1.6 Mechanical Systems

Transcription

1 1.6 Mechnicl Systems 1.6 Mechnicl Systems 45 Mechnics provies n excellent clss of systems for both motivting the ies of ynmicl systems n to which the ies of ynmicl systems pply. We sw some simple exmples of Newtonin mechnicl systems in the introuctory section. To illustrte n motivte the introuction of itionl structures governing mechnicl systems, we will evelop two exmples s we go long. Exmple 1. Consier prticle moving in Euclien 3 spce, R 3, subject to potentil forces. As vrint of this exmple, lso consier n object whose mss is time-vrying n is subject to externl forces (e.g. rocket burning fuel n generting propulsion). Exmple 2. Consier circulr hoop, rotting long the z xis with certin fixe ngulr velocity ω, s shown in Figure Consier prticle with mss m moving in this hoop. R Figure A prticle moving in rotting hoop. Configurtion Spce Q. The configurtion spce of mechnicl system is spce whose points etermine the sptil positions of the system. Although this spce is generlly prmetrize by generlize coorintes, enote (q 1,..., q n ), the spce Q itself nee not be n Euclien spce. It cn be rther thought s configurtion mnifol. Since we will be working

2 46 Introuction with coorintes for this spce, it is ok to think of Q s Euclien spce for purposes of this section. However, the istinction turns out to be n importnt generl issue. Exmple 1. Here Q = R 3 since point in spce etermines where our system is; the coorintes re simply stnr Euclien coorintes: (x, y, z) = (q 1, q 2, q 3 ). Exmple 2. Here Q = S 1, the circle of rius R since the position of the prticle is completely etermine by where it is in the hoop. Note tht the hoop s position in spce is lrey etermine s it hs prescribe ngulr velocity. The Lgrngin L(q, v). The Lgrngin is function of 2n vribles, if n is the imension of the configurtion spce. These vribles re the positions n velocities of the mechnicl system. We write this s follows L(q 1,..., q n, v 1,..., v n )) = L(q 1,..., q n, q 1,..., q n ). (1.6.1) At this stge, the q i s re not time erivtives yet (since L is just function of 2n vribles, but s soon s we introuce time epenence so tht the q i s re functions of time, then we will require tht the v i s to be the time erivtives of the q i s. In mny (but not ll) of our exmples we will set L = K E P E, i.e. s the ifference between the kinetic n potentil energies. The sign in front of the potentil energy in this efinition of L is very importnt. For instnce, consier prticle with constnt mss, moving in potentil fiel V, which genertes force F = V. We will see shortly tht the eqution F = m is prticulr cse of the bsic equtions ssocite to L, nmely the Euler Lgrnge equtions, therefore necessrily we nee the minus sign before the P E. There re other eeper resons why the minus sign tht involve reltivistic invrince. 4 Recll lso tht the kinetic energy of prticle with mss m moving in R 3 is given by K E = 1 2 m v 2. This efinition comes bout becuse the kinetic energy is relte to the work one by the force in simple wy s follows: if the prticle governe by F = m, then 1 t 2 m v 2 = mv F n so the funmentl theorem of clculus shows tht chnge in its kinetic energy from point to point b is given by: K E = b F (t) v t, 4 For instnce, the wve eqution φ tt φ xx = 0 hs Lgrngin lso given by the kinetic minus the potentil energy: L(φ, φ t) = 1 R `φ2 2 t φ 2 x x. Note tht this expression is invrint uner the Poincré group (the bsic group of specil reltivity), but with the plus sign (giving the totl energy) it woul not be invrint.

