Class Notes for Advanced Dynamics (MEAM535)

Transcription

1 Class Notes f Avance Dynamics MEAM535 Michael A. Carchii December 9, 9 Equilibrium Points, Small Perturbations Linear Stability Analysis The following notes were initially base aroun the text entitle: They of Vibrations with Applications 5th Eition by William T. Thomson Marie Dillon Dahleh.. Lagrange s Equations of Motion We have seen in lecture that if a system has n egrees-of-freeom escribe by the generalize coinateq,q,q 3,...,q n, then the kinetic energy of the system can be expresse as a function of the q k s q k s so that T T q,q,q 3,...,q n, q, q, q 3,..., q n if the fces, acting on the system escribe by the n generalize coinates q, q, q 3,..., q n, are conservative, which is the case with freely vibrational systems without amping, then we can fin a total potential function the Lagrangian of the system is then efine as V V q,q,q 3,...,q n. L T V Lq,q,q 3,...,q n, q, q, q 3,..., q n 3 the equations of motion become à t q j! q j 4

2 f j,, 3,..., n.. These n secon-er ifferential equations are couple in the q k s.. Example: A Cart-Ro System A cart of mass M is allowe to slie without friction along a hizontal tabletop. Connecte to the left of this cart is an ieal spring of stiffness constant k, which is also connecte to a fixe suppt to the left of the spring. A rigi ro of mass m is allowe to pivot without friction about the center of the cart as shown in the figure below. The Ro-Cart System Using the generalize coinates q x as measure along the tabletop from where the spring is at its natural length q θ istheangleshowninthe figure above, we see that the kinetic energy of the system is given by T T Cart + T Ro where T Cart Mẋ M being the mass of the cart T Ro mẋ cm + ẏ cm+ I cm θ. But x cm x + L sinθ y cm L cosθ

3 so that ẋ cm ẋ + L θ cosθ ẏ cm L θ sinθ I cm ml where L is the length of the ro m is the mass of the ro. Putting these into the expression f T Ro,weget T Ro µ m ẋ + L θ µ L cosθ + θ sinθ + µ ml θ which reuces to T Ro m ẋ + Lẋ θ cosθ+ L 4 θ + 4 ml θ simply T Ro mẋ + mlẋ θ cosθ+ 6 ml θ. Thus we see that the total kinetic energy of the cart-ro system is given by T Mẋ + mẋ + mlẋ θ cosθ+ 6 ml θ simply T m + Mẋ + mlẋ θ cosθ+ 6 ml θ. In aition, we the total potential energy of the cart-ro system is given by V kx + mgy cm kx mgl cosθ, using the pivot point O as the zero point f weight potential energy. From these, we can now construct the Lagrangian f the system as L T V L m + Mẋ + mlẋ θ cosθ+ 6 ml θ kx + mgl cosθ. Then ẋ m + Mẋ + ml θ cosθ, 3 x kx

4 hence leas to t Ã! ẋ x µ m + Mẋ + t ml θ cosθ + kx m + Mẍ + ml θ cosθ ml θ sinθ+kx as one of the equations of motion f the cart-ro system. We also have θ mlẋ cosθ+ 3 ml θ, θ mlẋ θ sinθ mgl sinθ hence leas to t t à µ mlẋ cosθ+ 3 ml θ θ! θ + mlẋ θ cosθ+ mgl sinθ mlẍ cosθ mlẋ θ sinθ+ 3 ml θ + mlẋ θ sinθ+ mgl sinθ which reuces to mlẍ cosθ+ 3 ml θ + mgl sinθ as the secon equation of motion f the cart-ro system. Stability Analysis About The Equilibrium Points f This Example From the expression f V we see that V x kx V θ mgl sinθ 4

5 give the equilibrium points at x e, θ e, at x e, θ e, π. It is interesting to consier small oscillations about the equilibrium at x e, θ e, f this problem by replacing cosθ cosε θ + θ e cosε θ ' sinθ sinε θ + θ e sinε θ ' ε θ thereby removing all ers higher than one. Uner these assumptions, the equations of motion m + Mẍ + ml θ cosθ ml θ sinθ+kx mlẍ cosθ+ 3 ml θ + mgl sinθ to first er become m + M ε x + ml ε θ + k ε x ml ε x + 3 ml ε θ + mglε θ where ε x x x e ε θ θ θ e. These equation reuce to m + M ε x + ml ε θ +kε x 3 ε x +L ε θ +3gε θ which in matrix fm can be expresse as m + M ml 3 L # εx ε θ + k 3g # εx ε θ. By setting εx ε θ A A e iωt 6 5

