Physcs 106a, Caltech 11 October, 2018 Lecture 4: Constrants, Vrtual Work, etc. Many, f not all, dynamcal problems we want to solve are constraned: not all of the possble 3 coordnates for M partcles (or 3M coordnates, total) can vary ndependently. There are (a potentally very large) number of constrants, expressble as forces of constrant; and usually we are not nterested n all the forces that mantan the constrants; we re only nterested n the unconstraned moton. For example, any macroscopc body contans of order Avogadro s number of pont-lke atoms (and even these may be broken down further), whereas often n mechancs we want to consder rgd body dynamcs n whch the sze and shape of the macroscopc body do not change. We wsh to study the constraned dynamcs n terms of sx coordnates (poston and orentaton of body) not n terms of Avogadro s number of coordnates movng under atomc constrant forces. Even f the sze changes, as when the body also vbrates, we want a macroscopc descrpton, not an atomc one. Fnte-element analyss s another example of a large (but not Avagadro-large) number of nteractng degrees of freedom. To make the problem manageable, we would lke to re-cast Newton s Second Law ( F = p, wth = 1... M) so that we consder only the subset of those equatons that correspond to drectons of moton that are unconstraned. We want to have the freedom to choose those k = 1... N unconstraned drectons of moton (degrees of freedom) n a way that fully satsfes (most or all of) the constrants, gve convenent descrptons of the resultng moton, and be able to treat them as ndependent varables. We want to project Newton s Law nto only those unconstraned degrees of freedom. The method of vrtual dsplacement, vrtual work, vrtual generalzed forces, and D Alembert s prncple, wll get us there. Wth these deas, we wll also go back and deduce the Lagrangan equatons of moton from the Newtonan formulaton, when the forces are conservatve. Along the way, we learn a number of other prncples, formulate the dea of constrants carefully, and learn more about when the Lagrangan approach s vald. Constrants We magne startng wth a prmtve descrpton of the dynamcs n terms of Newtonan equatons of moton for M elementary objects (pont partcles, or small peces of a sold body that can be treated as ponts, etc.) wth poston vectors r, = 1... M. The confguraton of the system may be defned by the 3M Cartesan components of the partcle postons. Newton s equatons contan all the forces actng, external and nterpartcle. Constrants may restrct the dynamcs, so that the dynamcs can be descrbed n terms of fewer than 3M varables. Ths leads to the noton of the number of degrees of freedom: The number of degrees of freedom s the number of coordnates that can be ndependently vared n a small dsplacement. Another way of sayng ths s the number of ndependent drectons n whch the system can move from any gven ntal condton. For constraned dynamcs the number of degrees of freedom wll usually be less than 3M. I take ths defnton from Taylor s book. Hand and Fnch and Goldsten et al. are less precse n ther usage of ths term. 1
For constraned dynamcs we can often descrbe any confguraton of the system allowed by the constrants usng a reduced number N of generalzed coordnates {q k }, k = 1... N. A key dstncton s between holonomc and nonholonomc constrants. For a holonomc constrant, we can fnd a reduced set of N generalzed coordnates such that the coordnates unquely defne any confguraton of the system allowed by the constrants, and so we can fnd an expresson for the postons of all the elementary components n the form r = r (q 1, q 2... q N, t), = 1... M 3, (1) the N coordnates can each be vared ndependently. For holonomc constrants, the number of degrees of freedom s the same as ths reduced number N of generalzed coordnates. Holonomc constrants are further classfed as tme ndependent or scleronomc f tme does not appear n Eq. (1), and tme dependent or rheonomc f tme does appear. Ths dstncton s most mportant when we consder the Hamltonan and the conservaton of energy. A nonholonomc constrant s one for whch the above cannot be done,.e. anythng else. The dynamcs of systems wth nonholonomc constants s often hard to treat. One sort of nonholonomc