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Center of mass

  1. May 3, 2009 #1
    1. The problem statement, all variables and given/known data

    Find the center of mass of the triangle with vertices at (0,0), (6,6) and (-6,6) if the density at (x,y) is equal to y.

    2. Relevant equations
    3. The attempt at a solution

    I am having trouble with these centroid/center of mass problems, and I can't even figure out the bounds for sure. Any help/tips?
     
  2. jcsd
  3. May 4, 2009 #2
    Center of mass coordinates,
    x= (My/m)
    y=(Mx/m)

    where My = Moment of y, and Mx = moment of x

    My = double integral of x times density
    Mx = double integral of y times density

    Your given the density = y, so for My its the double integral of xy over the given region
    for Mx its the double integral of y^2 over the given region

    to find m, mass, you just do the double integral of the density over the region.

    your region given to you is a triangle, if you draw this on paper it will be a triangle. Looking at it your going to have to split it into two integrals since the bounds of Y will change since it goes from a -x to x for its bottom bound.
    OR a much simpler way is that you can just multiply the integral by 2, and just doing the integral with the bounds of dy dx, and the bounds of dy will be from x to 6, and dx will from 0 to 6.
    Multiply this answer by 2 to get your answers to all of the integrals you have to do.

    Once you solve My, Mx, and m (mass) you can then plug them into the equations to get the location of the center of mass. which is,
    x = My/m
    y= Mx/m


    I basically went through the whole problem, to moreover cover your specific question about the bounds. You want to integrate the enclosed region as that is the object your finding. Since the bottom part of the DY is not the same throughout (it changes from -x to x), you cant do it as one whole integral, but in this case you know both sides (left of the Y axis, and right of the Y axis) are the same, you can just integrate one side and multiply it by 2.

    I hope this helps you out.
     
  4. May 4, 2009 #3
    Thank you very much! I am pretty sure I understand it now!
     
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