centre of mass of polygon

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 centre of mass of the polygon and $ O$ the origin. Then the position vector of $ M$ with respect to $ O$ is

$\displaystyle \overrightarrow{OM} = \frac{1}{n}\sum_{i=1}^n\overrightarrow{OA_i}.$ (1)

We can of course take especially $ O = A_1$ , and thus

$\displaystyle \overrightarrow{A_1M} = \frac{1}{n}\sum_{i=1}^n\overrightarrow{A_1A_i} =
\frac{1}{n}\sum_{i=2}^n\overrightarrow{A_1A_i}.$

In the special case of the triangle $ ABC$ we have

$\displaystyle \overrightarrow{AM} = \frac{1}{3}(\overrightarrow{AB}+\overrightarrow{AC}).$ (2)

The centre of mass of a triangle is the common point of its medians.

Remark. An analogical result with (2) concerns also the homogeneous tetrahedron $ ABCD$ ,

$\displaystyle \overrightarrow{AM} = \frac{1}{4}(\overrightarrow{AB}+\overrightarrow{AC}+\overrightarrow{AD}),$

and any $ n$ -dimensional simplex (cf. the midpoint of line segment: $ \overrightarrow{AM} = \frac{1}{2}\overrightarrow{AB}$ ).



Contributors to this entry (in most recent order):

As of this snapshot date, this entry was owned by pahio.