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Center of mass of an equilateral triangle (Kleppner)

  1. Dec 8, 2014 #1
    1. The problem statement, all variables and given/known data
    Find the center of mass of an equilateral triangle with side ##a##

    2. Relevant equations
    ## \vec R = \frac{1}{M} \int \vec r \ dm ##

    ## dm = \frac{M}{A} dx dy ##

    ## A = \frac{\sqrt{3}}{4}a^2 ##

    3. The attempt at a solution

    I set a pair of orthogonal axis ##(\vec x,\vec y)## so that one side of the triangle lies on the ##x## axis, and one vertex is at the origin.
    I find the following position for the center of mass:
    ##R_x = \frac{1}{A} (\int_0^{\frac{a}{2}}\int_0^{\sqrt{3}x} x \ dy\ dx + \int_\frac{a}{2}^{a}\int_0^{\sqrt{3}(a-x)} x \ dy\ dx ) = \frac{a}{2}##
    ##R_y = \frac{1}{A} (\int_0^{\frac{a}{2}}\int_0^{\sqrt{3}x} y \ dy\ dx + \int_\frac{a}{2}^{a}\int_0^{\sqrt{3}(a-x)} y \ dy\ dx ) = \frac{a}{2\sqrt{3}}##

    Do you think it is correct?
     
  2. jcsd
  3. Dec 8, 2014 #2

    PeroK

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    That's the correct answer. It would have been easier to put the apex of the triangle on the y-axis: then, by symmetry, R_x = 0.
     
  4. Dec 8, 2014 #3
    I did not think about that! Thank you!
     
  5. Mar 19, 2015 #4
    Please let me ask: I think I understood why the upper term from integration interval is equal to [itex]x\sqrt{3}[/itex], is the height in function of x, but what is the idea behind the upper term from [itex]\int_0^{\sqrt{3}(a-x)} x \ dy[/itex]?
     
  6. Mar 19, 2015 #5

    SteamKing

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    What are the equations of the two lines, which intersect at the apex, forming the sides of this triangle?
     
  7. Mar 20, 2015 #6
    Oh, of course, it would be something like a piecewise function where [itex]x \sqrt{3}[/itex] if [itex]0 \leq x \leq \frac{a}{2}[/itex] and [itex]a \sqrt{3} - x \sqrt{3}[/itex] if [itex]\frac{a}{2} < x \leq a[/itex]. Thank you for your answer.
     
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