continuing the assumption that the center of mass and the centroid are one and the same, yes, each and every triangular frame formed by three rods each, with each rod of the same thickness may not have the centroid at 1/3rd of the median's length, because the lengths of each rod can differ and hence their mass (Does this imply that the 1/3 shortcut does apply to every triangle formed by 3 equal-mass rods?). So please, if you answer, please differentiate b/w the situation of having three rods of the same mass and three rods of the same thickness.

    Thanks again.
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