6 in this equation we get # m + M ml ω A k e 3 L iωt + A 3g à # m + M ml k ω + 3 L 3g k ω m + M mlω # A 3ω 3g Lω A in er f A 6 A we must require that k ω et m + M mlω 3ω 3g Lω this leas to # A #! A A # A m +4MLω 4 kl +3m + Mgω +6kg e iωt so that the natural angular frequencies of the cart-ro system are given by ω± kl +3m + Mg ± q 4kL +3m + Mg 4kgLm +4M. m +4ML simply ω ± kl +3m + Mg m +4ML ± v u tã kl +3m + Mg m +4ML! 6kg m +4ML. To check this result, let us consier some limiting cases. The first is when L. This leas to m +4MLω 4 kl +3m + Mgω +6kg which f L reuces to simply 6m + Mgω +6kg 6

7 which leas to s k ω m + M, which is expecte, since we may treat this cart-ro system as a simple block of mass m + M connecte to a spring of constant k. The next limiting case is when M. This leas to ω ± 3g 4L ± s µ s 3g 3g ω + 4L L ω, which crespons to no motion, hence we igne this. The result f ω + is expecte, since we may treat the system as a compoun penulum with moment of inertia I O 3 ml about point O on the cart which is not moving since M. The center-ofmass of the ro is a istance D L/ frompointo.usingtheresult s mgd ω from basic physics putting in our expressions f D I O,wethenget which agrees with ω + above. v u I O u ω t mgl/ ml /3 s 3g L Let us also consier small oscillations about the equilibrium point at x e, θ e π, f this problem by replacing cosθ cosε θ + π cosε θ ' sinθ sinε θ + π sinε θ ' ε θ 7

8 thereby removing all ers higher than one. Uner these assumptions, the equations of motion, m + Mẍ + ml θ cosθ ml θ sinθ+kx mlẍ cosθ+ 3 ml θ + mgl sinθ to first er become m + M ε x ml ε θ + k ε x ml ε x + 3 ml ε θ mglε θ where ε x x x e ε θ θ θ e. These equation reuce to m + M ε x ml ε θ +kε x 3 ε x L ε θ +3gε θ which in matrix fm can be expresse as # m + M ml εx k + 3 L ε θ 3g # εx ε θ. By setting in this equation we get εx ε θ m + M ml ω 3 L # A A A A e iωt + e iωt 6 k 3g à # m + M ml k ω + 3 L 3g k ω m + M mlω 3ω 3g +Lω 8 # A #! A # A A A A e iωt

9 in er f A 6 A we must require that k ω et m + M mlω 3ω 3g +Lω # this leas to m +4MLω 4 kl 3m + Mgω 6kg so that the natural angular frequencies of the cart-ro system are given by ω± kl +3m + Mg ± q 4kL +3m + Mg +4kgLm +4M. m +4ML simply ω ± kl +3m + Mg m +4ML ± v u tã kl +3m + Mg m +4ML! + 6kg m +4ML. Note then that ω < showing that ω is not real so εx ε θ A A e iωt will not be boune, hence this equilibrium point is unstable. 3. Example: The Spherical Penulum The figure below shows a spherical penulum which consists of a mass m connecte to a pivot point O locate at the point,,l by a c of fixe 9

10 length L. The Spherical Penulum Let θ be the polar angle that the cs makes with the negative vertical z axis let ϕ be the azimuthal angle that the line connecting the projection of the mass on the xy plane the igin the otte line in the figure above makes with the positive x axis. Using the generalize coinates: q θ q ϕ, we see that the xyz position of the mass is given by Thus we have x L sinπ θcosϕ L sinθcosϕ y L sinπ θsinϕ L sinθsinϕ z L L cosπ θ L + L cosθ. ẋ L θ cosθcosϕ L ϕ sinθsinϕ ẏ L θ cosθsinϕ+l ϕ sinθcosϕ ż L θ sinθ. The kinetic energy of the penulum is then given by T mẋ + ẏ + ż

