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Transcript
Euler’s equation for rigid bodies∗ perucho† 2013-03-21 23:04:05 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 α21 + ω 21 × (IQ ω 21 ) + mQC × aQ2 1 , (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 orthogonal axes embedded in 2 and origin at Q2 1 , and aQ2 1 , ω 21 , α21 , are the acceleration of Q2, 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 C2. In that case we have Q = C, and therefore MC = IC α21 + ω 21 × (IC ω 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 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 translation, 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) ∗ hEulersEquationForRigidBodiesi created: h2013-03-21i by: hperuchoi version: h39490i Privacy setting: h1i hTopici h70G45i † This text is available under the Creative Commons Attribution/Share-Alike License 3.0. You can reuse this document or portions thereof only if you do so under terms that are compatible with the CC-BY-SA license. 1 That is possible because the kinematical concept of frame extension. 1
