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Lecture 10 Reprise and Generalized Forces

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Lecture 10 Reprise and Generalized Forces Lecture 10 Reprise and generalized forces The Lagrangian Holonomic constraints Generalized coordinates Nonholonomic constraints we haven’t done this, so let’s start with it Generalized forces Euler‐Lagrange equaons Hamilton’s equaons 1 Generalized forces are based on the principle of virtual work i δW = F ⋅δr + τ ⋅δΦ = Qiδq convert to me derivaves € δW δW = δt ⇒ W˙ δ t, Qδqi ⇒ Q q˙ iδt δt i i ˙ i ˙ i Wδ t = Qiq˙ δt ⇒ W = Qiq˙ € ∂W˙ Q = i ∂q˙ i € € so what we need is W˙ 2 € We need to look at the forces and torques applied to each link W˙ = F ⋅ r˙ + τ ⋅ Φ˙ The translaonal part is obvious: F ⋅ r˙ = F ⋅ v € The rotaonal part € δΦ = δφk + δθI1 + δψK 3 ˙ ˙ ˙ δΦ = φδ tk + θδ tI1 +ψ δtK 3 = ωδt = Ωδt € Φ˙ = ω € W˙ = F ⋅ v + τ ⋅ ω € € 3 So my total rate of work will be K ˙ W = ∑(Fi ⋅ vi + τ i ⋅ ω i ) i=1 The velocies and rotaons must be consistent with the constraints and everything must be expressed in ineral coordinates € I can believe that this looks mysterious, but it is actually simple, which we can see by looking at some examples REMEMBER THAT THESE FORCES AND TORQUES DO NOT INCLUDE THE POTENTIAL FORCES AND TORQUES 4 This is the three link robot that we have seen before It will have torques from the ground to link 1 from link 1 to link 2 from link 2 to link 3 5 The torque from the ground to link 1 τ 01k € The torque from link 1 to link 2 τ12I1 € The torque from link 2 to link 3 τ 23I2 € There is a force from the ground that balances the weight of the robot but it does no work because link 1 does not move up and down 6 The torque on link 1 τ 01k − τ12I1 € The torque on link 2 τ12I1 − τ 23I2 € The torque on link 3 τ 23I2 These will give us the generalized forces as we’ll see shortly € 7 There are a bunch of constraints affecng the rotaon rates φ1 = 0 = θ1 φ2 =ψ1 = φ3 € ψ2 = 0 =ψ3 € There are also enough connecvity constraints to reduce the system to three degrees of freedom € we’ll look at these shortly 8 All three I vectors are equal to cos ψ 1 sinψ1 0 The rotaon rates are € ˙ ˙ 0 cosψ1θ 2 cosψ1θ 3 ˙ ˙ ω1 = 0 , ω 2 = sinψ1θ 2 , ω 3 = sinψ1θ 3 ˙ ˙ ˙ ψ 1 ψ1 ψ1 € 9 Build the work terms for each link ˙ ω1 ⋅ (τ 01k − τ12I1) = τ 01ψ1 ˙ ω2 ⋅ (τ12I1 − τ 23I2 ) = (τ12 − τ 23 )θ 2 € ˙ ω3 ⋅ τ 23I2 = τ 23θ 3 € so we have € ˙ ˙ ˙ ˙ W = τ 01ψ1 + (τ12 − τ 23)θ 2 + τ 23θ 3 € 10 The three generalized coordinates ψ 1 ˙ q = θ 2 ⇒ W = τ 01q˙1 + (τ12 − τ 23)q˙ 2 + τ 23q˙ 3 θ 3 Differenate to get the generalized forces € Q1 = τ 01, Q2 = (τ12 − τ 23), Q3 = τ 23 € 11 We get a beer idea of the compability issues by looking further Apply a horizontal force to the end of link 3 Let it be directed in the –I direcon 12 We have to take the connecvity constraints in to account to handle this We see intuively that the only work is can do is to rotate the enre robot about k. It will contribute a torque to Q1; can we get that using the rubric? We have a force applied to link 3, and it is not at the center of mass, so there is an accompanying torque 1 1 1 τ = r K × F = r K × I F = r sinθ Fψ˙ 3 2 3 3 2 3 3 1 2 3 3 1 € 13 The connecvity constraint 1 1 p = p + r K + r K ⇒ v = r K˙ + r K˙ 3 1 2 2 2 3 3 3 2 2 2 3 3 The force