11 which reuces to T ml θ cosθcosϕ L ϕ sinθsinϕ + ml θ cosθsinϕ+l ϕ sinθcosϕ + m L θ sinθ T ml θ cos θcos ϕ+l ϕ sin θsin ϕ + ml θ cos θsin ϕ+l ϕ sin θcos ϕ + ml θ sin θ which further reuces to T ml θ + ϕ sin θ. The potential energy is given by V mgz mgl + cosθ so that the Lagrangian of the system is L T V, Then we see that L ml θ + ϕ sin θ mgl + cosθ. θ ml θ, θ ml ϕ sinθcosθ+mgl sinθ resulting in the one equation of motion t à θ! θ which reuces to ml θ ml ϕ sinθcosθ mgl sinθ, θ ϕ sinθcosθ g sinθ. L

12 Next we see that ϕ ml ϕ sin θ, ϕ resulting in the secon equation of motion t Ã! ϕ ϕ ³ ml ϕ sin θ t which says that ml ϕ sin θ equals a constant in time. Setting this equal to C, we get C ϕ ml sin θ putting this into the first equation of motion leas to θ Ã! C ml sin sinθcosθ g sinθ θ L simply θ C cosθ m L 4 sin 3 θ g sinθ. L Asacheck,wenotethatifthereisnomotioninϕ irection so that ϕ,then C the θ equation becomes θ g sinθ L which leas to a stable equilibrium at θ π, then the equation of motion becomes Setting ε θ π, wethenget sinθ ' sinπ+θ πcosπ θ π θ + g θ π. L ε + g L ε

13 which is the equation of a simple penulum with angular frequency ω r g L. We also note from the equation of motion θ C cosθ m L 4 sin 3 θ g sinθ. L that so that θ C cosθ m L 4 sin 3 θ + g sinθ Gθ L θ θ θgθ θ θ Gθθ θ Z Gθθ constant in time. This leas to θ Z Ã C cosθ m L 4 sin 3 θ + g! L sinθ θ constant θ C + m L 4 sin θ + g cosθ constant L θ + Ã Of course this is simply C ml sin θ! sin θ+ g cosθ constant L ml θ + ϕ sin θ + mg cosθ constant. T + V constant thereby showing that energy is conserve. 3

14 4. The Equations of Motion - General Fm Consier a system of N particles with positions given by the rectangular coinates x i,y i,z i,fi,, 3,...,N, having n egrees of freeom with generalize coinates q, q, q 3,..., q n generalize spees q, q, q 3,..., q n, so that x i x i q,q,q 3,...,q n y i y i q,q,q 3,...,q n z i z i q,q,q 3,...,q n f i,, 3,...,N. We know that the Lagrangian of the system can be written as L T V ingeneral,weshallalwaysfin that the kinetic energy of the system will have the fm T T jk q j q k 5a k with NX xi x i T jk m i + y i y i + z i z i i q j q j q j being only a function of the q k s not the q k s. Thus we have L k it shoul be clear from Equation 5a that T jk 5b T jk q j q k V, 6 T q j q k 7 so T jk T kj. Now all of T jk V are functions of only the generalize coinates q j they are not functions of the generalize spees q j. From Lagrange s equations t à q i! 4

15 we have Since we see that In aition we see that T q i But t k à T V q i! T V Ã! T T + V. 8 t q i T T k k T jk q j q k q i q j q i q k q i T jk q j q k T jk q j q k. 9a k, f j 6 i, f j i, f k 6 i, f k i so we fin that T T jk δ ji q k + δ ki q j q i k qj q k T jk q k + q j. q i q i δ ji δ ki T ik q k + T ji q j T ij q j + k ij + T ji q j T since T ij T ji,sowefin that T q i T ij + T ij q j 5 T ij q j. k T jk δ ji q k + T ji q j k T jk δ ki q j 9b

16 Putting Equations 9a 9b into Equation 8, we get t T ij q j t T ij q j k k T jk T jk q j q k + V q j q k + V. But Ã! Ã! t T Tij ij q j t q q j Tij j + T ij t t q j + T ij q j. However, using the chain rule from Calculus, we have T ij t k T ij q k t n X k T ij q k so t T ij q j k putting this into Equation, we have T ij q k q j + T ij q j k T ij q k q j + T ij q j k T jk q j q k + V just T ij q j + Since T ij T ji,wemaywrite k à Tij T ij T ij + T ji! T jk q j q k + V thenwehave à Tij T ij q j + k + T ji T! jk q j q k + V. 6

