mgz = mgz. mg mgz (mv) is conservative. His answer is the Lagrangian formulation of mechanics. The potential for d dt

Transcription

1 Ã Ò Ø Ò Ö Ý ÈÓØ ÒØ Ð The Idea: A force s conservatve f t s the (negatve gradent of a scalar-valued functon For example, the gravtatonal force actng on a partcle of mass m near the surface of the earth s f g mgz mg Φ mgz x mgz y mgz z The scalar-valued functon Φ(x,y,z mgz s called a potental (energy for f g The gravtatonal potental Φ(x,y,z mgz s a multple of the alttude z, so the hgher the partcle, the greater ts potental energy Gradents pont n the drecton of greatest ncrease of the potental up, n the case of the gravtatonal potental The mnus sgn n f g Φ says the force pulls n the opposte drecton down, n the case of the gravtatonal potental Not all forces are conservatve Non-conservatve forces nclude forces of frcton and external stmul lke parents pushng a swng Any force that dsspates or supples energy s lkely to be non-conservatve Newton s second law, f d (mv, (1 dt saysthat d dt (mv s the net forcef actng onthe partcle Lagrangeasksn what sense d dt (mv s conservatve Hs answer s the Lagrangan formulaton of mechancs The potental for d dt (mv s the knetc energy These notes derve (non-varatonal Lagrangan mechancs from Newtonan mechancs The dervaton clarfes how the sometmes-confusng gradents are nterpreted The proofs requre only the smplest tools of analyss: the lnearty of the dervatve, the product and chan rules, some elementary lnear algebra, and the equalty of mxed partal dervatves Newtonan framework: Newtonan mechancs s governed by Newton s second law, Equaton (1 The net force f actng on a partcle of mass m changes the momentum p mv The velocty v ẋ s the tme dervatve of poston x the dot means dervatve wth respect to tme When m s constant, the second law s f equals ma, but the general law s f equals ṗ Suppose, n an effort to solve a partcular problem, we make a change of varables x G(q Physcsts call q the generalzed coordnates and x the system vector Example: Asphercalpendulum s apont massm, called the bob, suspended by a strng (or rgd but massless rod of fxed length r The strng restrcts the bob s moton to a sphere, suggestng a change of varables to sphercal coordnates x rcos(θsn(ϕ r sn(θ sn(ϕ rcos(ϕ G( θ ϕ θ The parameter q s -dmensonal; the system vector x ϕ s 3-dmensonal The chan rule says that the bob s velocty s X Z θ ϕ Fgure 1: Sphercal Pendulum v ẋ rsn(θ θsn(ϕ+rcos(θcos(ϕ ϕ rsn(θsn(ϕ rcos(θcos(ϕ rcos(θ θsn(ϕ+rsn(θcos(ϕ ϕ r cos(θ sn(ϕ r sn(θ cos(ϕ θ ( ϕ rsn(ϕ ϕ r sn(ϕ Õ Ë(Õ 1 r m Y

2 The matrx S S(q s the Jacoban matrx of partal dervatves of G: x x θ ϕ rsn(θsn(ϕ rcos(θcos(ϕ y y S(q D Õ G(q r cos(θ sn(ϕ r sn(θ cos(ϕ r sn(ϕ θ z θ ϕ z ϕ Under the change of varables, v S q and Newton s second law, Equaton (1, becomes Knetc Energy as Potental f d (MS q, (3 dt where M s the mass matrx The sphercal pendulum, wth ts sngle pont mass, has M mi 3, the scalar m tmes the 3 3 dentty matrx M need not be dagonal, but, n Newtonan mechancs, t s symmetrc The key 1 to the Newtonan framework s to left-multply both sdes of Equaton (3 by S T : S T f S T d (MS q (4 dt The pont s to elmnate the forces of constrant on the left sde The argument goes lke ths: 1 S S(q maps generalzed veloctes q to system veloctes v ẋ S q, so The range (or column space of S(q les n the tangent plane to the constrant surface at x G(q, so 3 The columns of S(q are orthogonal to any vector normal to the constrant surface at x G(q In matrx language, S T n for any vector n normal to the constrant surface at x G(q In Newtonan mechancs, constranng forces are normal to the constrant surface For example, the sphercal pendulum s strng pulls up n a drecton parallel to x normal to the sphere Snce S T x rsn(θsn(ϕ rcos(θsn(ϕ r cos(θ cos(ϕ r sn(θ cos(ϕ r sn(ϕ and snce the force of constrant f c s parallel to x, S T f c for the force of constrant f c rcos(θsn(ϕ r sn(θ sn(ϕ rcos(ϕ, Ths assumpton s called the prncple of vrtual work, or Newton s thrd law, or Postulate A Fnally, forces add vectorally 3 n Newtonan mechancs We decompose the net force actng on the partcle nto appled forces f a, such as gravty, and constrant forces f c, such as the tenson n the strng: f f a +f c so S T f S T f a +S T f c S T f a + Only the appled force remans the force of constrant s elmnated Equaton (4 becomes S T f a S T d dt ( MS q, (5 wth (the known f a replacng (the unknown f on the left sde It would be nce to see the dervatve of somethng rather than S T tmes the dervatve of somethng on the rght sde of Equaton (5 The product rule s just the tcket: S T f a d dt ( S T MS q ṠT MS q (6 1 see Newton VarPrncpdf Cornelus Lanczos, The Varatonal Prncples of Mechancs, 4th ed, Dover, 197, p 76 3 See the notes n footnote 1 for a dscusson