constrant s a constrant n the form of an nequalty: a partcle must be outsde a sold object for example. Rollng dynamcs also often gves nonholonomc constrants. For example, consder a vertcal bcycle wheel rollng on a horzontal plane (Hand and Fnch, Appendx 1A). We need 4 varables to defne any confguraton, e.g. x, y coordnates of the center, and two angles: φ gvng the angle of a reference spoke, and θ gvng the drecton the wheel s pontng. On the other hand there are only 2 degrees of freedom, snce an nfntesmal change of the two angles prescrbes the change of all 4 varables. Four varables are needed to defne any confguraton, but there are only two degrees of freedom. Ths s actually an example of a nonntegrable dfferental nonholonomc constrant, snce we we can specfy nfntesmal changes n the r n terms of changes n a reduced set of the coordnates (e.g. θ, φ for the wheel) δ r = a k ({q k }, t) δq k (2) wth a lk some coeffcents that may depend on {q k } and tme. Equvalently, consderng the changes over an nfntesmal tme δt, ths s a relatonshp between the veloctes and generalzed veloctes r = a k ({q k ), t) q k. (3) Our methods can be appled to such nonholonomc systems wth a lttle more effort. Sometmes the dynamcs of a system wth strctly nonholonomc constrants can be treated as holonomc over part of the dynamcs. For example, consder a partcle sldng wthout frcton on the outsde surface of a sphere n gravty. Startng at the top, the partcle wll slde down on the surface holonomcally constraned dynamcs. However, the strct statement of the constrant s smply that the partcle must be outsde the sphere, a nonholonomc constrant. And ndeed, at some tme the partcle wll fly off the surface and can no longer be treated n terms of the surface-constraned moton. 2
Hand and Fnch descrbe sldng wth frcton as a nonholonomc constrant. However, the constrant, that the body s on the surface, s holonomc. The constrant force ncludes the frctonal force whch s parallel to the surface, and ths causes complcatons (energy dsspaton) later. Examples to llustrate these deas: 1. The sldng block-on-ramp. 2. A partcle of mass m free to slde on a frctonless ellptcal wre placed vertcally n a gravtatonal feld as n Assgnment 1, but now the sem-axes a(t), b(t) may be tme dependent, prescrbed by me. 3. A rgd body of M partcles Vrtual work I now consder the dynamcs of a system wth constrants. The key dea s to project Newton s 2nd law of moton onto drectons n the 3M dmensonal coordnate space that correspond to partcle dsplacements that are consstent wth the constrants. In ths way most types of constrant forces are elmnated, and we may be able to derve a reduced number of equatons of moton for the generalzed coordnates n terms of only the external or appled forces. It s useful to ntroduce the noton of a vrtual dsplacement whch s a coordnated dsplacement δ r of all the partcles wth tme and veloctes held fxed that s consstent wth the constrants. The projecton of Newton s law ( F = p) along the vrtual dsplacement s then F δ r = p δ r D Alembert s prncple (4) Note that ths equaton s true for any choce for the δ r t s a smple dentty multplyng the left and rght hand sdes of an equaton by the same thng. We choose to use the vrtual dsplacement rather than the actual dsplacement n the dynamcs r (t + δt) r (δt) = r δt because then most types of constrant forces are elmnated. Ths s a very mportant pont. Study the example of a bead on an ellptcal wre wth tme dependent rad to convnce yourself that the vrtual dsplacement elmnates the constrant force normal to the wre from Eq. (4), but that the dsplacement that occurs n tme δt n the dynamcs does not. Also, of course, we do not know the dsplacements occurrng n an nfntesmal tme n the actual dynamcs untl we have solved the problem! Equaton (4) can be repeated for a varety of ndependent vrtual dsplacements (for example correspondng to a small change n each degree of freedom). The quantty δw = F δ r for a vrtual dsplacement δ r s called the vrtual work. It s not necessarly the work done n the actual dynamcs. For an equlbrum (tme ndependent) problem, the rght hand sde of Eq. (4) vanshes, and ths gves us the prncple of vrtual work method for solvng constraned equlbra F δ r = 0 Prncple of vrtual work for equlbra (5) For the case of holonomc constrants the vrtual dsplacement comng from an nfntesmal change of a sngle generalzed coordnate q k s δ r = r /, and the correspondng vrtual work s 3
