16.61 Aerospace Dynamics Spring 2003 Lecture #9 Virtual Work And the Derivation of Lagrange?s Equations Massachusetts Institute of Technology ? How, Deyst 2003 (Based on Notes by Blair 2002) 1 16.61 Aerospace Dynamics Spring 2003 Derivation of Lagrangian Equations Basic Concept: Virtual Work Consider system of N particles located at ( ) 123 3 ,,, N xxx x… ) with 3 forces per particle ( 123 3 ,,, N F FF F… , each in the positive direction. F 3 F 2 F 1 (x 1 , x 2 , x 3 ) F 5 F 6 F 4 (x 4 , x 5 , x 6 ) x i x j x k F N F N-1 F N-2 (x N-2 , x N-1 , x N ) Assume system given small, arbitrary displacements in all directions. Called virtual displacements - No passage of time - Applied forces remain constant Massachusetts Institute of Technology ? How, Deyst 2003 (Based on Notes by Blair 2002) 2 16.61 Aerospace Dynamics Spring 2003 The work done by the forces is termed Virtual Work. 3 1 N jj j WFxδ δ = = ∑ Note use of δx and not dx. Note: ? There is no passage of time ? The forces remain constant. In vector form: 3 1 ii i Wδ δ = = ? ∑ Fr Virtual displacements MUST satisfy all constraint relationships, ! Constraint forces do no work. Massachusetts Institute of Technology ? How, Deyst 2003 (Based on Notes by Blair 2002) 3 16.61 Aerospace Dynamics Spring 2003 Example: Two masses connected by a rod l 1 R 2 R ? r e Constraint forces: 12 ? r Re 2 = ?=?RR Now assume virtual displacements 1 δr , and 2 δr - but the displacement components along the rigid rod must be equal, so there is a constraint equation of the form 21 rere rr δδ ?=? Virtual Work: () 11 2 2 212 22 1 ?? ? 0 RrRr rr r rr r W Re Re RRe =?+? =? ? + ? =? ? = 2 δ δδ δ δ δ So the virtual work done of the constraint forces is zero Massachusetts Institute of Technology ? How, Deyst 2003 (Based on Notes by Blair 2002) 4 16.61 Aerospace Dynamics Spring 2003 This analysis extends to rigid body case ? Rigid body is a collection of masses ? Masses held at a fixed distance. ! Virtual work for the internal constraints of a rigid body displacement is zero. Example: Body sliding on rigid surface without friction Since the surface is rigid and fixed, 0, 0 s rWδ δ= →= For the body, 1 W 1 δ δ=?R W r, but the direction of the virtual displacement that satisfies the constraints is perpendicular to the constraint force. Thus 0δ = . Massachusetts Institute of Technology ? How, Deyst 2003 (Based on Notes by Blair 2002) 5 16.61 Aerospace Dynamics Spring 2003 Principle of Virtual Work i m = mass of particle i i R = Constraint forces acting on the particle i F = External forces acting on the particle ! For static equilibrium (if all particles of the system are motionless in the inertial frame and if the vector sum of all forces acting on each particle is zero) 0 ii + =RF The virtual work for a system in static equilibrium is () 1 0 N ii i i Wδδ = =+? ∑ RF r= But virtual displacements must be perpendicular to constraint forces, so 0 ii δ? =Rr , which implies that we have 1 0 N ii i δ = ? = ∑ Fr Principle of virtual work: The necessary and sufficient conditions for the static equilibrium of an initially motionless scleronomic system which is subject to workless bilateral constraints is that zero virtual work be done by the applied forces in moving through an arbitrary virtual displacement satisfying the constraints. Massachusetts Institute of Technology ? How, Deyst 2003 (Based on Notes by Blair 2002) 6 16.61 Aerospace Dynamics Spring 2003 Example: System shown consists of 2 masses