# Center of Mass of Hemisphere

1. Jun 27, 2015

1. The problem statement, all variables and given/known data
Find the center of mass of a hemisphere of radius R.

2. Relevant equations
R_cm = (rho/M) Integral[ (vector r) dv]

dv = r^2 Sin[theta] dr d[theta] d[phi]

3. The attempt at a solution

In spherical polar coordinates, vector r = r r-hat, so the argument of the integral should be:
r^3 Sin[theta] dr d[theta] d[phi]

Well, that's what I would expect... I keep ending up with 3R/4, but other sources have it as 3R/8, and I see that the volume element dv is written with an extra Cos[theta] which I do not understand..........

I have no problem integrating over volumes in E&M problems, so I do not get why this is such an issue for me.

2. Jun 27, 2015

Ok, I started to figure it out, but I'm not all the way there yet:

Obviously, by symmetry the CM is along the vertical Z-axis if the flat part of the hemisphere lies in the XY plane.

The height at any random r is: z = r Cos[theta]

If I plug THIS value into the integral, in lieu of the extra "r" I put in earlier, the integral comes out to 3R/8, exactly as I have seen from other sources.

What I don't understand is:
Why on earth do you replace "r" (from the position vector r) with "r Cos[theta]" or "z"??? Why doesn't it work without this?

3. Jun 27, 2015

### SteamKing

Staff Emeritus
The extra factor of cos θ is included because in calculating the moment of dV, you are interested in the perpendicular distance of dV above the x-y plane, i.e., z, not the total distance r. This goes back to the definition of what a moment is.

4. Jun 27, 2015

### HallsofIvy

Staff Emeritus
The mathematical 'centroid' of a three dimensional figure, which is the same as the center of mass of a solid with constant density, is the given by $\overline{x}= \\int\int\int x dV$/V, $\overline{y}= \int\int\int y dV/V$, $\overline{y}= \int\int\int z dV/V$ where V is the volume, $\int\int\int dV$. (The density, being constant, can be factored out of the integral in the numerator, as well as the mass integral in the denominator, and canceled).

Since this figure is symmetric around the z axes, it is immediate that $\overline{x}= \overline{y}= 0$. To find $\overline{z}$, we can imagine the hemisphere divided into thin layers, each layer a circle with center at (0, 0, z), radius $r= \sqrt{x^2+ y^2}= \sqrt{R^2- z^2}$ and thickness dz. The area of such a disk is $\pi (R^2- z^2)$ and volume $\pi (R^2- z^2)dz$. So $\int\int\int z dV$ reduces to $\int_0^R (R^2z- z^3)dz$.. The volume is, of course, $\frac{2}{3}\pi R^3$.

5. Jun 27, 2015

I'm going through John Taylor's Classical Mechanics, and in the next section he talks about moments of inertia, but this has not been introduced yet where this problem arises. Is it possible to understand this problem without the use of the definition of the moment?

I understand it's easier to hand-wave away the horizontal components and see that the problem has symmetry along the XY plane, but I would rather be able to do it rigorously through math, rather than by assumption. Assumptions and hand-waving are all well and good once you've got a good hold of the problem, but I'd like to be able to work it out completely at least once before jumping to that part.

6. Jun 27, 2015

### TSny

You left out the r-hat unit vector in the integrand.

7. Jun 27, 2015

### SteamKing

Staff Emeritus
The moment of inertia is ∫ z2 dV, which is sometimes called the second moment, but this is irrelevant to the calculation of the centroid.

For calculating the centroid, you are interested in finding the first moment, which is ∫ z dV.

This distinction about moments should have been encountered in your mechanics classes at some point, and possibly also covered in a calculus class somewhere, as well.

8. Jun 27, 2015

Thank you for the tip. I did indeed forget to include r-hat, which is not constant. I put in r-hat = sincos(etc)...xhat, yhat, zhat, and indeed the xhat and yhat integrals evaluate to 0, and the z-hat integral gives exactly the right answer :)

I'm sure everything regarding the moments is true, but I really wanted to be able to understand this on a more fundamental level (mathematically) before including additional physics.

9. Jun 27, 2015

### SteamKing

Staff Emeritus
There's not much physics to include here.

The mathematical definition of moment and the physics definition are pretty much the same. If you study statistics to any depth, you'll come across these same concepts in figuring out things like expectation and variance and whatnot, which is why these concepts are often taught in calculus courses: Their utility is not confined to just one subject.