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 $z_{\rm cm} = R/2.$