3 1.6 Mechnicl Systems 47 tht is, the line integrl of F long the pth tken by the prticle, or the work one by F long the pth of the prticle. In prticulr, if the force is given by F = V for potentil V, then or K E = b F (s) s = V () V (b), (K E + P E ) = 0, which gives conservtion of energy. This will be lso verifie below. Exmple 1. From the bove iscussion, we see tht in Exmple 1, we shoul hve L(q, v) = 1 2 m v 2 V (q) Exmple 2. With reference to Figure 1.6.2, in n inertil (or lbortory) frme, the velocity is: v(t) = R θe θ + ωr sin θe φ. Where the vectors e θ, e φ n e R re n orthonorml bsis of vectors in R 3, ssocite with sphericl coorintes. Obviously, since the mss is moving insie the hoop, its component long e R is zero. The component R θe θ is the velocity long e θ since the istnce of the mss m from the stright own position long the circle is given by Rθ. The kinetic energy of this system is K E = 1 2 m v 2 = 1 2 m(r2 θ2 + ω 2 R 2 sin 2 θ), while the potentil energy is given by P E = mgh = mgr cos θ (with the zero of potentil energy tken to be the plne z = 0 n with θ mesure, s bove, from the ownwr position of the mss). Therefore the Lgrngin for this system is L(θ, θ) = 1 2 m(r2 θ2 + ω 2 R 2 sin 2 θ) + mgr cos θ. The Euler-Lgrnge Equtions. The first step in the escription of Lgrngin system ws giving the configurtion spce n the secon ws giving the Lgrngin. Now we come to the thir step, which is writing own the Euler-Lgrnge equtions: L t q i L q i = 0 (1.6.2)

4 48 Introuction R h = 0 R R cos θ m mg Figure A prticle moving in rotting hoop with the ttche orthonorml frme n sie view. Historicl Note. This eqution ws introuce by Lgrnge (25 Jn 1736, Turin Itly 10 April 1813, Pris, Frnce), n it correspons to Newton s Secon Lw F = m. Lgrnge writes this eqution in generlize coorintes (q i, q i ), the point being tht chnge of coorintes oes not lter the form of the Euler Lgrnge equtions, wheres chnging coorintes in Newton s lw is tricky s one hs to trnsform ccelertions. To this en, the efinition of L = K E P E is crucil, n mkes the Euler Lgrnge equtions inepenent of the chosen coorinte system). It is interesting lso to note how Lgrnge originlly nme wht we cll L toy s H = K E P E, fter the utch scientist Huygens, fmous n mire t tht time for his works on geometric optics. Let us go bck to our exmples: Exmple 1. Here the Euler Lgrnge equtions become L t q i L q i = (mv) + V = 0; t tht is, (mv) = V, t which is the sme s F = m. Exmple 2. Here we first compute L θ = mr2 θ, L θ = mr2 ω 2 sin θ cos θ mgr sin θ,

5 1.6 Mechnicl Systems 49 n so the Euler Lgrnge equtions become t (mr2 θ) (mr 2 ω 2 sin θ cos θ mgr sin θ) = 0. (1.6.3) Derivtion from F = m (for those who might hve oubts). We will now write the equtions of the motion of this mss in Euclien coorintes, n use Newton s secon lw to erive these equtions (1.6.3). A key point is to relize tht the hoop will exert forces of constrint on the prticle. It is lso importnt to trnsform between the ccelertions written in Euclien ccelertions to sphericl coorintes (not plesnt, but strightforwr tsk). We will show tht this proceure results in the sme equtions of motion. The position of the prticle in spce is specifie by the ngles θ n ϕ, s shown in Figure We cn tke ϕ = ωt, so the position of the prticle becomes etermine by θ lone. Let the orthonorml frme long the coorinte irections e θ, e ϕ, n e r be s shown. The forces cting on the prticle re: 1. Friction, proportionl to the velocity of the prticle reltive to the hoop: νr θe θ, where ν 0 is constnt Grvity: mgk. 3. Constrint forces in the irections e r n e ϕ to keep the prticle in the hoop. The equtions of motion re erive from Newton s secon lw F = m. To get them, we nee to clculte the ccelertion ; here mens the ccelertion reltive to the fixe inertil frme xyz in spce; it oes not men θ. Reltive to this xyz coorinte system, we hve x = R sin θ cos ϕ, y = R sin θ sin ϕ, z = R cos θ. (1.6.4) Clculting the secon erivtives using ϕ = ωt n the chin rule gives ẍ = ω 2 x θ 2 x + (R cos θ cos ϕ) θ 2Rω θ cos θ sin ϕ, ÿ = ω 2 y θ 2 y + (R cos θ sin ϕ) θ + 2Rω θ cos θ cos ϕ, z = z θ 2 + (R sin θ) θ. (1.6.5) 5 This is lw of friction tht is more like viscous flui friction thn sliing friction in which ν is the rtio of the tngentil force to the norml force; in ny ctul experimentl setup (e.g., involving rolling spheres) relistic moeling of the friction is not trivil tsk; see, for exmple,?.

