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Center of mass of a rigid body-sphere

  1. May 28, 2007 #1
    1. The problem statement, all variables and given/known data
    I know this sounds silly, but I need to understand where I'm going wrong here.

    Prove the center of mass of a spherical rigid-body (solid sphere) is found at the origin explicitly, using spherical coordinates (zenith, etc.).

    2. Relevant equations

    Centroid center of mass equation.

    3. The attempt at a solution

    I originally tried using my own spherical coordinates, then found the conventional ones. However I'm not sure what the value of dV would be. For example, why wouldn't every infinitesimal dimension of dV be the square root of (dx+dy+dz)?

    thanks for reading this and any help you can offer.
     
    Last edited: May 28, 2007
  2. jcsd
  3. May 28, 2007 #2
    Can you show us your work so far?

    dV is the volume of infinitesimal of the sphere.
    sqrt(dx+dy+dz)...I really don't know what's that supposed to mean.

    Volume != sqrt(Length)
     
  4. May 28, 2007 #3

    Doc Al

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    Staff: Mentor

    You probably mean: dV = dx*dy*dz. But those aren't spherical coordinates.
     
  5. May 28, 2007 #4
    explanation

    Sorry about that, all I can see dV as is dxdydz. However I think that would convert to r^2*sin(phi)*dr*d(phi)*d(theta) due to the Jacobian determinant, which I saw on wiki.
    But why would dV be dxdydz? All I know is the transformation, but I just can't get my head round how you would determine what the spherical dV would be. In ares it would be polar and easy, even a cylinder most likely. But a sphere?
     
  6. May 29, 2007 #5

    Doc Al

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    I assume that you're having difficulty picturing a volume element dV in spherical coordinates. dV in cartesian coordinates is trivial--just a cube of size dxdydz. But dV in spherical coordinates is a chunk of a sphere and thus a bit harder to visualize; This picture might help: Volume Element in Spherical Coordinates
     
  7. May 29, 2007 #6
    By symmetry, if the COM is a point, it can only be at the center. You don't even need to know the eqn. for the COM to know that must be true.
     
  8. May 29, 2007 #7

    Doc Al

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    I suspect that the OP is well aware of that, but is charged with carrying out the excercise nonetheless. (See the first post.)
     
  9. May 29, 2007 #8
    OK, but why? The symmetry argument is a complete proof.
     
  10. May 31, 2007 #9
    Thanks guys, I think that is enough material for me to work with. Thank you very much for your help.
     
  11. May 31, 2007 #10
    Actually, I have a little question, if you would care to answer it. Is it possible to prove formally the limits on an integral? Generally even?
     
  12. May 31, 2007 #11

    Doc Al

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    I don't understand the question. Perhaps you can rephrase it. The limits of the integrals needed in this exercise are well defined.

    Did you do the exercise of setting up the integrals representing the x, y, and z coordinates of the center of mass in spherical coordinates? Once you do that, it's trivial to do the integration and show that the center of mass is at the center.
     
  13. Jun 1, 2007 #12
    Guys to tell you the truth I just wanted to prove sophisticatedly Newton's shell theorem but i got over my head a little. I have tried repeatedly to get the value of rho, the origin to Point distance to zero through the supposedly simple proof, but to no avail. I integrate phi between zero and pi radians, theta between zero and 2 pi radians and rho between zero and R, the radius of the shell. I add phi, theta or rho in depending on which coordinate I want to find my COM in, and I get something ridiculous like 1/4*pi*R every time.
     
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