Moving coordinate system

Chapter Movng coordnate system Introducton Movng coordnate systems are mportant because, no materal body s at absolute rest As we know, even galaxes are not statonary Therefore, a coordnate frame at absolute rest s hypothetcal, hypotheszed by Newton, where hs laws of moton hold In realty, we have the movng frames, prme example beng Earth tself We therefore need to know how the Newton s laws operate n a movng frame lke a rotatng frame (eg Earth) Rectlnear movng frame Let us consder a smplest case of a rectlnear moton Let the frame S move relatve to S For smplcty, let us assume that, the axes of S and S are parallel The nstantaneous poston of O relatve to the orgn O s depcted by R(t) n fgure Consder a rectlnear moton of S along the drecton of R(t) Then vector r n S wll be r n S such that, r = r R(t) () z z P S r r S V O R y O x x y Fgure : Frame S, movng relatve to S along vector R(t) The tme dervatve of the equaton s r = r R(t) () v s the velocty n S and s related wth velocty v n S as, v = v V (3)

CHAPTER MOVING COORDINATE SYSTEM where V = R(t) s velocty of S relatve to S Let us assume that, the Newton s second law s vald n S Then, the external force F ext s F ext = m r = m d v ( ) ( ) d v = m + d V = m r + d V (4) If S moves relatve to S wth constant velocty, e, dv/ = 0, then S and S are ndstngushable F ext = m r = m r (5) Thus, S and S are nertal We call the nertal force measured n the fxed frame S as F = m r Then, the nertal force measured n S s F = m r The F ext s ndependent of moton of S and S If V 0, F = m r m R = F m R (6) It must be noted that f S s movng wth respect to S wth constant velocty, the law of force has a same form (In techncal jargon, t s covarant) However, f S s an accelerated frame, then t can be dstngushed from S by vrtue of the term m R Some tmes, the term m R of S s called fcttous force Ths s a msnomer because, we get a jerk when a tran starts 3 Rotatng frame of reference Consder two frame of reference S and S wth unt vectors n = (, j, k) and n = (, j, k ) and wth a common orgn S rotates wth some axs wth angular velocty ω Let us consder a poston vector r The components of r n S are (x, y, z) and n S are (x, y, z ) Then, r = îx + ĵy + ˆkz r = î x + ĵ y + ˆk z (7) (8) z z S S P r O x y x y Fgure : Vector r n S and S As S rotates, unt vectors n = (, j, k ) are functons of tme Let us wrte d/ as a tme dervatve operator n S and d / as a tme dervatve operator n S We do not wrte d / because we assume t = t, a non relatvstc case Snce S s a fxed frame, (, j, k) do not change wth tme Clearly, v = d r = îẋ + ĵẏ + ˆkż (9)

3 ROTATING FRAME OF REFERENCE 3 The tme dervatve of the equaton 8 s where d r d r = î dx + dî x + ĵ dy + dĵ y dz + ˆk + dˆk z = î dx + ĵ dy v = v + x dî + ˆk dz + y dĵ + x dî + z dˆk + y dĵ + z dˆk (0) v = î dx + ĵ dy + ˆk dz = d (î x + ĵ y + ˆk z ) s the velocty n the frame S We need to fnd the value of dn / n a fxed frame n terms of the known quanttes lke ω and r We must prove a followng mportant lemma Lemma: Let there be a general drecton OM around whch a vector A of constant magntude rotates wth a constant angular velocty ω n a fxed frame Then d A = ω A Proof: Let the tal of the vector A be at O and let t rotate about OM wth angular velocty ω as shown n M ω R ωδt δa A(t) α A(t+δt) O Fgure 3: Vector A s rotatng around a fxed drecton OM the fgure 3 The poston of the vector A at tme t and t + δt s shown n the fgure when t rotates n tme δt As the drecton of A changes, A(t) changes to A(t + δt) The magntude of A(t) and A(t + δt) s same but, A(t + δt) = A(t) + δ A () From the fgure 3, δa = R ω δt () Agan form the fgure 3, R = A sn α (3)

4 CHAPTER MOVING COORDINATE SYSTEM δ A = ωa sn α δt δ A = ( ω A) δt δ A δt = ω A When δt 0, the equaton 4 gves (4) d A = ω A (5) Ths dervatve s taken n the frame S In the frame where A s fxed (e, S frame), d A/ = 0 And, n S, OM rotates wth angular velocty ω e, n a counter clockwse drecton Equaton (5) can be wrtten n an operator form as, d( ) = ω ( ) (6) where, ( ) contan any vector operator Thus, dî dĵ dˆk = ω î = ω ĵ = ω ˆk Usng equatons 7, 8 and 9 n equaton 0, d r v = v + x ( ω î ) + y ( ω ĵ ) + z ( ω ˆk ) v = v + ω (î x + ĵ y + ˆk z ) v = v + ω r = d r + ω r From equaton, the general operator equaton can be wrtten as, d( ) = d ( ) (7) (8) (9) (0) () + ω ( ) () We use equaton to get acceleraton n a rotatng frame n terms of ts value n the fxed frame We denote a and a as acceleratons n S and S Thus by operatng v to the equaton, d v = d v + ω v (3) dv = d ( v + ω r) + ω ( v + ω r) a = d v + ω d r + d ω r + ω v + ω ω r a = a + ( ω v ) + ω ω r + d ω r (4) It must be noted that, the tme dervatve of a vector parallel to ω s same n both S and the S frames Thus, d ω = d ω If F ext s the external force, accordng to Newton s second law of moton, vald n the fxed frame as, F ext = ma Usng equaton 4, (5) F ext = ma = ma + m( ω v ) + m( ω ω r) + m( ω r) (6) As everyone s on the surface of Earth, everyone s n a rotatng frame, he measures v and a and not v and a Equaton 6 for an observer n rotatng frame as, ma = F ext m( ω v ) m( ω ω r) m( ω r) (7)

4 APPLICATIONS 5 Equaton 7 gves a moton of a partcle n the rotatng frames If ω s constant, and t s the case n many stuatons, the last term m( ω r = 0), then, ma = F ext m( ω v ) m( ω ω r) (8) F = F ext + F c + F r = F ext + F o (9) where F c = m( ω v ) s called Corols force and F r = m( ω ω r) s called centrfugal force In a fxed frame, the term m( ω ω r) s called a centrpetal force, drected towards the center and ts magntude s mω r = mv /r In a rotatng frame, we have a mnus sgn n front of the centrpetal force, drected outwards and s called centrfugal force F o = F c + F r s not real force It s called fcttous force It s not present n a fxed coordnate system We can treat a rotatng coordnate system as f t were fxed by addng a centrfugal force and a Corols force If we fx a coordnate system wth the rotatng partcle, then t s at rest n ths frame The centrpetal force s balanced by centrfugal force n ths frame Corols force depends on the velocty of the partcle and t acts n a drecton perpendcular to v Therefore, Corols force does no work but only changes ts drecton of moton Exercse: Estmate the magntude of Angular velocty of earth Centrfugal acceleraton at equator 3 Corols acceleraton at lattude of 45 o wth velocty 0 3 ms 4 Applcatons 4 Effect of centrfugal force on acceleraton due to gravty The acceleraton due to gravty vares wth lattude φ, beng about 05% smaller at equator than ts value at the poles Ths s the reason why our Earth has oblate shape e, Earth s sphere of flattened at the poles Consder a pont P wth lattude φ as shown n fgure 4 If Earth was not rotatng, g at P would act along PO Centrfugal acceleraton s to be superposed on g to get g eff Thus, g eff = g ω ( ω r) (30) N ω P g geff ω ω r O S Fgure 4: Drecton of effectve acceleraton due to gravty Drecton of g eff does not pass through O, the centre of Earth The drecton of plumb lne s along g eff as shown n fgure 4 Earth s surface s flattened to such an extent that s always perpendcular to g eff 4 Effect of Corols force on atmospherc ar flow It s observed that, there s not much of an ar flow n vertcal drecton as compared wth a horzontal drecton Ths s so because, n a vertcal drecton, pressure decrease as one moves upwards and ths force s balanced by weght of the ar parcel Ths leads to a persstent long range moton of ar n the form of wnds