3 Equaton (6 s the Newtonan framework The dfferentated varable S T MS q s the generalzed momentum, and s confusngly denoted p, the same notaton as the (non-generalzed momentum The components of S T f a are called generalzed forces, even when they are not measured n unts of force Example (sphercal pendulum, contnued: The forces actng on the bob are domnated by the (appled force of gravty pullng down, and the (constrant force of the strng pullng up, toward the pvot The downward force of gravty s easly modeled by f a, mg where m s the mass of the bob and g 98 m s The generalzed force s therefore S T f a rsn(θsn(ϕ rcos(θsn(ϕ r cos(θ cos(ϕ r sn(θ cos(ϕ r sn(ϕ mg Physcsts recognze mgr sn(ϕ as the magntude of the torque The sphercal pendulum s (generalzed momentum s mgr sn(ϕ (7 p S T MS q rsn(θsn(ϕ rcos(θsn(ϕ m rsn(θsn(ϕ rcos(θcos(ϕ m r cos(θ sn(ϕ r sn(θ cos(ϕ r cos(θ cos(ϕ r sn(θ cos(ϕ r sn(ϕ m r sn(ϕ mr sn mr ϕ The last term n Equaton (6 nvolves the tme dervatve of S, θ ϕ Ṡ rcos(θ θsn(ϕ rsn(θcos(ϕ ϕ rsn(θ θcos(ϕ rcos(θsn(ϕ ϕ rsn(θ θsn(ϕ+rcos(θcos(ϕ ϕ rcos(θ θcos(ϕ rsn(θsn(ϕ ϕ (9 r cos(ϕ ϕ The last term s therefore Ṡ T MS q mr sn(ϕcos(ϕ ϕ mr sn(ϕcos mr sn(ϕcos θ ϕ mr sn(ϕcos Equatons (7, (8, and (1 ft nto Equaton (6 s framework to gve the equatons of moton d mgr sn(ϕ dt mr sn mr ϕ mr sn(ϕcos (8 (1 (11 The pont s that the framework provdes equatons of moton from whch further analyss begns Lagrange smplfes the framework by recognzng each of the three vectors n Equaton (11 as a gradent Gradents: We have already seen that the constant force of gravty (near the surface of the earth s a gradent of the gravtatonal potental Φ(x, y, z mgz All constant forces are gradents, of course: c 1 c (c 1x 1+c x + x 1 (c 1x 1+c x + x Ü(c 1 x 1 +c x + 3 } {{ } Φ(Ü

