Center of mass of a rigid body-sphere

AI Thread Summary
The discussion revolves around proving that the center of mass of a solid sphere is at the origin using spherical coordinates. Participants clarify that the volume element dV in spherical coordinates is not simply the square root of (dx+dy+dz), but rather dV = r^2*sin(φ)dr dφ dθ. The symmetry of the sphere suggests that the center of mass must be at the center, simplifying the proof. The original poster expresses confusion about setting up integrals for calculating the center of mass and struggles with the integration limits. Overall, the conversation emphasizes the importance of understanding spherical coordinates and the symmetry argument in proving the center of mass location.
lewis198
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Homework Statement


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.).

Homework Equations



Centroid center of mass equation.

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.
 
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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)
 
lewis198 said:
For example, why wouldn't every infinitesimal dimension of dV be the square root of (dx+dy+dz)?
You probably mean: dV = dx*dy*dz. But those aren't spherical coordinates.
 
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?
 
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: http://www.spsu.edu/math/Dillon/VolumeElementSphericalCoordinates.htm"
 
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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.
 
I suspect that the OP is well aware of that, but is charged with carrying out the excercise nonetheless. (See the first post.)
 
Doc Al said:
I suspect that the OP is well aware of that, but is charged with carrying out the excercise nonetheless. (See the first post.)

OK, but why? The symmetry argument is a complete proof.
 
Thanks guys, I think that is enough material for me to work with. Thank you very much for your help.
 
  • #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?
 
  • #11
lewis198 said:
Is it possible to prove formally the limits on an integral? Generally even?
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.
 
  • #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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