6 CHAPTER MOVING COORDINATE SYSTEM In the absence of Corols force, the ar flow takes place from hgher pressure to lower pressure Let ω be perpendcular to the plane The Corols acceleraton ( ω v) deflects ar current as shown n fgure 5 The wnd current then crculates around a low pressure zone n clockwse drecton Low Pressure Low Hgh Pressure Fgure 5: Cyclone formaton due to Corols force 43 Foucault pendulum French physcst Foucault realzed that, the Corols force would rotate the plane of oscllaton of pendulum Foucault pendulum s an ordnary pendulum wth a large length as much as more than 0 meter Let us assume that a pendulum s suspended from the orgn O of a frame S at a place of lattude φ, as shown n fgure 6 The Z axs of S s vertcally upward, the X axs due south and the Y axs due east, but wth the orgn at a heght above the surface of the earth at P The poston vector of the bob B s r (x, y, z) and O B = l = r s the length of the pendulum The tenson n the strng s the appled force F a and snce t s along BO, we may wrte, F a = k m r r = k r (3) where k = g/l s a constant of proportonalty Therefore, the equaton of moton of the pendulum s, r k r = 0 (3) The equaton 3 can be satsfed only when there s no other force actng on the pendulum except gravtatonal force If the rotaton of Earth s taken nto account, g > g eff and n adon a Corols force act on the pendulum Thus, the equaton of pendulum becomes, r k r = ω v (33) where k = g eff /l Snce x-axs due south and the Y-axs due to east, the vector ω s n the ZX-plane, so that W O N ω P φ O z x y B E Fgure 6: Poston of the Foucault s pendulum on the surface of the Earth

4 APPLICATIONS 7 ω y = 0 and we have ω x = ωcosφ, ω y = 0 ω z = ωsnφ, (34) Snce moton of the pendulum s n the XY-plane, so that v z = 0 Then clearly ω = ( ωcosφ, 0, ω snφ) (35) r = (ẍ, ÿ, 0) (36) Resolvng the equaton of moton n components, ẍ k x = ( ω v) x = ωẏ snφ (37) ÿ k y = ( ω v) y = ω(ẋ snφ + ż cosφ) (38) z k z + g = ( ω v) z = ωẏ cosφ (39) For small oscllatons, z = l and therefore ż = z = 0 Agan, snce g 000 cms and ω = 73 0 5 rad/s, the term contanng ω n equaton 39 may be neglected n comparson wth g and therefore we obtan k = g/l Then equatons 37 and 38 may be wrtten as ẍ = g x + ωẏ snφ l ÿ = g y ωẋ snφ l (40) (4) Introducng the complex varable z = x + y and wrtng u = ωsnφ, equatons 40 and 4 may be wrtten as a sngle equaton z + uż + g l z = 0 (4) Soluton of the equaton 4 s n the form z = e αt, where α s a constant and on substtuton to the equaton 4, α + uα g l = 0 (43) The roots of the equaton 43 are, α = u + u + g α = u u l + g l (44) Thus general soluton of the equaton 4 s z = Ae αt + Be αt ż = α Ae αt + α Be αt (45) (46) where A and B are constants to be determned by the ntal conons If the pendulum s drawn due south (along the X-axs) at dstance a and let go wth zero ntal velocty n the ZX-plane, we have, as ntal conons, z = (x + y) = (a + 0) = a, ż = 0 at t = 0 Equaton 45 and 46 yelds, a = A + B (47) 0 = α A + α B, α A = α B (48) Then from equaton 45, ż = Aα (e αt e αt ) = Aα [ e ( u+ u + g l )t e ( u u + g l )t ] = [ Aα e ut e ( u + g l )t e ( ] u + g l )t ż = Aα e ut sn ( u + gl ) t (49) z = x + y, ż = ẋ + ẏ, z = ẍ + ÿ = g l x + ωẏ snφ g y ωẋ snφ l z = g l (x + y) ω snφ(ẋ + ẏ) = g z ω snφ ż l

8 CHAPTER MOVING COORDINATE SYSTEM As we know, the velocty of the bob becomes zero every tme the pendulum reverses ts drecton and at these values of t, we have ż = 0, and ths happens when the sne functon s zero and therefore for values of t for whch u + g t = nπ; (n = 0,,, 3, ) l For one oscllaton t = T and n =, u + g l T = π; T = π u + g l where u = ω snφ If ω s neglgble, u 0 and we obtan T = π If u 0, g/l u + g l = π T Usng equaton 5 n 44, α = u + π T α = u π T Rewrtng the equaton 45 usng 5, z(t) = Ae ( u+ π T )t + Be ( u π T )t ( ) z(t) = e ut Ae πt T + Be πt T (50) (5) (5) (53) When the pendulum s ntally set n moton by drawng t southward along the X-axs to the poston P wth a dstance a, we have t = 0 and therefore, we see from equaton 53 that z(0) = A + B = a (54) After half an oscllaton, t = T/, the bob s at the northern end at P and we have z(t/) = e ut/ ( A B) = ae ut/ = ae (π ut/) (55) Equaton 55 shows that, the vector Z(0) representng the pont P has been rotated nto the poston P through an angle (π ut/) Ths means that the trajectory of the pendulum, n the frst half oscllaton s the curve from P to P as shown n fgure 7 Smlarly, when t = T, we have z(t ) = e ut (A + B) = ae ut = z(0)e ut (56) N ut/ P t = T/ W O E P 3 t = T ut S P t = 0 Fgure 7: Rotaton of the plane of oscllaton of Foucault s pendulum Equaton 56 shows that after one complete oscllaton, n tme T, the poston of the bob of the pendulum s agan at the southern end but rotated through an angle ut nto poston P 3 Thus the trajectory n the second half oscllaton s the curved path from P to P 3 as shown n the fgure We thus see that the moton of the bob

