Revewng, what we have dscussed so far: Generalzed coordnates Any number of varables (say, n) suffcent to specfy the confguraton of the system at each nstant to tme (need not be the mnmum number). In general, let q, q,, qn be generalzed coordnates. Then, the poston vectors are descrbed by r r ( q, q,..., q ),,,..., N n
In general, we also have some geometrc constrants ( r,, r N, t) 0,,,, d. These are d equatons n 3N (scalar) varables. Let q, q,, qn be the generalzed coordnates or varables.e., r r( q, q,..., qn, t),,,..., N so that the geometrc constrants are satsfed.
If we elmnate all geometrc constrants n n 3 N d. o where n number of generalzed coordnates n n o. Sometmes, one may not want to solve for all the geometrc constrants Then q,..., qn are more than the mnmum needed and not all are ndependent. Now consder the work done by effectve forces n any vrtual dsplacement N W F r 3
Now, consder the poston vector, and ts vrtual dsplacement: r r ( q, q,..., q, t) n n r r q,,..., N q where q vrtual dsplacement n Then, the vrtual work s r N n n N W F q F q q q n W Q q Q F N r r q q. 4
Asde: consder the poston vector: r r( q,..., q, t) Dfferentatng, we get r r r r q q q q t n r r rdt q dt dt q t n r r d r dq dt q t r n r q q n 5
Here Q s generalzed force correspondng to the th generalzed coordnate. q Thus, N W F r Q q n 6
Ex 8: Consder the compound pendulum. Let, x(t) moton of slder. It s a specfed functon of tme Let, be O generalzed coordnates. Fnd: generalzed forces Q, Q (correspondng to the generalzed coordnates y x(t), ). A B x C F(t) 7
Now, to fnd the generalzed forces, we need to frst defne the poston of C n terms of generalzed coordnates: or r [ x Lsn Lsn( )] C L[cos cos( )] r L[cos cos( )( )] C L[ sn sn( )( )] r L[{cos cos( )} {sn C sn( )} ] L[cos( ) sn( ) ] 8
Asde: frst fnd the velocty to fnd vrtual dspalcement: r c [ x L cos L( )cos( )] L[ sn ( )sn( )] r [ x L cos c L( )cos( )] L[ sn ( )sn( )] 9
Now, the force actng at C s: F Thus, the vrtual work done s: or W FL[{cos cos( )} Q Q cos( )} ] Q Q FL{cos cos( )} FLcos( ) Note: f the forces are conservatve, the con generalzed forces are: Q V / q where the potental functon s F W F r C V q q n (,..., ) 0
The potental functon can be found as below: V FL{sn( ) sn( )} h ( ) V FLcos( ) Q or V FL{cos cos( )} Q V FLsn( ) h ( ) V (, ) FL{sn sn( )} h ( ) h ( )
6.6 Lagrange s Equatons (Important: ths dervaton s dfferent from the one n the text). The startng pont s the D Alembert s prncple: N ( F m r ) r 0 Recall that there are also d fnte and g knematc constrants to be satsfed by any vrtual dsplacement of the system:
N ( r, r,..., r N, t) 0,,,..., d (fnte constrants) l ( r,..., r, t) r D 0,,,..., g N (knematc constrants) Let q,..., qn generalzed coordnates. (need not be all ndependent;.e., need not satsfy fnte constrants dentcally) or n n0 3 N d. 3
r r ( q,..., q, t),,,..., N Then n Now, we calculate the dfferent terms n D Alembert s prncple: N N n r F r F ( q ) q n N r n ( F ) q Q q q effectve forces N N n m r m r q q n N r ( m r ) q q r accleraton 4
We now defne the knetc energy of the system of partcles to be N but r n r q q r t T m r r r r r r T m ( q ) ( q ) N n n k q t k qk t or r r T m q q N n n { ( ) k k q q k 5
