Peeter Joot's (OLD) Blog.

Math, physics, perl, and programming obscurity.

Center of mass for a toroidal segment

Posted by peeterjoot on May 15, 2010


[Click here for a PDF of this post with nicer formatting]Note that this PDF file is formatted in a wide-for-screen layout that is probably not good for printing.

After seeing Iron Man II with Lance earlier, a movie with bountiful toroids, and now that the kids are tucked in, finishing up the toroidal center of mass calculation started earlier seems like it is in order. This is a problem I’d been meaning to try since reading this blog post for the center of mass of a toroidal wire segment

Center of mass.

With the prep done, we are ready to move on to the original problem. Given a toroidal segment over angle \theta \in [-\Delta \theta/2, \Delta \theta/2], then the volume of that segment is

\begin{aligned}\Delta V = r^2 R \pi \Delta \theta.\end{aligned} \hspace{\stretch{1}}(3.13)

Our center of mass position vector is then located at

\begin{aligned}\mathbf{R} \Delta V &= \int_{\rho=0}^r \int_{\theta=-\Delta \theta/2}^{\Delta \theta/2} \int_{\phi=0}^{2\pi} e^{-j\theta/2} \left( \rho \mathbf{e}_1 e^{ i \phi } + R \mathbf{e}_3 \right) e^{j \theta/2} \rho \left( R + \rho \sin\phi \right) d\rho d\theta d\phi.\end{aligned} \hspace{\stretch{1}}(3.14)

Evaluating the \phi integrals we loose the \int_0^{2\pi} e^{i\phi} and \int_0^{2\pi} \sin\phi terms and are left with \int_0^{2\pi} e^{i\phi} \sin\phi d\phi = i \pi /2 and \int_0^{2\pi} d\phi = 2 \pi. This leaves us with

\begin{aligned}\mathbf{R} \Delta V &= \int_{\rho=0}^r \int_{\theta=-\Delta \theta/2}^{\Delta \theta/2} \left( e^{-j\theta/2} \rho^3 \mathbf{e}_3 \frac{\pi}{2} e^{j \theta/2} + 2 \pi \rho R^2 \mathbf{e}_3 e^{j \theta}  \right) d\rho d\theta \\ &= \int_{\theta=-\Delta \theta/2}^{\Delta \theta/2} \left( e^{-j\theta/2} r^4 \mathbf{e}_3 \frac{\pi}{8} e^{j \theta/2} + 2\pi \frac{1}{{2}} r^2 R^2 \mathbf{e}_3 e^{j \theta}  \right) d\theta \\ &= \int_{\theta=-\Delta \theta/2}^{\Delta \theta/2} \left( e^{-j\theta/2} r^4 \mathbf{e}_3 \frac{\pi}{8} e^{j \theta/2} + \pi r^2 R^2 \mathbf{e}_3 e^{j \theta}  \right) d\theta.\end{aligned} \hspace{\stretch{1}}(3.15)

Since \mathbf{e}_3 j = -j \mathbf{e}_3, we have a conjugate commutation with the e^{-j \theta/2} for just

\begin{aligned}\mathbf{R} \Delta V &= \pi r^2 \left( \frac{r^2}{8} + R^2 \right) \mathbf{e}_3 \int_{\theta=-\Delta \theta/2}^{\Delta \theta/2} e^{j \theta} d\theta \\ &= \pi r^2 \left( \frac{r^2}{8} + R^2 \right) \mathbf{e}_3 2 \sin(\Delta \theta/2).\end{aligned} \hspace{\stretch{1}}(3.18)

A final reassembly, provides the desired final result for the center of mass vector

\begin{aligned}\mathbf{R} &= \mathbf{e}_3 \frac{1}{{R}} \left( \frac{r^2}{8} + R^2 \right) \frac{\sin(\Delta \theta/2)}{ \Delta \theta/2 }.\end{aligned} \hspace{\stretch{1}}(3.20)

Presuming no algebraic errors have been made, how about a couple of sanity checks to see if the correctness of this seems plausible.

We are pointing in the z-axis direction as expected by symmetry. Good. For \Delta \theta = 2 \pi, our center of mass vector is at the origin. Good, that’s also what we expected. If we let r \rightarrow 0, and \Delta \theta \rightarrow 0, we have \mathbf{R} = R \mathbf{e}_3 as also expected for a tiny segment of “wire” at that position. Also good.


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