Solid hemisphere center of mass in spherical coordinates

[Ledamien]

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 \int \rho r^3 sin \theta dr d\theta d\varphi

M = ρ V → 1/M = 1/ρV

V = 2 \pi R^3 / 3

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

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

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

So, what do I do wrong?

 

[SteamKing]

How did you derive your equation for the Cm?

 

[haruspex]

\int \rho r^3 sin \theta dr d\theta d\varphi

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.

 

[Ledamien]

1/M \sum mi ri
1/M\int \rho(r) r dV - this is a volume integral.
1/M\int [\rho(r) r] r^2 dr sin \theta) d\theta d\varphi - 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 \pi/2 and 0 to 2\pi would be sufficient. As the density is constant I can pull it out of the integral, and \rho/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?

 

[haruspex]

1/M \sum mi ri
1/M\int \rho(r) r dV - this is a volume integral.

What exactly does r_i stand for in the first line? What is the appropriate formula to replace it in the second?

 

[Ledamien]

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\theta - sin\theta). That does not make much sense, and it does not work.

 

[haruspex]

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\theta - sin\theta). 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.

 

[Ledamien]

OK I got it, I skipped a step:
Cm = 1/V\int r (r hat) dV
As I know it's on the z axis:
Cm = 1/V\int z (z hat) dV

z = r cos\theta
dV = r^2 sin\thetadr d\theta d\varphi

Cm = 1/V\intr cos\theta (z hat) r^2 sin\thetadr d\theta d\varphi

I can pull out z hat from the integral.

With V = (2\piR^3)/3, this gives me 3R/8 (z hat).

Well, that was silly...

Thank you guys for your help.