or r r r r ( ) N n n m q q k q t k qk t r r N m t t n n N r r T ( m ) q q k q q k k r r r r n N N ( m ) q m q t t t T T ( q, q ) T ( q ) T k 0 6
Revew: Dervaton of Lagrange s Equatons N N ( F m r ) r 0 (,...,, ) 0,,,..., r r t d N l r D 0,,,..., g Now express r ( q, q,..., qn, t) 7
N n N q N where r F r F q n d T T m r r q dt q q Q N T m r r 8
Algebrac manpulatons N n r r qk,,,..., N k q r n k k r q k q 0,,,..., d k n N k r r q k q k 0 k n a ( q, t) q 0,,,..., d k k 9
constrants n dfferental form N r r 0,,,..., d N l r 0,,,..., g These are d + g relatons n 3N vrtual dspl. D. O. F. = 3N (d + g) = ( n0 g) n = number of generalzed coordnates D. O. F. 0
One can then show (wth some manpulaton) that N N r d T T m r,,,..., n q dt q q n d T T ( F m r ) r [ Q ] q 0 dt q q Case : Number of generalzed coordnates n n0 3N d degrees of freedom of the system,.e., all holonomc (geometrc) constrants are automatcally satsfed by the choce of generalzed coordnates; and there are no knematc or velocty constrants.
The generalzed coordnates (hence ndependent. Then n d T T dt q q Q q 0 q ) are d T T dt q q Q,,,..., n Ths s one form of Lagrange s equatons for a holonomc n degrees of freedom mechancal system. These are a system of n equatons. ( nd order dfferental equatons)
Lagrange s equatons are nd -order nonlnear ordnary dfferental equatons for n generalzed coordnates q We need to specfy q (0), q (0),,,..., n (n ntal condtons) Ther soluton gves One can then fnd the postons r ( t),,,..., N and the constrant forces R F m r,,,..., N q ( t),,,..., n. 3
Ex. 9: Consder a plane pendulum wth oscllatng support: Wth the coordnate O system shown, the poston: r f ( t) l(cos sn ) P l sn ( l cos f ( t)) The velocty s: r l cos ( l sn f ( t)) P θ - generalzed coordnate no constrant. B y f(t) l P x g m mg 4
The knetc energy s: T mr P r P m [ l f ( t ) l f ( t )sn ] Need to fnd generalzed force Q. W F r Q ; F mg P r d r wth tme frozen ( set t 0) P P df d r P l cos d ( l sn d dt) dt r ( l cos l sn ) P 5
W mgl sn Q mgl sn Thus, the Lagrange s d T T equaton s: dt Computng the varous terms: d dt T T T ml ml f ( t )sn ml ml f sn ml f cos ml f cos ml ml f sn ml f cos ml f cos mgl sn Q 6
or Ex 9 (text): [{ g f ( t)}sn ]/ l 0 x g μ=0 45 m m x μ=0 msldes on m; m sldes on the horzontal surface: 0. All surfaces n contact are smooth. 7
-m g x x absolute poston of m -m g x poston of m relatve to m Fnd: acceleraton of m usng Lagrange s equatons. There are two generalzed coordnates x, x n = degrees of freedom =. (no constrants on x, x ) 45 m m x 8
We proceed step by step and develop the varous quanttes, startng wth poston vectors: r x ; r x ( x x ) / v x ; v ( x x / ) x / T mx m{( x x / ) ( x / ) } ( m m ) x m{ x x x } We need to fnd generalzed forces Q, Q? 9
F m g ; r x ; F mg ; r ( x ) W F r m g x Q x Q x x x Q 0 ; Q mg / Then, the equatons of moton are: T m x : ( m m ) x x x 30
d T m ( ) ( m m ) x x dt x T 0 ; Q 0 x m ( m m) x x 0 T m x : m x x x d T m T ( ) m x x ; 0 dt x x Q mg 3
m x m m g x Solvng the two equatons for x x m g /( m m ) 3
Another form of Lagrange s equatons: Suppose that some forces actng on the system are conservatve,.e., the correspondng forces (as well as the generalzed forces) are dervable from a potental functon then F V Q( t, q, q ) Q( t, q, q ) q potental part nonconservatve part V ( q,..., q n ), 33