piece of the work term is € 1 1 W˙ = F ⋅ v = FI ⋅ r K˙ + r K˙ = r sinθ + r sinθ Fψ˙ F 3 1 2 2 2 3 3 2 2 2 3 3 1 combine that with the torque term and we have € ˙ ˙ W = (r2 sinθ2 + r3 sinθ3 )Fψ1 for an addional term in accord with intuion € Q1 = (r2 sinθ2 + r3 sinθ3 )F 14 € ?? We’ll look at this in general once again 15 The Lagrangian in physical variables K L = ∑Ti −Vi i=1 1 1 2 T = m x˙ 2 + y˙ 2 + z˙ 2 + A θ˙ cosψ + φ˙ sinθ sinψ 2 ( ) 2 ( ) € 1 2 1 2 + B −θ˙ sinψ + φ˙ sinθ cosψ + C ψ˙ + φ˙ cosθ 2 ( ) 2 ( ) € We’ve mostly been doing gravity potenals 16 Simple holonomic constraints Nonsimple holonomic constraints These remove some number of variables from the problem At this point you can assign the remaining variables to a set of generalized coordinates 17 Nonholonomic constraints: linear in the derivaves of q Write them in matrix form i j C jq˙ = 0 ⇔ Cq˙ = 0 C is an M x N matrix — number of constraints € x number of generalized coordinates 18 Generalized forces Idenfy the outside forces and torques on each link Form W˙ ∂W˙ Then Qi = € ∂q˙ i € Choose between Euler‐Lagrange and Hamilton 19 Euler‐Lagrange d ∂L ∂L j i = i + Qi + λ jCi dt ∂q˙ ∂q Convert to first order system € the first set of equaons is really simple q˙ i = ui We need to develop the symbolic representaon of the second set € 20 Here’s the Lagrangian we want its contribuon to the le and right hand sides 1 L = q˙ i M q˙ j −V qk 2 ij ( ) ∂L 1 ∂M ∂V right hand side = q˙ m mn q˙ n − € ∂qi 2 ∂qi ∂qi substute q˙ m = um, q˙ n = un € right hand side ∂L 1 ∂M ∂V = um mn un − € ∂qi 2 ∂qi ∂qi € 21 le hand side ∂L j j d ∂L j dMij j i = Mijq˙ = Mij u ⇒ i = Mij u˙ + u ∂q˙ dt ∂q˙ dt dMij ∂Mij € M u˙ j + u j = M u˙ j + u j uk ij dt ij ∂qk and we can combine the two and add back € the constraint contribuon and the generalized forces 22 ∂Mij ∂L M u˙ j + u j uk = + λ C j + Q ij ∂qk ∂qi j 1 i ∂L 1 m ∂Mmn n ∂V i = u i u − i € ∂q 2 ∂q ∂q € ∂Mij 1 ∂M ∂V M u˙ j + u j uk = um mn un − + λ C j + Q ij ∂qk 2 ∂qi ∂qi j 1 i € 23 ∂Mij 1 ∂M ∂V M u˙ j + u j uk = um mn un − + λ C j + Q ij ∂qk 2 ∂qi ∂qi j 1 i 1 ∂M ∂Mij ∂V M u˙ j = um mn un − u j uk − + λ C j + Q € ij 2 ∂qi ∂qk ∂qi j 1 i € q˙ i = ui and we can (in principle) solve for the individual components of u € j ki 1 m ∂Mmn n ∂Mij j k ∂V j u˙ = M u i u − k u u − i + λ jC1 + Qi 2 ∂q ∂q ∂q € 24 We did Hamilton the last me I have the pair of sets of odes j ji q˙ = M pi ∂L j p˙ i = + λ jC1 + Qi € ∂qi where all the q dots in the second set have been replaced by their expressions in terms of p € 25 We can write this out in detail, although it looks prey awful 1 L = q˙ i M q˙ j −V qk 2 ij ( ) ∂L 1 ∂M ∂V right hand side = q˙ m mn q˙ n − € ∂qi 2 ∂qi ∂qi substute m mr n ns q˙ = M pr, q˙ = M ps € ∂L 1 ∂M ∂V right hand side = M mr p mn M ns p − € ∂qi 2 r ∂qi s ∂qi € 26 ∂L p˙ = + λ C j + Q i ∂qi j 1 i ∂L 1 ∂M ∂V = M mr p mn M ns p − ∂qi 2 r ∂qi s ∂qi € € 1 ∂M ∂V p˙ = M mr p mn M ns p − + λ C j + Q i 2 r ∂qi s ∂qi j 1 i € i ij q˙ = M p j 27 € Both of these are of the same general form a form that we will see in the other methods we will address i ij q˙ = M p j 1 ∂M ∂V p˙ = M mr p mn M ns p − + λ C j + Q i 2 r ∂qi s ∂qi j 1 i Euler‐Lagrange € € Hamilton q˙ i = ui j ki 1 m ∂Mmn n ∂Mij j k ∂V j u˙ = M u i u − k u u − i + λ jC1 + Qi 2 ∂q ∂q ∂q € € 28 What you have to do in general — whether you use