connected by a massless bar. Determine the coefficient of friction on the floor necessary for static equilibrium. (Wall is frictionless.) Virtual Work: 12 Wmgx Nx 2 δ δμδ= ? Constraints and force balance: 122 ,2xxNmgδ δ= = Substitution: ( ) 12mg x 0μ δ? = Result: 1 2 μ = Massachusetts Institute of Technology ? How, Deyst 2003 (Based on Notes by Blair 2002) 7 16.61 Aerospace Dynamics Spring 2003 So far we have approached this as a statics problem, but this is a dynamics course!! Recall d?Alembert who made dynamics a special case of statics: () () i 1 1 0 0 RF r r Frr N ii i i N iii i i Wm m = = =+?? ? ? = ∑ ∑ "" "" δδ δ = ? ! So we can apply all of the previous results to the dynamics problem as well. Comments: " Virtual work and virtual displacements play an important role in analytical dynamics, but fade from the picture in the application of the methods. " However, this is why we can ignore the calculation of the constraint forces. Massachusetts Institute of Technology ? How, Deyst 2003 (Based on Notes by Blair 2002) 8 16.61 Aerospace Dynamics Spring 2003 Generalized Forces Since we have defined generalized coordinates, we need generalized forces to work in the same ?space.? Consider the 2 particle problem: (x 1 , x 2 ) (x 3 , x 4 ) x i x j l θ ()() 1234 22 2 13 24 Coordinates: , , , Constraint: DOF: 4 1 3 xxxx xx xx l? +? = ?= ? Select n=3 generalized coordinates: q xx q xx q xx xx 1 13 2 24 3 1 42 31 22 = + = + = ? ? ? () () tan () () ? Can also write the inverse mapping: ( ) 123 ,,, , ii n xfqqq qt= … Massachusetts Institute of Technology ? How, Deyst 2003 (Based on Notes by Blair 2002) 9 16.61 Aerospace Dynamics Spring 2003 Virtual Work: 3 1 N j j j WFxδ δ = = ∑ Constraint relations: 1 n j ji i i x x q q = ? ?? δ =δ ?? ? ?? ∑ Substitution: 3 11 Nn j ji ji i x WF q == ? ?? qδ =δ ?? ? ?? ∑∑ Define Generalized Force: 3 1 N j ij j i x q = QF ? ?? = ?? ? ?? ∑ ! Work done for unit displacement of q i by forces acting on the system when all other generalized coordinates remain constant. ?= = ∑ δδWQ ii i n 1 q n ? If is an angle, is a torque i q i Q ? If is a length, is a force i q i Q ? If the ?s are independent, then for static equilibrium must have: i q 0, 1, 2, i Qi= = … Massachusetts Institute of Technology ? How, Deyst 2003 (Based on Notes by Blair 2002) 10 16.61 Aerospace Dynamics Spring 2003 Derivation of Lagrange?s Equation ? Two approaches (A) Start with energy expressions Formulation Lagrange?s Equations (Greenwood, 6-6) Interpretation Newton?s Laws (B) Start with Newton?s Laws Formulation Lagrange?s Equations (Wells, Chapters 3&4) Interpretation Energy Expressions (A) Replicated the application of Lagrange?s equations in solving problems (B) Provides more insight and feel for the physics Massachusetts Institute of Technology ? How, Deyst 2003 (Based on Notes by Blair 2002) 11 16.61 Aerospace Dynamics Spring 2003 Our process 1. Start with Newton 2. Apply virtual work 3. Introduce generalized coordinates 4. Eliminate constraints 5. Using definition of derivatives, eliminate explicit use of acceleration ? Start with a single particle with a single constraint, e.g. o Marble rolling on a frictionless sphere, o Conical pendulum θ φ r = L x y z m F Free body Diagram Massachusetts Institute of Technology ? How, Deyst 2003 (Based on Notes by Blair 2002) 12 16.61 Aerospace Dynamics Spring 2003 1. Newton: m=Fa ? For the particle: ,, xyz F mx F my F mz==="" "" "" where axes x, y, z describe an inertial frame ? Note that ,, x yz F FF are the vector sum of all forces acting on the particle (applied and constraint forces) 2. Apply Virtual Work: ? Consider δs, which is an arbitrary displacement for the system, then the virtual work associated with this displacement is: xyz WFxFy Fzδ δδδ= ++ ? Note that δs may violate the applied constraints, because F contains constraint forces ? Combine Newton and Virtual Work x y z Fx mxx Fymy Fz mzz y δ δ δ δ δ δ = = = "" "" "" ? Add the equations ( ) xyz mxx y y z z F x F y F zδ δδ δ δ++ = + + "" "" "" δ Massachusetts Institute of Technology ? How, Deyst 2003 (Based on Notes by Blair 2002) 13 16.61 Aerospace Dynamics Spring 2003 ? Called D?Alembert?s equation: () xyz mxx y y z z F x F y F zδ δδ δ δ++ = + +"" "" "" δ ? Observations: o Scalar relationship o LHS ≈ kinetic energy o RHS ≈ virtual work term 3. Introduce generalized coordinates o Assumed motion on a sphere ! 1 stationary constraint o DOF = 3 - 1 = 2 generalized coordinates ( ) ( ) ( ) 112 212 312 ,, ,, ,x f qq y f qq z f qq=== ? Define virtual displacements in terms of generalized coordinates: 1 n j ji i i x x q q = ? ?? δ= δ ?? ? ?? ∑ ! 12 12 12 12 12 11 xx xq qq yy yq qq zz zq qq q q q ? ? =+ ?? ?? =+ ?? ?? =+ ?? δ δδ δ δδ δ δδ ? Note: these virtual displacements conform to the constraints, because the mapping of the generalized coordinates conforms to the surface of the sphere. Massachusetts Institute of Technology ? How, Deyst 2003 (Based on Notes by Blair 2002) 14 16.61 Aerospace Dynamics Spring 2003 ? Substitute virtual displacements into D?Alembert?s equation 12 12 12 12 12 11 xx xq qq yy yq qq zz zq qq q q q ? ? =+ ?? ?? =+ ?? ?? =+ ?? δ δδ δ δδ δ δδ () xyz mxx y y z z F x F y F zδ δδ δ δ++ = + +"" "" "" δ 12 111 2 22 111 2 22 xyz x y z xyz xyz mx y z q mx y z q qqq qqq xyz xyz F FF qFFF qqq qqq ??? ??? ??? ++ + ++ ??? ??? ??? ??? ??? ??? =++ +++ ??? ??? "" "" "" """" ""δδ q ? ? ? δ δ ? Facts: o Virtual displacements 1 qδ and 2 qδ conform to constraints o Virtual work Wδ is work that conforms to constraints o 1 qδ and 2 qδ are independent and can be independently moved without violating constraints Massachusetts Institute of Technology ? How, Deyst 2003 (Based on Notes by Blair 2002) 15 16.61 Aerospace Dynamics Spring 2003 ? Conclusion: o Force of the constraint has been eliminated by selecting generalized coordinates that enforce the constraint (Reason 2 for Lagrange, pg 24) o Further, we can split the equation into two equations in two unknowns due to independence of 1 qδ and 2 qδ . 111 1 1 1 222 2 2 xyz xyz xyz x y z mx y z F F F qqq q q q xyz x y mx y z F F F qqq q q q ??? ??? ? ? ? ++ = + + ??? ??? ? ? ? ??? ??? ??? ? ? ? ++ = + + ??? ??? ? ? ? ??? "" "" "" "" "" "" 2 z ? ? ? ? ? ? 5. Finally, eliminate acceleration terms ? Consider the total derivative 11 dx xdx xxx dt q q dt q ?? ? ?? ? =+ ?? ? ?? ? ?? ? """" 1 ? ? ? ? Rearrange 11 xd x dx xxx qdt q dtq ?? ? ?? =? ?? ? ?? ? "" " " 1 ? ? ? ? (1) Massachusetts Institute of Technology ? How, Deyst 2003 (Based on Notes by Blair 2002) 16 16.61 Aerospace Dynamics Spring 2003 ? Recall () () 112 112 ,, d x f qq x f qq dt ? ?