The quantty s called the generalzed force. δw = F k = r F δq k. (6) r F = δw (7) δq k For the common case where the constrant forces gve no contrbuton to the vrtual work (remember the vrtual dsplacements are consstent wth the constrants) Eqs. (4-7) can be evaluated from the non-constrant ( external ) forces alone F F nc,.e. what we know. For holonomc constrants the equaton of moton Eq. (4) for the vrtual dsplacement gven by changng the generalzed coordnate q k can be wrtten p r = F k. (8) Note that ths s the equaton of moton and not the defnton of the generalzed force, whch s gven by Eq. (7). After some unnformatve algebra (see Appendx 1 or the textbooks) the left hand sde of these equatons can be transformed, agan assumng holonomc constrants, to a form that can be calculated from the knetc energy T = 1 2 m r r, leadng to the generalzed equatons of moton (or as Hand and Fnch call t, Golden Rule #1) ( d T dt q k ) T = F k. (9) These are equatons of moton, bascally gvng the acceleratons q k n terms of the forces F k just as Newton s equatons do. As n Eq. (8), they are not the defntons of F k : n general you know the forces and want to calculate the dynamcs. The generalzed forces are defned and calculated by Eq. (7). The equatons also do not n general express the conservaton of energy (snce they were derved consderng vrtual dsplacements not real ones). As we have seen, for tme dependent constrants the energy need not n fact be conserved. The scheme for usng Eq. (9) s: calculate the knetc energy T = 1 2 m r r n terms of the q k and q k ; T = 1 2 jk A jk({q})q j q k, where A jk depends on masses, (potentally all) coordnates {q k }, but not veloctes { q k }. Usually ths s easy snce t s easy to see what veloctes a tme dervatve of each q k gves; calculate the generalzed force F k : you can use the formal expresson gven by the frst equalty n Eq. (7), but usually t s easer to make a small change n one q k, calculate δw for ths change and use the second equalty n Eq. (7). The concept of vrtual work s rather trcky. I present some more detals and examples n Appendx 2. Equaton (9) s true n general for holonomc constrants (or, of course, for no constrants at all). If the constrant forces gve no contrbuton to the vrtual work they do not contrbute to F k and so are elmnated from the equatons of moton. If they do contrbute to the vrtual work (e.g. frcton) they must be ncluded n F k. Note that Eq. (9) s more general than the Lagrangan equatons I wll now derve, snce the generalzed forces may contan both constrant forces and nonconservatve forces such as frcton. 4
Conservatve forces For conservatve forces F k = r F = ( V r ) r = V (10) For such forces the generalzed equaton of moton can be wrtten as the Euler-Lagrange equaton d L L = 0 (11) dt q k wth L({q k }, { q k }, t) the Lagrangan L = T V. Thus we have derved the Lagrangan formulaton of mechancs startng from Newton s 2nd law. Appendx 1: Dervaton of the Generalzed Equaton of Moton We start from Eq. (8) p r = F k δw δq k. (12) The task s to evaluate the left hand sde n terms of the knetc energy T = 1 2 m r r T ({q k }, { q k }, t) (13) for the case of holonomc constrants. Lemma: Dot Cancellaton: Frst we prove the lemma for holonomc constrants r q k = r. (14) The proof of the lemmas s as follows. For holonomc constrants we have r = r ({q k }, t) so that r d r dt = k r q k + r t. (15) Ths mples r = r ({q k }, { q k }, t) and we can magne changng q k and q k ndependently. But r / and r / t do not depend on { q l } so ( r ) r =. (16) q k {q l k },t { q l k },{q l },t Note that t s mportant to be careful about what s held constant n the partal dervatves! Now we use ths lemma to manpulate Eq. (12) p r = F k, T = 1 2 m r r T ({q k }, { q k }, t). (17) 5
