# Eulerβs equation for rigid bodies

Let $1$ be an inertial frame body (a rigid body) and $2$ a rigid body in motion respect to an observer located at $1$. Let $Q$ be an arbitrary point (fixed or in motion) and $C$ the center of mass of $2$. Then,

${\mathrm{\pi \x9d\x90\x8c}}_{Q}={\mathrm{I}}^{Q}\beta \x81\u2019{\mathrm{\pi \x9d\x9c\u0386}}_{21}+{\mathrm{\pi \x9d\x9d\x8e}}_{21}\Gamma \x97({\mathrm{I}}^{Q}\beta \x81\u2019{\mathrm{\pi \x9d\x9d\x8e}}_{21})+m\beta \x81\u2019\stackrel{{\rm B}\u2015}{\mathrm{\pi \x9d\x90\x90\pi \x9d\x90\x82}}\Gamma \x97{\mathrm{\pi \x9d\x90\x9a}}_{1}^{Q\beta \x81\u20192},$ | (1) |

where $m$ is the mass of the rigid body, $\stackrel{{\rm B}\u2015}{\mathrm{\pi \x9d\x90\x90\pi \x9d\x90\x82}}$ the position vector of $C$ respect to $Q$, ${\mathrm{\pi \x9d\x90\x8c}}_{Q}$ is the moment of forces system respect to $Q$, ${\mathrm{I}}^{Q}$ the tensor of inertia respect to orthogonal^{} axes embedded in $2$ and origin at $Q\beta \x81\u20192$ ^{1}^{1}That is possible because the kinematical concept of frame extension., and ${\mathrm{\pi \x9d\x90\x9a}}_{1}^{Q\beta \x81\u20192}$, ${\mathrm{\pi \x9d\x9d\x8e}}_{21}$, ${\mathrm{\pi \x9d\x9c\u0386}}_{21}$, are the acceleration of $Q\beta \x81\u20192$, the angular velocity and acceleration vectors respectively, all of them measured by an observer located at $1$.

This equation was got by Euler by using a fixed system of principal axes with origin at $C\beta \x81\u20192$. In that case we have $Q=C$, and therefore

${\mathrm{\pi \x9d\x90\x8c}}_{C}={\mathrm{I}}^{C}\beta \x81\u2019{\mathrm{\pi \x9d\x9c\u0386}}_{21}+{\mathrm{\pi \x9d\x9d\x8e}}_{21}\Gamma \x97({\mathrm{I}}^{C}\beta \x81\u2019{\mathrm{\pi \x9d\x9d\x8e}}_{21}).$ | (2) |

Euler used three independent scalar equations to represent (2). It is well known that the number of degrees of freedom associate to a rigid body in free motion in ${\mathrm{\beta \x84\x9d}}^{3}$ are six, just equal the number of independent scalar equations necessary to solve such a motion. (Newtonβs law contributing with three)

Its is clear if $2$ is at rest or in uniform and rectilinear translation^{}, then ${\mathrm{\pi \x9d\x90\x8c}}_{Q}=\mathrm{\pi \x9d\x9f\x8e}$, one of the necessary and sufficient conditions for the equilibrium of the system of forces applied to a rigid body. (The other one is the force resultant $\mathrm{\pi \x9d\x90\x85}=\mathrm{\pi \x9d\x9f\x8e}$)

Title | Eulerβs equation for rigid bodies |
---|---|

Canonical name | EulersEquationForRigidBodies |

Date of creation | 2013-03-22 17:10:36 |

Last modified on | 2013-03-22 17:10:36 |

Owner | perucho (2192) |

Last modified by | perucho (2192) |

Numerical id | 8 |

Author | perucho (2192) |

Entry type | Topic |

Classification | msc 70G45 |