6 50 Introuction If i, j, k, enote unit vectors long the x, y, n z xes, respectively, we hve the esily verifie reltion e θ = (cos θ cos ϕ)i + (cos θ sin ϕ)j + sin θk. (1.6.6) Now consier the vector eqution F = m, where F is the sum of the three forces escribe erlier n = ẍi + ÿj + zk. (1.6.7) The e ϕ n e r components of F = m tell us only wht the constrint forces must be; the eqution of motion comes from the e θ component: Using (1.6.6), the left sie of (1.6.8) is F e θ = m e θ. (1.6.8) F e θ = νr θ mg sin θ, (1.6.9) while from (1.6.5), (1.6.6), n (1.6.7), the right sie of (1.6.8) is m e θ = m{ẍ cos θ cos ϕ + ÿ cos θ sin ϕ + z sin θ} = m{cos θ cos ϕ[ ω 2 x θ 2 x + (R cos θ cos ϕ) θ 2Rω θ cos θ sin ϕ] + cos θ sin ϕ[ ω 2 y θ 2 y + (R cos θ sin ϕ) θ + 2Rω θ cos θ cos ϕ] + sin θ[ z θ 2 + (R sin θ) θ]}. Using (1.6.4), this simplifies to m e θ = mr{ θ ω 2 sin θ cos θ}. (1.6.10) Compring (1.6.8), (1.6.9), n (1.6.10), we get θ = ω 2 sin θ cos θ ν m θ g sin θ, (1.6.11) R which grees with our erlier equtions (1.6.3). Hmilton s Principle. Next, we will show the equivlence of Hmilton s vritionl principle (1830) n the Euler Lgrnge eqution: δ b L(q, q) = 0 L t q i L q i = 0. The vritionl principle tells us tht the ction integrl is sttionry to perturbtions of the curve going from to b, s in Figure Wht Hmilton s Principle mens precisely is tht the curve q(t) is such tht for ny fmily of curves q(t, ɛ) stisfying the conitions q(t, 0) = q(t) q(, ɛ) = q() q(b, ɛ) = q(b),

7 1.6 Mechnicl Systems 51 q(t) δq(t) q(b) q() Figure The Euler Lgrnge equtions re equivlent to Hmilton s Principle: the ction integrl is sttionry uner vritions of the curve q(t). we hve b ɛ L(q ɛ (t), q ɛ (t)) = 0. ɛ=0 As inicte in the preceing Figure, we write: ɛ q(t, ɛ) = δq(t). ɛ=0 Using this nottion, n ifferentiting uner the integrl sign n using integrtion by prts, Hmilton s Principle becomes 0 = b L(q ɛ (t), q ɛ (t)) = ɛ ɛ=0 = b b L δq + + Lδ q t q q [ ] L L δq + q q ( δq) t, where we use equlity of mixe prtils in interchnging the time erivtive n the ɛ erivtive (tht is, interchnging the overot n the δ opertions. Integrtion by prts then gives 0 = b [ L t q i L ] q i δq t = 0, where we notice tht there is no bounry term since δq vnishes t the enpoints becuse of the conitions q(, ɛ) = q() n q(b, ɛ) = q(b). Since this is zero for rbitrry δq(t), the integrn must vnish. In summry, this rgument shows inee tht Hmilton s Principle is equivlent to the Euler Lgrnge equtions.