=∴= ? ? " ? Perform the derivative (chain rule): 1 12 xx xq qq 2 q ? ? =+ ?? "" " (2) ? Partial derivative of (2) with respect to q gives 1 " 11 x qq x? ? = ? ? " " (3) ? Since () ( 112 112 1 ,, , x xfqq gqq q ) ? == ? is a ftn of both q 1 and q 2 the time derivative of 1 x q ? ? gives (chain rule again) 1 111 21 dx x x q dt q q q q q ?? ?? ?? ??? ?? =+ ?? ?? ?? ??? ?? ?? ?? ?? " 2 " (4) ? Partial derivative of (2) with respect to gives x" 1 q 1 111 12 xx x q qqq qq ?? ?? ??? ?? =+ ?? ?? ??? ?? ?? ?? " "" (5) Massachusetts Institute of Technology ? How, Deyst 2003 (Based on Notes by Blair 2002) 17 16.61 Aerospace Dynamics Spring 2003 ? Note RHS of 4 and 5 are the same, thus ! 11 xdx qdtq ?? ? ? = ?? ? ? ?? " (6) ? Now, insert (3) and (6) into (1): 11 xd x dx xxx qdt q dtq ?? ? ?? =? ?? ? ?? ? "" " " 1 ? ? ? ? (1) 11 x qq ?? = ?? " " x 11 dx dt q q ?? x? ? = ?? ? ? ?? " (3 and 6) ? Results in: 11 xd x xx qdt q q ?? ?? =? ?? ?? " "" " " " 1 x?" (7) ? Note that 2 11 2 x x x qq ?? ? ?? ? ? = ?? " " " "" ? and 2 11 2 x x x qq ?? ? ?? ? ? = ?? " " " ? Massachusetts Institute of Technology ? How, Deyst 2003 (Based on Notes by Blair 2002) 18 16.61 Aerospace Dynamics Spring 2003 ? Finally 22 11 2 xx xd x qdt q q ?? ?? ?? ?? ?? ?? ?? ? ?? ? =? ?? ?? ?? "" "" " 1 2 ? ? (8) ? The above process is identical for y and z. ? Recall our virtual work equation for : 1 q 111 1 1 xyz x 1 y zxy z mx y z F F F qqq q q q ??? ??? ? ? ? ++ = + + ??? ??? ? ? ? ??? "" "" "" ? ? ? ? Insert equations (8) for x, y and z and collect terms to eliminate acceleration terms. (Reason 3 for Lagrange, pg 24) 222 222 11 111 xyz dxy zxy z mm dt q q xyz FFF qqq ?? ?? ?++ ?++ ? ?? ?? ?? ???? ?? ??? =++ ?? ??? ?? "" """" " Massachusetts Institute of Technology ? How, Deyst 2003 (Based on Notes by Blair 2002) 19 16.61 Aerospace Dynamics Spring 2003 ? Observe that: () 222 11 22 Tmv mxyz== ++ """ 2 which is the kinetic energy of the particle ? Finally: 11 11 xyz dx TTFFF dt q q q q q ?? ? ?? ??? ?= + + ?? ? ?? ??? ?? ? " 1 y ? ? ? ? Similarly: 22 22 xyz dx 2 y z TTFFF dt q q q q q ?? ? ?? ??? ?= + + ?? ? ?? ??? ?? ? " ? ? ? ? The general form of Lagrange?s equation is thus: r q rr d TT dt q q ?? ?? ?= ?? ?? ?? " Q r qx y z rrr xyz QF F F qqq ? ?? =++ ??? Massachusetts Institute of Technology ? How, Deyst 2003 (Based on Notes by Blair 2002) 20 16.61 Aerospace Dynamics Spring 2003 ? Some observations: o One Lagrange equation needed for each DOF o Easily extendable for a system of particles o T ? Expression of system kinetic energy o All inertial forces contained in the LHS o only contains external forces q r Q ? How to use this ?. 1. Determine number of DOF and constraints 2. Identify generalized coordinates and equations of constraint a. Iterate on 1 and 2 if needed 3. Write expression for T a. v inertial velocity that can be written in terms of the coordinates of any frame b. Find required derivatives of T 4. Find generalized forces Q r q a. If forces are known in inertial coordinates, transform them to generalized coordinates b. Apply generalized force equation for each force r qx y z rr x r y z QF F F qq ?? q ? ?? =++ ?? ??? ?? 5. Substitute into Lagrange?s equation 6. Solve analytically or numerically Massachusetts Institute of Technology ? How, Deyst 2003 (Based on Notes by Blair 2002) 21 16.61 Aerospace Dynamics Spring 2003 Example: Projectile Problem: z x y 1, 2, 3. DOF = 3, no constraints 4. () 222 11 22 Tmv mx 2 y z== ++""" 5. ,, TTT mx my mz xyz ??? === ??? """ """ ,, dT d T dT mx my mz dt x dt y dt z ????? ?? ?? == ???? ?? ?? ?? ?? "" "" "" "" " = 0 TTT xyz ??? === ??? Massachusetts Institute of Technology ? How, Deyst 2003 (Based on Notes by Blair 2002) 22 16.61 Aerospace Dynamics Spring 2003 6. Generalized forces: r qx y z rr x r y z QF F F qq ?? ??? =++ ?? ??? ?? q ?mg z=?F 0, xy z qq qz z QQ QF mg z ? == = =? ? 