The kinetic energy ( ) T. 1 2 n n ( ) n n0. m11 m12 m1n. m21 m22 m2n n n T

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1 The ietic eergy 1 T T= Ttqq (,, ) = ( M + Mq 1 + qmq ) M = M(, tq) = m ( ) T q= q q q 1 1 ( ) M = M(, tq) = m + m m + m m + m m11 m1 m1 m1 m m T M M( tq, ) = =, M = M m1 m m 1

2 Mass matrix elemets where the matrix elemets, m, l, = 1,,...,, are defied by l χq χq m = m (, t q) = ρ dv( X ) t t B χq χq m = m( t, q) = dv( X ) m, 1,..., ρ = = t q B χq χq ml = ml ( t, q) = dv( X ) ml,, l 1,..., l ρ = = q q B

3 Mass matrix elemets 1 T T = T (, t q, q ) = q Mq T ( ) ( ) q = q q q, q = t M 1 m m1 m M M1 m1 m11 m1 = Mtq (, ) = = 1 M T 1 M m m1 m 3

4 The ietic eergy 4

5 The ietic eergy Propositio 9.8 The trasport iertial force may be writte Q m 1 m ta i = i= 1 t q 5

6 The ietic eergy 6

7 Lagrage s equatios d T T ta ga i e ( ) + Q + Q Q Q =, = 1,..., dt q q 1 1 T T tqq mqq qmq l T = (,, ) = l = l, = 1 7

8 Lagrage s equatios Mq = Q + Q + Q cif T it et 8

9 The complemetary iertia force i a scleroomic coordiate system cif T M 1 M t T M M1 1 M1 M Q = q ( ( ) ) q+ q ( + sew( )) + ( ) = q q t q t q q M =, M = T 1 1 M 1 M t ( ( )) q q q q M cif = Q = 1 9

10 Summary 1

11 Regular coordiates Propositio 9.1 The system of geeralized coordiates is regular if ad oly if the mass matrix M is positive defiite. 11

12 1

13 Example 9.4 Uderlyig, i our discussio, is the assumptio that we may use differet sets of geeralized coordiates for the cofiguratio space of the multibody. The choice of coordiates must be decided by the egieer. For istace, for a sigle particle P we may use Cartesia coordiates ( x, y, z ) or spherical coordiates (, r θϕ, ) x = r siθcosϕ y = r siθsiϕ z = r cosθ Let ( ex ey ez) deote the RON-basis correspodig to the Cartesia coordiate system with origi O. The trasplacemet of the particle may be writte x= ( x,y,z; P) = x + e x + e y + e z, ( x, y, z) Ω= χq O x y z 3 The χ q x = e x χ q y = e y χ q z = e z ad thus the Cartesia coordiate system is regular. 13

14 I spherical coordiates the trasplacemet is: χ( r, θϕ, ; P) = x + e rsiθcosϕ+ e rsiθsiϕ+ e rcos θ, ( r, θϕ, ) Ω q O x y z χq = exsiθcosϕ+ ey siθsiϕ+ ezcosθ r χq = exrcosθcosϕ+ eyrcosθsi ϕ+ ez ( rsi θ) θ χq = ex( rsiθsi ϕ) + eyrsiθcosϕ ϕ χq χq χq = r si θ, ( r, θϕ, ) Ω r θ ϕ This spherical coordiate system is thus a regular coordiate system. 14

15 Example 9.5 The coordiate system adopted for the double pedulum i Example 9. is regular sice, for fixed ( θφ, ) Ω, χq w θ χ + w = φ 1 q 1 1 ( icosθ + jsi θ) Xw =, X l1, ( for ) 1 icosθ + jsi θ) l1w + i(cosφ+ jsi φ) Xw =, X l, ( for ) 1 w = w = 15

16 Coordiate chages 16

17 Coordiate chage 1 ( ) T..., q = q q q 1 (... ) q = q q q T x= χq( tqt, ( ); X) = χq ( tq, ( t); X) X B Coordiate chage: q q q= qtq ˆ(, ) q = qˆ ( t, q), qˆ = qˆ 1 ˆq 17