8 52 Introuction Externl Forces. We hve shown tht Hmilton s Principle is equivlent to the Euler Lgrnge equtions. In the cse of externl forces F ext (such s frictionl forces or control forces), we moify Hmilton s principle to the Lgrnge- Alembert principle: δ L(q, q)t + F ext δq t = 0 Forces tht come from potentil re conventionlly put into the Lgrngin. Hence, by (externl) we men forces tht re not erive from potentil (such s friction or the propulsion force of rocket). However, if potentil forces were to be inclue s externl forces, then the system remins consistent. The sme rgument tht ws use to show tht Hmilton s Principle is equivlent to the Euler Lgrnge equtions shows tht the Lgrnge Alembert principle is equivlent to the Euler Lgrnge equtions with externl forces: L t q L q = F ext(q, q) For instnce in the rocket exmple, these equtions give: t (m q) = V + F ext These re lso the correct equtions when the mss of the rocket is llowe to chnge with time. Note tht one cnnot just pull the m out of the erivtive sign s one might guess if one nively uses F = m. For the bll in the hoop exmple, if we friction term tht is proportionl to the velocity (s ws expline in the F = m erivtion, this is bit hoc; in fct moeling friction relisticlly is subtle business), we get mr 2 θ = mr 2 ω 2 sin θ cos θ mgr sin θ νr θ, where we regr ν s the coefficient of friction. The Energy Eqution. Next we iscuss the energy eqution for mechnicl system. First suppose tht there re no externl forces, so tht the Euler Lgrnge equtions hol. Define the energy to be E(q, q) = n k=1 L q k qk L(q i, q i ), or, for short, E(q, q) = L q L(q, q). q

9 1.6 Mechnicl Systems 53 Using the Euler Lgrnge equtions, we compute the time erivtive of E with the help of the prouct rule n the chin rule s follows. t E = = n ( L t q k qk + L ) q k qk n ( q k L t q k L ) q k k=1 k=1 = 0. n j=1 L q j qj n j=1 L q j qj Thus, A similr clcultion shows tht when there is n externl force present, then t E = F ext q = Power of externl forces. Notice tht for the simple system with Lgrngin L(q, v) = 1 2 m v 2 V (q), the energy switches the sign of V so we get E(q, v) = 1 2 m v 2 +V (q). Bll in the Hoop n the Simple Penulum. We now look t the bll in the hoop bit more closely. First consier the cse when the hoop is not rotting (tht is, ω = 0) n there is no friction (tht is, ν = 0). In this cse, the eqution of motion becomes tht of the simple penulum: θ + g R sin θ = 0 Note tht for smll oscilltions, in which cse sin θ θ, the eqution of motion simplifies to θ+ g Rθ = 0 which is simple hrmonic oscilltor whose ngulr frequency is g ω pen = R. The phse portrit of the simple penulum is shown in Figure Phse Portrits for the Bll in the Hoop. We write the bll in the hoop s usul s first orer ynmicl system s follows. θ = v v = g (α cos θ 1) sin θ βv, R where α = (R/g)ω 2 n β = ν/m. The equilibrium points of this system re obtine by setting ẋ n v equl to zero. Thus, the equilibrium points correspon to zeros of (α cos θ 1) sin θ. If sin θ = 0, then either θ = 0 or θ = π (plus multiples of 2π). The other equilibrium points occur when α cos θ = 1. There re no other equilibri if 1/α > 1, 2 solutions if 1/α < 1 n one solution if α = 1.