7. EOMs: 0, 0,mx my mz mg== = =?"" "" "" 8. Solve differential equations. ? Comments: o Method is overkill for this problem o Inspection shows agreement with Newton Massachusetts Institute of Technology ? How, Deyst 2003 (Based on Notes by Blair 2002) 23 16.61 Aerospace Dynamics Spring 2003 Example: Mass moving along a frictionless track. θ φ X Y Z ρ z ? Track geometry defined such that: azρ = , and bzφ =? DOF = 3 ? 2 =1 ? Constraint equations: azρ = , and bzφ = ? ? Generalized coordinate: z ? Find 2 1 2 m=T , what is v? v Massachusetts Institute of Technology ? How, Deyst 2003 (Based on Notes by Blair 2002) 24 16.61 Aerospace Dynamics Spring 2003 ? Define rotating coordinate frame such that mass remains in plane. ? ?x? z φ X Y Z ρ ?x ?y ?x z ρ z ?z r x zz azx zz=+= +ρ # # # # and ω φ= = ? " ## z bzz " "# "# ( # ) # # "# " # "# r azx zz bzz azx zz azx abzzy zz =++?×+ =? + ! vrr az abzz z aabzz 2 22 2222 1 =? =+ + =+ + "" ( ")(" ) " ()" 2 2 T m aabzz=++ 2 1 2222 ()" 2 and 2 222 (1 ) T maabz z z ? =++ ? " " Massachusetts Institute of Technology ? How, Deyst 2003 (Based on Notes by Blair 2002) 25 16.61 Aerospace Dynamics Spring 2003 () 2 222 22 2 (1 ) 2 dT maabzzmabzz dt z ? ?? =++ + ?? ? ?? "" " " 22 2 () T mabzz z ? = ? " ? External force is gravity r qx y z rr xy QF F F qq ?? r z q ? ?? =++ ?? ??? ?? ?mg z= ?F 0, xy z qq qz z QQ QF m z g ? == = =? ? ? Equation of Motion: ( ) 2 222 22 2 1aabz zabz+++ ="" " g? ? Comments: o Solution highly nonlinear o ?Trick? was finding inertial velocity o Still need to use FARM approach Massachusetts Institute of Technology ? How, Deyst 2003 (Based on Notes by Blair 2002) 26 16.61 Aerospace Dynamics Spring 2003 Extending Lagrange?s Equation to Systems with Multiple Particles ? Assume a system of particles and apply Newton?s laws: 111 11 ,, ,, ppp xyz1 x py pz p F mx F my F mz F mx F my F mz === === "" "" "" $$$ "" "" "" ? As before, the F?s contain both external and constraint forces. ? Multiply both sides of each equation by the appropriate virtual displacement and add all the equations together. () () 11 iii pp ii ii ii x i y i z i mx x y y z z F x F y F zδ δδ δ δ δ == ++ = + + ∑∑ "" "" "" ? Recall that this is D?Alembert?s equation ? Assume the system has n DOF, np3≤ ? Select generalized coordinates, that enforce the constraints: i q ( ) () 12 12 12 ,,,, ,,,, ,,,, ii n ii n ii n xfqq qt ygqq qt zhqq qt = = = … … … Massachusetts Institute of Technology ? How, Deyst 2003 (Based on Notes by Blair 2002) 27 16.61 Aerospace Dynamics Spring 2003 ? Express virtual displacements in terms of generalized coordinates: 12 12 12 12 12 12 ii i in n ii i n ii i in n xx x xqq qq q yy y yqq qq q zz z zqq qq q ?? ? =+++ ?? ? ?? ? =+++ ?? ? ?? ? =+++ ?? ? … … … q q q δ δδ δ δ δδ δ δ δδ δ ? Substitute the relations into D?Alembert?s equation 1 1 iii p iii iiir i rrr p iii x yz i rrr xyz mx y z q qqq xyz r F FF qqq = = ?? ??? ++ ?? ??? ?? ??? =++ ??? ∑ ∑ "" "" "" δ qδ ? As before, have used fact that the generalized coordinates automatically enforce the constraints. o Sum over the entire system of particles decouples for each of the generalized coordinates. o This leaves us n such equations. ? Using the calculus relations (chain rule), one can show that r q rr d TT dt q q ?? ?? ?