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## Center of mass for a circular wire segment

Posted by peeterjoot on May 16, 2010

As a check for the torus segment center of mass calculation, there should be agreement in the limit where the radius of the torus goes to zero (with a non-zero correction otherwise).

# Center of mass for a circular wire segment.

As an additional check for the correctness of the result above, we should be able to compare with the center of mass of a circular wire segment, and get the same result in the limit $r \rightarrow 0$.

For that we have

\begin{aligned}Z (R \Delta \theta) = \int_{\theta=-\Delta \theta/2}^{\Delta \theta/2} R i e^{-i\theta} R d\theta\end{aligned} \hspace{\stretch{1}}(4.21)

So we have

\begin{aligned}Z &= \frac{1}{{\Delta \theta}} R i \frac{1}{{-i}} (e^{-i\Delta \theta/2} - e^{i\Delta\theta/2}).\end{aligned} \hspace{\stretch{1}}(4.22)

Observe that this is

\begin{aligned}Z &= R i \frac{\sin(\Delta\theta/2)}{\Delta\theta/2},\end{aligned} \hspace{\stretch{1}}(4.23)

which is consistent with the previous calculation for the solid torus when we let that solid diameter shrink to zero.

In particular, for $3/4$ of the torus, we have $\Delta \theta = 2 \pi (3/4) = 3 \pi/2$, and

\begin{aligned}Z = R i \frac{4 \sin(3\pi/4)}{3 \pi} = R i \frac{2 \sqrt{2}}{3 \pi} \approx 0.3 R i.\end{aligned} \hspace{\stretch{1}}(4.24)

We are a little bit up the imaginary axis as expected.

I’d initially somehow thought I’d been off by a factor of two compared to the result by The Virtuosi, without seeing a mistake in either. But that now appears not to be the case, and I just screwed up plugging in the numbers. Once again, I should go to my eight year old son when I have arithmetic problems, and restrict myself to just the calculus and algebra bits.