Notes on Analytical Dynamics

Transcription

1 Notes on Analytcal Dynamcs Jan Peters & Mchael Mstry October 7, 004 Newtonan Mechancs Basc Asssumptons and Newtons Laws Lonely pontmasses wth postve mass Newtons st: Constant velocty v n an nertal frame (e, no acceleraton) Newtons nd: F = d(mv)=dt Newtons 3rd: Acton has reacton solated partcles: m v + m v = p =) F + F = 0 Practcal Pendulum Equatons We have a base acceleraton a 0 = x 0 + y 0 j and angular acceleraton k = _! () _k =! Ths mples that a B = a 0 + a Bj0 ; wth a Bj0 = _! R +! (! R) : 3 Energy We de ne T = Z xf x 0 F (x)dx = Z xf x 0 rv (x)dx = mv f mv 0; where we use T = mv = 0 as the knetc energy, and have F (x) = r x V (x) for the potental energy for conservatve systems From ths we also realze that V (x 0 ) + T (x 0 ) = V (x f ) + T (x f ) = const Furthermore, we have mx = F (x)

2 4 Langrangans from Energy From Knetc Energy, we can derve the generalzed equatons of moton Assume we have T = 0:5 P n = m _x as knetc energy for a system of partcle, and the postons as functons of generalzed coordnates, e, x = x(q) In ths case, we also = = _q k m _x = X m _q k = X _x ; ; where the later part s only true for holomonc constrants (e, constrants do only depend on the generalzed postons and tme or, equvalently, x = x (q ; : : : ; q n ; t)) for whch we have dot-cancellaton" When d erentatng the later of the two wth respect to tme, we obtan d = _q k + X _x : When addng up these equatons, we realze = _p = X _q k = k ; where s generalzed force We can repeat the same excercse for any V (q), where we obtan k k _q k = 0 When de nng the Lagrangan ths mples that d dt L = _q = 0, whch s equvalent to sayng that the force derved from V s equal to the one derved from T 5 Duraton of Moton From dx m + V (x) = h; dt we can nfer the duraton of movement for two ponts, e, Z tf r Z m xf dx T = dt = p t 0 h V (x) If T = g(x 0 ; x f ; h) s nvertable, we have x = g (x 0 ; t t 0 ; h) Its usually ntractable x 0

3 An addtonal representaton s gven by V (q) = V 0 (q)+"v (q), whch mples that T (h) = T 0 (h) + "T (h) also has 6 Small Oscllatons Z q0 T 0 (h) p m d p h V0 (q)dq; dh q 0 T (h) p m d Z q0 p dh h V0 (q) dq: q 0 When lnearzng 0:5m _x + V (x) = h 0 around an equlbrum pont x 0 wth a small devaton x, we obtan 0:5m _x +V (x 0 +x) = h 0 +h, and subtractng the two yelds m _x x = h: x0 When comparng ths equaton to a lnear oscllator, t yelds the frequency and ampltude! V ; x0 s h A = V 00 (x 0 ) : 7 Phase planes We can draw phase plane motons usng the potental functon We have have the followng rules, llustrated n Fgure Endponts on the x-axs have V (x) = h These are the ampltudes of the oscllaton Endponts on the _x-axs have T ( _x; x) = h A hll s a seperatrx on the axs, a valley an attractor Hamltonan Dynamcs Basc Idea The basc dea of Hamltonan Dynamcs s to replace the varables x by the generalzed coordnate q and _x by the conjugate momentum p, where x = q, p = m _x: 3

4 Fgure : Ths gure demonstrates how to obtan a phaseplane moton from a potental functon n a conservatve system For example, for the Hamltonan H (q; p; t) = p =(m) + V (q; t), where we see that _q = (q; p; t) = (q; p; t) _p = F (q; t) = When d erentatng a tme-nvarant H (q; p) wth respect to tme, we see that dh (q; p) (q; p) (q; p) dp = dt (q; (q; p; @H (q; p; = 0; e, that all trajectores have constant Hamltonans or are on the soclnes Understandng Hamltonans A quantty s consdered conserved" f t s constant over tme whle the moton proceed, e, t s ndependent of the generalzed veloctes _q k For example, the 4

5 energy E = V + T s conserved We can express ths n general as the d erence of the part of the Langrangan whch depends explctly on _q k and the whole langrangan, e, H _q k _q k k By d erentatng, we obtan dh ; whch mples that H s only conserved f tme does not appear explctly n the Lagrangan Addtonally, we have dh : 3 Area Preservaton When lnearzng the resultng equatons from the Hamltonan, we obtan q(t) = q(t 0 + t) = q(t 0 ) + dq dt t + O(t ) = q(t 0 ) t + O(t ); {z } f(q;p;t) p(t) = p(t 0 + t) = p(t 0 ) + dp dt t + ) = p(t 0 ) t + O(t ): {z } g(q;p;t) Now, we can also lnearze dp and dq, and obtan dp(t) = df df dq + dq dp dp = dq(t) = dg dg dq + dq dp dp dq + dq dp: When denoted as dp(t) = dq(t) " @q # {z } A dp(t0 ) ; dq(t 0 ) we realze that det A = + O(t ), e, that the area of any pont n state space s constant and we have area preservng mappng Ths area conservaton s equvalent to Energy conservaton, see Fgure 5

