Classcal Mechancs Vrtual Work & d Alembert s Prncple Dpan Kumar Ghosh UM-DAE Centre for Excellence n Basc Scences Kalna, Mumba 400098 August 15, 2016 1 Constrants Moton of a system of partcles s often constraned, ether geometrcally or knematcally. Constrants reduce the number of degrees of freedom of a gven body. Consder the moton of a sngle partcle n space. For free, unconstraned moton, t has three degrees of freedom whch are usually expressed by three coordnates such as x, y, z or r, θ, ϕ etc. If, however, the partcle s constraned to move on the surface of a sphere, we must have (takng Cartesan coordnates), x 2 + y 2 + z 2 = R 2 whch reduces the number of degrees of freedom by one. Consder two masses connected by a rgd rod, lke a dumbbell. Two partcles have 6 degrees of freedom, Snce the dstance between the two bodes remans constant, we have the constrant (x 1 x 2 ) 2 + (y 1 y 2 ) 2 + (z 1 z 2 ) 2 = d 2 whch reduces the degree of freedom to 5. These are examples of geometrc or holonomc constrants whoch are expressble as algebrac equatons nvolvng the coordnates. There are other constrants whch restrct the moton of bodes. Some of these are expressble as dfferental equatons whch constran the coordnates and components of veloctes. These are called knematc constrants. Non-ntegrable knematc constrants whch cannot be reduced to holonomc constrants are called non-holonomc constrants. Thus, f m s the dmenson of the confguraton space (.e., the number of generalsed coordnates), holonomc constrants are expressble as equatons of the form f (t, q 1, q 2,..., q m ) = 0, 1 k
c D. K. Ghosh, IIT Bombay 2 where k s the number of constrants. Holonomc constrants are called scleronomc f they do not explctly depend on tme. Tme dependent constrants are called rheonomc. Knematc constrants are expressed as equatons n the phase space f (t, q 1, q 2,..., q m ; q 1, q 2,..., q m ) = 0, 1 k Both the constrants are classfed as rheonomc f they explctly depend on tme. Sometmes a constrant may appear to be knematc but may be n realty holonomc. For nstance, a constrant of the type Aẋ + B = 0 may actually be holonomc f there exsts a functon f such that A = f/ x and B = f/ t. We then have, df = f df x + f t whch gves a holonomc constrant f = constant. Example 1: Consder two masses connected by pulleys, as shown. In general two partcles have 6 degrees of freedom. However, m 1 can only move along the x drecton and m 2 along the z drecton. Thus y 1 = z 1 = 0 and x 2 = y 2 = 0. We are left wth two degrees of freedom. However, f m 1 moves along the x drecton by a dstance d, m 2 would have to move along z drecton by 2d,.e., we get another holonomc constrant, z 2 2x 1 = 0 whch reduces the degree of freedom further by one. The problem s essentally a one dmensonal problem. m 1 P 1 z x P 2 m 2 00000000000000000 11111111111111111 00000000000000000 11111111111111111 00000000000000000 11111111111111111 Non-holonomc constrant could be n the form of dfferental equatons or algebrac nequaltes. For nstance, a partcle constraned to move nsde a sphere of radus R satsfes x 2 + y 2 + z 2 < R 2. Consder a dsk rollng on a horzontal surface, on x-y plane.
c D. K. Ghosh, IIT Bombay 3 z y P φ θ x If the dsk s rollng wthout slppng, keepng ts plane vertcal, we need four coordnates to descrbe the poston of the dsk. These are the x and y coordnates of the centre of the dsk, the angle ϕ by whch a fxed pont on the rm of the dsk has rotated about the axs of rotaton and the angle θ that the axs of the dsk makes wth the x axs. If R s the radus of the dsk, the velocty of the dsk s gven by v = Rω = R ϕ (1) Snce the dsk remans vertcal, the components of the velocty of the centre of the dsk are gven by Usng eqs. (1) to (3) we get dx = v sn θ (2) dy = v cos θ (3) dx dy = R sn θ dϕ = R cos θ dϕ the mnus sgn s due to the sense of rotaton beng opposte to the postve angle. Combnng these two we get the followng par of dfferental equatons: dx = R sn θdϕ dy = R cos θdϕ These equatons cannot be further reduced and we cannot connect x, y, θ and ϕ by an algebrac equalty, showng that the constrant s non-holonomc.
