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Electromechanical Dynamics (Part 1).0067

# Electromechanical Dynamics (Part 1).0067 - the mass were...

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Lumped Electromechanical Elements Thus the infinitesimal element of mass p dV at point p will have an accelera- tion force d ' r df, = p dV . (2.2.13) dt 2 The force df, consists of two components: df from sources external to the body and dfT from sources within the body. Thus (2.2.13) can be written as dar df + df = p dV d (2.2.14) dt 2 The mass density associated with each point p in the rigid body is constant. Hence, when we integrate this expression throughout the volume of the body and recognize that the internal forces integrate to zero,* we obtain the result f = M d (2.2.15) dt 2 where f is the total external force applied to the body, M = f p dV is a constant and is the total mass of the body, _ prdV rm, = is the position vector of the center of mass of the body. M From the result of (2.2.15) it is clear that the translational motion of a rigid body can be described completely by treating the body as if
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Unformatted text preview: the mass were concentrated at the center of mass. Consequently, (2.2.10) and (2.2.11) * To illustrate that the internal forces integrate to zero consider an ensemble of N inter-acting particles. An internal force is applied between two particles: fij = -fii, where ft is the force on the ith particle due to the source connected between the ith and jth particles. The total internal force on the ith particle is N fi= fij' 5=1 i-i The total internal force on the ensemble is N NN i=1 i=1 1=1 Because f,, = --fi, we conclude that each term in this summation is canceled by an equal and opposite term and the net internal force is zero. If we let N -oo, the nature of the result is unchanged; thus we conclude that the integral of internal forces over the volume of a rigid body must be zero. A-PDF Split DEMO : Purchase from www.A-PDF.com to remove the watermark...
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