18 Coordiate chages Virtual velocity fields w = w (, tqwx, ; ) = q χq q = 1 w w = wq (, tq, w; X) = q χq j w j j= 1 ( ) T w= w w w 1 ( ) T w = w w w 1 Compoet chage: w = ˆ q q j j= 1 w j w ˆq = q w Jacobia: qˆ q qˆ = j q 18

19 Coordiate chages Propositio 9.13 If the q - coordiate system is regular the the q - coordiate system is regular if ad oly if qˆ det( ) q 19

20 Coordiate chages qˆ qˆ j qt ( ) = qtq ˆ(, ( t)) q = + q, 1,..., j = t q j= 1 q qˆ = + t qˆ q q χ q χq χ e q qˆ Q j = Pda( X ) + ρdv( X ) = da( X ) + j j j q b q P q q B B B = 1 χq qˆ χq χq qˆ e qˆ bρdv( X ) = ( Pda( X ) + bρdv( X )) = Q q q q q q q B j j j = 1 = 1 = 1 B B

21 Covariat ad cotravariat vectors Covariat vector: * ˆq Q = Q q a i e a i e qˆ Q Q Q = ( Q Q Q ) q Cotravariat vector: w ˆq = q w q ˆq = q q qˆ ( = 1) t Lagrage s equatios are o covariat form: a i e Q Q Q = 1

22 Ivariace of Lagrage s equatios Theorem 9. Lagrage s equatios are ivariat uder a regular coordiate trasformatio, i.e. a i e a i e Q Q Q = Q Q Q = Lagrage s equatios are ivariat uder regular chages of the cofiguratio coordiates!

23 Summary 3

24 9.5.1 The rigid part Assume that the motio of the rigid part is give χq (, tqx ; ) = xa(, tq) + R (, tq)( X X A), X B x = x + ω ( x x ) A A x A = = x A, q ω = ω; q = A is a reductio poit. x A, x = q def A def R T ω; = ax( R ), =, 1,..., q 4

25 The rigid part x A = = x A, q ω = ω = ; q Corollary 9.3 x x A A, = x A ; : = q ω ω; =, = 1,,..., q 5

26 The rigid part: The mass matrix A = m l = ; A ; l x ω I ω Iertia tesor: I = (( x x ) 1 ( x x ) ( x x )) ρdv( x) A A A A B t 6

27 The rigid part: The mass matrix m = x x m + ω I ω l c; c; l ; c ; l 7

28 The double pedulum x A ; x = q A ω ω = q ; m = ( x x + x ω p + x ω p ) m + ω I ω, l, = 1,,..., l A; A; l A; ; l Ac A; l ; Ac ; A ; l 8

29 The double pedulum x = O ω 1 = θ ω = φ m = ( x x + x ω p + x ω p ) m + ω I ω = ω I ω l O; O; l O; ; l Oc O; l ; Oc ; O ; l ; O ; l m = x,,, ; x ; m + ω; Iω; l= θφ l c c l c l 9

30 3

31 31

32 Summarizig, 1 1 ml 11 1 l mθθ = mθθ + mθθ = + ml 1, mθφ = mθφ + mθφ = m1 l cos( φ θ), 3 m ml 3 = φφ ad M ml 11 mll 1 + ml 1 cos( φ θ) 3 = M ( θφ, ) = mll 1 ml cos( φ θ) 3 3

33 The rigid part: The iteral force Q i E = S q B q dv( X ) Propositio 9.16 For a rigid part the geeralized iteral force is zero, i.e. Q =, = 1,...,, i 33

34 The rigid part: The exteral force 34

35 Poit force e t(, tqx ; ) = f(, tq) δ ( X X) B ec Geeralized force from cotact χq(, tqx ; ) χq(, tqx ; ) ec Q = (, ) ec f tqδ ( X X) ( ) (, ) dax tq ec q = f B q B 35

36 Example 9.6 Assume that the plaar double pedulum is subjected to a follower poit force ad a couple. The correspodig geeralized force is Q = x ; f + ω; M, = θφ, e, e, e, B B 36

