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Center of mass of a curved surface

  1. Dec 26, 2013 #1
    1. The problem statement, all variables and given/known data
    Let ##\Sigma## be surface given as ##f(x,y)=\frac{1}{2a}(x^2+y^2)## above circle ##K((0,0),a)## for ##a>0##. Calculate the center of mass.


    2. Relevant equations

    If you are not familiar with ##K((0,0),a)## , that is a circle in point ##(0,0)## with radius a.

    3. The attempt at a solution

    I would attempt to get a solution if I would be able to imagine ##\Sigma##. If I understand correctly, ##f(x,y)## is also a circle with radius ##\sqrt{2a}## ? or...?

    My question here is...: What does the fact, that the ##f(x,y)## is above a circle tell me? o_O
     
  2. jcsd
  3. Dec 26, 2013 #2

    mfb

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    I think the surface is meant as 3-dimensional object, so you can write $$z=f(x,y)=\frac{1}{2a}(x^2+y^2)$$ That's a [strike]cone[/strike] paraboloid. The center of mass is trivial for the x- and y-coordinate, and the z-coordinate is interesting.

    Edit: fixed, thanks.
     
    Last edited: Dec 26, 2013
  4. Dec 26, 2013 #3

    LCKurtz

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    Actually, it's a paraboloid.
     
  5. Dec 26, 2013 #4
    Ok, if the surface is 3-dimensional than i can write ##x=rcos\varphi ## and ##y=rsin\varphi ## for ##r\in \left [ 0,a \right ]## and ##\varphi \in \left [ 0,2\pi \right ]##

    Therefore ##f(r,\varphi )=\frac{1}{2r}r^2=\frac{r}{2}##

    So ##z=\frac{\int_{0}^{2\pi }\int_{0}^{a}\frac{r^2}{4}drd\varphi }{\int_{0}^{2\pi }\int_{0}^{a}\frac{r}{2}drd\varphi }=\frac{a}{3}##

    Or is this not so easy? :D
     
  6. Dec 26, 2013 #5

    mfb

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    Why did you replace a by r in the denominator for f?
    And I think there are factors of r missing in the integration (r dr dφ).
     
  7. Dec 26, 2013 #6
    Blah, I don't know. I shouldn't have done that!

    ##x=rcos\varphi ## and ##y=rsin\varphi ## for ##r\in \left [ 0,a \right ]## and ##\varphi \in \left [ 0,2\pi \right ]##

    Therefore ##f(r,\varphi )=\frac{1}{2a}r^2##

    So ##z=\frac{\int_{0}^{2\pi }\int_{0}^{a}\frac{r^4}{4a^2}drd\varphi }{\int_{0}^{2\pi }\int_{0}^{a}\frac{r^2}{2a}drd\varphi }=\frac{3}{10}a##
     
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