c D. K. Ghosh, IIT Bombay 4 Tacklng problems wth non-holonomc constrants s more dffcult and no general prescrpton can be provded for ther soluton. Constrants ntroduce two new elements nto problem solvng. Snce the generalsed coordnates are no longer ndependent, the equatons of moton are not ndependent ether. Constrants arse from forces between elements whose nature s unknown. These forces are known only by the effect they have on moton of the system. 2 Vrtual Dsplacement A real dsplacement of partcles consttutng a system happens over a fnte tme. Such a dsplacement of, say, the -th partcle s, n general, functon of all the generalsed coordnates as well as of tme. If the poston of the -th partcle s wrtten as r = r (q 1, q 2,..., q s ) The total dfferental of the poston vector s then wrtten as d r = r dq j + r (4) q j t j=1 A vrtual or an magned dsplacement s nstantaneous and s consstent wth the constrants on the system. The dsplacements, whether real or vrtual, result n admssble geometrcal confguraton of the system. In case of a vrtual dsplacement, we have, d r = j=1 r q j δq j (5) where we have use δq to ndcate a vrtual dsplacement whle reservng dq for real dsplacement. For a real dsplacement, the forces of constrants may change. The velocty s gven by so that we have, ṙ = d r = ṙ q k = j=1 r q j + r q j t (Note the structure of the above equaton - as f the dots cancel!) j=1 (6) r q j δ jk = r q k (7) 2.1 Prncple of Vrtual Work Consder a system of N partcles under tme dependent holonomc constrants. If q 1,..., q s be a set of generalzed coordnates, the vrtual dsplacement of the -th partcle s gven
c D. K. Ghosh, IIT Bombay 5 by eqn. (5). Suppose the system, under the acton of appled forces as well as those of constrants, s n equlbrum. The total force actng on each partcle s then zero, F = 0; ( = 1,..., N) We then have, for the vrtual work done by F n the dsplacement δ r s so that the total work done s F δ r = 0 δw = F δ r = 0 The total force actng on any partcle can be splt nto two: an appled part and a part due to the constrants, F = F a + F c then we have δw = F a δ r + F c δ r = 0 (8) The forces of constrants (e.g. normal reacton, tenson, rgd body constrants etc.) do not do any work. Ths s general true of scleronomc holonomc constrants and ths statement s central hypothess n the prncple of vrtual work. Two examples llustrate the hypothess. Consder two types of dsplacements consstent wth the constrants on a rgd body. A dsplacement along the lne jonng two partcles does not do any net work because the reactons are equal and opposte. In order to be consstent wth rgd body constrants, for a par of partcles j and k, we must have δr k = δr j. The work done s f kj δr j + f jk δr k = (f kj + f jk ) δr j = 0 where we have used δr j = δr k. Thus there s no work done. The second type of dsplacement s along the arc of a crcle normal to the lne jonng the partcles.as the forces of constrants are normal to the drecton of dsplacement, the work done s once agan zero. Consder the pulley arrangement n Example 1. When m 1 moves by an amount δx to the rght, m 2 moves by 2δx downwards n order to keep the sum of the lengths of the two ropes constant. The only appled forces are the gravty and frcton. Thus by the prncple of vrtual work, we have, µmgδx + mg2δx = 0 whch shows that for statc equlbrum m 2 = µm 1 /2. Example 2: Two frctonless blocks of mass m each are connected by a massless rgd rod. The system s constraned to move n the vertcal plane.
