Finding the centre of mass of a hemispherical shell

[clementc]

Homework Statement

Hey everyone,
I'm studying for my physics and came across a question for the COM of a hemisphere. However, I was wondering what methods there are to find the COM of a hemispherical shell instead. Any insight would be very much appreciated! =)
After much of tearing Google apart in order to find a solution, I found 2, but was wondering if there were any more methods (I always like to find as many solutions as possible because it really helps me understand).

Homework Equations

I just learned integration in cylindrical and spherical coordinates, so I kinda know that:
x_\textrm {COM}=\frac{\int x dm}{\int dm}
dA = r^2 \sin\theta\ d\theta\ d\phi
dV = r^2 \sin\theta\ dr\ d\theta\ d\phi
where \theta is the polar angle, and \phi is the angle about the equator (I learned it this way, but I think most people use \phi as the polar angle instead)

The Attempt at a Solution

The 2 methods I found were:
1. Expressing dA as a ratio of the total SA, and this would be equal to the ratio of dm/M; i.e. that \frac{dA}{2\pi r^2} = \frac{dm}{M} (OMG this TeXnology is so cool!)
ANYWAY so with this, you would be able to

• find dm

• find that z=r \cos \theta, where z is the axis passing through the apex and centre of the hemisphere.

and plug it all into the equation, giving R/2. YAY!

2. A bit more complicated but really nifty way was to call big radius R, and the one on the closer side of the shell \kappa R, where 0\leq k\leq 1.
Then
\textrm {Total Volume} = \frac{2}{3}\pi R^3 (1-{\kappa}^3)
Once again, z=r \cos \theta and dV = r^2 \sin\theta\ dr\ d\theta\ d\phi \Rightarrow dm = \rho r^2 \sin\theta\ dr\ d\theta\ d\phi where \rho is density. From here, you can place into the formula, and end up with
\frac{3R(1-{\kappa}^4)}{8(1-{\kappa}^3)}.
Now, as 1-{\kappa}^4 = (1-\kappa)(1+\kappa+{\kappa}^2+{\kappa}^3) and 1-{\kappa}^3 = (1-\kappa)(1+\kappa+{\kappa}^2), you can simplify this.
For a spherical shell, you just take \lim_{\substack{x \rightarrow 1}}, and that gives you R/2 again!

 

[TSny]

Those methods look good.

For fun, here's another way. Consider the figure below:

Let $dA_{xy}$ be an element of area under the hemisphere on the x-y plane. If you project that area up onto the curved surface of the hemisphere, you get a patch on the hemisphere with area $dA_H = \large \frac{dA_{xy}}{\cos \theta}$, where $\theta$ is the angle between the normal vector $\hat n$ on the hemisphere and the z-axis. The z-coordinate of the patch is $z = R \cos \theta$.

Let $\sigma$ denote the uniform surface mass density of the hemisphere. Then, $$z_{\rm cm} = \frac{1}{M} \int{z \, dm} = \frac{1}{M} \int {z \, \sigma \, dA_H} = \frac{\sigma}{M} \int(R \cos \theta) \frac{dA_{xy}}{\cos \theta} = \frac{\sigma}{M} R \int dA_{xy} = \frac{\sigma}{M} R \left(\pi R^2\right)$$

Using $\sigma = \large \frac{M}{2 \pi R^2}$ gives the result $z_{\rm cm} = R/2.$