ME 242 Lecture 30 04.17

ME 242 Lecture 30 04.17 - ME 242 Dynamics Today's Key...

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1 ME 242 Dynamics April 17, 2009 Eric Wang Today’s Key Concepts Chapter 7: Kinetics of Rigid Bodies 2D Motion Example of 2D general motion Rigid Body General 2D Motion Kinematics Kinetics a B = a A + α × r B / A + ω × ( × r B / A ) v B = v A + × r B / A F = ma G M G = I G ˙ ˙ θ b 3 M A = I G ˙ ˙ b 3 + r G / A × G M A = I A ˙ ˙ b 3 + r G / A × A Recall: Degrees of Freedom • DOF = number of independent directions of motion possible 1-DOF How Many DOF’s? How Many DOF’s? 1 2 3
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2 How Many DOF’s? Example (prob. 7.3.11) Given : cart mass = m c , Rod mass = m a rod length = 2L , inertia about O = I O Cart applies a moment Mk on rod at O Find : equations of motion for system A d N 1 + N 2 m c g + R y = m c ˙ ˙ y FBD=IRD m c g F m c ˙ ˙ x N 1 m c ˙ ˙ y M N 2 R x R y F + R x = m c ˙ ˙ x O N 1 d + N 2 d M = 0 R x FBD=IRD m a g m a a A x m a a A y M R y R x i ( R y + m a g ) j = m a a A O A I ˙ ˙ θ a A = a O + ˙ ˙ k × r A / O + ˙ k × ( ˙ k × r A / O ) = ˙ ˙ x i + ˙ ˙ k × Lb 1 + ˙ k × ( ˙ k × Lb 1 ) b 1 b 2 = ˙ ˙ x i + ˙
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ME 242 Lecture 30 04.17 - ME 242 Dynamics Today's Key...

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