Euler‐Lagrange or Hamilton Set up the physical Lagrangian Put in the simple holonomic constraints Put in the nonsimple holonomic constraints Define the generalized coordinates Find the nonholonomic constraints and the constraint matrix Define the Lagrange mulpliers Find the rate of work and the generalized forces 29 For Hamilton’s equaons i ij Define the momentum and use it to eliminate q˙ = M p j (These are half the differenal equaons) € The other half are the momentum equaons 1 ∂M ∂V p˙ = M mr p mn M ns p − + λ C j + Q i 2 r ∂qi s ∂qi j 1 i € 30 Solve some of the momentum equaons for the Lagrange mulpliers Combine the remaining momentum equaons with the differenated constraints These are the new momentum equaons i ij Solve the new momentum equaons simultaneously with q˙ = M p j € 31 An example link 1, the frame link 2, the fork link 2 link 3, the rear wheel link 4, the front wheel link 1 link 4 link 3 32 The (physical) Lagrangian There are four idencal Lagrangians — we’ve seen them before I’m not going to write them out I am going to hold this system erect, so that gravity does nothing I can set g = 0 33 Simple holonomic constraints I want the system to be erect Let the frame and fork rotate only about the vercal φ1 = 0 = φ2, θ1 = 0 = θ 2 minus 4 Define the difference between their k = K rotaons € ψ2 =ψ1 + α no change € 34 Simple holonomic constraints The heights of all the links are fixed z1 = h1, z2 = h2, z3 = rW = z4 minus 4 € 35 Simple holonomic constraints I want the wheels to be upright π θ = − = θ minus 2 3 2 4 The rear wheel points in the same direcon as the frame the front wheel in the same direcon as the fork € φ3 =ψ1, φ4 =ψ1 + α minus 2 The horizontal locaons of the centers of mass are connected the front connecon is simple € x4 = x3, y4 = y3 minus 2 36 € Nonsimple holonomic constraints The other horizontal locaons of the centers of mass are also connected x3 = −l1 cosψ1, y3 = −l1 sinψ1 minus 4 x2 = l2 cosψ1, y2 = l2 sinψ2 At this point we have introduced eighteen constraints, leaving six variables € 37 Assign the generalized coordinates x1 y 1 ψ q = 1 α ψ 3 ψ 4 € We sll have rolling constraints 38 Nonholonomic constraints rolling constraints v3 = ω 3 × r3, v4 = ω 4 × r4 r3 = k = r4 vercal components are already in place, so there are but four of these € x˙ 3 −€r Wψ˙ 3 cosφ3 = 0 = y˙ 3 − rWψ˙ 3 sinφ3 x˙ 4 − rWψ˙ 4 cosφ4 = 0 = y˙ 4 − rWψ˙ 4 sinφ4 constraint matrix € 3 3 1 0 l2 sinq 0 −rW cosq 0 3 3 0 1 −l2 cosq 0 −rW sinq 0 C = 3 3 4 1 0 −l1 sinq 0 0 −rW cos(q + q ) 3 3 4 0 1 l1 cosq 0 0 −rW sin(q + q ) 39 € Nonholonomic constraints in physical coordinates 1 0 l2 sinψ1 0 −rW cosψ1 0 0 1 −l cosψ 0 −r sinψ 0 C = 2 1 W 1 1 0 −l1 sinψ1 0 0 −rW cos(ψ1 + α) 0 1 l1 cosψ1 0 0 −rW sin(ψ1 + α) There will be four Lagrange mulpliers € j We’ll add λ jCi to the momentum equaons € 40 Generalized forces There is torque from the frame to the fork τ12k and from the frame to the rear wheel τ13K 3 € The rate of work will be € ˙ W = ω1 ⋅ (−τ12k − τ13K 3) + ω 2 ⋅ τ12k + ω 3 ⋅ τ13K 3 from which we can find € Q = {0 0 0 τ12 τ13 0} € 41 At this point we have the structure of the problem It’s me to go to Mathemaca to see how to do this in pracce And to see how to convert this to numerical form for integraon And some simple examples 42 .
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