4 APPLICATIONS 9 s as f the plane of vbraton of the pendulum rotated clockwse through an angle ut n one perod Thus the tme requred for one complete rotaton of the plane, e, though an angle π s τ = T ut π π = ω snφ = 4 hours (57) snφ snce π ω = day = 4 hours Thus, at a place of lattude φ = 45o, the tme τ for one complete rotaton s 4 34 hours We note that the curved path from P to P of the bob of the pendulum s due to the force of Corols actng to the rght of the velocty vector as the bob moves from south to north For the same reason, the force of Corols actng to the rght of the velocty as the bob moves from north to south accounts for the part of the trajectory from P to P 3 We recall that the other nertal force, namely the centrfugal force has already been taken care of n defnng, the weght of a materal pont so that the trajectory P to P to P 3 of the bob s solely due to the force of Corols Foucault carred out hs experments n 85 n Pars, but hs experments only confrmed the rotaton of the earth qualtatvely Quanttatve confrmaton came only n the year 879 by the work of Kamerlngh Onnes of low temperature fame Reference books: Classcal Mechancs - P V Panat Classcal Mechancs - K N Srnvasa Rao

0 CHAPTER MOVING COORDINATE SYSTEM

Chapter MECHANICS OF A SYSTEM OF PARTICLES Conservaton of lnear momentum and angular momentum Consder a system of n partcles of masses m, m, m 3,, m n at respectve postons r (t), r (t), r 3 (t),, r n (t) at a tme t Then, the total momentum P of the system s, ( n n P = p = m v = d n ) m r () = = = The force actng on the th partcle of the system has two parts () external force F (e), (from our sde) and () nternal force, F j (nternal force on the th partcle due to the j th partcle) Thus, the equaton of moton (Newton s second law) for the th partcle s wrtten as n d p = j= F j + F (e) () j, because a partcle cannot exert any force on tself Clearly, from equaton, the total force actng on the system s d P = d p = F j + j F (e) (3) If the Newton s thrd law s vald n the system, F j = F j, then j F j = 0, the equaton 3 becomes, d P = d p = d m v = d m r = (4) F (e) = F ext (5) We defne a vector R as the average of the rad vectors of the partcles, weghted n proporton to ther mass, R = m r + m r + m 3 r 3 + + m n r n = m r (6) m + m + m 3 + + m n M The vector R defnes a pont known as the center of mass The equaton 5 reduces to d P = M d R = F (e) = F ext (7) The equaton 5 states that the center of mass moves as f the total external force were actng on the entre mass of the system concentrated at the center of mass If F ext = 0, the total lnear momentum of the system,

CHAPTER MECHANICS OF A SYSTEM OF PARTICLES m j Center of mass r j R m r Fgure : The center of mass of a system of partcles d P = M d R = F (e) = 0 (8) P = M d R = Constant That s total lnear momentum of the system (total mass of the system tmes the velocty of the center of mass) s constant Thus the conservaton theorem for the lnear Momentum of a system of partcles s stated as, f the total external force actng on the system s zero, the total lnear momentum s conserved The angular momentum of the system of partcles s L = r p (9) The torque actng on the system s dl ( ) = d r p = = d r p + r F (e) + r d p n r j F j (0) If the Newton s thrd law s vald n the system, F j = F j, then j F j = 0, the equaton 0 becomes, d L = r F (e) d L = N (e) = N ext () If N ext = 0, the total angular momentum of the system s d L = N (e) = 0 L = Constant That s total angular momentum of the system s constant Thus the conservaton theorem for the angular momentum of a system of partcles s stated as the angular momentum of the system s constant n tme, f the appled (external) torque s zero Theorem: Angular momentum of a system of partcles about a general orgn O s equal to the angular momentum of the system concentrated at the CM plus the angular momentum of the system about ts CM Proof: Consder a partcle whose coordnate s r wth respect to O and r wth respect to the CM Thus

ENERGY 3 z m r CM r R o y x Fgure : Poston of th partcle r = R + r v = V CM + v m v = m VCM + m v p = m VCM + p () (3) (4) where, p = m v s a velocty of th partcle wth reference to the CM Then, L = r p = ( R + r ) (m VCM + p ) L = R V CM m + R p + (m r ) V CM + ( r p ) (5) If CM s taken taken as the orgn, (m r ) = 0 = p (6) The equaton 5 reduces to, L = R V CM m + ( r p ) L = R P + ( r p ) (7) R P s an angular momentum of the centre of mass wth respect to O, and ( r p ) s an angular momentum of the system of partcles wth respect to the CM From the above equaton t s obvous that, L depends upon the choce of orgn If the CM of the system of partcles s statonary, then L s ndependent of the choce of orgn Energy The work done by all forces n movng the system from an ntal confguraton, to a fnal confguraton s W dw = W dw = F d r F (e) d r + n j F j dr j (8)

4 CHAPTER MECHANICS OF A SYSTEM OF PARTICLES If the Newton s thrd law s vald n the system, F j = F j, then j F j = 0, the equaton 8 becomes, T T = F (e) d r = d p d r = d p m p = p d p m [ ] [ ] p p T T = m m T = m v In center of mass coordnates, v = V CM + v and v = ( V CM + v )( V CM + v ), T = m ( V CM + v )( V CM + v ) = m V CM + V CM m v + m ( v ) (9) If CM s taken taken as the orgn, p = m v = 0, T = MV CM + m ( v ) (0) Thus the knetc energy of a system of partcles s equal to the sum of the knetc energy of the CM plus the knetc energy of the system about ts CM If the partcle moves from ntal confguraton, to a fnal confguraton under the acton of a conservatve force, then the external forces are dervable n terms of the gradent of a potental, the frst term of the equaton 8 can be wrtten as F (e) d r = V d r () The nternal forces are also conservatve, then the mutual forces between the th and j th partcles, F j and F j can be obtaned from a potental functon V j The seond term of the equaton 8 can be wrtten as F j d r j = ( V j d r + j V j d r j ) () j j To satsfy the strong law of acton and reacton, F j = j V j = V j = F j Then, equaton can be wrtten as, F j d r j = V j (d r d r j ) = j V j d r j (3) j j j Where r j = d r d r j and j stands for the gradent wth respect to r j The term j V j d r j n the equaton 3 s for the j par of partcles The total work arsng from nternal forces then reduces to F j d r j = j j jv j d r j (4) Combnng the equaton 4, equaton and equaton 8, we see that, W dw = V d r j jv j d r j V = [( V ) ( V ) ] [( j V j ) ( j V j ) ] j V = V + j V j (5) j If F s conservatve, F = 0 then F = V

ENERGY 5 The second term on the rght n equaton 5 wll be called the nternal potental energy of the system In general, t need not be zero and, more mportant, t may vary as the system changes wth tme Only for the partcular class of systems known as rgd bodes the nternal potental always be constant Reference books: Classcal Mechancs - P V Panat Classcal Mechancs, 3rd Ed - HGoldsten, CPoole and JSafko