6 Fgure : Ths gure llustrates the conservaton of the area n (q; p) space whch s equvalent to the conservaton of energy 4 Hamltonans and Lagrangans The connecton between fx; _xg and fq; pg s not always a smple scalng operaton but are n fact de ned by some operatons We de ne the _q = u, and p _q k From H X k _q k L = X k u k p k L: Ths yelds for a one partcle system the equaton L (q; u; t) = pu H(q; p; t): We realze that u = (L + H)=p Ths transformaton s llustrated n Fgure 3 Furthermore, we realze _q = : The Lagrangan s de ned as an functon L(q; u; t) n terms of poston and the slope of the path wth respect to q The Hamltonan s de ned as a functon H(q; p; t) n terms of the poston and the momentum Ths s llustrated n Fgure 4 If you are gven a Hamltonan such as H(p) = p =(m), you can always compute the Lagrangan whch would be L(u) = pu H(p) = mu = = m _q = as u = p=m () p = mu n ths example 6

7 Fgure 3: Ths Fgure llustrates the transformaton of a Lagrangan nto a Hamltonan 3 Systems of Partcles 3 Momentum The moton of a system or partcles can be descrbed n terms of the moton X c of the center of mass (e, the translaton) and the rotaton around the center of mass If we de ne y = x X c, and M = P m, we can obtan L 0 = MX c _X c + X m y _y = X m x _x ; where P m x = MX c 3 Torques The torques around the the center of mass s gven by nx c = y F = d dt =! X m y _y = _L c ; 7

8 Fgure 4: Ths gure llustrates the transformaton of a Hamltonan nto a Lagrangan and by z0 = r M X c + L _ c ; where M X c = P n = F, and r = z 0 X c Ths also yelds z0 = Mr z 0 + _L z0 ; where L z0 represents the angular moment around z 0 Mr z 0 s zero f z 0 = 0, or f r = 0, or f r k z 0 33 Knetc Energy The knetc energy s gven by T = X m _x _x = M _X c _X c + X m _y _y Note that 0:5 P m _y _y = 0:5 P m r! = 0:5I!, and therefore T = M _X c + I! : The rotatonal part s gven by T rot = 0:5!L 0 = 0:5! P m r v 8

9 34 General Representaton The general representaton of a system of partcle s gven by Mx(t) = F(x; _x; t) + F c ; where F represents the external forces whch create work, and F c represents the constrant forces Ths gves us 3n equatons but we have equally many new unknowns as we do not know the constrant forces Let us assume that we are gven some (even non-holomonc) m constrants ' (x; _x; t) = 0; then we can d erentate these and obtan where A s a 3n by m matrx 35 General Soluton A(x; _x; t)x = b(x; _x; t); In general, we the unconstraned acceleraton Ma = F, and the constrant forces F c = M += AM = + (b Aa) combned, ths yelds the fundamental equaton of moton x(t) = M F + M += AM = + b AM F : We can nd M = by ful llng the followng three condtons () M = M = = I, () M = M = = M, and () M = MM = = I For M = dag(m ; : : : ; m n ), we also have M = = dag(= p m ; : : : ; = p m ) For a system wth M = mi, we can smplfy the fundamental equaton to x(t) = m F + b m A m AF : Note that e = (b Aa) s lke an error n a control equaton 36 Interpretaton of the General Soluton We have two nterpretatons for ths result Least-Squares Soluton The general soluton can be nterpreted as the soluton to the mnmzaton problem under the constrant Ax = b mn x (x a) T M (x a) ; 9

10 Nature as a controller We realze that nature s bascally a controller x = x a = M = AM = + (b Aa) = K e; wth K = M = AM = + The motor command s F c = M += AM = + (b Aa) = K e; wth K = M += AM = + 37 Langrange s Vew of Lfe Determne the unconstraned moton from Mx u = F, where M s a n by n matrx and x, F are n vectors Determne the constrants n the form Ax = b where A s a n by m Matrx wth rank A = r 3 Several acceleratons x 0,, x n r are possble acceleratons as they ful ll Ax = b where exactly n r of the n r + are lnear ndependent 4 When subtractng the x 0 we obtan n r vrtual dsplacements x = x x 0 ; for whch Ax = 0 These dsplacements do not create work, e, we have F T c x = 0 (D Alemberts prncple) These can also we obtaned by rewrtng Ax = 0 as creatng vectors lke 3 v (v r+ ; : : : ; v n ) v = v r (v r+ ; : : : ; v n ) v r+ ; v n 0

11 and then obtanng n r vectors as 3 v (0; : : : ; 0; v r+ ; 0; : : : ; 0) v r (0; : : : ; 0; v r+ ; 0; : : : ; 0) 0 x = : Ths gves us Mx = F + F c ; Ax = b; F T c x = 0; whch allows us to determne the n unknowns n x, and the n n F c 38 Gauss vew of Lfe (Le Chatelers Prncple) Determne the unconstraned moton from Mx u = F, where M s a n by n matrx and x, F are n vectors Nature consders all possble acceleratons A ~x = b whch comply wth the constrants 3 Nature pcks the one possble acceleraton whch mnmzes the Gaussan T G(~x) = ~x x u M ~x a, e, Nature s solvng a global mnmzaton problem at each nstant of tme or Nature takes the mnmum devaton from the unconstraned moton