10 54 Introuction θ homoclinic orbit θ = π θ θ = π Figure Phse portrit of the simple penulum. Therefore criticl vlue is when α = 1; i.e., Rω 2 = g. Thus, bifurction occurs when ω = g/r; i.e., interestingly, when the hoop rottes with the sme ngulr velocity s frequency of oscilltions of the simple penulum. The chnge in the phse portrit s α crosses the criticl vlue α = 1 is shown in Figure Aing bit of issiption oes n(t)chnge this picture too much from lrge scle perspective, in the sense tht the bifurction from one to three equilibri still occurs t the sme criticl vlue even in the presence of issiption, but it oes turn the centers into sinks. Off-centere Hoop. So fr we hve consiere the problem when the xis of rottion of the hoop is concentric with its xis of symmetry. When, the rottion xis is slightly shifte by n ɛ the phse portrit will be s shown in Figure Like the symmetric cse, one chnges from one equilibrium insie the penulum homoclinic loop, but unlike the symmetric cse, the one equilibrium oes not splt into three, but rther two new equilibri pper through center-sle bifurction s ω increses. The Legenre Trnsformtion. Now we show how to rewrite the Euler Lgrnge equtions into wht is clle Hmiltonin form. To o this, we introuce the Legenre trnsformtion; nmely we efine the conjugte momentum by p i = L q, introuce the chnge of vribles i (q i, q i ) ( q i, p i ) n ssume (for exmple, vi the implicit function theorem) tht this is legitimte chnge of vribles; tht is, tht it efines, implicitly q i s (smooth) function of (q i, p i ).

11 1.6 Mechnicl Systems 55 θ θ θ θ α = 0.5, β = 0 θ α = 1.5, β = 0 θ α = 1.5, β = 0.1 Figure Phse portrits for the bll in the hoop. No mping, but incresing rottion rte s α increses from α =.05 to α = 1.5 n β = 0 n ing bit of mping when β = 0.1. ω Figure The bll in the off centere hoop n the chnges in its phse portrit s the ngulr velocity is increses. Using this chnge of vribles, introuce the Hmiltonin by n H(q i, p i ) = p i q i L(q i, q i ) i=1

12 56 Introuction Hmilton s equtions for given Hmiltonin re t qi = H p i = H tṗi q i The key link between Lgrngins n Hmiltonins is given in the following theorem: Theorem. The Euler Lgrnge equtions for (q i, q i ) re equivlent to Hmilton s equtions for q, p. Proof. First ssume tht the Euler Lgrnge equtions hol n we will show tht Hmilton s equtions hol (the converse is shown in similr mnner). We compute crefully using the chin rule s follows: H n = q i q + (p i j L q j ) p i p j q j p i = q i, j=1 the two terms cncelling by virtue of the efinition of the momentum. Similrly we clculte the other prtil erivtive of H s follows H n q i = q (p j j L q j q i q j q i L ) δq i j=1 = L t q i = p i t. In going from the first line to the secon, we gin use the efinition of the conjugte momentum long with the Euler Lgrnge equtions n in going to the lst equlity we gin use the efinition of the momentum. Thus we hve estblishe Hmilton s equtions. Exmple (Bll in the hoop) Recll tht for the bll in the hoop, L(θ, θ) = 1 2 mr2 ( θ 2 + ω 2 sin 2 θ) + mgr cos θ From this we clculte p = L/ θ = mr 2 θ n so H(θ, p) = p θ L = mr 2 θ2 L = p2 2mR 2 mgr cos θ 1 2 mr2 ω 2 sin 2 θ

13 1.6 Mechnicl Systems 57 From this Hmiltonin one cn check irectly tht Hmilton s equtions give the sme equtions tht we h before. Note, however, tht this Hmiltonin is not the kinetic plus potentil energy! If one use tht, then one woul get the incorrect equtions. Conservtion of Energy. Lets hve nother look t conservtion of energy from Hmiltonin point of view. Assuming Hmilton s equtions hol, we get using the Chin rule, t H = = = 0 n ( H q i + H ) ṗ i q i p i n ( H H H ) H q i p i p i q i i=1 i=1 Dirichlet s Stbility Theorem. Becuse of conservtion of energy, one cn sometimes use the energy s Lipunov function. Theorem. Consier Hmiltonin H(q i, p i ) n the corresponing Hmiltonin ynmicl system: t qi = H p i = H tṗi q i Let (q i, p i ) be n equilibrium point; i.e., criticl point of H. If the secon erivtive of H s symmetric mtrix, evlute t this equilibrium point, is positive efinite, then the equilibrium is Lipunov stble. This result follows from using conservtion of H: H hs strict minimum t the equilibrium n is conserve. We will return to stbility results using Lipunov functions more generlly lter on. Eigenvlue Theorem. We hve seen vi Lipunov s spectrl theorem the importnt role plye by the istribution of eigenvlues of the lineriztion of system ẋ = f(x) t n equilibrium point. For Hmiltonin (n lso Lgrngin) systems, there is severe restriction on how the eigenvlues cn pper. In fct, the next theorem shows tht the eigenvlues must be istribute symmetriclly not only with respect to the rel xis (lwys true for ny rel system), but lso with respect to the imginry xis. Theorem. As in the previous theorem, consier Hmiltonin system with n equilibrium point. Then,the spectrum of the lineriztion t (q i, p i ) is symmetric with respect to the imginry xis. In prticulr, if the conitions of the preceing theorem hol, tht is, if this criticl point is