c D. K. Ghosh, IIT Bombay 6 R mg δx 1 mg θ 00000000000000000 11111111111111111 00000000000000000 11111111111111111 00000000000000000 11111111111111111 N δx 2 F If the block on the vertcal track undergoes a vrtual dsplacement δx 1 and that on the horzontal plane has a vrtual dsplacement δx 2, we have δx 1 sn θ = δx 2 cos θ whch s the constrant whch keeps the rod length constant. The gravty does work on the mass on the vertcal track whle the appled force F 2 s responsble for work on the horzontally movng block, mgδx 1 + F 2 δx 2 = 0. Thus we have F 2 = mg δx 1 δx 2 = mg cot θ Example 3: Consder an Atwood s machne n equlbrum. In ths case we have the constrant y 1 + y 2 = l = constant. 00000000000 11111111111 000 111 00 11 00 11 m m 2 1 Thus we have δy 1 = δy 2. The appled forces are m 1 g and m 2 g, both actng downwards. The vrtual work done s δw = m 1 gδy 1 + m 2 gδy 2 = (m 1 m 2 )gδy 1 = 0, whch gves the condton for equlbrum to be m 1 = m 2. 2.2 Generalzed Force We recall that n the presence of holonomc constrants, all the components of generalsed coordnates can be made ndependent. If the system has s degrees of freedom (s 3N), the poston vector of the -th partcle ( = 1, N) can be wrtten as
c D. K. Ghosh, IIT Bombay 7 r = r (q 1, q 2,..., q s ) The velocty of the -th partcle s gven by ṙ = d r = j=1 r q j + r q j t { q } are known as the generalsed veloctes. (If the constrants are scleronomc, we would have r t = 0. ) Let us return to the expresson for vrtual work and express t n terms of the generalsed coordnates. (9) δw = F a δ r = F a r δq α q α=1 α Q α δq α (10) α=1 where we have defneed a generalsed force correspondng to the coordnate q α by the relaton Q α = F a r q α (11) Note that the generalsed force depends only on appled forces and not on forces of constrants. In equlbrum, we have δw = α Q α δq α = 0 Snce the generalsed coordnates are ndependent, we have Q α = 0. However, ths does not apply that appled forces vansh. The condton for vanshng of F a δ r = 0 s applcable only to statc stuatons. d Alembert extended ths to nclude general moton of the system. 3 d Alembert s Prncple d Alembert s prncple, developed from an dea orgnally due to Bernoull, s to use the fact that accordng to Newton s law, the force appled on a partcle results n a rate of change of ts momentum, F = ṗ
c D. K. Ghosh, IIT Bombay 8 ṗ s known as the nertal force or pseudo force actng on the partcle. One can then thnk of brngng the body to equlbrum by applyng a pseudo force ṗ on the -th partcle of the system (1 N), F ṗ = 0 Note that F contans both appled and constrant forces actng on the -th partcle. Once agan, we can splt F nto two parts and wrte the above equaton for vrtual dsplacements as ( F a + F c ṗ ) δ r = 0 Snce the force of constrants do not do any work, we get ( F a ṗ ) δ r = 0 (12) Equaton (12) s the statement of d Alembert s prncple. It may be noted that the only force appearng n ths equaton s the appled force. Further, F a refers to the force on the -th partcle and the sum n (12) s over all partcles and not over the ndependent generalsed coordnates. Consequently, (12) does not mply F a ṗ = 0. An nterestng consequence follows f the dsplacement n (12) happens to be real dsplacement nstead of vrtual ones. In that case the dsplacement can be wrtten as d r = r If the force s conservatve and s dervable from a potental V,.e. F = V, can rewrte (12) as follows: ( F a ṗ )d r = [ V m r ] ṙ = [ V d r d (1 2 mṙ2 )] = d( T + V ) = 0 3.1 Example 4: Consder a mass restng on a frctonless nclne. The mass would slde down wth an acceleraton when released. A horzontal acceleraton s appled on the mass to keep the mass from sldng. We need to fnd the acceleraton. The problem s
