Virtual Work and the Derivation of Lagrange’s Equations

Lecture #9 Virtual Work And the Derivation of Lagrange’s Equations 1 Derivation of Lagrangian Equations Basic Concept: Virtual Work Consider system of N particles located at ( x1 , x2 , x3 , … x3 N ) with 3 forces per particle ( F1 , F2 , F3 , … F3 N ) , each in the positive direction. xk F3 F1 F5 F6 F4 F2 (x4, x5, x6) (x1, x2, x3) xj FN xi (xN-2, xN-1, xN) FN-2 FN-1 Assume system given small, arbitrary displacements in all directions. Called virtual displacements - No passage of time - Applied forces remain constant 2 The work done by the forces is termed Virtual Work. 3N δ W = ∑ F jδ x j j =1 Note use of δx and not dx. Note: • There is no passage of time • The forces remain constant. In vector form: 3 δ W = ∑ Fi • δ ri i =1 Virtual displacements MUST satisfy all constraint relationships, ! Constraint forces do no work. 3 Example: Two masses connected by a rod l R1 R2 eˆr Constraint forces: R1 = − R 2 = − R2eˆr Now assume virtual displacements δ r1 , and δ r2 - but the displacement components along the rigid rod must be equal, so there is a constraint equation of the form er • δ r1 = er • δ r2 Virtual Work: δ W = R1 • δ r1 + R 2 • δ r2 = − R2eˆr • δ r1 + R2eˆr • δ r2 = ( R2 − R2 ) eˆr • δ r1 =0 So the virtual work done of the constraint forces is zero 4 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, δ rs = 0, → δ W = 0 For the body, δ W = R1 • δ r1 , but the direction of the virtual displacement that satisfies the constraints is perpendicular to the constraint force. Thus δ W = 0 . 5 Principle of Virtual Work mi = mass of particle i R i = Constraint forces acting on the particle Fi = 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) R i + Fi = 0 The virtual work for a system in static equilibrium is N δ W = ∑ ( R i + Fi ) • δ ri = 0 i =1 But virtual displacements must be perpendicular to constraint forces, so R i • δ ri = 0 , which implies that we have N ∑F •δ r = 0 i =1 i i 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. 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: δ W = mgδ x1 − µ N 2δ x2 Constraints and force balance: δ x1 = δ x2 , N 2 = 2mg Substitution: Result: mg (1 − 2 µ ) δ x = 0 µ = 12 7 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: N δ W = ∑ ( R i + Fi − mi""ri ) • δ ri = 0 i =1 N ⇒ ∑ ( Fi − mi"" ri ) • δ ri = 0 i =1 ! 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. 8 Generalized Forces Since we have defined generalized coordinates, we need generalized forces to work in the same “space.” Consider the 2 particle problem: xj (x3, x4) θ l (x1, x2) xi Coordinates: x1 , x2 , x3 , x4 Constraint: ( x1 − x3 ) DOF: 4 −1 = 3 2 + ( x2 − x4 ) = l 2 2 • Select n=3 generalized coordinates: ( x1 + x3 ) ( x + x4 ) ( x − x2 ) q2 = 2 q3 = tan −1 4 2 2 ( x3 − x1 ) • Can also write the inverse mapping: q1 = xi = fi ( q1 , q2 , q3 , … qn , t ) 9 3N δ W = ∑ F jδ x j Virtual Work: j =1  ∂x j δx j = ∑  i =1  ∂qi n Constraint relations:  δqi   ∂x j δW = ∑∑ F j  j =1 i =1  ∂qi 3N n Substitution:  δqi   ∂x j  Qi = ∑ Fj   q ∂ j =1  i 3N Define Generalized Force: ! Work done for unit displacement of qi by forces acting on the system when all other generalized coordinates remain constant. n ⇒ δW = ∑ Qi δqi i =1 • If qi is an angle, Qi is a torque • If qi is a length, Qi is a force • If the qi ’s are independent, then for static equilibrium must have: Qi = 0, i = 1, 2, … n 10 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 11 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 z m r=L φ θ x y F Free body Diagram 12 1. Newton: F = ma • For the particle: Fx = mx"", Fy = my"", Fz = mz"" where axes x, y, z describe an inertial frame • Note that Fx , Fy , Fz 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: δ W = Fxδ x + Fyδ y + Fzδ z • Note that δs may violate the applied constraints, because F contains constraint forces • Combine Newton and Virtual Work Fxδ x = mx""δ x Fyδ y = my""δ y Fzδ z = mz""δ z • Add the equations m ( "" xδ x + "" yδ y + "" zδ z ) = Fxδ x + Fyδ y + Fzδ z 13 • Called D’Alembert’s equation: m ( "" xδ x + "" yδ y + "" zδ z ) = Fxδ x + Fyδ y + Fzδ z • 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 x = f1 ( q1 , q2 ) , y = f 2 ( q1 , q2 ) , z = f 3 ( q1 , q2 ) • Define virtual displacements in terms of generalized coordinates: ∂x ∂x δx= δ q1 + δ q2 ∂q1 ∂q2 n  ∂x j δx j = ∑  i =1  ∂qi  δqi  ! δy= ∂y ∂y δ q1 + δ q2 ∂q1 ∂q2 δz = ∂z ∂z δ q1 + δ q2 ∂q1 ∂q1 • Note: these virtual displacements conform to the constraints, because the mapping of the generalized coordinates conforms to the surface of the sphere. 14 • Substitute virtual displacements into D’Alembert’s equation δx= ∂x ∂x δ q1 + δ q2 ∂q1 ∂q2 δy= ∂y ∂y δ q1 + δ q2 ∂q1 ∂q2 δz = ∂z ∂z δ q1 + δ q2 ∂q1 ∂q1 m ( "" xδ x + "" yδ y + "" zδ z ) = Fxδ x + Fyδ y + Fzδ z  ∂x  ∂x ∂y ∂z  ∂y ∂z  "" "" "" δ m  "" x y z q m x y z + "" + "" + + +  1   δ q2 ∂q1 ∂q1  ∂q2 ∂q2   ∂q1  ∂q2  ∂x  ∂x ∂y ∂z  ∂y ∂z  δ q F F F =  Fx + Fy + Fz + + +  1  x  δ q2 y z q q q q q q ∂ ∂ ∂ ∂ ∂ ∂ 1 1 1  2 2 2    • Facts: o Virtual displacements δ q1 and δ q2 conform to constraints o Virtual work δ W is work that conforms to constraints o δ q1 and δ q2 are independent and can be independently moved without violating constraints 15 • 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 δ q1 and δ q2 .  ∂x ∂y ∂z   ∂x ∂y ∂z  + "" + "" = + + m  "" x y z F F F   x  y z ∂ ∂ ∂ ∂ ∂ ∂ q q q q q q   1 1 1  1 1 1   ∂x ∂y ∂z   ∂x ∂y ∂z  + "" + "" = + + m  "" x y z F F F   x  y z ∂ ∂ ∂ ∂ ∂ ∂ q q q q q q   2 2 2  2 2 2  5. Finally, eliminate acceleration terms • Consider the total derivative d  ∂x  ∂x d  ∂x  " "" " x = x + x     dt  ∂q1  dt  ∂q1  ∂q1 • Rearrange "" x ∂x d  ∂x  d  ∂x  " =  x" − x    ∂q1 dt  ∂q1  dt  ∂q1  (1) 16 • Recall x = f1 ( q1 , q2 ) ∴ x" = d  f1 ( q1 , q2 )  dt  • Perform the derivative (chain rule): x" = ∂x ∂x q"1 + q"2 ∂q1 ∂q2 (2) • Partial derivative of (2) with respect to q"1 gives ∂x" ∂x = ∂q"1 ∂q1 (3) ∂x = g1 ( q1 , q2 ) is a ftn of both q1 and q2 ∂q1 ∂x the time derivative of gives (chain rule again) ∂q1 • Since x = f1 ( q1 , q2 ) , d  ∂x  ∂  ∂x  ∂  ∂x  " q = +     1   q"2 dt  ∂q1  ∂q1  ∂q1  ∂q2  ∂q1  (4) • Partial derivative of x" (2) with respect to q1 gives ∂x" ∂  ∂x  ∂  ∂x  " q = +   1   q"2 ∂q1 ∂q1  ∂q1  ∂q1  ∂q2  (5) 17 • Note RHS of 4 and 5 are the same, thus ! ∂x" d  ∂x  =   ∂q1 dt  ∂q1  (6) • Now, insert (3) and (6) into (1): "" x d  ∂x  ∂x d  ∂x  " x =  x" −    ∂q1 dt  ∂q1  dt  ∂q1  ∂x ∂x" = ∂q1 ∂q"1 d  ∂x  ∂x"  = dt  ∂q1  ∂q1 (1) (3 and 6) • Results in: "" x ∂x d  ∂x"  ∂x" " x =  x" −  ∂q1 dt  ∂q"1  ∂q1 (7) • Note that  x" 2  ∂  2 " ∂x =   x" ∂q"1 ∂q"1  x" 2  ∂  2 " ∂x and x" =   ∂q1 ∂q1 18 • Finally   x" 2    x" 2   ∂   ∂  2 ∂x d   2   "" x = −   ∂q1 dt  ∂q"1  ∂q1     (8) • The above process is identical for y and z. • Recall our virtual work equation for q1 :  ∂x ∂y ∂z   ∂x ∂y ∂z  m  "" x + "" y + "" z = F + F + F   x  y z ∂ q ∂ q ∂ q ∂ q ∂ q ∂ q 1 1 1  1 1 1    • Insert equations (8) for x, y and z and collect terms to eliminate acceleration terms. (Reason 3 for Lagrange, pg 24) d  ∂  x" 2 + y" 2 + z" 2   ∂  x" 2 + y" 2 + z" 2  m m  −     2 2 dt  ∂q"1    ∂q1    ∂x ∂y ∂z  =  Fx + Fy + Fz  ∂ ∂ ∂ q q q  1 1 1  19 • Observe that: 1 2 1 T = mv = m ( x" 2 + y" 2 + z" 2 ) 2 2 which is the kinetic energy of the particle • Finally:  ∂x d ∂  ∂ ∂y ∂z  T =  Fx + Fy + Fz  " T −  ∂ ∂ ∂ dt  ∂q1  ∂q1 q q q 1 1 1   • Similarly:  ∂x ∂y ∂z  d ∂  ∂ + Fy + Fz T =  Fx  " T −  dt  ∂q2  ∂q2 ∂ q ∂ q ∂ q  2 2 2  • The general form of Lagrange’s equation is thus: d ∂  ∂ T = Qqr  " T − dt  ∂qr  ∂qr  ∂x ∂y ∂z  Qqr =  Fx + Fy + Fz  ∂ q ∂ q ∂ q r r r   20 • 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 Qq only contains external forces r • 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 Qqr a. If forces are known in inertial coordinates, transform them to generalized coordinates b. Apply generalized force equation for each force  ∂x ∂y ∂z  Qqr =  Fx + Fy + Fz  ∂ q ∂ q ∂ q r r r   5. Substitute into Lagrange’s equation 6. Solve analytically or numerically 21 Example: Projectile Problem: z y x 1, 2, 3. DOF = 3, no constraints 4. 5. T= 1 2 1 mv = m ( x" 2 + y" 2 + z" 2 ) 2 2 ∂T ∂T ∂T = mx" , = my" , = mz" ∂x" ∂y" ∂z" d  ∂T  d  ∂T  d  ∂T  "" "" mx , my , = =       = mz"" dt  ∂x"  dt  ∂y"  dt  ∂z"  ∂T ∂T ∂T = = =0 ∂x ∂y ∂z 22 6. Generalized forces:  ∂x ∂y ∂z  Qqr =  Fx + Fy + Fz  q q q ∂ ∂ ∂ r r r   F = − mg zˆ Qqx = Qq y = 0, Qqz = Fz ∂z = −mg ∂z 7. EOMs: mx"" = 0, = my"" = 0, mz"" = −mg 8. Solve differential equations. • Comments: o Method is overkill for this problem o Inspection shows agreement with Newton 23 Example: Mass moving along a frictionless track. Z ρ θ z Y φ X • Track geometry defined such that: ρ = az , and φ = −bz DOF = 3 – 2 =1 • Constraint equations: ρ = az , and φ = −bz • Generalized coordinate: z • Find T = 1 2 mv , what is v? 2 24 • Define rotating coordinate frame such that mass remains in xˆ − zˆ plane. ẑ Z ρ ρ z x̂ z ŷ Y x̂ φ X r = ρx# +zz# = azx# + zz# and ω = φ" z# = −bzz# v 2 = r" • r" " # + zz "# + ( −bzz# ) × azx# + zz# r" = azx ! " # − abzzy " # + zz "# = azx " )2 + z" 2 = (az")2 + (abzz = (1 + a 2 + a 2 b 2 z 2 )z" 2 ∂T m 2 2 2 2 2 = m(1 + a 2 + a 2b2 z 2 ) z" T = (1 + a + a b z )z" and ∂z" 2 25 d  ∂T  2 2 2 2 z + 2m a 2b2 z z" 2  "  = m(1 + a + a b z )"" dt  ∂z  ( ) ∂T = m(a 2b2 z ) z" 2 ∂z • External force is gravity  ∂x ∂y ∂z  Qqr =  Fx + Fy + Fz  ∂ q ∂ q ∂ q r r r   F = −mg zˆ Qqx = Qq y = 0, Qqz = Fz ∂z = −mg ∂z • Equation of Motion: 2 2 2 2 2 2 2 "" " a + a b z + 1 z + a b z z = −g ( ) • Comments: o Solution highly nonlinear o “Trick” was finding inertial velocity o Still need to use FARM approach 26 Extending Lagrange’s Equation to Systems with Multiple Particles • Assume a system of particles and apply Newton’s laws: Fx1 = mx""1 , Fy1 = my""1 , Fz1 = mz""1 $ $ $ Fx p = mx""p , Fy p = my""p , Fz p = mz""p • 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. p ∑ m ( ""x δ x + ""y δ y i =1 i i i i p ( + "" ziδ zi ) = ∑ Fxi δ xi + Fyi δ yi + Fzi δ zi i =1 ) • Recall that this is D’Alembert’s equation • Assume the system has n DOF, n ≤ 3 p • Select generalized coordinates, qi that enforce the constraints: xi = f i ( q1 , q2 , …, qn , t ) yi = gi ( q1 , q2 , …, qn , t ) zi = hi ( q1 , q2 , …, qn , t ) 27 • Express virtual displacements in terms of generalized coordinates: δ xi = ∂xi ∂x ∂x δ q1 + i δ q2 + … + i δ qn ∂q1 ∂q2 ∂qn δ yi = ∂yi ∂y ∂y δ q1 + i δ q2 + … + i δ qn ∂q1 ∂q2 ∂qn ∂zi ∂z ∂z δ q1 + i δ q2 + … + i δ qn ∂q1 ∂q2 ∂qn • Substitute the relations into D’Alembert’s equation δ zi =  ∂xi ∂y ∂z m xi yi i + "" zi i + "" ∑  "" ∂qr ∂qr i =1  ∂qr p  δ qr   ∂x ∂y ∂z  = ∑  Fxi i + Fyi i + Fzi i  δ qr ∂qr ∂qr ∂qr  i =1  • 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. p • Using the calculus relations (chain rule), one can show that d ∂  ∂ T = Qqr  " T − dt  ∂qr  ∂qr • Once again, one Lagrange equation for each DOF. 28 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 29 • General Expression for V, the potential energy A P ( V = − ∫ ∑ Fxi dxi + Fyi dyi + Fzi dzi B i =1 ) • 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: ∂V Fxi = − ∂xi ∂V Fyi = − ∂yi ∂V Fzi = − ∂zi (C1) • Observe that: ∂Fx3 ∂  ∂V  ∂ 2V = − =− ∂y4 ∂y4  ∂x3  ∂x3∂y4 ∂Fy4 ∂  ∂V  ∂ 2V = − =− ∂x3 ∂x3  ∂y4  ∂x3∂y4 • Thus, in general ∂Fxi ∂yr = ∂Fyr ∂xi (C2) • Equation (C1) represents a necessary condition for a force to be conservative, Equation (C2) is a sufficient condition. 30 • Recall expression for generalized forces:  ∂xi ∂yi ∂zi  Qqr = ∑  Fxi + Fyi + Fzi  ∂ ∂ ∂ q q q i =1  r r r  p o Separate forces into conservative and non-conservative  ∂V ∂xi ∂V ∂yi ∂V ∂zi  N Qqr = −∑  Q + + +  qr ∂yi ∂qr ∂yi ∂qr  i =1  ∂xi ∂qr ∂V =− + Q N qr ∂qr p • Lagrange’s Equation: d ∂  ∂ T = Qqr  " T − dt  ∂qr  ∂qr • Substitute in generalized force: d ∂  ∂ ∂V N T − T = − + Q   qr dt  ∂q"r  ∂qr ∂qr ⇒ d ∂  ∂ N T T V Q − − = ( )   qr dt  ∂q"r  ∂qr • Since conservative forces are not functions of ∂ velocities: V =0 ∂q"r 31 • Thus, can define the Lagrangian L = T − V to obtain the final form of Lagrange’s equation: d  ∂L  dt  ∂q"r  ∂L = Fqr − ∂ q r  Example: Planar pendulum with an inline spring. y k r θ m x • • • • DOF = 3 – 1 = 2 Constraint equation: z = 0 Generalized coordinates: r, θ Coordinate mapping: x = r cosθ , y = r sin θ 32 • Kinetic energy 1 2 1 mv = m ( x" 2 + y" 2 ) 2 2 • Derivatives of coordinates: T= x" = r" cosθ − rθ" sin θ , y" = r" sin θ + rθ" cosθ • Substitute into kinetic energy equation T= • Potential energy V= • Lagrangian 1 m r" 2 + r 2θ" 2 2 ( ) 1 2 k ( r − ro ) − mgr cosθ 2 1 1 2 m r" 2 + r 2θ" 2 − k ( r − ro ) + mgr cosθ 2 2 • Derivatives of L (note need to do this for each GC) L = T −V = ∂L = mr", ∂r" ∂L = mr 2θ", ∂θ" ( d  ∂L   "  = mr"", dt  ∂r  ) ∂L = mrθ" 2 − k ( r − ro ) + mg cosθ ∂r d  ∂L  2 "" θ + 2mrr"θ", = mr  " dt  ∂θ  ∂L = −mgr sin θ ∂θ • Substitute into Lagrange’s Equation: mr"" − mrθ" 2 + k ( r − ro ) = mg cosθ mr 2θ"" + 2mrr"θ" − mgr sin θ = 0 33