Snce we want the tme dervatve of p we look at what we get when we evaluate dervatves of T T = T q k = m r r = m r r q k = p r, p r, usng dot cancellaton n the second expresson. Now dfferentate the second expresson wth respect to t along path of dynamcs {q k (t)} d T = p r + p d r. (18) dt q k q k dt q k The frst term s what we want, so we look at the second term, rememberng r = r ({q k }, t) so that Ths gves and so the result we want d r = dt l = p d r = dt 2 r q l + 2 r, q l t r q l + r 1 q l t l = r, p r = T. (19) p r = d T T, (20) dt q k d T T = F k δw. (21) dt q k δq k Appendx 2: More on Vrtual Work To understand vrtual work better, lets consder the settng of M partcles or elementary objects of the mechancal system wth postons r, = 1... M. For a system wth holonomc constrants these postons are determned by a reduced number N of generalzed coordnates q k, k = 1... N. Holonomc means we can wrte expressons for the postons r n terms of the generalzed coordnates q k and, maybe, tme. r = r (q 1, q 2,... q N, t), = 1... M. (22) The vrtual work correspondng to some vrtual dsplacement n whch each object moves by the dsplacement δ r s M δw = F δ r. (23) =1 Here F s the total physcal (Newtonan) force actng on object. 1 A vrtual dsplacement s a coordnated move of the objects such that the constrants reman satsfed. It s an magnary 1 An mportant pont s that F only ncludes the real physcal forces, due to gravty, the contact between the objects (.e. the constrant forces) etc., and not fcttous (also known as vrtual) forces such as the centrfugal force. 6
set of dsplacements n whch tme and veloctes are held fxed, not the actual dsplacements for the dynamcs over a tme δt (whch we don t know untl we solve the problem n any case). The dea s that for such vrtual dsplacements, the constrant forces usually do not contrbute to the vrtual work. To fnd the generalzed equatons of moton for the coordnates q k we want to fnd the vrtual work for the vrtual dsplacements correspondng to changng each generalzed coordnate q k separately. (We can do ths, snce the q k can be vared ndependently, whle stll mantanng the constrants.) For example, frst consder just a change δq 1 n the coordnate q 1. Ths wll correspond to a coordnated set of dsplacements δ r of the M objects, each proportonal to δq 1. The vrtual work s calculated from Eq. (23): t too wll be proportonal to δq 1. The generalzed force correspondng to q 1 s then F 1 = δw δq 1. (24) Then repeat for generalzed coordnate q 2, etc. We mght calculate δw for the coordnate change δq 1 n two ways: Geometrcally: Draw pctures showng the vector dsplacements δ r for the change δq 1. Form force dot dsplacement F δr for each object, and sum over the objects. Algebracally: Calculate δ r for the coordnate change δq 1 usng the explct equatons Eq. (22) Now calculate the vrtual work, etc. Some other thoughts: δ r = r q 1 δq 1. (25) Varyng a sngle coordnate δq 1 leads, n general, to dsplacements of all the objects. The dsplacements wll be consstent wth the constrants. There s a generalzed force F k s assocated wth each generalzed coordnate q k The generalzed force does not act on one object: t s compled from the forces actng on all the objects. The Golden Rule No.1 d T T = F k (26) dt q k s the equaton of moton, gvng q k n terms of the generalzed force F k calculated as above. It s equvalent to Newtons m a = F. It s not the defnton of F k. Snce Eq. (26) s more complcated than Newton s second law, we need to work out the consequences of the equaton. For example, F k = 0 does not alway mean q k = 0 (an example s equaton for the X generalzed coordnate for the block-on-wedge problem). The vrtual work can be defned for any set of object dsplacements that satsfy the constrants, ncludng general nonholonomc constrants. In these more general cases, the vrtual dsplacement cannot be assocated wth a sngle generalzed coordnate, and so the noton of a vrtual force cannot be extracted from the vrtual work. Also, n these cases t s usually not possble to wrte the other sde of the equaton n a smple form such as the left hand sde of Eq. (26). Thus, although the vrtual work can be calculated n these cases, t may not be useful to solve for the dynamcs. 7