14 58 Introuction nonegenerte minimum of H, then the spectrum of the lineriztion must lie on the imginry xis. For exmple, this mens tht in Hmiltonin system, the spectrum t n equilibrium point cn never be totlly in the left hlf plne. The stuent shoul clculte the eigenvlues for the bll in the rotting hoop to verify tht this symmetry oes hol in tht cse. Proof. follows where Let z = (q i, p i ) n write Hmilton s equtions in mtrix form s ż = JDH(z) ( q ż = ṗ) n the 2n 2n mtrix J is given by ( ) 0 1 J =, 1 0 where the ones stn for the n n ientity mtrix. Also, DH(z) = Note these esily verifie properties of J: 1. First of ll, J T J = I (tht is, J is n orthogonl mtrix; in prticulr, in the cse n = 1, it is rottion by 90 o ). 2. Secon, J 2 = Thirly, note tht et J = 1. The lineriztion of the equtions ż = JDH(z) t criticl point z = (q, p) is given by ż = JD 2 H(z)z =: Az, where A = JD 2 H(z) n where D 2 H(z) is the mtrix of secon prtil erivtives of H evlute t the equilibrium point z. We clim tht the ientity JAJ = A T hols. To see this, we use the property J 2 = I to give H δq i H δp i JAJ = J ( JD 2 H(z) ) J = D 2 H(z)J = ( JD 2 H(z) ) T = A T where we hve lso use the fct tht D 2 H(z) is symmetric ue to equlity of mixe prtils, n skew symmetry of J.

15 1.6 Mechnicl Systems 59 The chrcteristic polynomil p of the mtrix A is given by p(λ) = et(a λi) = et(j(a λi)j) = et(jaj + λi) where in going from the first line to the secon, we use the fct tht et J = 1 n in going from the secon line to the thir we use the fct tht J 2 = I. From our clcultion bove, JAJ = A T n so et(jaj + λi) = et(a T + λi) = et(a + λi) = p( λ). Thus, p(λ) = p( λ); tht is, p is n even function of λ. Thus, if λ is n eigenvlue, so is λ, which proves the result. Homoclinic Orbit of the Simple Penulum. Finlly we return to the simple penulum. Here our gol is to clculte the eqution of the hetroclinic orbit. In generl the trjectories cn be foun using elliptic functions. However, it is interesting tht the homoclinic orbit cn be written in terms of elementry functions. Here is the proceure: First of ll, recll tht the equtions (with g/r = 1) re which hve the energy integrl which we cn rewrite s θ + sin θ = 0 E = 1 2 θ 2 cos θ θ = 2(E + cos θ) n so θ = 2(E + cos θ) t = t + c The vlue of E on the homoclinic trjectory equls the vlue of E t the sle point (π, 0), nmely E = 1; in this cse, the integrl is vilble (eg, in tbles of integrls) n is given by θ = 1 ( ) 1 + sin(θ/2) 2(1 + cos θ) 2 log 1 sin(θ/2) We cn lso fix the integrtion constnt by choosing t = 0 to correspon to the point θ = 0 (note tht the point (0, 1), which hs energy E = 1 lies on the homoclinic orbit); from these clcultions, one conclues tht the homoclinic orbit is given by θ(t) = ±2 tn 1 (sinh t).