Lagrangian Dynamics: Virtual Work and Generalized Forces

Admissible Variations/Virtual Displacements 1 2.003J/1.053J Dynamics and Control I, Spring 2007 Paula Echeverri, Professor Thomas Peacock 4/4/2007 Lecture 14 Lagrangian Dynamics: Virtual Work and Generalized Forces Reading: Williams, Chapter 5 L =T −V � ∂L d ∂L − = Qi dt ∂q̇i ∂qi � All qi are scalars. qi : Generalized Coordinates L: Lagrangian Qi : Generalized Forces Admissible Variations/Virtual Displacements Virtual Displacement: Admissible variations: hypothetical (not real) small change from one geometri­ cally admissible state to a nearby geometrically admissible state. Bead on Wire Figure 1: Bead on a wire. 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]. Admissible Variations/Virtual Displacements 2 Both δx and δy are admissible variations. Hypothetical geometric configuration displacement. δ �= d δx �= dx dx implies t involved. y = f (x) df · dx dx df (x) δy = · δx dx dy = Generalized Coordinates Minimal, complete, and independent set of coordinates s is referred to as complete: capable of describing all geometric configurations at all times. s is referred to as independent : If all but one coordinate is fixed, there is a continuous range of values that the free one can take. That corresponds to the admissible system configurations. Example: 2-Dimensional Rod Figure 2: 2D rod with fixed translation in x and y but free to rotate about θ. 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]. Virtual Work 3 If we fix x and y, we can still rotate in a range with θ. # degrees of freedom = # of generalized coordinates: True for 2.003J. True for Holonomic Systems. Lagrange’s equations work for Holonomic systems. Virtual Work � W = f i · dr i ← Actual Work i i = forces act at that location δW = � f i · δr i ← Virtual Work i f i = f applied + f constrained i i Constrained: Friction in roll. Constraint to move on surface. Normal forces. Tension, rigid body constraints. δw = � f app · δr i = 0 at equilibrium i i No work done because no motion in direction of force. No virtual work. � fi = 0 i 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]. Virtual Work 4 Example: Hanging Rigid Bar Figure 3: Hanging rigid bar. The bar is fixed translationally but is subject to a force, F . It therefore can displace itself rotationally about its pivot point. Figure by MIT OCW. Displacement: δy A = −aδθĵ δy B = −lδθĵ Forces: F = −F ĵ R = Rĵ Two forces applied: i = 2 δw = F lδθ − Raδθ = 0 Fl at equilibrium R= a Could also have taken moments about O. 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]. Virtual Work 5 Example: Tethered Cart Figure 4: Tethered cart. The cart is attached to a tether that is attached to the wall. Figure by MIT OCW. δw = F δyB − Rδxc = 0 yB = l sin θ Using δy = df (x) dx δxc δyB = l cos θδθ δxc = −2l sin θδθ (−F l cos θ + 2R sin θ)δθ = 0 −F l cos θ + 2R sin θ = 0 ⇒ R = F at equilibrium 2 tan θ Figure 5: Application of Newton’s method to solve problem. The indicated extra forces are needed to solve using Newton. 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]. Generalized forces for Holonomic Systems 6 Generalized forces for Holonomic Systems In an holonomic system, the number of degrees of freedom equals the number of generalized coordinates. δw = � f i · δr i = i � Qi δqj i = number of applied forces: 1 to n j = number of generalized coordinates r i = ri (q1 , q2 , . . . qj ) ri : Position of point where force is applied δri = m � ∂ri δqj ∂qj j Substitute: n � fi i � m m �� n � � ∂ri ∂r i · δqj = ·∂qj fi ∂qj ∂qj j j i Qj = n � i fi · ∂ri Generalized Forces ∂qj + f iCONS. f i = f NC i fiCONS. : Gravity, Spring, and Buoyancy are examples; Potential