Example: Frctonless wedge and block As an example, consder the frctonless sldng block on wedge problem we dscussed n class. d X α Label the wedge (nclned plane at angle α) as = 1, and the small block as = 2. The Cartesan coordnates n ths 2-dmensonal system wth two objects are r 1 = (X, Y ) and r 2 = (x, y); that s, four functons of tme. However, t s clear that there s no moton n Y because the wedge plus small block s pressng down on the table and the table s pushng back up an equal but opposte amount. Y s thus an gnorable coordnate. It s also clear that the small block experences no moton perpendcular to the top surface of the wedge, for the same reason. Also, the moton parallel to the top surface of the wedge s most easly descrbed wth respect to the wedge (d), not the observer at rest wth respect to both blocks. So out of these four Cartesan coordnates, we choose two generalzed coordnates d, X, shown n the fgure. Note that a change n X moves both the wedge and the block, snce the constrant block s on wedge must be mantaned. In terms of these coordnates, we can wrte down the knetc energy T = 1 2 m1ẋ2 + 1 ) 2 2 m 2 (Ẋ + d 1 ( cos α + 2 m 2 d 2 sn α), (27) where the second and thrd terms correspond to m 2 (v 2 x + v 2 y)/2 of the small block. The potental energy due to gravty has no effect on X, and Y s gnorable. For the small block, along d, we have V = m 2 gd sn α. To apply the method of vrtual work, we can use ether Eq. (8)or Eq. (9). We need to evaluate the terms r / and F k. Apply the two general methods mentoned above. 8
Geometrcal method Frst consder a change δx n the generalzed coordnate X. N F c δx F c δx α mg α Mg Draw the free body dagrams as you were taught n freshman physcs. The forces actng are gravty, and acton and reacton forces normal to the contact surfaces. In the fgures, the black arrows denote the drectons of the forces, and the symbols the magntudes. The dsplacements δ r of the two objects due to the change δx are the red arrows. Now calculate the vrtual work for ths δx. For object 1 (the wedge) the contrbuton to the vrtual work s δw 1 = F c sn α δx (gravty and the normal surface force N don t contrbute snce F δr s zero for these forces and the horzontal dsplacement.). For object 2 (the block) the contrbuton to the vrtual work s δw 2 = F c sn α δx. The total vrtual work for the change δx s δw = δw 1 + δw 2 = 0: the contrbutons from the constrant forces vansh as promsed! Ths s of course because the forces are equal and opposte, whereas the dsplacements of the two bodes are the same to mantan the constrant. Snce the vrtual work s zero, ths means the generalzed force F X correspondng to X s zero. 2 Now consder a change δd n the generalzed coordnate d. F c mg α δd There s not dsplacement of the wedge and so I haven t shown the wedge n the fgure. The vrtual work s δw = mg sn αδd. The constrant force does not contrbute, snce t s perpendcular to the dsplacement. The generalzed force for the generalzed coordnate d s F d = δw/δd = mg sn α. 2 Ths does not mean that Ẍ = 0 check the generalzed equaton of moton. 9
Algebrac method The relatonshp between the postons of the objects and the generalzed coordnates s, usng the notaton (u, v) for a vector wth x-component (horzontal, to the rght) u and y-component (vertcal, up) v, (wth h the heght of the wedge). The forces on the objects are r 1 = (X, 0), (28) r 2 = (X + d cos α, h d sn α), (29) F 1 = F c (sn α, cos α) + (0, N Mg), (30) F 2 = F c (sn α, cos α) + (0, mg). (31) Now use F k = F r /. For the generalzed coordnate X we have r 1 / X = (1, 0) and r 2 / X = (1, 0). Summng over wedge and block F r = 0, (32) X and so F X = 0. For the generalzed coordnate d we have r 1 / d = (0, 0) and r 2 / d = (cos α, sn α). Summng gves F r = mg sn α, (33) d so that F d = mg sn α. The two methods agree, as they should. The geometrc method may seem longer, snce we went through t n some detal. But t s probably easer once you get famlar wth t. Pluggng the knetc energy n Eq. 27, and the generalzed forces gven above, nto Eq. (9), we have two equatons of moton, whch may look a bt complcated but are relatvely easy to solve. We note that T/ q k = 0 for both q 1 = X and q 2 = d, and that F X = 0. (for q 1 = X) : m 1 Ẍ + m 2 Ẍ + m 2 d cos α 0 = 0, (34) ) (for q 2 = d) : m 2 (Ẍ + d cos α cos α + m 2 d sn 2 α = m 2 g sn α. (35) The second equaton smplfes to Ẍ cos α + d = g sn α. You can solve that for d, plug nto frst equaton, solve for Ẍ. Wth ntal condtons for the poston and velocty of X and d, the problem s solved. 10