Solid hemisphere center of mass in spherical coordinates

  • Thread starter Ledamien
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Hello,

I am struggling with what was supposed to be the simplest calc problem in spherical coordinates. I am trying to fid the center of mass of a solid hemisphere with a constant density, and I get a weird result.
First, I compute the mass, then apply the center of mass formula. I divide both and voila, obviously wrong result. What is wrong here?

Cm = 1 / M [itex]\int[/itex] [itex]\rho[/itex] r^3 sin [itex]\theta[/itex] dr d[itex]\theta[/itex] d[itex]\varphi[/itex]

M = ρ V → 1/M = 1/ρV

V = 2 [itex]\pi[/itex] R^3 / 3

Cm = 3/ (2 [itex]\pi[/itex] R^3) [itex]\int[/itex] r^3 sin [itex]\theta[/itex] dr d[itex]\theta[/itex] d[itex]\varphi[/itex]

Integrated over:
r → 0 to R
[itex]\theta[/itex] → 0 to [itex]\pi[/itex]/2
[itex]\varphi [/itex] → 0 to 2 [itex]\pi[/itex]

Cm = 3/ (2 [itex]\pi[/itex] R^3) * [itex]\pi[/itex] R^4 / 2
Cm = 3R / 4

So, what do I do wrong?
 
Last edited:

Answers and Replies

  • #2
SteamKing
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How did you derive your equation for the Cm?
 
  • #3
haruspex
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[itex]\int[/itex] [itex]\rho[/itex] r^3 sin [itex]\theta[/itex] dr d[itex]\theta[/itex] d[itex]\varphi[/itex]
Isn't there a trig term missing in that integral? There should be one from the Jacobian and another for the fact that you're interested in the displacement in only one Cartesian coordinate.
 
  • #4
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1/M [itex]\sum[/itex] mi ri
1/M[itex]\int[/itex] [itex]\rho[/itex](r) r dV - this is a volume integral.
1/M[itex]\int[/itex] [[itex]\rho[/itex](r) r] r^2 dr sin [itex]\theta[/itex]) d[itex]\theta[/itex] d[itex]\varphi[/itex] - this is a volume integral in spherical coordinates.
My idea was that a simple volume integral for a hemisphere, going from 0 to r, 0 to [itex]\pi[/itex]/2 and 0 to 2[itex]\pi[/itex] would be sufficient. As the density is constant I can pull it out of the integral, and [itex]\rho[/itex]/M is simply 1/V, one over the total volume - the same integral but with r^2 instead of r^3. I seem to be missing a 1/2 somewhere, but where?
 
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haruspex
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1/M [itex]\sum[/itex] mi ri
1/M[itex]\int[/itex] [itex]\rho[/itex](r) r dV - this is a volume integral.
What exactly does ri stand for in the first line? What is the appropriate formula to replace it in the second?
 
  • #6
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ri is the vector position of a mass in some direction. I know that by symmetry it should be on the z axis. Should I multiply the whole thing be z hat in spherical (cos[itex]\theta[/itex] - sin[itex]\theta[/itex]). That does not make much sense, and it does not work.
 
  • #7
haruspex
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ri is the vector position of a mass in some direction.
Yes, it's a vector. But you have replaced it by r, the magnitude of the vector. You cannot do that because in the integral the components of the vector in the x and y directions would cancel.
Should I multiply the whole thing be z hat in spherical (cos[itex]\theta[/itex] - sin[itex]\theta[/itex]). That does not make much sense, and it does not work.
Multiplying by sin or cos of theta (depending on which way it's measured) makes sense to me.
 
  • #8
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OK I got it, I skipped a step:
Cm = 1/V[itex]\int[/itex] r (r hat) dV
As I know it's on the z axis:
Cm = 1/V[itex]\int[/itex] z (z hat) dV

z = r cos[itex]\theta[/itex]
dV = r^2 sin[itex]\theta[/itex]dr d[itex]\theta[/itex] d[itex]\varphi[/itex]

Cm = 1/V[itex]\int[/itex]r cos[itex]\theta[/itex] (z hat) r^2 sin[itex]\theta[/itex]dr d[itex]\theta[/itex] d[itex]\varphi[/itex]

I can pull out z hat from the integral.

With V = (2[itex]\pi[/itex]R^3)/3, this gives me 3R/8 (z hat).

Well, that was silly...

Thank you guys for your help.
 
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