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Center of Mass w/ variable density

  1. Nov 12, 2013 #1
    1. The problem statement, all variables and given/known data
    A box with one corner at the origin and the opposite corner at (a,b,c). The density is:##\rho=\rho_o\frac{z^2}{c^2}##


    2. Relevant equations
    ##r_{cm}=\frac{\int\vec{r}dm}{\int dm}##


    3. The attempt at a solution
    I calculated the mass:

    ##\int_{0,0,0}^{a,b,c}\rho_o\frac{z^2}{c^2}=\frac{abc\rho_o}{3}##

    But I'm not sure what the top of the equation is suppose to mean. Am I suppose to multiply the density by the direction it is changing (z) then integrate? I can't find an example anywhere that explains it well enough so that I can understand.

    Thanks for the help.
     
  2. jcsd
  3. Nov 12, 2013 #2

    haruspex

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    Top (numerator?) of which equation?
    It asks for the centre of mass, so that requires three co-ordinates. In vectors, ∫ρ(r)r.dv/mass
    E.g. to get the x-co-ordinate, ∫ρ(x,y,z).x.dxdydz/mass.
     
  4. Nov 12, 2013 #3
    so I would need to do:

    ##\int_0^a\rho_o\frac{z^2}{c^2}x dx##
    ##\int_0^b\rho_o\frac{z^2}{c^2}y dy##
    ##\int_0^c\rho_o\frac{z^2}{c^2}z dz##

    Am I understanding correctly then?

    edit: also, by top equation I meant the position integral. I should just solve these 3 integrals and then divide by the mass then.
     
  5. Nov 13, 2013 #4

    haruspex

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    No, each one is a triple integral, dxdydz.
     
  6. Nov 13, 2013 #5

    HallsofIvy

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    That is
    [tex]\int_{x=0}^1\int_{y=0}^1\int_{z= 0}^1\rho_0\frac{z}{c}dzdydx[/tex]
    By "Fubini's theorem" that can be written as a product of integrals:
    [tex]\dfrac{\rho_0}{c}\left(\int_{x=0}^1 dx\right)\left(\int_{y= 0}^1 dy\right)\left(\int_{z= 0}^1 z dz\right)[/tex]
     
  7. Nov 13, 2013 #6
    This has me confused. It appears like my first equation but with (a,b,c) replaced with (1,1,1) and you've dropped the squares for z and c. Is this suppose to be another method to find the CM?

    haruspex,

    So, I should have something that looks like:

    ##\int_0^a\int_0^b\int_0^c\rho_o\frac{z^2}{c^2}x dxdydz##
    ##\int_0^a\int_0^b\int_0^c\rho_o\frac{z^2}{c^2}y dxdydz##
    ##\int_0^a\int_0^b\int_0^c\rho_o\frac{z^2}{c^2}z dxdydz##

    which would give me ##\hat{x}+\hat{y}+\hat{z}## all divided by the mass I found?

    If I understand it correctly now:

    ##\int_0^a\int_0^b\int_0^c\rho_o\frac{z^2}{c^2}x dxdydz=\frac{\rho_o}{c^2}\frac{a^2c^2b}{4}=\frac{\rho_oa^2b}{4}##
    so for ##\hat{x}## I would get ##\frac{\frac{\rho_oa^2b}{4}}{\frac{\rho_0abc}{3}}##

    Which is just## \frac{3a}{4c}\hat{x}##

    Thanks for the help.
     
    Last edited: Nov 13, 2013
  8. Nov 13, 2013 #7

    haruspex

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    Yes.
    I don't understand the plus signs. Each integral gives you one co-ordinate of a vector.
    That doesn't look right. The dz should still be giving you a factor 1/3.
     
  9. Nov 13, 2013 #8
    Ack, I forgot to square the z's for integration. Don't know why I did that. But, I get the gist so I'll be good from here. I just did plus instead of commas.

    Thanks for all the help!
     
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