6 CHAPTER MECHANICS OF A SYSTEM OF PARTICLES

Chapter 3 The Lagrangean method 3 Constrants Constrants means restrctons; constraned moton means restrcted moton Most of the moton that we encounter, s constraned moton Most physcal realzatons of constraned moton nvolve surfaces of other bodes, for example, Moton of a bllard ball on the table: Moton of a bllard ball s restrcted by the boundares of the table, and t moves on the surface of the table If the centre of mass of a bllard ball of radus R movng on a bllard table of length a and breah b, must satsfy the relaton a + R x a R, b + R y b R, z = R assumng that the orgn of the coordnate axes s at the centre of the rectangular table and x and y axes are parallel to length and breah respectvely e, a set of one equaton and two nequaltes, defnes the moton of a bllard ball at all nstants of tme The moton of a smple pendulum: The bob of the pendulum moves n a vertcal plane (say zx plane) Its dstance from the fulcrum s fxed Thus, f (x, y, z) s coordnate of the bob then, y = constant, z + x = l are the restrctons on the coordnates of the bob Physcally, constraned moton s realsed by the forces whch arse when the object n moton s n contact wth the constranng surfaces or curves These forces, called constrant forces, are usually stff elastc forces at the contact If there are no constrants, moton of the partcle s descrbed by the trajectory r(t) = x + jy + kz and by ts momentum p(t) = p x + jp y + kp z Thus the poston of the partcle s specfed by three coordnates If there are N partcles, 3N ndependent coordnates are necessary for the poston specfcaton of the system at a tme t Presence of constrants may reduce the number of ndependent varables 3 Classfcaton of Constrants (a) Scleronomc: constrant relatons do not explctly depend on tme, (b) Rheonomc: constrant relatons depend explctly on tme, (a) Holonomc: conons of constrant can be expressed as equatons connectng the coordnates of the partcles, (b) Non holonomc: constrant relatons are not holonomc, 3 (a) Conservatve: total mechancal energy of the system s conserved whle performng, the constraned moton Constrant forces do not do any work, (b) Dsspatve: constrant forces do work and total mechancal energy s not conserved 4 (a) Blateral: at any pont on the constrant surface both the forward and backward motons are possble Constrant relatons are not n the form of nequaltes but are n the form of equatons, (b) Unlateral: at some ponts no forward moton s possble Constrant relatons are expressed n the form of nequaltes 7

8 CHAPTER 3 THE LAGRANGEAN METHOD 3 Holonomc and non holonomc constrants If one can wrte the equatons of constrants as f ( r, r, r 3, r n ; t) = 0 f ( r, r, r 3, r n ; t) = 0 f ( r, r, r 3, r n ; t) = 0 (3) f + ( r, r, r 3, r n ; t) = 0 f k ( r, r, r 3, r n ; t) = 0 where k < n, then such constrants are known as holonomc constrants The constrants whch cannot be expressed n the form of algebrac equatons are non holonomc constrants, however, they could be expressed as nequaltes 33 Examples of constrants Rgd body: A rgd body s a system of partcles such that the dstance between any par of partcles remans constant n tme Thus the moton of a rgd body s constraned by the equatons r r k = const (3) where the par of subscrpts (, k) run over all dstnct pars of partcles formng the body Obvously ths constrant s scleronomc The constrant s also holonomc and blateral The constrant relatons 3 can be wrtten as r r k = const Takng dfferentals ( r r k ) ( r r k ) = 0 (33) Work done by the system s W = ( F k r + F k r k ) (34) k Let the nternal force of constrant on the th partcle due to the k th partcle be represented by F k By Newton s thrd law we have, F k = F k (35) Thus we have for the work done by F k due to a dsplacement r of the th partcle, F k r = F k r k (36) On combnng equatons 34 and 36 we can wrte the total work done by the system W = F k ( r r k ) (37) k Snce all Fk are the nternal forces whch arse purely due to nteracton between all possble pars of partcles, t s only natural that F k wll act parallel to the lne jonng the th and k th partcles Thus we can wrte, F k = C k ( r r k ) (38) where C k s are real constants and symmetrc n and k Substtutng n the above expresson for the total work, we have W = C k ( r r k )( r r k ) (39) k In equaton 39 each ndvdual term of the summand s zero Thus the constrant of rgy s conservatve n nature, apart from ts beng scleronomc, holonomc and blateral

3 GENERALIZED COORDINATES 9 Deformable bodes: Suppose that the deformaton of the body s changng n tme accordng to a certan prescrbed functon of tme Then the moton of such a body s constraned by the equaton r r k = f(t) (30) where r and r k are poston vectors and the par of subscrpts (, k) runs over all dstnct pars of partcles n the body These constrant relatons cannot gve the total work W = 0 Hence ths s a rheonomc, holonomc, blateral and dsspatve constrant 3 Gas n a sphercal contaner of radus R Here f r s a poston vector of the th gas molecule (orgn s at the centre of the sphere) then x + y + z R (3) Thus, we have a constrant equaton gven by an nequalty and hence s non holonomc constrant 4 Rollng wthout sldng: Suppose a sphercal ball s rollng on a plane wthout sldng We assume that the surfaces n contact are perfectly rough Thus the frctonal forces are not neglgble Snce the pont of contact s not sldng, the frctonal forces do not do any work, and hence the total mechancal energy of the rollng body s conserved Thus the constrant s conservatve To obtan the constrant equaton we note that rollng wthout sldng means that the relatve velocty of the pont of contact wth respect to the plane s zero Then the velocty v of any pont P n the rollng body, as seen from a fxed frame of reference, s gven by v = V CM + ω r (3) where V CM s the velocty of the centre of mass and r s measured from the CM to the pont P under consderaton Thus the velocty of the pont of contact s obtaned by puttng r = rˆn n equaton 3, where ˆn s the unt vector along the outward normal to the plane and r s the radus of the sphere Snce there s no sldng of ths pont we must have the nstantaneous velocty v at the contact v = V CM r( ω ˆn) = 0 (33) For a sphere ths constrant s non ntegrable because ω s generally not expressble n the form of a total tme dervatve of any sngle coordnate Thus the constrant s non holonomc However, for a cylnder, ω = dθ/ where θ s the angle of rotaton of the cylnder about ts axs Therefore ths equaton of constrant can be ntegrated and reduced to a holonomc form, gvng a relaton between r and the coordnates of the centre of mass 3 Generalzed coordnates The problem of system of n partcles can be solved when the number of constrant equatons are less than 3n Let there be k equatons of constrants k < 3n, f ( r, r, r 3, r n ; t) = 0 f ( r, r, r 3, r n ; t) = 0 f k ( r, r, r 3, r n ; t) = 0 (34) e, 3n k coordnates may be regarded as free and whch defne the poston of the system at any moment of tme t Then the number of ndependent coordnate to specfy the moton at a gven tme t s 3n k These ndependent coordnates are called degrees of freedom In the case of a free materal partcle, for nstance, n = and k = 0 so that t has 3n k = 3 degrees of freedom If the partcle s constraned to move on a surface whose equaton may be taken as f(x, y, z) = 0(z = 0), we clearly have k = and therefore t would have 3n k = degrees of freedom On the other hand, for a dumb-bell shaped structure, wth two partcles connected by a rod of length l; a constrant equaton becomes f(x, y, z) = (x x ) + (y y ) + (z z ) l = 0