c D. K. Ghosh, IIT Bombay 9 reasonably smple and s readly solved by Newton s laws. N 90 θ a N = mg cos θ + ma sn θ mg ma cos θ = mg sn θ θ whch gves a = g tan θ Let us look at the problem from d Alembert s prncple. Suppose the mass has an nstantaneous (vrtual) dsplacement δl along the nclne. We then have δx = δl cos θ and δy = δl sn θ. The only appled force s mg along the y axs: F = mgŷ. From the prncple of vrtual work t follows that F x δx + F y δy ma x δx ma y δy = 0 We only apply a horzontal acceleraton so that we have, a y = 0. Snce F x = 0, we get mgδy ma x δx = 0.e., mgδl sn θ maδl cos θ = 0. Thus we get, a = g tan θ. Example 5: Consder the arrangement shown n the fgure. The pulley s fxed on the fxed wedge. Fnd the acceleraton of the masses when released. l l 1 2 m 1 m 2 α m g m g 2 1 β From d Alembert s prncple we have Snce l 1 + l 2 = constant, we have ( F 1 ṗ 1 ) δ l 1 + ( F 2 ṗ 2 ) δ l 2 = 0 (13) δl 1 = δl 2 l1 = l 2 (14a) (14b) The nertal forces are ṗ 1 = m 1 l1 and ṗ 2 = m 2 l2 = m 2 l1 and the only appled forces are the weght of the masses. Takng the components of (13) along the nclne, we have, Usng (14a) and (14b) we get, (m 1 g sn α m 1 l1 )δl 1 + (m 2 g sn β m 2 l2 )δl 2 = 0 (m 1 g sn α m 1 l1 m 2 g sn β m 2 l1 )δl 1 = 0
c D. K. Ghosh, IIT Bombay 10 so that l1 = m 1g sn α m 2 g sn β m 1 + m 2 4 Euler- Lagrange Equaton We wll now derve the Euler-Lagrange equaton from d Alembert s prncple. Recall eqn. (7) where we showed ṙ r = δ jk = r (15) q k q j q k j=1 we also had, from (6), for scleronomc constrants, ṙ = j=1 r q j q j (16) Usng chan rule dfferentaton, we can wrte, d ( ) r = q j 2 r dq k q j q k ( r q k q k k=1 = q j k=1 ) = q j ṙ (17) where n the last lne we have used (16) and have used the fact that q s ndependent of q. Let us return to d alembert s prncple k=1 N ( F ṗ ) δ r = 0 The nertal force term may be smplfed as follows: ( ) m r δ r = m r r δq k q k=1 k [ ( ) d = m ṙ r q k=1 k [ ( ) d = m ṙ ṙ q k =1 ( d r m ṙ q k ] m ṙ ṙ q k where n the last lne we have used the dot cancelaton relatonshp (17). ) ] δq k δq k (18)
c D. K. Ghosh, IIT Bombay 11 The rght hand sde of the above expresson can be smplfed and the equaton can be wrtten as follows: [ ( ) ( )] d 1 m r δ r = q k 2 m ṙ 2 1 q k=1 k 2 m ṙ 2 δq k [ ( ) d T = T ] δq k (19) q k q k =1 k=1 where T s the knetc energy of the system of partcles. Substtutng (19) n d Alembert s equaton, we have N ( ( ) d T F a δ r T ) δq k = 0 (20) q k q k In terms of generalzed coordnates, we could wrte the frst term as s k=1 Q kδq k. Thus we have, [ ( ( ) d T Q k T )] δq k = 0 (21) q k q k k=1 Snce the generalzed coordnates are ndependent, we may vary each coordnate ndependently and get ( ) d T T = Q k (22) q k q k Equaton (22) s the form of Euler-Lagrange equaton whch s derved from d Alembert s prncple. If, however, the external forces actng on the system are conservatve,.e., f we can express F = V, where mples gradent taken wth respect to the coordnates of the -th partcle, we have N Q k = F r q =1 k N = V r q k =1 = V q k (23) If the potental s a functon only of the poston q k, we have V = 0 whch enables us q k to wrte the generalzed force as Q k = V + d ( ) V (24) q k q k On brng ths expresson for the generalzed force to the left hand sde of (23) and recognzng that L = T V, we recover the Euler-Lagrange equaton d ( ) L q k L q k = 0 (25)
