Lagrangian Dynamics: Derivations of Lagrange’s Equations

Constraints and Degrees of Freedom 1 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, m1 and m2 connected by a spring and dashpot in parallel. Figure by MIT OCW. 2 degrees of freedom If we prescribe the motion of m1 , the system will have only 1 degree of freedom, only x2 . For example, x1 (t) = x0 cos ωt x1 = x1 (t) is a constraint. The constraint implies that δx1 = 0. The admissible variation is zero because position of x1 is determined. For this system, the equation of motion (use Linear Momentum Principle) is mẍ2 = −k(x2 − x1 (t)) − c(ẋ2 − ẋ1 (t)) mẍ2 + cẋ2 + kx2 = cẋ1 (t) + kx1 (t) 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]. Lagrange’s Equations 2 cẋ1 (t) + kx1 (t): known forcing term differential equation for x2 (t): ODE, second order, inhomogeneous Lagrange’s Equations For a system of n particles with ideal constraints Linear Momentum ṗi = f ext + f constraint i i N � (1) (f ext + f constraint − ṗi ) = 0 i i (2) i=1 � f constraint =0 i i=1 D’Alembert’s Principle N � − ṗi ) · δri = 0 (f ext i (3) i=1 Choose ṗi = 0 at equilibrium. We have the principle of virtual work. Hamilton’s Principle Now ṗi = mi r̈i , so we can write: N � (mi r̈ i − f ext ) · δr i = 0 i (4) i=1 δW = N � f ext · δr i , i (5) i=1 which is the virtual work of all active forces, conservative and nonconservative. N � mi r̈ i · δr i = i=1 (6) is obtained by using N � i=1 d dt (ṙ mi � d (ṙ · δr i ) − ṙ i · δṙ i dt i � (6) · δr) = r̈δr + ṙδṙ 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]. Lagrange’s Equations 3 ṙi · δṙ i can be rewritten as 21 δ(ṙ · r) by using δ(ṙ · ṙ) = 2ṙδṙ. Substituting this in (6), we can write: N � mi r̈ i · δr i = i=1 N � mi i=1 N � d 1 (ṙ i · δr i ) − δ m(ṙ i · ṙ i ) dt 2 i=1 (7) The second term on the right is a kinetic energy term. δ N � 1 i=1 2 m(ṙ i · ṙ i ) = δ(Kinetic Energy) = δT Now we rewrite (4) as: N � mi r̈ i · δri − i=1 N � f ext · δri = 0 i (8) i=1 Substitute (5) and (7 into (8) to obtain: N � i=1 Rearrange to give mi d (ṙ · δr i ) − δT − δW = 0 dt i δT + δW = N � i=1 mi d (ṙ · δri ) dt i (9) 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]. Lagrange’s Equations 4 Integrate (9) between two definite states in time r(t1 ) and r(t2 ) Figure 2: Between t1 and t2 , there are admissible variations δx and δy. We are integrating over theoretically admissible states between t1 and t2 that satisfy all constraints. Figure by MIT OCW. � t2 N t2 � d (ṙ i · δri )dt dt t1 i=1 � N � �t2 � mi ṙ i · δri � = (δW + δT )dt = t1 � mi i=1 (10) (11) t1 �t2 � � i=1 mi ṙ i · δr i � = 0. �N The right hand side, t1 �t2 � Why? ṙ i · δr i �� = 0, because at a particular time, δr i (ti ) = 0. Also, we know t1 the initial and final states. It is the behavior in between that we want to know. The result is the extended Hamilton Principle. � t2 (δW + δT )dt = 0 (12) t1 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]. Lagrange’s Equations 5 Generalized Fores and the Lagrangian δW = δW conservative + δW nonconservative = −δV + m � Qj δqj j=1 Conservative δW : δW = f cons · δr i i f cons =− i δW = − ∂V ∂ri ∂V · δri = −δV ∂ri Nonconservative δW : Qj δqj m � Qj δqj j=1 m: Total number of generalized coordinates Qj = Ξj : Generalized force for nonconservative work done qj = ξj : Generalized coordinate Substitute for δW in (12) to obtain: � t2 (δT − δV + t1 m � Qj δqj )dT = 0 (13) j=1 Define Lagrangian L =T −V The Lagrangian is a function of all the generalized coordinates, the generalized velocities, and time: L = L(qj , q̇j , t) where j = 1, 2, 3 . . . , m (13) can now be written as � t2 � t1 δL(qj , q̇j , t) + m � j=1 � Qj δqj dt = 0 (14) 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]. Lagrange’s Equations 