0 CHAPTER 3 THE LAGRANGEAN METHOD we have n = and k =, therefore t has 3n k = 5 degrees of freedom The degrees of freedom are represented by 3n k varables, q, q,, q n k The old coordnates r, r, r 3, r n are expressed n terms of q s as, r = r (q, q,, q n k ; t) r = r (q, q,, q n k ; t) r n = r n (q, q,, q n k ; t) (35) 3 Vrtual dsplacement A vrtual (nfntesmal) dsplacement of a system refers to a change n the confguraton of the system as the result of any arbtrary nfntesmal change of the coordnates δ r, consstent wth the forces and constrants mposed on the system at the gven nstant of tme t The dsplacement s called vrtual to dstngush t from an actual dsplacement of the system occurrng n a tme nterval, durng whch the forces and constrants may be changng 3 Vrtual work Total work done by the external forces when vrtual dsplacements are made n n partcle system, s known as vrtual work If F (a) be the appled force and f be the constrant force actng on th partcle, the net force actng on the system s F = F (a) + f (36) If δ r s the vrtual dsplacement, the work done on the system s W = F δ r = F (a) δ r + f δ r (37) When system s n equlbrum W = F δ r = 0 (38) Thus equaton 37 reduces to F (a) δ r + f δ r = 0 (39) The vrtual dsplacements δ r are such that the constrant forces do no work ( f δ r = 0) Thus, F (a) δ r = 0 (30) e, The conon for statc equlbrum s that the vrtual work done by all the appled forces should vansh, provded the vrtual work done by all the constrant forces vanshes Ths s called the prncple of vrtual work 33 D Alembert s Prncple Consder the moton of n partcle system Then, by Newton s law, F = p (3) Combnng equatons 3 and 36, F (a) + f = p F (a) + f p = 0 (3)

34 LAGRANGE S EQUATIONS OF THE SECOND KIND The equaton 3 states that the partcles n the system wll be n equlbrum under a force equal to the actual force plus a reversed effectve force p The work done now can be wrtten as ( F (a) + f p )δ r = 0 (33) The vrtual dsplacements δ r are such that the constrant forces do no work ( f δ r = 0) Thus, ( F (a) p )δ r = 0 (34) The equaton 34 s called D Alembert s Prncple D Alembert s prncple does not nvolve forces of constrant e, any dynamcal problem could be converted nto an effectve statc problem 34 Lagrange s Equatons of the second knd In D Alembert s prncple, the vrtual dsplacements δ r are not ndependent Therefore, the D Alembert s equaton s a sngle equaton If the constrants are holonomc, we use ndependent set of varables {q } When ths s done, we get, n equatons, one each for each q These equatons are Lagrange s equatons For n partcle system the D Alembert s equaton s, ( F (a) p ) δ r = 0 (35) If we have n partcles at r, r, r 3, r n and there are k equatons of holonomc constrants, then there are 3n k = m generalzed coordnates They are denoted by q, q,, q m Thus we have, r = r (q, q,, q m ; t) = r (q j, t) (36) Velocty of the th partcle, v s gven as (on dfferentatng the equaton 36), d r m δ r δq j = δq j δt + δ r δt v = j= m j= δ r q j + δ r δq j δt (37) δv = δ r δ j δq j The vrtual work of the system s W = F δ r = F δ r δq j (by usng equaton??) δq j j F δ r = Q j δq j j where the Q j are called the components of the generalzed force, defned as (38) (39) Q j = F δ r δq j (330) Note that q s need not have the dmensons of length, so the Q s do not necessarly have the dmensons of force, but Q j δq j must always have the dmensons of work For example, Q j mght be a torque N j and dq j a dfferental angle dθ j, whch makes N j dθ j a dfferental of work Agan consder the equaton 39, Q j δq j = F δ r j = p δ r = = Q j δq j j = m r δ r m r δ r δq j δq j j [ ( d m r δ r δq j j ) m r d ( δ r δq j )] δq j (33)

CHAPTER 3 THE LAGRANGEAN METHOD We can see from the equaton 33, r s dfferentable wth respect to both t and q j, we can nterchange the dfferentaton wth respect to t and q j n equaton 33 Q j δq j = [ ( d m r δ r ) m r δ ( )] d r δq j δq j j j δq j = [ ( d m r δ r ) m r δ v ] δq j (33) δq j j δq j From the equaton 38, we can wrte δv = δ r δ j δq j Equaton 333 n equaton 33 gves, Q j δq j = [ ( d m v δv δ j j j = { [ ( d δ δ j j Q j δq j = [ ( ) d δt δt δ j j j δq j ) m v δ v δq j m v ] δq j ] δq j )] ( δ δq j where T = m v s the knetc energy of the system [ ( ) d δt δt ] Q j δq j = 0 δ j δq j j )} m v δq j (333) Snce all δq j are ndependent whereas, δ r were not Thus, ( ) d δt δt Q j = 0 (334) δ j δq j When the forces F are dervable from a scalar potental functon V F = V Equaton 335 n 330 gves Q j = V δ r δq j ( Q j = δv + j δv + k δv ) ( δx + j δy + k δz ) δx δy δz δq j δq j δq j The equaton 336 s exactly the same expresson for the partal derves of a functon V (r l, r,, r n ; t) wth respect to q j Q j = δv δq j (335) (336) (337) On combnng the equatons 334 and 337 ( ) d δt δt + δv = 0 δ j δq j δq j ( ) d δt δ(t V ) = 0 (338) δ j δq j The equatons of moton n the form 338 are restrcted to conservatve systems, only when V s ndependent of tme Hence the potental V s a functon of generalzed coordnates q j only, and does not depend upon q j We defne a new functon, the Lagrangan L, as L(q j, j ) = T (q j, j ) V (q j ) (339) The partal derves of the equaton 339 are δl = δt δ j δ j δl = δt δv = δ(t V ) δq j δq j δq j δq j (340) (34)

34 LAGRANGE S EQUATIONS OF THE SECOND KIND 3 Equatons 340 and 34 n equaton 338, ( ) d δl δl = 0, j =,, 3,, m (34) δ j δq j Equaton 34 gves set of m equatons These m equatons, one for each ndependent generalzed coordnates, are known as Lagrange s Equatons of moton of the second knd n a potental feld 34 Smple applcatons of the Lagrangan formulaton Knetc energy T s gven by T = m v = m m δ r q j + δ r δq j= j δt = m m δ r m δ r m q j q k + δq j δq k where j= k= j= by usng equaton 38 δ r δq j q j δ r δt + ( ) δ r δt = δ r δ r m q j q k + δ r δ r m δq j k j δq k δq j j δt q j + ( ) δ r m δt T = m jk q j q k + m j q j + m 0 (343) j j k m jk = m δ r δq j δ r δq k (344) m j = m 0 = δ r δ r m δq j δt m ( δ r δt ) Thus, the knetc energy of a system can always be wrtten as the sum of three homogeneous functons of the generalzed veloctes, T = T + T + T 0 (345) where T 0 s ndependent of the generalzed veloctes, T, s lnear n the veloctes, and T s quadratc n the veloctes If the transformaton equatons do not contan the tme explctly, as may occur when the constrants are ndependent of tme (scleronomous), m 0 = 0 and m j = 0 = T 0 = 0 and T = 0 Then T = T = m jk q j q k j k Thus, T s always a homogeneous quadratc form n the generalzed veloctes Moton of a sngle partcle (346) (a) Usng Cartesan coordnates: If (x, y, z) are the Cartesan coordnates at tme t of a free materal pont of mass m movng n a potental feld V (x, y, z), we may take q = x, q = y, q 3 = z as there are no equatons of constrant The appled force F on the partcle has the components δv δx, δv δy, δv δz, whle the knetc energy T s gven by T = m(ẋ + ẏ + ż ) Thus the Lagrangan for the partcle s T V = L = m(ẋ + ẏ + ż ) V (x, y, z) (347) δl δẋ δl δx = mẋ = δv δx (348) (349)