4 In the language of matrx multplcaton, Ü ( c T x c, d whch s just the multdmensonal verson of dxmx m It wll be convenent to wrte Φ(x nstead of Φ(x 1,x,, and to use vectors as the subscrpt n Ü Lagrangan mechancs uses two gradents: Ü Φ(x,y and Ý Φ(x,y The frst means the gradent wth respect to the x varables, holdng the y varables constant, and the second s smlarly defned The mportant gradent for Lagrangan mechancs s the gradent of a quadratc potental Quadratc functons appear n lnear algebra n second-dervatve tests for extrema, and n least-squares problems Example: If A s a symmetrc matrx, constant n x, then Φ(x 1 xt Ax 1 x A j x j s a quadratc form Its partal dervatve wth respect to x 1 s, by the product rule, Φ(x 1 x 1 j,j A 1j x j + x A 1 The frst sum s the frst rowof Ax and the second s the frst rowof A T x Snce A T A, the partal dervatve s just the frst row of Ax The gradent assembles all the partal dervatves n a vector, so ( 1 Ü xt Ax Ax (1 The pont s that Ax s the gradent wth respect to x of the quadratc potental 1 xt Ax whenever A s symmetrc Ths s smply the mult-dmensonal verson of ( d 1 dx mx mx Remark: The gradent and the dervatve are transposes of each other: D Ü Φ(x x 1 Φ(x x Φ(x ( T x n Φ(x Ü Φ(x, and they serve dfferent purposes The dervatve must satsfy the chan rule d dt Φ(x(t D ÜΦ(x(t d ẋ 1 (t dt x(t D ÜΦ(x(t ẋ (t, so D Ü Φ(x must be a row vector The gradent s a drecton n the system space, so t must be a column vector Remark: The chan rule for gradents s awkward, and not smply because transposes reverse the order of multplcaton If x G(q, then the chan rule for dervatves says D Õ Φ(G(q D Ü Φ(xD Õ G(q Take transposes of both sdes for the chan rule for gradents: Õ Φ(G(q D Õ G(q T Ü Φ(x 4

5 Note that the matrx D Õ G(q T s not a gradent because gradents are vectors In other words, whle the chan rule for dervatves nvolves only dervatve matrces, the chan rule for gradents nvolves both (vector gradents and (matrx dervatves Ths techncal detal suggests a strategy for provng theorems about gradents: prove the correspondng theorem about dervatves, then take transposes Remark: The dervatve verson of the gradent of a quadratc form s ( 1 D Ü xt Ax x T A, (13 provded A s symmetrc and constant n x In ths case, we ve reversed the strategy, takng the transpose of the gradent n Equaton (1 to get the dervatve Lagrangan Framework: Our goal s to dentfy each of the three vectors S T f a, p S T MS q, and ṠT MS q from Equaton (6 as a gradents Consder frst the generalzed momentum p S T MS q Snce M s symmetrc, so s A S T MS If x q, then by Equaton (1, the ant-gradent of Ax s S } T {{ MS } q Ü ( 1 Õ qt S T MS q, provded A s constant n x But A S T (qms(q s constant n q provded M s Indeed, the pont of factorng the chan rule d dtx(g(q S(q q was to separate the q s from the q s We therefore have Theorem: If the (symmetrc mass matrx M s ndependent of the generalzed veloctes q, then p S T 1 MS q Õ qt S T MS q T(Õ, Õ In ths dfferentaton, all components of q wthout dots are held constant Remark: The quadratc form T T(q, q s the knetc energy of the system Snce v S q, T 1 vt Mv In Englsh: The knetc energy s one-half m v squared Example (sphercal pendulum, contnued: Equaton (8 says the sphercal pendulum s generalzed mr momentum S T MS q sn mr The knetc energy of the sphercal pendulum s therefore ϕ T 1 qt ( S T MS q 1 θ ϕ The gradent wth respect to q Õ T θ 1 1 ϕ mr sn mr 1 ( mr sn +mr ϕ ϕ θ, treatng ϕ (and θ as constant, s ndeed ϕ (mr sn +mr ϕ (mr sn +mr ϕ 5 mr sn mr S T MS q ϕ