The equatons of moton for an n degree of freedom holonomc system take the form: d T T V ( ) Q, dt q q q Let L T V, Then =,,.,n the Lagrangan functon. d L L ( ) Q ( t, q, q ),,,..., n dt q q Ths s the standard form of Lagrange s equatons. 34
Explct form of the equatons of moton The dea here s to express all the quanttes n terms of the generalzed coordnates and ther tme-dervatves Now: N T m r r where d T T V ( ) Q,,,..., n dt q q q r r ( t, q, q,..., q ),,,..., n r n n r q q r t 35
r r r r T m ( q ) ( q ) or N n n k q k t qk t n n N k r r ( m ) q q q q k k r r r r n N N ( m ) q m q t t t n n n T m k ( q, t) q q k a ( q, t) q T0 ( q, t) k 36
T T T T 0 quadratc lnear n ndependent of n q q generalzed velocty q If the constrants are ndependent of tme, (that s, the system s schleronomc), r r r( q) 0 and t n n T T m ( q, t) q q k k k 37
n n n T m ( q, t) q q a ( q, t) q T0 ( q, t) T n Then m ( q, t) q a ( q, t) q d T n n ( ) m ( q, t) q m ( q, t) q a ( q, t) dt q Now n m m ( q, t) q l q a ( q, t) n a q l q l m a t t 38
T T T T q q q q T 0 m n n l qq l q l q T a n q q q T q 0 0 T q combnng all these expressons 39
n n n l l m ( q, t) q ( ) q q l l ql q q n n q T0 V Q ( t, q, q ),,,..., n q q Note that a q m a a a ( ) q t q q t a q m m m - lnear gyroscopc coeffcents 40
Ex 0: Dsk of radus l rotatng wth const.. A double pendulum attached at P (O ) on the dsk. Rods are massless, each partcle of mass m. Moton n horzontal plane. Ω O θ X l O q l y A m x l q Y B m 4
Fnd: T knetc energy; dentfy terms of dfferent type. XYZ fxed frame xyz movng frame attached to the dsk at O = P K k ; 0 va vo ( ) r vo roo k l (cos I sn J A k l l l(cosq sn q ) q, q ; T, T, T 0 4
( ) ( sn cos ) r lq q q A k l(cos q sn q ) l(cos q sn q) v ( l q l )sn q [( l q l )cos q l] A v v ( ) B O B B r B l(cosq cos q ) l(sn q sn q ) ( ) l( sn q q sn q q ) B r l(cosq q cos q q ) 43
Now vb [ l(sn q sn q ) l(sn qq sn q q )] [ l( cos q cos q ) l(cos qq cos qq )] T m ( v A v A v B v B ) [ cos( ) ] ml q q q q q q [ ( cos ) ( cos ) ml q q q q ( q q )cos( q q )] [5 4cos cos cos( )] ml q q q q 44
Lagrange s equatons: T q ml q q q q : [ cos( )] q d dt T ml q q q [( cos ) cos( )] ( ) ml [q q cos( q q ) q q ( q q )sn( q q )] [ sn ( )sn( )] ml q q q q q q 45
T q ml qq q q ( )sn( ) [ sn ( )sn( )] ml q q q q q q [ sn sn sn( )] ml q q q q The equaton of moton s d T T ( ) Q ( q, q, t) dt q q 46
[ cos( ) ml q q q q q ( q q )sn( q q )] [ sn ( )sn( )] ml q q q q q q ml q q sn( q q ) ml [q sn q ( q q )sn( q q )] ml [ sn q sn q sn( q q )] Q( q, q, t) 47
Smplfyng ml [q q cos( q q ) q sn( q q )] ml q sn( q q ) [ sn sn( )] ml q q q Q There s a smlar equaton for q : 48
T q : ml [ q cos( q q ) q ] d dt q ml q q q [( cos ) cos( )] [ T ( ) ml [ q cos( q q ) q ( q q ) q sn( q q ) q ] ml [ q sn q ( q q )sn( q q )] 49
T q ml qq sn( q q ) ml q q q q q q [ sn ( )sn( )] [ sn sn( )] ml q q q The fnal equaton s d T T ( ) Q( q, q, t) dt q q 50
[ cos( ) ( )sn( ) ] ml q q q q q q q q q [ sn ( )sn( )] ml q q q q q q ml q q sn( q q ) ml [ q sn q ( q q )sn( q q )] ml q q q Q [sn sn( )] 5