4 CHAPTER 3 THE LAGRANGEAN METHOD Equatons 348 and 349 n equaton 34, d δv (mẋ) + δx = 0 mẍ = δv δx = F x Smlarly mÿ = F y, m z = F z The Lagrangan equatons of moton are mẍ = F x, mÿ = F y, m z = F z (350) Equaton 350 gves Newton s equatons of moton (b) Usng cylndrcal polar coordnates: If (r, θ, z) are the cylndrcal coordnates at tme t of a free materal z ẑ ˆθ ˆr θ r y x Fgure 3: Cylndrcal coordnates; d r = dr ˆr + rdθ ˆθ + dz ˆk pont of mass m, we may take (q = r, q = θ, q 3 = z) as there are no equatons of constrant The appled force F on the partcle has the components n generalzed coordnates (Q r, Q θ, Q z ) The three Lagrangan equatons can be wrtten by usng the equaton 334, d d d ( ) δt δṙ ( ) δt δ ( θ δt δż δt δr Q r = 0 (35) δt δθ Q θ = 0 (35) ) δt δz Q z = 0 (353) The poston vector r n cylndrcal coordnates, r = rˆr + rθˆθ + zˆk where ˆr, ˆθ and ˆk are unt vectors n the r, θ and z drectons, respectvely The components of the force n generalzed coordnates can be obtaned from the equaton 330 as, Q r = δ r F r δr = F rˆr (354) Q θ = δ r F θ δθ = F θrˆθ (355) Q z = δ r F z δz = F zˆk (356) In cylndrcal coordnates x = rcosθ, y = rsnθ, z = z dx = rsnθ dθ + dr cosθ, dy dθ = rcosθ + dr snθ, dz = dz ẋ = rsnθ θ + ṙcosθ, ẏ = rcosθ θ + ṙsnθ, ż = ż

34 LAGRANGE S EQUATIONS OF THE SECOND KIND 5 The knetc energy T s δt δṙ δt T = m(ẋ + ẏ + ż ) = m( rsnθ θ + ṙcosθ) + (rcosθ θ + ṙsnθ) + ż T = m[ṙ + (r θ) + ż ] (357) δt = mṙ, δr = mr θ (358) = mr δt θ, δθ = 0 (359) δt = mż, δz = 0 (360) δ θ δt δż Equatons 354 and 358 n equaton 35, d (mṙ) mr θ F r = 0 m r mr θ = F r (36) If r s constant, F r = mr θ beng the centrpetal acceleraton Equatons 355 and 359 n equaton 35, d (mr θ) rfθ = 0 d L = rf θ = N l (36) where L = mr θ s the angular momentum and N l s the appled torque Equatons 356 and 360 n equaton 353, d (mż) F z = 0 m z = F z (363) (c) Usng sphercal polar coordnates: If (r, θ, φ) are the sphercal polar coordnates at tme t of a free materal pont of mass m, we may take (q = r, q = θ, q 3 = φ) as there are no equatons of constrant The appled force F on the partcle has the components n generalzed coordnates (Q r, Q θ, Q φ ) The three Lagrangan equatons can be wrtten by usng the equaton 334, z x θ φ y r z ˆr ˆθ ˆφ y x ( ) d δt δṙ ( d δt δ φ d ( δt δ θ Fgure 3: Sphercal coordnates; d r = dr ˆr + rsnθdφ ˆφ + rdθ ˆθ δt δr Q r = 0 (364) ) δt δφ Q φ = 0 (365) ) δt δθ Q θ = 0 (366)

6 CHAPTER 3 THE LAGRANGEAN METHOD The poston vector r n sphercal coordnates, r = r ˆr + rsnθφ ˆφ + rθ ˆθ where ˆr, ˆφ and ˆθ are unt vectors n the r, φ and θ drectons, respectvely The components of the force n generalzed coordnates can be obtaned from the equaton 330 as, Q r = δ r F r δr = F r ˆr (367) Q φ = δ r F φ δφ = F φrsnθ ˆφ (368) Q θ = δ r F θ δθ = F θr ˆθ (369) In sphercal coordnates x = rcosφ snθ, y = rsnφ snθ, z = rcosθ ẋ = ṙcosφ snθ rsnφ φ snθ + rcosφ cosθ θ ẏ = ṙsnφ snθ + rcosφ φ snθ + rsnφ cosθ θ ż = ṙcosθ rsnθ θ The knetc energy T s T = m(ẋ + ẏ + ż ) T = ( [ṙ m + rsnθ φ ) ( + r θ ) ] (370) δt δṙ = mṙ, δt δ φ = mr sn θ φ, δt δt ( δr = m rsn θ φ + r θ ) (37) δt δφ = 0 (37) = mr θ, δt δθ = mr φ cosθ (373) δ θ Equatons 367 and 37 n equaton 364, d ( (mṙ) m rsn θ φ + r θ ) F r = 0 ( rsn θ φ + r θ ) = F r (374) m r m Equatons 368 and 37 n equaton 365, d (mr sn θ φ) rsnθ F φ = 0 d (mr sn θ φ) = rsnθ F φ (375) Equatons 369 and 373 n equaton 366, d (mr θ) mr φ cosθ rf θ = 0 d (mr θ) mr φ cosθ = rf θ (376) Atwood s machne: Fgure 33 shows the schematc dagram of Atwood s machne whch s an example of a conservatve system wth holonomc, scleronomous constrant (the pulley s assumed frctonless and massless) Clearly there s only one ndependent coordnate y, the poston of the other weght beng determned by the constrant that the length of the rope between them s l The Lagrangan equaton for the moton can be wrtten by usng equaton34 as, d ( ) δl δl δẏ δy The potental energy s = 0 (377) V = M gy M g(l y) (378) The knetc energy of the system s

34 LAGRANGE S EQUATIONS OF THE SECOND KIND 7 y M (l y) M Fgure 33: Atwood s machne T = (M + M )ẏ (379) The Lagrangan L s L = T V = (M + M )ẏ + M gy + M g(l y) (380) L δl δẏ = (M + M )ẏ L δl δy = M g M g Equatons 38 and 38 n equaton 377, d [(M + M )ẏ] M g + M g = 0 (M + M )ÿ = (M M )g ÿ = (M M ) (M + M ) g (38) (38) Ths s the famlar result obtaned by more elementary means Ths emphaszes that the forces of constrant here the tenson n the rope appear nowhere n the Lagrangan formulaton Also the tenson n the rope can not be found drectly by the Lagrangan method 3 A bead (or rng) sldng on a unformly rotatng wre n a force-free space: Consder beads n a straght wre, and s rotated unformly wth angular velocty ω about some fxed axs perpendcular to the wre Ths example has been chosen as a smple llustraton of a constrant beng tme dependent, wth the rotaton axs along z and the wre n the xy plane Beads can move along wre, only one Lagrangan equaton as usng the equaton 34, d ( δl δṙ ) δl δr = 0 (383) The transformaton equatons explctly contan the tme, x = rcosωt, y = rsnωt ẋ = rsnωt ω + ṙcosωt, ẏ = rcosωt ω + ṙsnωt, The knetc energy T s T = m(ẋ + ẏ ) T = m[ṙ + (rω) ] (384)

