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Finding the center of mass of an arbitrary uniform triangle

  1. Jan 14, 2017 #1
    1. The problem statement, all variables and given/known data
    1. Show that for an arbitrary uniform triangle ABC, with A at (x1, y1), B at (x2,y2), C at (x3, y3), the CM (xcm, ycm), is simply defined by xcm=(x1+x2+x3)/3, and ycm =(y1+y2+y3)/3

    2. Relevant equations
    xcm = 1/M * ∫xdm

    ycm = 1/M * ∫ydm

    M = ∫dm = ∫δdA where δ = M/A = dm/dA

    for uniform mass distribution

    M = δ∫dA

    3. The attempt at a solution

    I'm not really too sure on how to set up the triangle for this problem. My professor had a picture of an arbitrary triangle with the vertices at the specified coordinates. Would it be alright to place my origin with the longest side of the triangle lying on the x-axis and then both end points of that side connect to a point on the y-axis? The coordinates of the vertices from left to right would then be (x1,0), (0,y2), and (x3, 0). Would this defeat the purpose of the exercise? I'm pretty lost here.
     
  2. jcsd
  3. Jan 14, 2017 #2
    If it's a uniform triangle, then the center of mass is the geometric centroid. You really don't need to use calculus for a uniform triangle.
     
  4. Jan 14, 2017 #3
    How would I prove it then?
     
  5. Jan 14, 2017 #4
    The triangle's coordinates are arbitrary. They don't matter. When you want to find the average value of a set of data, you are finding the average value of a set of discrete data. When you integrate, you are finding the average value of a continuous function. Imagine now a function of TWO variables. It's domain will not be a "line" (such as a subset of the x-axis) but a region of the x-y plane. And it's range will be the density. The triangle shape is a domain. But you don't have to jump into multi-variate calculus since f(x, y) = 1, or any other arbitrary constant. It's uniform. So first use integration to find the sum of all "x values times density at that location (constant here)" and the sum of all "y values times density at that location (constant here)" in the domain. Then use integration to find the "average x value" and "average y value" for every possible point. The domain itself is bounded by linear functions, so your result will be a simple average. I won't set up a single equation for you because you need to figure this out on your own. This is a very basic concept that, while hard to grasp at first, should be burned into your psyche.
     
  6. Jan 14, 2017 #5
    I did not really understand what you meant by this.
     
  7. Jan 14, 2017 #6
    This is the set up that I'm thinking about, not sure if its okay though.
     

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  8. Jan 14, 2017 #7

    ehild

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    It is all right, but you have to substitute the y(x) functions for both domains ( from 0 to x2 and from x2 to x3)
     
  9. Jan 14, 2017 #8
    so from 0 to x2 I can use y(x) = (y2/x2)x and from x2 to x3 y(x) = qx - qx3 Where q = -y2/(x3-x2) ?
     
  10. Jan 14, 2017 #9

    ehild

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    Right.
     
  11. Jan 14, 2017 #10
    So I tried to do the calculation but I think im doing something wrong cause im ending up with a long expression that I cant seem to simplify.
     

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  12. Jan 14, 2017 #11

    ehild

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    Yes, it will be a long expression, but it simplifies. Do the integral first, without M and ρ. When substituting the limits, the x3-x2 term appears both in the numerator and denominator, you can cancel it. Remember, that a2-b2=(a-b)(a+b) and
    a3-b3=(a-b)(a2+ab+b2).
     
  13. Jan 14, 2017 #12
    I keep evaluating the integral and I am really not seeing how the x3 - x2 cancels...
     
  14. Jan 15, 2017 #13

    TSny

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    upload_2017-1-15_0-17-18.png
    Find the common denominator and express this as a single fraction. Simplify.
     
  15. Jan 15, 2017 #14
    ah jeez after trying like a million times I finally got it lol I ended up with xcm = (x3 + x2)/3
    Do you feel this constitutes a proof of what was originally asked since I made x1 = 0 my defining my origin in the way that I did?
     
  16. Jan 15, 2017 #15

    ehild

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    You defined the horizontal coordinate with respect to x1. In principle, you used the variable u=x-x1. The integration happened with respect to u, and the limits were 0, u2 and u3. So the u coordinate of the CM is ucm=(u2+u3)/3.
    In terms of x, it is
    xcm-x1=[(x2-x1)+(x3-x1)]/3--->xcm=x1+[(x2-x1)+(x3-x1)]/3=(x1+x2+x3)/3
     
  17. Jan 15, 2017 #16
    Wouldn't that change of variables just slide the triangle so that vertex A lies on the y-axis? the way I defined my origin would be that change of variables plus something that slides vertex A up or down so that it is exactly at (0,0) and something else that rotates vertex C down to lies on the x-axis. Am I thinking about this correctly?
     
  18. Jan 15, 2017 #17

    ehild

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    I spoke about the x variable. You can do the same with y.
     
  19. Jan 15, 2017 #18
    So I can define u = x - x1 and v = y - y1 as you said and that would place vertex A at the origin, but how can I now rotate the triangle while holding vertex A fixed so that vertex C lies on the x-axis?
     
  20. Jan 15, 2017 #19
    Would I have to define a new variable such that r = v - v3?
     
  21. Jan 15, 2017 #20

    ehild

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    Make a picture and show what you mean.
    The integration can be performed for the general position of the triangle, integration between the sides, with respect to x (first picture) or y (second picture).

    upload_2017-1-16_5-2-25.png
     
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