37 Example

38 Example 9.6, cot. 38

39 9.5. The elastic part, elastic eergy Elastic potetial: ue ue = ue( X, E), X B, S = E Elastic eergy: ueq, = ueq, (, tqx ; ) = ue( X, Eq(, tqx ; )) Ueq, = Ueq, (, tq) = ueq, (, tqxdvx ; ) ( ) B q Liear elastic sprig: 1 Ue( q) = q 39

40 The elastic part, iteral geeralized force S = E u E u Q E dv( X ) E = = dv( X ) = dv( X ) =, i q ue q eq, S q Eq q q B B B U ueq, dv( X ) = q q B eq, Q U = q, i eq, 4

41 The elastic part 41

42 The liear isotropic elastic part Corollary 9.5 For a liear elastic part, i 1 Q = q ( X ) qdv( X ) q E E B where = ( X ) is the elasticity tesor for body. Corollary 9.6 For a liear isotropic elastic part Q 1 = ( λ (tr ) + µ ) dv( X ) q E E, i B 4

43 The elastic bar Referece ceter-lie : { X X X s, s l } C = B = +e O 1 q Preset ceter-lie : Ct = x Bt x= χq(, tqx ; ) = xo + e1() t s, s l l 43

44 The elastic eergy 1 ( q+ l ) Ue = Ue( q) = ( q l) 4l Sprig costat : A E 1 A = ν ( λ + µ ) l = 1 + ν 1 ν l 44

45 The elastic geeralized force i U e 1 q Qq = = (( ) 1q ) q l i 1 ( q l)( q+ l) Qq = q l l i q; l Q q ( l) q The elastic geeralized force, small elogatios : i Q q ( l) q 45

46 9.6. Body forces: Gravitatio g = e ( g) z c p Oc B t z c O g χ p U Q dv X dv X dv X q q q q, g q q gq, = gρ ( ) = gρ ( ) = ( pqρ ( ) g) = B B B Ugq, = Ugq, (, tq) = poc(, tq) gm = zc(, tqgm ) 46

47 Exteral forces: Body forces : Q U = Ugq, = Ugq, (, tq) = poc(, tq) gm = zc(, tqgm ) q, g gq, The potetial eergy for the multibody i the gravitatioal field: g = e ( g) z c p Oc B t z c N N U gq, = U gq, = z c(, t q) gm = zc(, t q) gm = 1 = 1 O g Q g U = q gq, 47

48 Double pedulum with torsio sprigs Torsio sprigs 1 1 U ( θ, φ) = U + U = κθ + κ ( φ θ) 1 eq, eq, eq, 1 48

49 Example 9.7 Let the plaar double pedulum i Example 9. be equipped with elastic torsioal sprigs at the joits at O ad A. The elastic potetial eergy of these iteral forces is the give by 1 1 U ( θ, φ) = U + U = κθ + κ ( φ θ) 1 eq, eq, eq, 1 where κ 1 ad κ are the sprig stiffess costats. The correspodig geeralized force is the give by U i eq, Q θ = = κ1θ + κ( φ θ) = ( κ1+ κ) θ + κφ θ U i eq, Q φ = = κ( φ θ) = κθ κφ φ ad ( ) i i i κ1+ κ κ Q = ( Qθ Qφ) = ( θ φ) κ κ 49

50 Example 9.9 The plaar double pedulum i Example 9. is subjected to the force of gravity with potetial eergy U ( θφ, ) = U + U = y ( θφ, ) gm + y ( θφ, ) gm 1 gq, gq, gq, c 1 c 1 where, see Figure 9.13, Thus y c 1 l1 l ( θφ, ) = cosθ, yc ( θφ, ) = l cos cos 1 θ φ l1 l Ugq, ( θφ, ) = ( m1cos θ+ m( l1cosθ + cos φ)) g j O i θ c 1 A φ c B 5

51 j O i θ c 1 A φ c Figure 9.13 The potetial eergy of the double pedulum i the gravitatioal field. The geeralized force correspodig to the gravitatioal potetial reads B U g gq, 1 Qθ = = ( m1+ m) lg 1 siθ, θ Q U l g gq, φ = = φ mgsiφ ad 1 l θ φ θ φ ( ) ( 1 ) 1 si si g g g Q = Q Q = m + m lg mg 51