8 CHAPTER 3 THE LAGRANGEAN METHOD The Lagrangan L s T V = L = m[ṙ + (rω) ] δl δṙ δl δr = mṙ = mrω Equatons 385 and 386 n equaton 383, d (mṙ) mrω = 0 r = rω (385) (386) (387) The equaton 387 s the famlar smple harmonc oscllator equaton wth a change of sgn The soluton of the equaton s r = e ωt shows that the bead moves exponentally outward because of the centrpetal acceleraton But the method cannot furnsh the force of constrant that keeps the bead on the wre 35 Velocty dependent potental Consder an electrc charge, q, of mass m movng at a velocty, v, n an electrc feld, E, and a magnetc feld, B, whch may depend upon tme and poston The electrc charge experences a force, called the Lorentz force, gven by F = q[ E + ( v B)] (388) Both E(t, x, y, z) and B(t, x, y, z) are contnuous functons of tme and poston dervable from a scalar potental φ(t, x, y, z) and a vector potental A(t, x, y, z) Faraday s law of electromagnetc nducton s E = δ B δt = δ δt ( A) = δ A δt E + δ A = 0 δt ( E + δ A δt ) = 0 Thus, E + δa/δt s gradent of some scalar functon φ e, ( E + δ A ) = φ δt E = φ δ A δt Equaton 389 n 388, [ F = q φ δ A ] δt + ( v A) (389) (390) Takng x component of F, [ F x = q δφ δx δa ] x + ( v A) δt x [ = q δφ δx δa ] x + v y ( A) δt z v z ( A) y [ = q δφ δx δa ( x δay + v y δt δx δa ) ( x δaz + v z δy δx δa )] x δz We can wrte da x δa x δt = δa x δt = da x + δa x δx δx δt + δa x δy δy δt + δa x δz δz δt (39) + δa x δx v x + δa x δy v y + δa x δz v z (39)

36 HAMILTON S PRINCIPLE 9 Equaton 39 n 39, [ F x = q δφ δx da x + δa x δx v x + δa x δy v y + δa x δz v z + v y [ = q δφ δx da x + δa x δx v x + δa y δx v y + δa ] z δx v z [ = q δφ δx da x + δ ] δx (A xv x + A y v y + A z v z ) [ F x = q δφ δx da x + δ ] δx ( A v) [ F x = q δ δx (φ A v) da ] x ( δay δx δa ) ( x δaz + v z δy δx δa )] x δz (393) We also can wrte da x = d δ ( A v) δv x da x = d δ (φ A v) δv (394) x Equaton 394 n 393, [ F x = q δ δx (φ A v) + d ] δ (φ A v) δv x F x = δu δx + d δu δv x (395) where U = q(φ A v) s the velocty dependent potental The Lagrange s equaton was wrtten as (equaton 334), ( ) d δt δt = Q j δ δq j d q j ( δt δẋ ) δt δx = Q x = F x On combnng equatons 395 and 396, (396) d δ δ (T U) (T U) δẋ δx = 0 d δl δẋ δl δx = 0 (397) Smlarly, d δl δẏ δl δy d δl δż δl δz = 0 (398) = 0 (399) where L = T U = mv q(φ A v) s the Lagrangan for a charged partcle n electromagnetc feld Thus, even electromagnetc forces can be accommodated n Lagrange s formulaton 36 Hamlton s prncple The moton of a conservatve system from ts confguraton at tme t to ts confguraton at tme t s such that the lne ntegral between the tme t and t of the Lagrangan of the system has a statonary value for the actual path of the moton I = t t L(q, q, t) (300) L = T V s the Lagrangan Snce L has the dmensons of energy tme called acton, the prncple s sometmes referred to as the prncple of least acton The ntegral s called the acton ntegral Then the varaton of the acton ntegral for fxed tme t and t must be zero e, δi = δ t t L(q, q, t) = 0 (30)

30 CHAPTER 3 THE LAGRANGEAN METHOD y (q, t ) (q, t ) x Fgure 34: Vared paths of the functon of q(t) n the one-dmensonal extremum problem I has a statonary value relatve to paths dfferng nfntesmally from the correct functon q(t) Let η(t) be a contnuous functon wth contnuous frst dervatve and η(t ) = η(t ) = 0 We construct an another curve q(t, α) as q(t, α) = q(t, 0) + αη(t) (30) Then, wth dfferent values of α we wll get dfferent paths For α = 0, equaton 30 gves the curve q(t, 0) = q(t) For smplcty, t s assumed that both the correct path q(t) and the auxlary functon η(t) are well-behaved functons and are contnuous and nonsngular between t and t, wth contnuous frst and second dervatves n the same nterval For any such parametrc famly of curves, equaton 300 can be wrtten as, I(α) = t t L{q(t, α), q(t, α), t} (303) and the conon for obtanng a statonary pont s, ( ) di = 0 dα α=0 Consder the equaton 300, I = δi δα dα = Snce = = = = δi δα dα = δi δα dα = At α = 0, δi = t t t t t t t t t t t t L(q, q, t) ( δl δq δl δq δl δq δl δq δl δq δq δα ) δl δ q dα + δ q δα dα δq dα + δα δq dα + δα δq dα + δα δq δα t t t t t t dα + δl δ q ( ) ( ) δq δq = = 0, δα t δα t t t t t t t ( δq δα δl δq δq δα ( δl δq d ) t dα δl δ q t ) δq dα = δq and 0 ( δl δq d ) δl δq δ q δl δ q δ q dα δα δl δ q dα δ q δαδt ( ) δl δ δq dα δ q δt δα t δq t δq d δα δα t δq d δα dα δα ) ( δi δα 0 t ( ) δl dα δ q dα = δi ( ) δl dα δ q

36 HAMILTON S PRINCIPLE 3 Accordng to Hamlton s prncple δi = 0, t ( δl δq d ) δl δq = 0 (304) δ q t Snce the q varables are ndependent, the varatons δq are ndependent In the equaton 304, the coeffcents of δq must vansh separately δl δq d δl δ q = 0 (305) The equaton 305 s the Lagrange equaton of moton Ths equaton s vald for any functon f(q, q, t) and s called the Euler-Lagrange dfferental equaton Reference books: Classcal Mechancs - P V Panat Classcal Mechancs, 3rd Ed - HGoldsten, CPoole and JSafko Classcal Mechancs - N C Rana and P S Joag

