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lec15 (3) - Constraints and Degrees of Freedom 1...

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Cite as: Thomas Peacock and Nicolas Hadjiconstantinou, course materials for 2.003J/1.053J Dynamics and Control I, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. 1 Constraints and Degrees of Freedom 2.003J/1.053J Dynamics and Control I, Spring 2007 Professor Thomas Peacock 4/9/2007 Lecture 15 Lagrangian Dynamics: Derivations of Lagrange’s Equations Constraints and Degrees of Freedom Constraints can be prescribed motion Figure 1: Two masses, m 1 and m 2 connected by a spring and dashpot in parallel. Figure by MIT OCW. 2 degrees of freedom If we prescribe the motion of m 1 , the system will have only 1 degree of freedom, only x 2 . For example, x 1 ( t ) = x 0 cos ωt x 1 = x 1 ( t ) is a constraint. The constraint implies that δx 1 = 0. The admissible variation is zero because position of x 1 is determined. For this system, the equation of motion (use Linear Momentum Principle) is mx ¨ 2 = k ( x 2 x 1 ( t )) c x 2 x ˙ 1 ( t )) mx ¨ 2 + cx ˙ 2 + kx 2 = cx ˙ 1 ( t ) + kx 1 ( t )
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Cite as: Thomas Peacock and Nicolas Hadjiconstantinou, course materials for 2.003J/1.053J Dynamics and Control I, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. 2 Lagrange’s Equations cx ˙ 1 ( t ) + kx 1 ( t ): known forcing term differential equation for x 2 ( t ): ODE, second order, inhomogeneous Lagrange’s Equations For a system of n particles with ideal constraints Linear Momentum p ˙ = f ext + f constraint (1) i i i N ( f ext + f constraint p ˙ ) = 0 (2) i i i i =1 f constraint = 0 i i =1 D’Alembert’s Principle N ( f ext p ˙ ) · δr i = 0 (3) i i i =1 Choose p ˙ i = 0 at equilibrium. We have the principle of virtual work. Hamilton’s Principle Now p ˙ = m i r ¨ i , so we can write: i N ( m i r ¨ i f ext ) · δr i = 0 (4) i i =1 N δW = f ext · δr i , (5) i i =1 which is the virtual work of all active forces, conservative and nonconservative. N N d m i r ¨ i · δr i = m i r i · δr i ) r ˙ i · δr ˙ i (6) dt i =1 i =1 d (6) is obtained by using dt r · δr ) = ¨ rδr + ˙ rδr ˙
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3 Lagrange’s Equations r ˙ i · δr ˙ i can be rewritten as 1 2 δ r · r ) by using δ r · r ˙) = rδr ˙. Substituting this in (6), we can write: N N N d 1 m i r ¨ i · δr i = m i r i · δr i ) δ m r i · r ˙ i ) (7) dt 2 i =1 i =1 i =1 The second term on the right is a kinetic energy term.
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  • Fall '08
  • MarkusBuehler
  • Lagrangian mechanics, DD Month YYYY, Thomas Peacock, 2.003J/1.053J Dynamics, Nicolas Hadjiconstantinou

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