52 9.6 Exteral forces: Cotact forces β = B = B UB β B B B ec ic = U Exterior cotact surface: ec B Iterior cotact surface: ic B ec B t Q Q ec q e = t ecb χ q da( X ) χq = t da( X ), t = P q ic i i ic B B t ic B t β B t Q = Q + Q c ec ic 5

53 Lagrage s equatios d T T ( ) Q Q = qm = Q + Q + Q dt q q i e T cif i e Q i U = q e, Q e = Q c + Q b, Q = q b U g d T T U c g ( ) e U + Q + = dt q q q q Potetial eergy: U = U + U e g d T T U c ( ) + Q = dt q q q 53

54 Lagrage s equatios The Lagragia: L= T U d L L U ( ) Q = qm = Q + Q dt q q q c T cif c Cotact forces: c ic ec Q = Q + Q U qm = Q + Q + Q q T cif ic ec 54

55 Exercise 4:14 The pedulum system, movig i a vertical plae, cosists of a small ball with mass m fixed, with a iextesible mass-less strig of legth l, to a carriage which may move, without frictio, alog a horizotal lie as show i the figure below. The carriage, with mass m c, is coected to a fixed poit O with a liear elastic sprig, with sprig costat, ad subjected to a exteral force F= Ft () alog its lie of motio. Use Lagrage s equatios to formulate the equatios of motio for the system, ( x=, correspods to ustressed sprig). 55

56 Exercise 4:14 Solutio e l x R O 1 A m c B N 1 N F e 1 ϕ l 3 C g m Cofiguratio coordiates: x,ϕ 56

57 Exercise 4:14 Solutio e l x R O 1 A m c B N 1 N F e 1 ϕ l 3 C g m X 1 xo + pop = xo + e1 ( l + x a), P, X l l Placemet: x= χq ( x, ϕ; P) = xo + pob + pbp = xo + e1 ( l + x) + pbp, P xo + pop = xo + e1 ( l + x + lsi ϕ) + e ( lcos ϕ), P 3 57

58 Exercise 4:14 Solutio Partial derivatives of the trasplacemet: χ t q ( x, ϕ; P) =, P χq X 1 ( x, ϕ; P) = e, P, X l x l 1 χ ( x, ϕ; P) = e1, P x q 3 1, P, X l χq ( x, ϕ; P) =, P ϕ e1lcosϕ+ elsi ϕ, P 3 58

59 Exercise 4:14 Solutio The material velocity: χq χq χq x = x( x, ϕ, x, ϕ; P) = + x + ϕ = t x ϕ X 1 + e 1 x +, P, X l l + e 1 x +, P = 3 + e 1x + e1l ϕcosϕ + e ϕlsi ϕ, P e e e X x, P, X l 1 1 l x, P ( x + l ϕcos ϕ) + e ϕlsi ϕ, P

60 Exercise 4:14 Solutio Kietic eergy: T = T( x, ϕ, x, ϕ) = vcmc + vbm= x mc + (( x + l ϕcos ϕ) + ( ϕlsi ϕ) ) m= 1 1 x m + mc ml cosϕ x ( x ( m + mc ) + ml x ϕcos ϕ+ ml ϕ ) = ϕ ml cosϕ ml ϕ T Mass matrix: M m + mc ml cosϕ = ml cosϕ ml M =, M = 1 1 6

61 Exercise 4:14 Solutio Calculatio of massmatrix elemets: q q m = χ χ ρ dv( X ) = dv( X ) = dm = m + m x x e e ρ xx 1 1 c B dm B dm 3 m χ χ = dv( X ) = ( 1l cos + l si ) ( 1l cos + l si ) dm = ml e e e e ϕϕ ρ ϕ ϕ ϕ ϕ ϕ ϕ B q q dm 3 χq χq mx = dv( X ) = 1 ( 1l cos + l si ) dm = ml cos x e e e ϕ ρ ϕ ϕ ϕ ϕ B dm 3 61

62 Exercise 4:14 Solutio Partial derivatives of ietic eergy: T = x T = ml x ϕsiϕ ϕ T = x ( m + mc ) + ml ϕcosϕ x T = ml x cosϕ + ml ϕ ϕ Lagrage s equatios: d T T i e ( ) Q x Q x = dt x x d T T i e ( ) Q ϕ Q ϕ = dt ϕ ϕ 6