3 CHAPTER 3 THE LAGRANGEAN METHOD

Chapter 4 Rgd body dynamcs A rgd body s defned as a system of mass ponts subject to the holonomc constrants that the dstances between all pars of ponts reman constant throughout the moton 4 The ndependent coordnates of a rgd body - Degrees of freedom A rgd body wth N partcles can at most have 3N degrees of freedom, but these are greatly reduced by the constrants, whch can be expressed as equatons of the form r j = c j (4) where r j s dstance between any th and j th partcle and c j s constant But all these relatons are not ndependent Consder three non collnear ponts, and 3 of rgd body, as shown n the fgure 4 If there s no constrant equatons, the number of degrees of freedom becomes 9 If we wrte the constrant equatons, Fgure 4: Rgd body wth three non collnear ponts r = c, r 3 = c 3 r 3 = c 3 (4) Then the number of degrees of freedom becomes 9 3 = 6 That s only sx coordnates are needed Of these 6 coordnates, 3 are the coordnates of CM of rgd body wth respect to the fxed axs and the other three generalzed coordnates are the angles that the axs of rotaton makes wth the fxed coordnate system A rgd body n space thus needs sx ndependent generalzed coordnates to specfy ts confguraton, no matter how many partcles t may contan 4 Rotaton about an axs, Orthogonal matrx Consder a rgd body n whch Cartesan coordnate system x, y, z fxed at the pont O r s the poston vector of a mass pont rotated spacally at an angle θ The coordnate system x, y, z s also rotates wth angles 33

34 CHAPTER 4 RIGID BODY DYNAMICS θ j as shown n the fgure 4 Note that the angle θ j s defned so that the frst ndex refers to the prmed system and the second ndex to the unprmed system Thus from the fgure 4 we can wrte, r = r x + j y + k z = x + jy + kz ( x + j y + k z ) = (x + jy + kz) x = ( ) x + (j ) y + (k ) z x = cosθ x + cosθ y + cosθ 3 z Smlarly y = cosθ x + cosθ y + cosθ 3 z (43) z = cosθ 3 x + cosθ 3 y + cosθ 33 z (44) z z z r r x x z z y y θ 33 x θ 3 θ 3 θ x y y y θ x Fgure 4: Rotaton of a rgd body Equatons 44 consttute a group of transformaton equatons from a set of coordnates x, y, z to a new set x, y, z wth drecton cosnes cosθ j as transformaton coeffcents They form an example of a lnear or vector transformaton, defned by transformaton equatons of the form x = a x + a y + a 3 z y = a x + a y + a 3 z z = a 3 x + a 3 y + a 33 z (45) where a j = cosθ j The equatons 45 can be wrtten n the form of matrx equaton as x y z = a a a 3 a a a 3 a 3 a 3 a 33 x y z (46) [X ] [A] [X] (47) where [X ] s a column matrx wthcomponents (x, y, z ); [X] s a smlar column matrx wth components a a a 3 (x, y, z) and [A] = a a a 3 s the rotaton matrx a 3 a 3 a 33 Consder a rgd body rotated n x y plane by an angle φ as shown n the fgure 43 Then a = cos φ a = sn φ a 3 = 0 a = sn φ a = cos φ a 3 = 0 (48) a 3 = 0 a 3 = 0 a 33 = The rotaton matrx A can be wrtten as cos φ sn φ 0 A = sn φ cos φ 0 = (49) 0 0

43 EULER ANGLES 35 z z y y x x x y Fgure 43: Rotaton of a rgd body n x y plane Aà = cos φ sn φ 0 sn φ cos φ 0 0 0 cos φ sn φ 0 sn φ cos φ 0 0 0 Aà = (40) From the equaton 40, t s clear that à = A That s, the rotatonal matrx A s orthoganal and t can be taken as an operator The matrx A operatng on the components of a vector n the unprmed system yelds the components of the vector n the prmed system Symbolcally, the process can be wrtten as, r = A r = r (4) That s, the transformaton matrx [A] affects rotaton of the rgd body wth one pont fxed has egenvalue + Ths s called Euler s theorem whch states that A general dsplacement of a rgd body wth one pont fxed s a rotaton about some axs A more general theorem than Euler s s proved by Chasles and t states that, The most general dsplacement of rgd body s a translaton of the rgd body plus rotaton of the rgd body It can be shown that, the orthogonal matrx whose determnant s - represents nverson and cannot represent physcal dsplacement of a rgd body 43 Euler angles Euleran angles are three rotatons about three ndependent axes chosen n a certan successve way Consder a rgd body wth ntal system of axes xyz Rotate the rgd body by by and angle φ counterclockwse about the z axs as shown n the fgure 43, and the resultant coordnate system s labeled as ξηζ Then, ζ z η φ φ y x ξ Fgure 44: Rotaton about z axs

36 CHAPTER 4 RIGID BODY DYNAMICS ξ η ζ ξ η ζ = = [A φ ] cos φ sn φ 0 sn φ cos φ 0 0 0 x y z x y z (4) (43) In the second stage, the ntermedate axes, ξηζ are rotated about the ξ axs counterclockwse by an angle θ to ζ z ζ η θ η φ y x ξ ξ Fgure 45: Rotaton about ξ axs produce another ntermedate set, ξ η ζ axes as shown n the fgure 43 The ξ axs s at the ntersecton of the xy and ξ η planes and s known as the lne of nodes The transformaton then s wrtten as, ξ η ζ ξ η ζ = = [A θ ] 0 0 0 cos θ sn θ 0 sn θ cos θ ξ η ζ ξ η ζ (44) (45) Fnally, ξ η ζ axes are rotated counterclockwse by an angle ψ about the ζ axs as shown n the fgure 46 to produce the desred x y z system of axes x y z x y z = cos ψ sn ψ 0 sn ψ cos ψ 0 0 0 = [A ψ ] ξ η ζ ξ η ζ (46) (47) ζ z z y θ φ ψ x y x Fgure 46: Rotaton about ζ axs

44 ANGULAR MOMENTUM AND KINETIC ENERGY OF MOTION ABOUT A POINT 37 Then, the transformaton of axes (xyz) to (x y z ) can be wrtten by usng the equatons 43, 45 and 47 as, x y z x y z = [A φ ] [A θ ] [A ψ ] = [A] x y z where [A] = [A φ ] [A θ ] [A ψ ] By usng equatons 4, 44 and 46, x y z (48) (49) A = A = 4 4 cos φ sn φ 0 sn φ cos φ 0 0 0 3 5 4 0 0 0 cos θ sn θ 0 sn θ cos θ 3 5 4 cos ψ sn ψ 0 sn ψ cos ψ 0 0 0 cos ψ cos φ cos θ sn φ sn ψ cos ψ sn φ + cos θ cos φ sn ψ sn ψ sn θ sn ψ cos φ cos θ sn φ cos ψ sn ψ sn φ + cos θ cos φ cos ψ cos ψ sn θ sn θ sn φ sn θ cos φ cos θ 3 5 3 5 (40) It can be shown that à = A, hence A s orthoganal 44 Angular momentum and knetc energy of moton about a pont Consder a rgd body moves wth one pont statonary, the total angular momentum about that pont s L = m ( r v ) (4) where r and v are the radus vector and velocty, respectvely, of the th partcle relatve to the gven pont Snce r, s a fxed vector relatve to the body, the velocty v, wth respect to the fxed frame arses solely from the rotatonal moton of the rgd body about the fxed pont Then, v = ω r where ω s the angular velocty of the rgd body The equaton 4 becomes z z z r r m y m y x x x y Fgure 47: Rotaton of a rgd body L = m [ r ( ω r )] (4)