Smplfyng ml [ q cos( q q ) q sn( q q ) q ] ml q sn( q q ) ml [sn q sn( q q )] Q ( q, q, t) In vector form cos( q q ) q ml cos( q q ) q 5
ml 0 sn( q q ) q sn( q q) 0 q m l 0 sn( q q ) q sn( q q ) 0 q ml sn q sn( q q ) Q sn q sn( q q ) Q 53
or, n compact notaton M q G q G q F( q) Q( q) Note: G G G G T T skew-symmetrc matrces 54
Readng Assgnment: Examples 6.3 6.7 Imp: Note the dscusson n Ex. 6.6 on page 77 (begns at the bottom of p. 76) and page 78. Specally, note the redefned T and V (the fcttous knetc and potental energes). We wll see ths when we study lnearzaton and stablty n the last week of the course. 55
6.7 Lagrange Multplers: (nonholonomc systems or systems wth n n o Recall the equatons for dynamcs of an N partcle system N (F m r ) r 0 Subect to fnte constrants: f (r,r,...,r N,t) 0,,,...,d knematc constrants: N l ( r,..., r, t) r D 0,,,..., g N 56
Note that there are only d geometrc constrants let n n0 (3 N d) be the number of generalzed coordnates. Note: degrees of freedom of the system are n 3 N ( d g). In terms of generalzed coordnates and vrtual dsplacements n the generalzed coordnates, the constrants can be wrtten as n a ( t, q) q 0,,,...,( d g) 57
Note that f n n, 0 then the d geometrc constrants are automatcally satsfed and only g constrant expressons reman. We now assume that the constrants are workless n any vrtual dsplacements permtted by the constrants. Let C,,,..., n be the constrant forces correspondng to the generalzed coordnates q,,,..., n 58
Then, the vrtual work done by constrant forces n any vrtual dsplacement s W R r C q (ths mples that C N n N R r q 0 ) Now, f the vrtual dsplacements q are all ndependent, C 0,,,..., n. (Note: ths does not mean that are zero). R 59
In the present case, we have constrant relatons nvolvng q ' s and, hence, C 0,,,..., n even though n W C q 0. Summarzng: the D Alembert s prncple wrtten n terms of the generalzed coordnates s n d T T dt q q Q q 0 () 60
subected to the assocated constrant relatons: We need some way of makng ndependent. Let,,,...,( d g) be new parameters (equal n number to the constrant relatons) that wll be utlzed to accomplsh ths task (makng n a ( t, q) q 0,,,...,( d g) () qs ' ndependent) qs ' Lagrange multplers 6
The dea s to manpulate the constrants and the workless constrant forces n some way. The constrant equatons are n a ( t, q) q 0,,,...,( d g) Addng these d g n a ( t, q) q 0 (3) Addtonally, the vrtual work relaton as n W C q 0 6
Addng these two Note: at ths pont we have (d + g) unknown multplers unknowns ; n C, the generalzed constrant forces; and the vrtual dsplacements are not ndependent they satsfy (d + g) constrants. n d g C a q qs ' 0 (4) 63
The trck now s to choose 's such that become ndependent assume that 's have values such that d g C a,,,..., n (5) Add the constrant relatons (3) to () qs ' n d T T d g Q a q dt q q 0 (6) Snce qs ' are ndependent wth the related to C ' s by (5), ' s 64
d T T d g Q a,,,..., n (6) dt q q We thus have varables,. q n equatons n n ( d g) The addtonal (d + g) equatons are the constrant relatons n a ( t, q) q d 0,,,...,( d g) 65
Readng Assgnment: Ex: 6-8, 6-9 66