63 Exercise 4:14 Solutio Geeralized iteral forces: Elastic potetial eergy: U e 1 = x Q i x U x e = = x U ϕ i e Qϕ = = Geeralized exteral forces: Gravitatioal potetial eergy: U = mgl cosϕ g U eg, g Qx = = x Q = = mgl siϕ ϕ U eg, g ϕ 63

64 Exercise 4:14 Solutio The exteral force F = e : 1 F e l x χ ef, q p OC Qx = F = F = e1 e1f = F x x R O 1 A m c B N 1 N F e 1 χ ef, q p OC Qϕ = F = F = e1 F = ϕ ϕ ϕ l 3 C g m The exteral force R: χq X χq X = ( x, ϕ; P) = = Q = = x e x R χq ( x, ϕ; ), ϕ er, 1 x l χ er, q P = X l Qϕ = R= ϕ 64

65 Exercise 4:14 Solutio The exteral force N 1 : e l x χ x q N1 = e1 N1 = R O 1 A m c B N 1 N ϕ F l e 1 χq N = = ϕ 1 N1 3 C g m Sum of iteral ad exteral forces: x x x i e Q + Q = mgl si i e Q + Q = + F ϕ ϕ ϕ 65

66 Exercise 4:14 Solutio The equatios of motio: d dt ( x( m + m ) + ml ϕcos ϕ) + x F = c d ( ml x cos ϕ+ ml ϕ) ml x ϕsiϕ+ mgl siϕ = dt x( m + mc ) + ml ϕcosϕ ml ϕ siϕ+ x F = ml x cosϕ+ ml ϕ+ mgl siϕ = m + m ml cosϕ = + + ml cosϕ ml cif i e c ( x ϕ) ( ml ϕ siϕ ) ( x ) ( F mgl siϕ) q Q Q Q M 66

67 Exercise 4:14 Solutio Kietic eergy of the sprig: l l 3 1 X X 1 X 1 x ρ X Ts = Ts( x, ϕ, x, ϕ) = 1 1 dx ( ) dx e x e xρ = = = l l x ρ l l 3 l 1 x l 1 m = 3 3 ρ x Kietic eergy of the rod: 1 1 Tr = Tr( x, ϕ, x, ϕ) = vcmr + ωr Iω c r = 1 l l 1 1 (( x + ϕcos ϕ) + ( ϕsi ϕ) ) mr + ϕ ml r 1 67

68 Lagrage s equatios U qm = Q + Q + Q q T cif ic ec ic Q? β t Bt xc g β β S β, t β Bt S β, t β t 68

69 Exteral forces: Cotact forces β =, B = B UB β, B B B ec ic = U, Exterior cotact surface: ec B Iterior cotact surface: ic B ec B t Q Q ec q e = t ecb χ q da( X ) χq = t da( X ), t = P q ic i i ic B B t ic B t β B t Q = Q + Q c ec ic 69

70 9.7 The iteractio betwee parts Parts:, β Cotact surface: S = B B B β β ic t t t t Two parts are i geometric cotact at poit C o the iteral boudary surface if: x = x = χ ( tqt, ( ); X ) = x = χ ( tqt, ( ); X ), X S, X S β β C C q C C q C C t, C t, β β β Relative velocity: χ χ β β x x x β, q q C : = C C = ( ) q = q q Impeetrability costrait: β x, β C Kiematical cotact: β x =, β C B t xc g β St β β B t Separatig cotact: β x <, β C 7

71 Costrait: Impeetrability A q-coordiate system is said to be compatible with the impeetrability costrait if: (, tqx ; ) β χ (, ; ), q tqx χ β β β C q C β C = ( ) q, ( tq, ) Τ Ω, q = q q x 71

72 β Lemma 9.1 If ad are i geometrical cotact at C, durig the time iterval Τ, ad if the q - coordiate system is compatible with impeetrability coditio, the for (, tq) Τ Ω, β β β χ (, ; ) (, ; ) q tqx χ C q tqxc β (, tqx, C ) ( ) t t β χ (, ; ) (, ; ) q tqx χq tqx C C β ( tqx,, C ) ( ) =, = 1,..., q q β β where xc = χq (, tqx ; C ) = χq (, tqx ; C ), XC St,, XC S t,. Costrait: Impeetrability β β 7

