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Center of Mass

  1. Feb 12, 2013 #1
    1. The problem statement, all variables and given/known data
    find the center of mass of a thin plate with constant density in the given region.
    region bounded by y-axis, x=y-y^3 ; 0<=y<=1


    2. Relevant equations

    x(bar) = (integral)(a to b) α(x) * x * (f(x) - g(x))
    ----------------------------------
    (integral)(a to b) α(x) * (f(x) - g(x))


    y(bar) = (integral)(a to b) (1/2) α(x) * x * (f(x)^2 - g(x)^2)
    ----------------------------------
    (integral)(a to b) α(x) * (f(x) - g(x))


    3. The attempt at a solution
    how do i solve this since it is f(y) and not f(x). would i replace the "X" in x(bar) integral with a "y"?
     
    Last edited: Feb 12, 2013
  2. jcsd
  3. Feb 12, 2013 #2

    HallsofIvy

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    Your "x= y= y^3" is "shorthand" (and not very good if you ask me) for "x= y" and x= y^3. Solve the latter for y: y= x^(1/3).
     
  4. Feb 12, 2013 #3
    are the limits of integration from 0 to 1 still?
     
  5. Feb 12, 2013 #4
    update: it is x=y-y^3

    sorry for the misunderstanding.
     
  6. Feb 12, 2013 #5

    Dick

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    Then you had the start of a good idea. Take the function y=x-x^3 bounded by the x-axis for 0<=x<=1. It looks the same as your original region, just has x and y reversed. So work out it's center of mass then exchange x and y again.
     
  7. Feb 12, 2013 #6
    not getting the right answer.
     
  8. Feb 12, 2013 #7
    made a mistake on one of the integrals. now i'm getting the correct answers.
    but i'm getting the two backwards the X value of center of mass is the y value for the center of mass on the answer key.
     
  9. Feb 12, 2013 #8

    Dick

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    Without seeing what you are doing, it's hard to say. But if you are getting say, (1/2,1/3) for y=x-x^3, then the answer for x=y-y^3 should be (1/3,1/2). You need to interchange the center of mass coordinates to go from one to the other.
     
  10. Feb 12, 2013 #9
    what do you mean?
     
  11. Feb 12, 2013 #10

    Dick

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    I mean that the regions bounded by x=y-y^3 and y=x-x^3 don't have the same center of mass but there is a simple relation between them. What did you get for the center of mass of y=x-x^3?
     
  12. Feb 12, 2013 #11
    i got (8/15, 16/105) whereas the answer is (16/105, 8/15).

    should i just note that whenever solving a center of mass with x= to switch the values at the end?
     
  13. Feb 12, 2013 #12

    Dick

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    You could make a note of it, but it would better if you clearly understood why. As I've said, switching x and y to turn x=y-y^3 into y=x-x^3 is going to switch the x and y coordinates of the center of mass. You need to switch them back.
     
  14. Feb 12, 2013 #13
    how would i solve this where it would not involve switching the coordinates?
     
  15. Feb 12, 2013 #14

    Dick

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    You can't solve x=y-y^3 to get y as a function of x in any useful way. So you can't use your formulas directly. Either get similar formulas for the case where you have x expressed as a function of y or set up the center of mass formulas as double integrals, if you've covered that.
     
  16. Feb 12, 2013 #15
    we haven't covered double integrals.
     
  17. Feb 12, 2013 #16

    Dick

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    Then stick with the switching approach.
     
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