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How would one find the center of mass of a circle? The center of mass of a rod is given by:

$$\frac{1}{M}\int^{L}_{0}\rho x dx$$

So, for a sphere, it would be an area integral, such as:

$$\frac{1}{M}\int^{?}_{?} \rho ? dA$$

The "?" mean I don't know what to put there.

This is an area integral, which means it needs to be taken over the two axises, correct?

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Suppose we are interested in the centroid of a sphere, meaning we ignore mass. We know that the centroid is at the center of the sphere. How do we know that? – John Douma Apr 20 '13 at 2:25
You have to know double integrals for circle and triple for sphere. – Belov Apr 20 '13 at 2:41

When one knows, or can divine, the answer in advance there is often a co-ordinate system that makes the problem trivial. For the Center of Mass of a sphere, integrate from r = 0 to R, the two semi-spheres above and below the x-y plane. That then trivially becomes the integral from 0 to R of ( (2 * pi * r^2) - (2 * pi * r^2) ) * dr. Amazingly, this integral evaluates to the origin!

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do not forget the density distribution, if the disk is not homogeneous the center of mass might not be at the center – Wolphram jonny Apr 20 '13 at 3:10
True, but the post stands; taking advantage of all symmetries to reduce the evaluation to a summation over a finite set of point masses is how physicists prefer to perform this calculation. – Pieter Geerkens Apr 20 '13 at 3:15
I am not sure to understand what you say, why is not the mass density included in your equations? – Wolphram jonny Apr 20 '13 at 3:19

Your first integral is correct. However, to find the center of mass of a sphere requires three integrals, one for each coordinate. Each of those integrals will look like the one you wrote down.

$$ \bar x = \frac{1}{M}\int_{S} \rho xdA $$ $$ \bar y = \frac{1}{M}\int_{S} \rho ydA $$ $$ \bar z = \frac{1}{M}\int_{S} \rho zdA $$

The limits of integration will have to cover the whole sphere so if you use spherical coordinates, which you should, the vertical angle will go from $0$ to $\pi$ and the azimuth will go from $0$ to $2\pi$. Of course, you'll have to convert $x,y$ and $z$ to whatever coordinate system you are using.

Let's set up $\bar x$ in spherical coordinates as an example. In spherical coordinates $$ dA = (Rd\theta)(R\sin \theta d\phi) = R^2\sin \theta d\theta d\phi $$

where I am using $\theta$ for the vertical angle and $\phi$ for the azimuth.

$$ \bar x = \frac{1}{M}\int_{S} \rho xdA = \frac{1}{M}\int_{0}^{2\pi}\int_{0}^{\pi}\rho xR^2\sin \theta d\theta d\phi = \frac{R^2}{M}\int_{0}^{2\pi}\int_{0}^{\pi}\rho x\sin \theta d\theta d\phi $$

Substituting $x = R\sin \theta \cos \phi$ gives us

$$ \bar x = \frac{R^2}{M}\int_{0}^{2\pi}\int_{0}^{\pi}\rho(R\sin \theta \cos \phi)\sin \theta d\theta d\phi = \frac{R^3}{M}\int_{0}^{2\pi}\int_{0}^{\pi}\rho \sin^2 \theta \cos \phi d\theta d\phi $$

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For a disk (assuming is centered at $(x_c,y_c)=(0,0)$: the x coordinate would be:

$\frac{1}{M}\int^r_{-r}\int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}} \rho (x,y) x dx dy$, where $r$ is the radius

and the y coordinate would be:

$\frac{1}{M}\int^r_{-r}\int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}} \rho (x,y) ydx dy$, where $r$ is the radius

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Are you calculating the center of mass of a circle or a disk? – John Douma Apr 21 '13 at 17:36
a disk of zero height (I assumed that you meant that when you said a circle). – Wolphram jonny Apr 21 '13 at 18:16
By circle, I mean the set of points in a plane equidistant from a fixed point. By disk, I mean a circle together with the area it encloses. Take $\rho$ to be $1$ when $y\ge 0$ and $0$, otherwise. You will get a different result for the two. – John Douma Apr 21 '13 at 18:23
ok, I missundertood you, my equations are for a disk. Let me know if you want a circle. – Wolphram jonny Apr 21 '13 at 18:39
I didn't ask the original question. Your first sentence starts, "For a circle". – John Douma Apr 21 '13 at 18:53

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