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Motivation
My friend’s son is an exceptional little guy. I think he’s about 12 or so, but he’s already half way through our old 1st year calculus book. When they visited, he was intensely dissatisfied with an “obvious” statement that the center of mass necessarily lies on the axis of symmetry (when there is such a symmetry). I didn’t take the time when he was here to reason through that statement in a systematic way. It turns out that it’s pretty simple to do so.
A uniform mass with symmetry
Suppose that we have equal density thin plate, with surface density \( \rho \), with rotational symmetry about some axis, as illustrated in fig. 1.
fig. 1. An flat mass with rotational symmetry
In the plot example, I’ve plotted a positive function \( g(x) = \Abs{f(x)} \), and rotated it, plotting \( -g(a-x) \) below the axis, with the whole things bounded, in this case by \( x \in [-2, 5] \). We don’t actually need those bounds if the function is integrable everywhere (we just need a finite mass, i.e.: a finite area).
The problem of computing the center of mass coordinates is to find the values
\begin{equation}\label{eqn:centerOfMassRotationalSymmetry:20}
\begin{aligned}
\bar{x} M &= \int \sigma x dA \\
\bar{y} M &= \int \sigma y dA,
\end{aligned}
\end{equation}
where \( M = \int \sigma dA \). We are also only interested in \( \bar{y} \), since we want to show that the center of mass lies on the axis of symmetry. In symbols, with the surface mass density \( \sigma \) constant, that is
\begin{equation}\label{eqn:centerOfMassRotationalSymmetry:40}
\bar{y}
= \inv{A} \int y dx dy.
\end{equation}
Let’s ignore the bounds in the example, and perform the integral for all \( x \). If the area is bounded, then that integral will be zero outside of some region. That is
\begin{equation}\label{eqn:centerOfMassRotationalSymmetry:60}
\begin{aligned}
\bar{y}
&= \inv{A} \int_{x = -\infty}^\infty dx \int_{y = -g(a-x)}^{g(x)} y dy \\
&= \inv{A} \int_{x = -\infty}^\infty dx \inv{2} \evalrange{ y^2 }{-g(a-x)}{g(x)} \\
&= \inv{2 A} \int_{x = -\infty}^\infty dx \lr{ g^2(x) – \lr{ -g(a-x) }^2 } \\
&= \inv{2 A} \int_{x = -\infty}^\infty g^2(x) dx – \inv{2 A} \int_{x = -\infty}^\infty g^2(a-x) dx
\end{aligned}
\end{equation}
A change of variables \( u = a – x \) in the second integral gives
\begin{equation}\label{eqn:centerOfMassRotationalSymmetry:80}
\begin{aligned}
\inv{2 A} \int_{x = -\infty}^\infty g^2(a-x) dx
&=
\inv{2 A} \int_{u = \infty}^{-\infty} g^2(u) (-du) \\
&=
\inv{2 A} \int_{u = -\infty}^{\infty} g^2(u) du,
\end{aligned}
\end{equation}
which is exactly the first integral (other than a change of dummy variables.) That means
\begin{equation}\label{eqn:centerOfMassRotationalSymmetry:100}
\bar{y} = 0.
\end{equation}
This wasn’t the most general function, as we used a well behaved function (not a relation), that isn’t multivalued for any value of \( x \), but this argument would still work (requiring superposition arguments to generalize it.)
