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Center of Mass using vectors

  1. Feb 20, 2008 #1
    I am looking for a way to find the center mass of an object whose area is enclosed by [tex]x^{2}[/tex] and [tex]\sqrt{x}[/tex] without computing the x and y seperately (a great deal of paperwork).

    So...

    [tex]M\overrightarrow{R_{cm}} = \int \overrightarrow{r} dm[/tex]

    [tex]\sigma = \frac{M}{A} = \frac{dm}{dA}[/tex]

    [tex]\sigma A \overrightarrow{R_{cm}} = \int \sigma \overrightarrow{r} dA[/tex]

    [tex]\overrightarrow{R_{cm}} = \frac{\int\int \sigma \overrightarrow{r} dy dx } {\int \sigma (f(x)-g(x))dx}[/tex]

    Because they are constants, the sigmas cancel and I eventually end up with...

    [tex]\overrightarrow{R_{cm}} = \frac{\int^1_0\int^{x^{2}}_{\sqrt{x}} (xi+yj) dydx}{\int^1_0 x^{2} - \sqrt{x} dx}[/tex]

    (The answer is [tex]\frac{9}{20}i + \frac{9}{20}j[/tex])

    But what happens when sigma/density is not constant, but is given a value say... xi or something like that? Obviously taking the dot product will not work, and I am unsure about the cross product (I haven't done a lot of vectors).

    -Ataman
     
  2. jcsd
  3. Feb 20, 2008 #2
    What do you mean that the dot product won't work? I don't see any dot product in the above derivation.
     
  4. Feb 20, 2008 #3
    In the above derivation, the density is constant, so it is not defined by a vector.

    What I am looking for is a case where there is a varying density within the region.

    -Ataman
     
  5. Feb 20, 2008 #4
    Why define it as a vector? Why not make it as a function of the position vector?
     
  6. Feb 20, 2008 #5
    That's what I meant. Excuse me.

    -Ataman
     
  7. Feb 20, 2008 #6
    Then don't you know the answer?
     
  8. Feb 21, 2008 #7

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Then it is still not a vector. "Density" is a numeric value, not a vector function. You don't need a "dot product", you just have a scalar product- multiply each component of the position vector by the density function.

    You should notice that you aren't really doing less work that if you did x and y as separate integrals. Since [itex]\int u\vec{i}+ v\vec{j} dx= \int u\vec{i}dx+ \int v\vec{j}dx[/itex] you are just writing two integrals as if they were one.
     
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