6 Remark: The dervaton and the example llustrate the formal computaton of the gradent Õ holdng all the undotted components of q constant Ths manpulaton s the source of endless confuson, but the dervaton exposes the correct computaton: Hold all varables wthout dots constant and dfferentate wth respect to the dotted varables Remark: The physcst wll very lkely see the knetc energy rather than compute t The component of the velocty n the ϕ-drecton s r ϕ, and the component n the θ-drecton s rsn The two drectons are orthogonal, so Pythagoras s theorem says 1 m v 1 ( m r sn(ϕ θ +r ϕ, wthout explctly computng S Ths shortcut s one of the ways the Lagrangan formulaton smplfes the Newtonan framework Once we ve commtted to lookng for gradents, t s easy to see that S T MS q s the gradent wth respect to q of the quadratc form T(q, q The next gradent n Lagrange s formulaton, ṠT MS q, s beyond my ablty to see If ts ant-gradent s obvous to you, please show me how to thnk about t Perhaps the best startng pont s to apply the strategy mentoned earler: take transposes and ask s q T S T MṠ a dervatve of some scalar? The crux of the computaton s Ṡ, whch we compute column-by-column so, by the chan rule, Column j of S q j G(q q j x, column j of Ṡ d x dtq j q 1 ( x q ( x q n q j q n q j q 1 q j ( x q q n q j ( x q n, provded the q are constant under dfferentaton wth respect to the q k By the equalty of mxed partal dervatves, column j of Ṡ x q x q n S q q j q 1 q n q j Assemble the columns of Ṡ nto the matrx Ṡ DÕ S q D Õ v (14 Example (sphercal pendulum, contnued From Equaton (, the pendulum s velocty s S q v rsn(θ θsn(ϕ+rcos(θcos(ϕ ϕ rcos(θ θsn(ϕ+rsn(θcos(ϕ ϕ, rsn(ϕ ϕ so θ D Õ S q ( rsn(θ θsn(ϕ+rcos(θcos(ϕ ϕ ϕ ( rsn(θ θsn(ϕ+rcos(θcos(ϕ ϕ θ ( rcos(θ θsn(ϕ+rsn(θcos(ϕ ϕ ϕ ( rcos(θ θsn(ϕ+rsn(θcos(ϕ ϕ θ ( rsn(ϕ ϕ ϕ ( rsn(ϕ ϕ ( rcos(θ θsn(ϕ rsn(θcos(ϕ ϕ ( rsn(θ θcos(ϕ rcos(θsn(ϕ ϕ ( rsn(θ θsn(ϕ+rcos(θcos(ϕ ϕ ( rcos(θ θcos(ϕ rsn(θsn(ϕ ϕ, r cos(ϕ ϕ 6

7 n agreement wth the pendulum s Ṡ n Equaton (9 Our search for ant-gradents therefore reduces to s q T S T MD Õ S q a dervatve of some scalar? Yes, by the chan rule: Equaton (13, the dervatve of a quadratc form, says so the chan rule says q T S T M D Ë Õ ( 1 T S q M S q, ( 1 T q T S T MD Õ S q ( 1 D Ë Õ S q M S q D Õ S q D Õ qt S T MS q Take transposes for the gradent theorem: Theorem: If (the symmetrc M s ndependent of q, and f S has contnuous partal dervatves (so the mxed partal dervatves are equal, then ( 1 Ṡ T MS q Õ qt S T MS q Remark: The computaton s subtle For example, the dervatve D Ë Õ s computed wth respect to the components of the vector S q, but the expresson D Ë makes no sense because S s a matrx, not a vector Even f q s constant wth respect to dfferentaton, t doesn t factor out of the expressons Lagrange s gradents turn the Newtonan framework n Equaton (6 nto S T f a d dt ÕT(q, q Õ T(q, q (15 Equaton (15 s correct no matter what forces f a are appled to the system If, however, f a happens to be conservatve, then the framework smplfes consderably: Theorem: If f a Ü Φ(x, then S T f a Õ Φ(G(q Remark: In Englsh, the theorem says that f f a s a gradent, then so s S T f a More precsely, f f a s a gradent wth respect to x, then S T f a s a gradent wth respect to q Proof: Snce x G(q, the chan rule says D Õ Φ(G(q D Ü Φ(xD Õ G(q Take the negatve transpose of both sdes to complete the proof //// Example: The gravtatonal potental Φ(x mgz s, n sphercal coordnates, Φ(G(θ, ϕ mgr cos(ϕ The gradent wth respect to the generalzed coordnates q s n agreement wth Equaton (7 θ Õ Φ(q mgrcos(ϕ ϕ mgrcos(ϕ 7 Ë mgr sn(ϕ,

8 Suppose we decompose the appled forces f a nto conservatve and non-conservatve contrbutons: f a Õ V(q+f n, where V(q Φ(G(q s the potental (energy expressedn the generalzedcoordnatesq, and f n s whatever force s left over Accordng to our conventon, Õ V(q because V(q s ndependent of q Consequently, Lagrange s framework, Equaton (15, s S T f n d dt Õ ( ( T(q, q V(q Õ T(q, q V(q (16 Defne the Lagrangan as L(q, q T(q, q V(q (17 Lke the knetc energy, the Lagrangan s a scalar functon of the generalzed coordnates and veloctes Lagrange s framework smplfes to S T f n d dt ÕL(q, q Õ L(q, q (18 Examples: The followng examples llustrate Lagrange s technque for some smple systems Example: The sphercal pendulum s Lagrangan, Equaton (17, s so the gradents are ( θ θ L, 1 ( mr sn +mr ϕ mgrcos(ϕ, (19 ϕ ϕ Õ L(q, q Õ L(q, q L θ L ϕ L θ L ϕ mr sn(ϕcos +mgrsn(ϕ mr sn mr ϕ If f n no one s pushng on the pendulum the equatons of moton are d dt mr sn mr ϕ mr sn(ϕcos +mgrsn(ϕ n agreement wth Equaton (11 (Note that Equaton (11 s gravtatonal force has moved from the left sde to the rght sde here If f n µv s a frctonal force proportonal to the velocty, but n the opposte drecton, then S T f n S T ( µv µs T S q µ In ths case, Lagrange s formulaton, Equaton (18, s µ r sn r d ϕ dt mr sn mr ϕ r sn r ϕ, mr sn(ϕcos +mgrsn(ϕ 8