c D. K. Ghosh, IIT Bombay 12 5 Velocty Dependent Potental - Lorentz Force The expresson for the generalzed force n the expresson (24) can be used to defne a potental functon for velocty dependent forces as well. Only n such a stuaton, the second term on the rght hand sde of (24) s not dentcally zero but would depend on the exact dependence of the force on velocty. Thus n such stuatons, we could derve Euler-Lagrange equaton form an approprately defned Lagrangan functon. It s an mportant generalsaton for us because Lorentz force whch a charge partcle experences n an electromagnetc feld s a velocty dependent force. The force on an electrc charge q (not to be confused wth generalzed coordnate whch s also usually wrtten by the same notaton), s gven by F = q( E + v B) (26) Usng the expressons for the electrc and the magnetc felds n terms of the scalar and the vector potentals ϕ and A respectvely, we have E = ϕ A t B = A we can wrte the expresson for the force as [ F = q ϕ A t + v ( A) ] (27) The scalar potental depends only on poston and we wll need to deal only wth the remanng two terms. Snce v does not depend on poston, we have the followng dentty, Proof of (28) v ( A) = ( v A) ( v ) A (28) Proof. To prove the dentty (28), we consder the x-component of ts left hand sde. [ v ( A)] x = v y ( A) z v z ( A) y ( = v y x A y ) ( y A x v z z A x ) x A z ( ) ( ) A y = v y x + v A z A x z v y x y + v A x z z ( ) ( A x = v x x + v A y y x + v A z A x z v x x x + v y where n the last lne we have added and subtracted v x A x x. A x y + v z ) A x z
c D. K. Ghosh, IIT Bombay 13 Snce v or ts components do not depend on poston coordnates, we can wrte the frst term as x ( v A) and the second term as ( v )A x. Addng three components of v ( A) then yelds the dentty (28). We wll now rewrte the second term of (28) usng a smart trck. Note that the total dervatve of any component of A can be wrtten as follows: da x = A x x dx + A x y = v x A x x + v y = ( v )A x + A x t Combnng three components, we can wrte, dy + A x x A x y + v z dz + A x t A x x + A x t (29) da = ( v ) A + A t Usng (28) and substtutng the expresson for A from (30) nto (27), we get t [ F = q (ϕ v A) d A ] (30) (31) Now, we can wrte usng the fact that A does not depend on velocty components da x = d ( ) (A x v x ) v x = d [ ] (A x v x + A y v y + A z v z ) v x = d ( ) ( v v A) x = d ( ) ( v v A ϕ) x where, n the last step, we have added a term ϕ/ v x whch s zero because the scalar potental does not depend on the velocty ether. Thus da = d ( v ( v A ) ϕ) where v = î + ĵ + v x v ˆk s gradent wth respect to the velocty vector. Substtutng ths n the expresson (31), we get y v z [ F = q (ϕ v A) + d ( A)) ] v (ϕ v (34) (32) (33)
c D. K. Ghosh, IIT Bombay 14 Thus f we defne a velocty dependent potental as U = ϕ v A (35) the component of Lorentz force would be gven by F j = U + d ( ) U q j q j (36) whch s of the form (24). Wth ths modfcaton, the Euler-Lagrange equaton s stll vald for velocty dependent potental for whch the Lagrangan s gven by. L = T U