Function Exists. f CONS. = − ∂V ∂r Example: Vg = mgz, r = zĵ f g = −mg ∂z dz ĵ = −mgĵ ficons. · ∂r i ∂V ∂r ∂V =− =− ∂q ∂r ∂qj ∂qj The conservative forces are already accounted for by the potential energy term in the Lagrangian for Lagrange’s Equation 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: Cart with Pendulum, Springs, and Dashpots C QN = j n � C fN · i i 7 ∂r i ∂qj � � d ∂L ∂L C − = QN j dt ∂q̇j ∂qj Lagrange’s Equation C QN = nonconservative generalized forces j ∂L ∂qj contains ∂V ∂qj . Example: Cart with Pendulum, Springs, and Dashpots Figure 6: The system contains a cart that has a spring (k) and a dashpot (c) attached to it. On the cart is a pendulum that has a torsional spring (kt ) and a torsional dashpot (ct ). There is a force applied to m that is a function of time F = F (t) We will model the system as 2 particles in 2 dimensions. 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: Cart with Pendulum, Springs, and Dashpots 8 4 degrees of freedom: 2 constraints. Cart moves in only 1 direction. Rod fixes distance of the 2 particles. Thus, there are a net 2 degrees of freedom. For 2.003J, all systems are holonomic (the number of degrees of freedom equals the number of generalized coordinates). q1 = x q2 = θ Figure 7: Forces felt by cart system. Figure by MIT OCW. F 1 : Damper and Spring in −x direction −(kx + cẋ)ı̂ F 2 : Two torques: τ = −(kt θ + ct θ̇k̂ F 3: F 3 = F0 sin ωtı̂ rA = xı̂ = q1 ı̂ ← r 1 rB = r A + rB/A = (x + l sin θ)ı̂ − l cos θĵ ← r 3 r 2 = θk̂ (Torque creates angular displacement) = q2 k̂ 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: Cart with Pendulum, Springs, and Dashpots 9 Q1 : ∂r1 ∂q1 = 1ı̂, ∂r 2 ∂q1 = 0, ∂r 3 ∂q1 = 1ı̂ Q1 = −cq̇1 + F0 sin ωt ∂r1 ∂q2 = 0, ∂r2 ∂q2 = 1k̂, ∂r 3 ∂q2 = l cos q2 ı̂ + l sin q2 ĵ Q2 = −ct q̇2 + F0 sin ωt · l cos q2 With the generalized forces, we can write the equations of motion. Kinematics M: rM = xı̂ ṙM = ẋı̂ r̈M = ẍı̂ m: r m = (x + l sin θ)ı̂ − l cos θĵ ṙ m = (ẋ + l cos θ · θ̇)ı̂ + l sin θθ̇ĵ r̈ m = (ẍ + l(cos θ)θ¨ − l(sin θ)θ̇2 )ı̂ + (l(sin θ)θ¨ + l(cos θ)θ̇2 )ĵ Generalized Coordinates: q1 = x and q2 = θ. Lagrangian L =T −V T = TM + Tm TM = 1 1 M (ṙ M · ṙM ) = M ẋ2 2 2 1 m(ṙ m · ṙm ) 2 1 = m(ẋ2 + 2lẋ cos θθ̇ + l2 θ̇2 ) 2 Tm = (1) (2) 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: Cart with Pendulum, Springs, and Dashpots T = 10 1 1 M ẋ2 + m(ẋ2 + 2lẋ cos θθ̇ + l2 θ̇2 ) 2 2 V = VM,g + MM,k + Vm,g + Vm,kt 1 1 = M g(0) + k(ṙ M · ṙ M ) + mg(−l cos θ) + kt θ2 2 2 Symbol VM,g VM,k Vm,g Vm,kt V = (3) (4) Potential Energy Gravity on M Spring on M Gravity on m Torsional Spring on m 1 1 2 kx + (−mgl cos θ) + kt θ2 2 2 Substitute in L = T − V L= 1 1 1 1 M ẋ2 + m(ẋ2 + 2lẋθ̇ cos θ + l2 θ̇2 ) − kx2 + mgl cos θ − kt θ2 2 2 2 2 Equations of Motion Use d dt � ∂L ∂ q̇i � � � ∂L − ∂q = Ξi to derive the equations of motion. Ξi = Qi . i From before, Ξx = F0 sin ω0 t − cẋ and Ξθ = F0 (sin ωt)l cos θ − ct θ̇. For Generalized Coordinate x δx �= 0 and δθ = 0. Units of Force. ∂L = −kx ∂x ∂L = (M + m)ẋ + ml(cos θ)θ̇ ∂ ẋ � � d ∂L = (M + m)ẍ + ml θ¨ cos θ + mL(− sin θ)θ̇2 dt ∂ ẋ � � ∂L d ∂L − = (M + m)ẍ + ml θ̈(cos θ) + ml(− sin θ)θ̇2 + kx = F0 sin ωt − cẋ dt ∂ẋ ∂x 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: Cart with Pendulum, Springs, and Dashpots 11 For Generalize Coordinate θ δx = 0 and δθ �= 0. Units of Torque. ∂L = mlẋθ̇(− sin θ) − mgl sin θ − kt θ ∂θ ∂L = mlẋ cos θ + ml2 θ̇ ∂θ̇ � � d ∂L = mlẋ(− sin θ)θ̇ + mlẍ cos θ + ml 2 θ¨ dt ∂θ̇ � � ∂L d ∂L − = mlẋθ̇(− sin θ)+mlẍ cos θ+ml 2 θ̈−mlẋθ̇(− sin θ)+mgl sin θ+kt θ dt ∂θ̇ ∂θ � � ∂L d ∂L − = mlẍ cos θ + ml 2 θ¨ + mgl sin θ + kt θ = F0 (sin ωt)l cos θ − ct θ dt ∂θ̇ ∂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].