9 Example: A seesaw s loaded wth masses m at (x,y Parameterze the (-dmensonal system usng polar coordnates, measurng the angle θ from the horzontal Take r < for masses to the left of the fulcrum, and r > for masses to the rght The system vector x and the system velocty v ẋ are x 1 y 1 x x n y n r 1 cos(θ r 1 sn(θ r n cos(θ r n sn(θ r 1 sn(θ r 1 cos(θ and v θ r n sn(θ Õ r n cos(θ Ë θ m (x m,y 1 (x 1,y 1 Fgure : Seesaw m 3 (x 3,y 3 The mass matrx s dagonal, so the knetc energy s r 1 sn(θ T m 1 r 1 sn(θ T 1 r 1 cos(θ m θ 1 r 1 cos(θ θ r n sn(θ m n r n sn(θ r n cos(θ m n r n cos(θ 1 ( m1 r1 + +m n r θ n The physcst, of course, computes the system s knetc energy by summng 1 m v 1 m r θ, wthout explctly parameterzng x The gravtatonal potental energy of mass s m gy m gr sn(θ The system s gravtatonal potental s the sum of the masses potentals Check that the gravtatonal force actng on the system s m 1 g f g m n g x 1 m gy y 1 m gy x n m gy y n m gy Ü m gy and that S T f g m gr cos(θ θ m gr sn(θ The Lagrangan for the seesaw s therefore ( ( 1 L T V m r θ m gr sn(θ Lagrange s equaton of moton s ( ( S T f n d m r θ m gr cos(θ dt Remark: The sums are named m M the total mass or the zeroth moment m r M 1 the frst moment m r r the center of mass M m r I the second moment or moment of nerta 9

10 Wth ths notaton, the equatons of moton are S T f n I θ M rcos(θ ( Remark: The system need not be dscrete If the seesaw has densty ρ(r, then the dscrete sums turn nto ntegrals: M ρ(rdr r 1 rρ(rdr I r ρ(rdr M Wth ths notaton, Equaton ( s stll the equaton of moton In terms of the pcture, the masses n Fgure ( may be spread out along the seesaw n any way we please The seesaw tself need not be massless, and the masses need not be downward-pontng trangles Systems n whch the mass s dstrbuted contnuously rather than dscretely are called contnuous meda Example: A double pendulum s a system wth one pendulum s bob beng the second pendulum s pvot A double pendulum swngng n a plane s parameterzed by the two angles made wth the vertcal: so x r 1 sn(θ 1 r 1 cos(θ 1 r 1 sn(θ 1 +r sn(θ r 1 cos(θ 1 r cos(θ r 1 cos(θ 1 r v 1 sn(θ 1 θ1 r 1 cos(θ 1 r cos(θ θ r 1 sn(θ 1 r sn(θ Õ The mass matrx s dagonal, so the knetc energy s Ë r 1 θ 1 m 1 r θ m Fgure 3: Double Pendulum r 1 cos(θ 1 T 1 θ r 1 θ 1 sn(θ 1 r 1 cos(θ 1 r cos(θ r 1 sn(θ 1 r sn(θ 1 θ 1 θ 1 T (m 1 +m r1 m r 1 r cos(θ θ 1 m r 1 r cos(θ θ 1 m r ( (m 1 +m r 1 θ 1 +m r 1 r cos(θ θ 1 θ 1 θ +m r θ The gravtatonal potental for the system s m 1 r 1 cos(θ 1 m 1 r 1 sn(θ 1 m r 1 cos(θ 1 r cos(θ m r 1 sn(θ 1 r sn(θ θ1 θ V m 1 gy 1 +m gy m 1 gr 1 cos(θ 1 m g(r 1 cos(θ 1 +r cos(θ (m 1 +m gr 1 cos(θ 1 m gr cos(θ θ1 θ If M m 1 +m, then the Lagrangan s L T V 1 ( Mr1 θ 1 +m r 1 r cos(θ θ 1 θ 1 θ +m r θ +Mgr 1 cos(θ 1 +m gr cos(θ Lagrange s equatons of moton are therefore S T f n d Mr 1 θ1 +m r 1 r cos(θ θ 1 θ dt m r 1 r cos(θ θ 1 θ 1 +m r θ 1 m r 1 r sn(θ θ 1 θ 1 θ Mgr 1 sn(θ 1 m r 1 r sn(θ θ 1 θ 1 θ m gr sn(θ