17 with T ij q j + k T ijk T ij T ij q j q k T ijk q j q k + V. T + T ji T jk. a b c Equation give the equations of motion f the n-egree of freeom system it clearly shows the explicit occurrences of the q j s the q j s. 5. Linear Stability Analysis About Static Equilibrium Points The static equilibrium points f the system are where q i q ie constant, resulting in q i q i putting this into Equation a, we fin that V q j q,q,q 3,...,q nq e,q e,q 3e,...,q ne 3 f j,, 3,...,n gives the set of equations that can be solve f the q ie s. Now suppose that q,q,q 3,...,q n q e,q e,q 3e,...,q ne. is one such static equilibrium point suppose that ε, ε, ε 3,...,ε n q q e,q q e,q 3 q 3e,...,q n q ne is a small perturbation about this static equilibrium point. Using the notation V V q,q,q 3,...,q n V e V q e,q e,q 3e,...,q ne exping V/ to first er, we have V V + q,q,q 3,...,q n q e,q e,q 3e,...,q ne V q j q,q,q 3,...,q n q e,q e,q 3e,...,q ne 7 q j q je

18 since If we efine then we get V V q j V V ije q,q,q 3,...,q nq e,q e,q 3e,...,q ne q,q,q 3,...,q n q e,q e,q 3e,...,q ne V V q j. q,q,q 3,...,q n q e,q e,q 3e,...,q ne V ije q j q je V ije ε j q j q je to first er in the ε j s. Setting q j q je in the T ij s the T ijk s, we have T ije T ij q e,q e,q 3e,...,q ne T q i q j q,q,q 3,...,q n q e,q e,q 3e,...,q ne 4 5 T ijke T ij + T ji T jk Putting these into Equation, we get q,q,q 3,...,q nq e,q e,q 3e,...,q ne 6 T ije q j + T ijke q j q k + V ije ε j k T ije ε j + T ijke ε j ε k + V ije ε j k since the mile term in this equation is secon er in the ε j s, we may rop it simply write T ije ε j + Writing this in matrix fm, we have V ije ε j. 7 [T e ] n n { ε} n +[V e ] n n {ε} n {} n 8a 8

19 where [T e ] n n [V e ] n n T q i q j V q j q,q,q 3,...,q n q e,q e,q 3e,...,q ne. q,q,q 3,...,q n q e,q e,q 3e,...,q ne 8b 8c we set To examine the stability of the equilibrium point at into Equation 8a resulting in q,q,q 3,...,q n q e,q e,q 3e,...,q ne {ε} n {μ} n e iωt 6 {} n ω [T e ] n n {μ} n e iωt +[V e ] n n {μ} n e iωt {} n ω [T e ] n n +[V e ] n n {μ} n {} n. In er f {μ} n 6 {} n,wemusthave et ω [T e ] n n +[V e ] n n 9 which shows that if all solutions to the equation are real, the equilibrium point at q,q,q 3,...,q n q e,q e,q 3e,...,q ne is stable if at least one solution to Equation is not real, then the equilibrium point at q,q,q 3,...,q n q e,q e,q 3e,...,q ne is unstable. If both [T e ] n n [V e ] n n are zero, then higher-er analysis is necessary. 9

20 6. The Cart-Ro Example: Revisite Consier the cart-ro system iscusse earlier. We saw that the total kinetic energy of the cart-ro system is given by T m + Mẋ + mlẋ θ cosθ+ 6 ml θ. In aition, we saw that the total potential energy of the cart-ro system was given by V kx mgl cosθ, From these, we can now construct V/ x V/ θ kx mgl sinθ setting this equal to zero gives the equilibrium points at x e, θ e, at x e, θ e, π. Nextwehave [T] T/ ẋ ẋ T/ θ ẋ T/ ẋ θ T/ θ θ # m + M ml cosθ ml cosθ 3 ml [V] V/ x x # # V/ θ x k V/ x θ V/ θ θ mgl cosθ so f x e, θ e,, wehave m + M [T e ] ml # k ml [V e ] 3 ml mgl # # f x e, θ e, π, wehave m + M [T e ] ml ml 3 ml Then f x e, θ e,, wehave # [V e ] k mgl # et ω [T e ]+[V e ]

21 which leas to et k m + Mω mlω mlω mgl 3 ml ω # m +4MLω 4 kl +3m + Mgω +6kg as befe, x e, θ e, π, wehave et ω [T e ]+[V e ] which leas to et k m + Mω mlω mlω mgl 3 ml ω # m +4MLω 4 kl 3m + Mgω 6kg as befe.