Ex. 6-9 Z Consder two wheels on a common axle of length l. Wheels are rollng on the horzontal plane wthout slppng, wheels reman normal to the ground. Wheel mass concentrated at the hubs. System defned by the postons ( x, y ) X m o G l ( x, y ) Y m ( x, y ) 67
z x o y G l m n : ( x, y) : ( x, y ) m vg n 0 68
vg x x y y n ( r r ) / r r vg n ( x x ) ( y y ) x x y y 0 {( ) ( ) } ( x x ) ( x x ) ( y y ) ( y y ) 0 69
Wheels roll wthout slppng G moves to the axle. Constrants: fnte ( x x ) ( y y ) l 0 () r knematc y y ( x x ) x x ( y y ) or y y y y x x x x 4 general coordnates, constrants n = degrees of freedom = 4 =. ( )( ) ( )( ) 0 () n0 N ( d) 4 3, n 4 70
We now use the generalzed coordnates (x, y, ): x x ( sn ) /, y y ( cos )/, x x ( sn ) /, y y ( cos )/ ths defnton allows the fnte constrant to be satsfed automatcally there are 3 generalzed coordnates, There s only constrant, d = 0, g =. n 3, x sn y cos 0 (3) ( another formofv G n 0) 7
Let q x, q y, q3. Then, the constrant can be wrtten n standard form as n 3 (3) a q d 0 constrant a sn, a cos, a3 0, d 0. The equatons of moton for the constraned system are: d T T d g Q a dt q q =,, 3. 7
T m x x y y etc. ( ) x x ( sn )/ x x cos / y y ( cos ) / y y sn / x x ( sn )/ x x cos / y y ( cos )/ y y sn / 73
The knetc energy (K.E. ) s T ( m)( x y ) m 4 K.E.of the C.M. K.E.of rotaton about the C.M. d q x : T / x mx ; ( T / x ) mx dt T / x 0 d T T d g Q a ; d g dt x x 74
Note that there s only one constrant there s only one Lagrange multpler., a sn, Q 0 (there s no external effectve force) mx sn (4) d q y : T / y my ; ( dt T / y ) my T / y 0, a cos my cos (5) 75
d T T q3 : Q3 a3 dt T / ml /; d ( T / ) ml / dt T/ 0, a 0 3 ml 0 (6) There are 3 equatons of moton + constrant equaton for the four varables x, y, and. Need ntal condtons to determne the moton. 76
example of a moton: Intal condtons: ( x, y ) 0 ( x, y ) ( v,0) 0 0 0 ( 0 0, 0 ) 0 0 mass center at the orgn; - mass center gven a velocty n x-drecton; - gven angular velocty about the z-axs m O G v 0 m y ntal condtons x 77
(6) ml / 0 0, t (7) (4) mx sn, (5) my cos Now, note that y x a t tangent acceler cos sn m m /m s te h force along e t 78
Now x ( sn t)/ m, y ( cos t)/m Thus, the acceleraton n e n drecton s a xcos ysn n sn cos cos sn 0 m m Integratng v = constant v ( t 0) v n n 0 x v cos v cos t 0 0 y v sn v sn t 0 0 velocty components 79
Integratng agan, we get v0 x( t) sn t v (8) 0 y( t) cos t constant ( t) or v0 (9) y( t) ( cos t) v 0 / Then (4) sn t mx m( v0 or mv 0 sn (0) t) Note: turned out to be constant here. 80
Clearly wth t x v v ( y ) ( ) 0 0 () = Path of G s a crcle wth center at (x, y) (0, v / ) and radus v / 0 0 8
v 0 / path of G O y v 0 x 8
6.8 Conservaton Laws: The Lagrange s equatons for a holonomc system wth n degrees of freedom and n generalzed coordnates are: d T T V ( ) Q,,,..., n dt q q q These equatons can be put n frst-order form by defnng generalzed momenta: L T p,,,..., n q q 83
Then, the equatons can be wrtten as: d T V ( p) Q, dt q q L p,,,..., n q Suppose that there s a system for whch a generalzed coordnate, say q s, s absent from the Lagrangan L although ts tme dervatve does appear,.e., L T ( q,..., q, q,.., q q, q,... q ) s s n, n V ( q,..., qs, qs,.., qn) Further, suppose that the generalzed force Q s 84 0.