= ?? ?? ?? " Q ? Once again, one Lagrange equation for each DOF. Massachusetts Institute of Technology ? How, Deyst 2003 (Based on Notes by Blair 2002) 28 16.61 Aerospace Dynamics Spring 2003 Lagrange?s Equation for Conservative Systems ? Conservative forces and conservative systems o Forces are such that the work done by the forces in moving the system from one state to another depends only on the initial and final coordinates of the particles (path independence). ? Potential Energy, V o Work done by a conservative force in a transfer from a general configuration A to a reference configuration B is the potential energy of the system at A with respect to B. o Note: V is defined as work from the general state to the reference state. ? Examples of conservative forces: o Springs (linear elastic) o Elastic bodies o Gravity force ? Non-conservative forces o Friction o Drag of a fluid o Any force with time or velocity dependence Massachusetts Institute of Technology ? How, Deyst 2003 (Based on Notes by Blair 2002) 29 16.61 Aerospace Dynamics Spring 2003 ? General Expression for V, the potential energy () 1 iii A P xi yi zi i B VFdxFdF = =? + + ∑ ∫ d ? Note the ??? sign since the path is from B to A. The sum is over the P particles in the system. ? For path independence, integrand must be an exact differential. Thus: iii xyz ii VV FFF x i V y z ?? =? =? =? ? (C1) ? Observe that: 3 4 2 44 3 3 2 33 4 3 x y F VV yyx x F VV xxy x ? ?? ?? ? =?=? ?? ??? ?? ?? ? ?? ?? ? =?=? ?? ??? ?? ?? 4 4 y y ? Thus, in general i r x y ri F F yx ? ? = ? ? (C2) ? Equation (C1) represents a necessary condition for a force to be conservative, Equation (C2) is a sufficient condition. Massachusetts Institute of Technology ? How, Deyst 2003 (Based on Notes by Blair 2002) 30 16.61 Aerospace Dynamics Spring 2003 ? Recall expression for generalized forces: 1 iii p ii qr x y z i rr xy QFFF qq = ?? i r z q ? ?? =++ ?? ??? ?? ∑ o Separate forces into conservative and non-conservative 1 p N iii qr qr i ir ir ir N qr r xyzVVV QQ xq yq yq V Q q = ?? ?????? =? + + + ?? ?? ?? ?? ?? ? =? + ? ∑ ? Lagrange?s Equation: r q rr d TT dt q q ?? ?? ?= ?? ?? ?? " Q ? Substitute in generalized force: () N qr rr r N qr rr dV TT Q dt q q q d TTV dt q q ?? ??? ?=?+ ?? ?? ? ?? ?? ?? ??= ?? ? ?? ? " " Q? ? Since conservative forces are not functions of velocities: 0 r V q ? = ?" Massachusetts Institute of Technology ? How, Deyst 2003 (Based on Notes by Blair 2002) 31 16.61 Aerospace Dynamics Spring 2003 ? Thus, can define the Lagrangian LTV= ? to obtain the final form of Lagrange?s equation: qr rr dL L F dt q q ?? ?? ?= ?? ?? ?? " Example: Planar pendulum with an inline spring. m θ y x k r ? DOF = 3 ? 1 = 2 ? Constraint equation: z = 0 ? Generalized coordinates: r, θ ? Coordinate mapping: cos , sinxr yrθ θ= = Massachusetts Institute of Technology ? How, Deyst 2003 (Based on Notes by Blair 2002) 32 16.61 Aerospace Dynamics Spring 2003 Massachusetts Institute of Technology ? How, Deyst 2003 (Based on Notes by Blair 2002) 33 ? Kinetic energy () 22 11 22 Tmv mxy== +"" 2 ? Derivatives of coordinates: cos sin , sin cosxr r yr rθ θθ θθ=? =+ "" "" "" θ ? Substitute into kinetic energy equation () 222 1 2 Tmrrθ=+ " " ? Potential energy () 21 cos 2 o Vkrrmgrθ=?? ? Lagrangian () () 2 222 11 cos 22 o LTV mr r krr mgrθ θ=?= + ? ? + " " ? Derivatives of L (note need to do this for each GC) () 2 ,, o LdL L mr mr mr k r r mg rdtr r cosθ θ ??? ?? ===??+ ?? ?? " """ "" 22 ,2, LdL L mr mr mrr mgr dt sinθ θθ θθ θ ?? ? ?? ==+=? ?? ?? """" " "" θ ? Substitute into Lagrange?s Equation: ( ) 2 2 cos 2sin0 o mr mr k r r mg mr mrr mgr ?+?= +? = " "" "" " " θ θ θθ θ