6 Lagrange’s Equations We would like to express δL(qj , q̇j , t) as (a function) · δqj , so we take the total derivative of L. Note δt is 0, because admissible variation in space occurs at a fixed time. δL = � � � � � � m �� � ∂L ∂L ∂L δqj + δq̇j + δt ∂qj ∂q̇j ∂t j=1 t2 � (δL)dt = � m �� t2 � t1 t1 j=1 � � � � ∂L ∂L δqj + δq̇j dt ∂qj ∂q̇j (15) To remove the δq̇j in (15), integrate the second term by parts with the following substitutions: � ∂L ∂q̇j � � d ∂L du = dt ∂q̇j u= � y = δqj dy = δq̇j � m � t2 � t1 j=1 � ∂L ∂q̇j m � � m � t2 � t1 � ∂L ∂q̇j � � ∂L δq̇j dt = δq̇j dt ∂q̇j j=1 t1 �� �t2 � t2 � � � � � � m � � ∂L d ∂L � = δqj � − δqj dt ∂q̇j dt ∂q̇j t1 t1 j=1 � � t2 j=1 �t2 � δqj �� = 0 t1 � � � � t2 � m ∂L d ∂L δq̇j dt = − δqj dt ∂q̇j t1 j=1 dt ∂q̇j (16) Combine (14), (15), and (16) to get: � m �� t2 � t1 j=1 � � � � � ∂L d ∂L δqj − δqj + Qj δqj dt = 0 ∂qj dt ∂q̇j m � t2 � t1 j=1 − � � � � � ∂L d ∂L + +Qj δqj dt = 0 dt ∂q̇j ∂qj 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]. Example 1: 2-D Particle, Horizontal Plane 7 dt has finite values. δqj are independent and arbitrarily variable in a holonomic system. They are finite quantities. Thus, for the integral to be equal to 0, � � � � ∂L d ∂L + +Qj = 0 − dt ∂q̇j ∂qj Equations of Motion (Lagrange): � � � � ∂L d ∂L Qj = − dt ∂q̇j ∂qj or: � � � � ∂L d ∂L Ξj = − dt ∂ξ̇j ∂ξj Where Qj = Ξj = generalized force, qj = ξj = generalized coordinate, j = index for the m total generalized coordinates, and L is the Lagrangian of the system. Although these equations were formally derived for a system of particles, the same is true for rigid bodies. Example 1: 2-D Particle, Horizontal Plane Figure 3: 2-D Particle on a horizontal plane subject to a force F . Figure by MIT OCW. 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]. Example 1: 2-D Particle, Horizontal Plane 8 Cartesian Coordinates q1 = x q2 = y r = xı̂ + yĵ ṙ = ẋı̂ + ẏĵ |v|2 = ṙ · ṙ = ẋ2 + ẏ 2 = q̇12 + q̇22 Q1 = F cos θ Q2 = F sin θ L =T −V 1 m(ṙ · ṙ) 2 1 = m(q̇12 + q̇22 ) 2 T = V =0 (in horizontal plane, position with respect to gravity same at all locations) For q1 or (x) ∂L =0 ∂q1 ∂L = mq̇1 ∂q̇1 d ∂L = mq̈1 dt ∂q̇1 mq̈1 − 0 = F cos θ mq̈2 = F sin θ 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]. Example 1: 2-D Particle, Horizontal Plane 9 Polar Coordinates Figure 4: 2-D Particle subject to a force F described by polar coordinates. Figure by MIT OCW. q1 = r q2 = φ F = Fr êr + Fφ êφ r = r(t)êr ṙ = ṙêr + rφ̇êφ |v|2 = ṙ2 + r2 φ̇2 L =T −V = 1 m(q̇12 + q12 q̇22 ) + 0 2 q1 : q2 : ∂L = mq1 q̇22 ∂q1 � � d ∂L = mq̈1 dt ∂q̇1 ∂L =0 ∂q2 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]. Example: Falling Stick 10 � � d d ∂L = (mq12 q̇2 ) = m(2q1 q̇1 q̇2 + q12 q̈2 ) dt ∂q̇2 dt q 1 (r): Q1 = Fr Q2 = Fφ · r: moment. m(2q̇1 q1 q̇2 + q12 q̈2 ) = Fφ · q1 m(2q̇1 q̇2 + q1 q̈2 ) = Fφ mq̈1 − mq1 q̈2 = Fr Example: Falling Stick Figure 5: Falling stick. The stick is subject to a gravitational force, mg. The frictionless surface causes the stick to slip. Figure by MIT OCW. G: Center of Mass l: length Constraint: 1 point touching the ground. 2 degrees of freedom q1 = xG q2 = φ Must find L and Qj . Look for external nonconservative forces that do work. None. Normal does no work. Gravity is conservaitve. Q1 = Q2 = 0 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]. Example: Falling Stick 11 Lagrangian L =T −V Rigid bodies: Kinetic energy of translation and rotation T = 1 1 m(ṙ G · ṙ G ) + IG (ω · ω) 2 2 l sin φ 2 l ẏG = cos φφ̇ 2 yG = ω = φ̇k̂ ṙ G = ẋG ı̂ + ẏG ĵ = ẋG ı̂ + l cos φφ̇ĵ 2 l2 cos2 φφ̇2 4 � � � � 1 1 1 2 l2 2 2 2 q̇ + cos q2 q̇2 + ml q̇22 T = 2 1 4 2 12 ṙG · ṙ G = ẋ2G + See Lecture 16 for the rest of the example. 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].