Eulerβs equation for rigid bodies
Let be an inertial frame body (a rigid body) and a rigid body in motion respect to an observer located at . Let be an arbitrary point (fixed or in motion) and the center of mass of . Then,
(1) |
where is the mass of the rigid body, the position vector of respect to , is the moment of forces system respect to , the tensor of inertia respect to orthogonal axes embedded in and origin at 11That is possible because the kinematical concept of frame extension., and , , , are the acceleration of , the angular velocity and acceleration vectors respectively, all of them measured by an observer located at .
This equation was got by Euler by using a fixed system of principal axes with origin at . In that case we have , and therefore
(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 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 is at rest or in uniform and rectilinear translation, then , 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 )
Title | Eulerβs equation for rigid bodies |
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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 |