Euler’s 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,

𝐌Q=IQ⁒𝜢21+𝝎21Γ—(IQ⁒𝝎21)+mβ’ππ‚Β―Γ—πš1Q⁒2, (1)

where m is the mass of the rigid body, 𝐐𝐂¯ the position vector of C respect to Q, 𝐌Q is the moment of forces system respect to Q, IQ the tensor of inertia respect to orthogonalMathworldPlanetmathPlanetmathPlanetmath axes embedded in 2 and origin at Q⁒2 11That is possible because the kinematical concept of frame extension., and 𝐚1Q⁒2, 𝝎21, 𝜢21, are the acceleration of Q⁒2, 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⁒2. In that case we have Q=C, and therefore

𝐌C=IC⁒𝜢21+𝝎21Γ—(IC⁒𝝎21). (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 ℝ3 are six, just equal the number of independent scalar equations necessary to solve such a motion. (Newton’s law contributing with three)
Its is clear if 2 is at rest or in uniform and rectilinear translationMathworldPlanetmathPlanetmath, then 𝐌Q=𝟎, 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 𝐅=𝟎)

Title Euler’s 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