Talk:PlanetPhysics/Centre of Mass of Polygon

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Let $A_1A_2{\ldots}A_n$ be an $n$-gon which is supposed to have a constant surface-density in all of its points, $M$ the \htmladdnormallink{centre of mass}{http://planetphysics.us/encyclopedia/CenterOfGravity.html} of the polygon and $O$ the origin. Then the \htmladdnormallink{position vector}{http://planetphysics.us/encyclopedia/PositionVector.html} of $M$ with respect to $O$ is \begin{align} \overrightarrow{OM} = \frac{1}{n}\sum_{i=1}^n\overrightarrow{OA_i}. \end{align} We can of course take especially\, $O = A_1$,\, and thus $$\overrightarrow{A_1M} = \frac{1}{n}\sum_{i=1}^n\overrightarrow{A_1A_i} = \frac{1}{n}\sum_{i=2}^n\overrightarrow{A_1A_i}.$$

In the special case of the triangle $ABC$ we have \begin{align} \overrightarrow{AM} = \frac{1}{3}(\overrightarrow{AB}+\overrightarrow{AC}). \end{align} The centre of mass of a triangle is the common point of its medians.\\

\textbf{Remark.} An analogical result with (2) concerns also the homogeneous tetrahedron $ABCD$, $$\overrightarrow{AM} = \frac{1}{4}(\overrightarrow{AB}+\overrightarrow{AC}+\overrightarrow{AD}),$$ and any $n$-dimensional simplex (cf. the midpoint of line segment:\, $\overrightarrow{AM} = \frac{1}{2}\overrightarrow{AB}$).

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