Is there a way to find the "center of mass" of an irregular polygon?

Question

Just like the medicenter of a triangle is its "center of mass" if we cut it. What about some irregular shaped object? My first thought is to divide it up in triangles, then find the medicenter of each triangle, then connect the medicenters and repeating the process until you have only 3 medicenters to connect. Does this give you the "center of mass" of the object, because it seems to me it does, but I cannot prove it.

Answer 1

Here's an algebraic way. By Stokes' theorem, the moment of inertia of any region $R$ about the $y$-axis is $$ \iint_R x\, dx\, dy = \oint_{\partial R} \frac{1}{2} x^2\, dy.$$

If one side $S$ of a polygon connects points $(h_1, k_1)$ and $(h_2, k_2)$, then parametrize it as $x = h_1 (1-t) + h_2 t, y = k_1 (1-t) + k_2 t$. Then the integral over the side $S$ is $$ \begin{align*} \int_S \frac{1}{2} x^2\, dy &= \frac{1}{2} \int_0^1 \left[ h_1^2 (1-t)^2 + 2 h_1 h_2 t (1-t) + h_2^2 t^2 \right]\, (k_2 - k_1)\, dt \\ &= \frac{k_2 - k_1}{2} \int_0^1 \left[(h_1 - h_2)^2 t^2 - 2(h_1^2 + h_1 h_2) t + h_1^2 \right]\, dt \\ &= \frac{k_2 - k_1}{2} \left[ \frac{(h_1 - h_2)^2}{3} - (h_1^2 + h_1 h_2) + h_1^2 \right] \\ &= \frac{(k_2 - k_1)(h_1^2 + h_1 h_2 + h_2^2)}{6}.\end{align*} $$

The moment of inertia of the whole polygon is just the sum of this quantity over every side, taken anticlockwise. Note that the $h_2^2 k_2$ term in one side cancels the $h_1^2 k_1$ term in the next, giving the formula at the Wikipedia link in sigmabe's answer. There's a formula for area that can be obtained in similar fashion from noting $$\int_R\, dx\, dy = \oint_{\partial R} x\, dy$$ and parametrizing each side of a polygon.

Answer 2

You can do it in two phases . . .

Phase $(1)\,$:

Arbitrarily triangulate the polygonal region.

For each triangle, create a mass-point pair $(m,P)$, where $m$ is the area of the triangle, and $P$ is the centroid.

Phase $(2)\,$:

Choose any two distinct mass-points $(m_1,P_1)$ and $(m_2,P_2)$, and replace them by $(m,P)$, where \begin{align*} m&=m_1+m_2\\[4pt] P &= \left({\small{\frac{m_1}{m_1+m_2}}}\right)P_1 + \left({\small{\frac{m_2}{m_1+m_2}}}\right)P_2\\[4pt] &\!\!\!\!\! \left(\text{i.e., $P$ is ${\small{\frac{m_2}{m_1+m_2}}}$ of the way from $P_1$ to $P_2$.}\right)\\[4pt] \end{align*} If the new point $P$ happens to coincide with a point from one of the other remaining mass-point pairs, merge the two mass-point pairs into a single mass-point pair, summing the two masses.

When only one mass-point $(m,P)$ remains, $m$ is the area of the original region, and $P$ is the center of gravity.