11 Addenda: Knetc Energy as Potental Conservaton Laws: Some conservaton laws follow mmedately from the Lagrangan framework For example, the top row of Equaton (11 reads so the quantty n parentheses s constant: d dt ( mr sn, mr sn constant The physcst recognzes the left sde as the angular momentum, so the sphercal pendulum obeys the law of conservaton of angular momentum whenever no non-conservatve forces push t In general, a (generalzed coordnate q s gnorable or cyclc f the Lagrangan L(q, q s ndependent of q In ths language, θ s an gnorable coordnate of the sphercal pendulum s Lagrangan, Equaton (19 If q s gnorable, then row of the equatons of moton are ( S T f n d dt L(q, q q L(q, q q }{{ } If, n addton, ( S T f n, then the system obeys the conservaton law L(q, q q constant Movng Coordnates: The Lagrangan framework apples even f the coordnate system s movng For example, a laboratory on the surface of the earth may feel statonary, but the earth s rotatng If system vector x G(q,t depends (explctly on tme, then by the chan rule, v ẋ D Õ G(q,t Ë(Õ,t q+d t G(q,t S q+s (Õ,t In ths notaton, the uppercase S s a matrx; the lowercase s s a vector Furthermore, both S and s are ndependent of q, although both may depend on q and t Newton s second law, Equaton (1 (wth the mass matrx M replacng the partcle mass m, s f d dt ( M(S q+s The only physcs argument n the dervaton s how to elmnate forces of constrant The correct technque s to left multply both sdes by S T Use the product rule to fnd the Newtonan framework S T f a d dt ( ( S T M(S q+s ṠT M(S q+s The knetc energy s T(q, q,t 1 vt Mv 1 (S q+st M(S q+s The part of Equaton (14 that reads Ṡ D Õv s stll true (exercse for the reader, but v has changed, so Ṡ D Õ v D Õ (S q+s 11

12 All that remans s to check that, by the (gradent chan and product rules, Knetc Energy as Potental Õ T(q, q,t ṠT M(S q+s, Õ T(q, q,t S T M(S q+s and Consequently, Lagrange s framework, Equaton (16, remans vald even f G(q, t depends explctly on t Example (rotatng earth: Let x l be coordnates n a laboratory fxed on the surface of the rotatng earth Then the coordnates n space (gnorng the moton relatve to the sun and stars are cos(ωt sn(ωt x sn(ωt cos(ωt x l } {{ 1 } Ê(t where ω π s 1 The Newtonan reference frame the coordnates n whch Newton s law holds s the frame of x (not x l The velocty s, by the product rule, The knetc energy of the system s v R(tẋ l +Ṙ(tx l sn(ωtω cos(ωtω R(tv l + cos(ωtω sn(ωtω x l cos(ωt sn(ωt R(tv l + sn(ωt cos(ωt ω R(t (v l +ωjx l T 1 Snce M mi and R(θ s orthogonal, Ê(t ( T v l +ωjx l R(t T MR(t (v l +ωjx l T 1 m ( v l +ωjx l T ( v l +ωjx l 1 m ( v l +ωv T l Jx l +ω x T l J T Jx l Take q x l and, therefore, q v l Then S R(t and s R(tωJx l and the equatons of moton are R(t T f a d dt m(v l +ωjx l m ( ωjv l +ω J T Jx l  x l d dt mv l +mωjv l +ω J T Jx l Physcsts wrte ωj, the cross product wth the vector, the vector angular velocty The ω ω frst term on the rght s the rate of change of momentum n the laboratory coordnates The second term on the rght s the Corols acceleraton, and the thrd s the centrpetal acceleraton It s straghtforward to derve the same equatons of moton from the Newtonan framework, but the pont s that Lagrange s formulaton s vald n movng coordnates, too Note that ω s 1 t s no wonder we don t feel the earth rotatng 1