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Center of mass of a uniform sphere!

  1. Nov 2, 2013 #1

    ShayanJ

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    We know,by symmetry,that the center of mass of a uniform sphere is at its center.So we expect the formula [itex] r_{com}=\frac{\int r \rho d\tau}{\int \rho d\tau} [/itex] to give us zero for this case.So lets see:
    [itex]
    r_{com}=\frac{\int_0^{R} \int_0^{\pi}\int_0^{2\pi} r^3 \sin{\theta} d\phi d\theta dr}{\int_0^{R} \int_0^{\pi}\int_0^{2\pi} r^2 \sin{\theta} d\phi d\theta dr}=\frac{\int_0^R r^3 dr \int_0^{2\pi} d\phi \int_0^{\pi} \sin{\theta} d\theta}{\int_0^R r^2 dr \int_0^{2\pi} d\phi \int_0^{\pi} \sin{\theta} d\theta}=\frac{\frac{1}{4}R^4\times 2\pi \times 2}{\frac{1}{3}R^3\times 2\pi \times 2}=\frac{3}{4}R

    [/itex]!!!!!!

    What's going on???
     
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  3. Nov 2, 2013 #2

    D H

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    What's going on is that that is not the equation for the center of mass.
     
  4. Nov 2, 2013 #3

    arildno

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    What's going on is that the center of mass formula in terms of coordinate choice doesn't transform in that simplistic manner.

    Basically, you should look at how the formula for the CM in terms of (x,y,z)-coordinates treansform to polar, and verify to yourself that a) They specify the origin as the C.M, and b) None of the formulae looks the slightest bit like your attempt.
     
  5. Nov 2, 2013 #4

    Simon Bridge

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    Why would you expect that formula to hold?

    The denominator is just the mass of the sphere.
    The numerator is supposed to be the sum of the moments due to the mass elements.
    What are you computing the moments about?

    [edit] arildno is telling it more directly ;)
     
  6. Nov 2, 2013 #5

    arildno

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    But, you might ask:
    What sort of weighted average do you get then by your approach?


    You are, as you readily can find out, averaging r over the function (rho)*4(pi)r^2, which is the mass per length of a spherical shell centered about the origin.
     
    Last edited: Nov 2, 2013
  7. Nov 2, 2013 #6

    ShayanJ

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    Can you tell me how to gain the center of mass formulas in spherical coordinates?
    Because I don't think transforming the cartesian formulas to spherical coordinates gives the right answer,because we would have e.g.
    [itex]
    \overline{r\sin{\theta}\cos{\phi}}=\frac{1}{M}\int r \sin{\theta}\cos{\phi} r^2 \sin{\theta}d\phi d\theta dr
    [/itex]!!!!!!
     
  8. Nov 2, 2013 #7

    arildno

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    It sure is the right formula.
    Take first the integral over the cos(phi) from 0 to 2*pi; what do you get?
     
  9. Nov 2, 2013 #8

    ShayanJ

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    I know the formula is right!
    The point is,by transforming the three cartesian formulas for center of mass coordinates to spherical coordinate,one will get three equations which give you not sphercial coordinates of the center of mass but three functions of those coordinates.Oh...so looks like then the three equations should be solved simultanously to find the spherical coordinates of the center of mass!
    I got it...thanks guys!
     
    Last edited: Nov 2, 2013
  10. Nov 2, 2013 #9

    ShayanJ

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    Still another point remains.
    Center of mass is in fact the weighed average of the position of mass elements of the body under consideration,with the mass of every element being the weight.So the formula [itex] r_{com}=\frac{\int r \rho d\tau}{\int \rho d\tau}[/itex] seems reasonble.Why isn't it right?
    And...What if you want to write the formula in spherical coordinates in the first place,without using cartesian ones?what would you do?

    Thanks everyone
     
  11. Nov 2, 2013 #10

    arildno

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    "Center of mass is in fact the weighed average of the position of mass elements of the body under consideration,with the mass of every element being the weight"

    It sure is.
    But, then again, your scalar "r" is no position vector is it?
    Nor is it the scalar factor in a variable-independent vector component, either, in contrast to "x" as in [itex]x\vec{i}[/itex], where [itex]\vec{i}[/itex] is independent of x,y and z.

    Thus, for position vectors [itex]\vec{r}[/itex], we have the coordinate independent formulation:
    [tex]\vec{r}_{C.M}=\frac{\int_{V}\vec{r}\rho(\vec{r})dV}{\int_{V}\rho(\vec{r})dV}[/tex]

    Try as much as you like, you won't be able to reduce it into what you wrote, and that is why what you write doesn't have anything to do with the position of C.M.

    Therefore, the correct formula for the C.M in polar coordinates for your uniform sphere is:
    [tex]\vec{r}_{C.M}=\frac{\int_{V}\vec{i}_{r}\rho{r}^{3}\sin\theta {drd\theta{d}\phi}}{\int_{V}\rho{dV}}=\frac{3}{16\pi}R\int_{\theta,\phi}\vec{i}_{r} \sin\theta{d}\theta{d}\phi[/tex]

    Your fallacy can then be seen to be reduced to the following misconception:
    [tex]\vec{i}_{r}=||\vec{i}_{r}||={1}[/tex]
    Or, something like that anyway.
     
    Last edited: Nov 2, 2013
  12. Nov 2, 2013 #11

    Simon Bridge

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    Yah - we normally "cheat" by picking a symmetry line to do the integration along.
    ##r_{com}## is then the distance along this line from some reference point, also on the line.
     
  13. Nov 3, 2013 #12

    arildno

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    True enough.
    But what I wrote highlights the particular conceptual flaw of OP rather well, I think.
    And that was the main issue to be dealt with in this thread, not particular integration techniques.
    But, and I think you are right that simplification techniques, like picking out a symmetry line, and where the vectorial formulation is thereby suppressed might lead to some misunderstandings that finding C.M is "essentially" a calculation technique involving scalars.
     
    Last edited: Nov 3, 2013
  14. Nov 4, 2013 #13

    Simon Bridge

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    Sure - the problem wasn't with the integration but with what it means.
     
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