Then Lagrange s equatons gve for the generalzed coordnate q s : d L ( ps) 0 or ps constant dt q Ths says that the generalzed momentum assocated wth the coordnate q s s conserved remans constant throughout moton. q s s called an gnorable coordnate. The term gnorable refers to the fact that the degree of freedom correspondng to the coordnate q s can be gnored from the formulaton of the problem. s 85
Ex 9 (text): x g μ=0 45 m m x μ=0 msldes on m; m sldes on the horzontal surface: All surfaces n contact are smooth. 0. 86
-m g x x absolute poston of m -m g x poston of m relatve to m Fnd: Equatons of moton of the system, gnorable coordnates, and the conserved quanttes. There are two generalzed coordnates x, x n = degrees of freedom =. (no constrants on 87 x, x ) 45 m m x
We proceed as before and develop the varous quanttes, startng wth poston vectors: r x ; r x ( x x ) / v x ; v ( x x / ) x / T mx m{( x x / ) ( x / ) } ( m m ) x m{ x x x } We need to fnd generalzed forces Q, Q? 88
F m g ; r x ; F mg ; r ( x ) W F r m g x Q x Q x x x Q 0 ; Q mg / Then, the equatons of moton are: T m x : ( m m ) x x x 89
Note that T does not depend on x. In addton, Q 0. So, x s an 'gnorable' coordnate, and the corrospondng generalzed momentum p x s conserved. T m ( ) ( m m) x x cons tant p x x T m x : m x x x d T m T ( ) m x x ; 0 dt x x mg Q m x x m m g 90
Lagrangan ndependent of tme: Suppose that L does not depend explctly on tme. Then, L t 0 Now, the Lagrangan s L=T-V, and t depends both on generalzed coordnates and generalzed veloctes. The total dervatve of Lagrangan s: n n dl L L q k dt k qk k q k Now, Lagrange's equatons gve d L L ( ) 0 ( for the case of Qk ' s 0) dt q q k k q k 9
So, we can wrte n n n dl d L L d L ( ) q q ( q ) dt dt q q dt q k k k k k k k k k n n d L dl d L or ( q ) 0 ( q k L) 0 dt q dt dt q k k k k Ths shows that the quantty remans constant durng moton,.e., n L ( q k L ) h Jacob ' s Integral q k k k n L ( q k L ) q k k We can further manpulate ths as follows. 9
Consder the expresson for a Lagrangan: n n n L T T T0 V q q q V rs r s s s r s s Snce the Lagrangan s ndependent of tme, the coeffcents,, and depend on generalzed rs coordnates only, and thus s n L h ( qk L) T T ( T T T0 V ) T T0 V q k k When the knetc energy s a homogeneous quadratc functon of the generalzed veloctes, 93
we have and thus n n T T q q r s rs r s n L h ( q k L) T ( T V ) T V q k k E ( total energy) In other words, n a natural system for whch the Lagrangan does not depend explctly on tme, the total energy of the system E s conserved. 94
Example: Sphercal pendulum Consder the sphercal pendulum. The knetc and potental energes are: The Lagrangan s [( ) ( sn ) ] T m L L V mgl( cos ) x O [( ) ( sn ) ] ( cos ) L m L L mgl z L y mg 95