# centre of mass of polygon

Let ${A}_{1}{A}_{2}\mathrm{\dots}{A}_{n}$ be an $n$-gon (http://planetmath.org/Polygon^{}) which is supposed to have a 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

$\overrightarrow{OM}={\displaystyle \frac{1}{n}}{\displaystyle \sum _{i=1}^{n}}\overrightarrow{O{A}_{i}}.$ | (1) |

We can of course take especially $O={A}_{1}$, and thus

$$\overrightarrow{{A}_{1}M}=\frac{1}{n}\sum _{i=1}^{n}\overrightarrow{{A}_{1}{A}_{i}}=\frac{1}{n}\sum _{i=2}^{n}\overrightarrow{{A}_{1}{A}_{i}}.$$ |

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

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

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

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

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

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

Title | centre of mass of polygon |

Canonical name | CentreOfMassOfPolygon |

Date of creation | 2013-03-22 17:33:13 |

Last modified on | 2013-03-22 17:33:13 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 11 |

Author | pahio (2872) |

Entry type | Result |

Classification | msc 51P05 |

Classification | msc 51M04 |

Classification | msc 26B15 |

Classification | msc 15A72 |

Synonym | centroid of polygon |

Related topic | ArithmeticMean |

Related topic | AreaOfPolygon |

Related topic | CentreOfMassOfHalfDisc |

Related topic | BarycentricSubdivision |

Related topic | CoordinatesOfMidpoint |