73 Costrait: Kiematically coected Two parts are said to be iematically Coected at the cotact poit C if they Are i geometrical cotact ad if:, β C = x A q-coordiate system is said to be compatible with the iematically coected costrait if: χ (, tqx ; ) χ β q (, tqx ; C ) β ( ) q =, ( tq, ) Τ Ω, q q q = q C 73

74 Costrait: Kiematically coected χ (, ; ) (, ; ) q tqx χ β C q tqxc β ( ) q =, ( tq, ) Τ Ω, q q q = χ (, ; ) (, ; ) q tqx χ β q tqx C C β =, ( tq, ) Τ Ω, = 1,,..., q q β Lemma 9. If ad are iematically coected at C, at time t Τ, ad if the q - coordiate system is compatible with the iematically coected costrait, the for (, tq) Τ Ω, χ (, ; ) (, ; ) q tqx χ β C q tqxc β =, = 1,,..., q q 74

75 Geeralized cotact forces Assume that geeralized force, ad Q β, from β β are i geometrical cotact over the surface t β o, is give by χ β q β Q = t da( x), β, = 1,..., β q S t S, at time t. The ad the correspodig geeralized force, Q β, from o β by β χ β q β Q = t da( x), β, = 1,..., β q S t where the cotact force t β is the tractio vector from β vector from o. β o ad β t is the tractio 75

76 The mechaical iteractio t β β t = t β Bt xc g β β S β, t β Bt S β, t β t The geeralized force fromiteral cotacts betwee parts of the multibody is: 1 1 Q Q ( Q Q ) I, N N N ic β β β β = = + = β, = 1 β, = 1 β, = 1 Q =, = 1,..., N We itroduce the mechaical iteractio: I = Q + Q β β β 76

77 The equatios of motio for the multibody 1 qm Q Q I Q Q N T cif i β ec b = β, = 1 1 N I β = N β, = 1 β, = 1 < β I β 77

78 The mechaical iteractio β χq χ β β β β q β I = Q + Q = t da( x) + da( x) = β q t β q S β t S t β χq χq β ( ) t da( x), q q S t t β = t β β χq χ β β q β I = I = ( ) t da( x), = 1,..., β q q S t 78

79 The cotact force t = T β β t = N + τ β β β β β β τ = Normal compoet: β β N = N ( x,) t C ormal force compoet Tagetial composat: τ β = τ β ( x,) t C frictio force Uilateral cotact: β β N ( x, t), x S, t Τ C C t I mechaical cotact: N β < β x, β C Out of mechaical cotact: β β β N = τ ( x,) t = t = C 79

80 The mechaical iteractio If the q-coordiate system is compatible with the impeetrability costrait the : β χq χq β ( ) = q q ad cosequetly: β β χq χq χq χ β β q β β I = ( ) t da( x) = ( ) ( N + ) da( x) = β q q τ β q q S t S t S β t β χq χq β ( ) τ da( x) q q 8

81 The mechaical iteractio Corollary 9.7 If the q - coordiate system is compatible with the impeetrability costrait at β all cotact poits o S t, the β χq χ β q β I = ( ) τ da( x), = 1,..., β q q S t β β No frictio: τ ( x,) t = I = C 81

82 Equlibrated iteractio I geeral: I β The mechaical iteractio is said to be equilibrated, with respect to the system of cofiguratio 1 coordiates q = ( q, q,..., q ) Ω if β I =, = 1,...,, β, = 1,..., N N 1 β β, I = β, = 1 I this case the equatios of motio for the multibody read qm = Q + Q+ Q + Q T cif i ec b q = qtq ˆ(, ) β β ˆq β β I = I I = I =, = 1,..., q The iteractio is a covariat tesor! 8

83 Equlibrated iteractio Kiematically coected:, β C = x β β Corollary 9.8 If ad are iematically coected at all cotact poits o S t, at time t Τ, ad if the q - coordiate system is compatible with these costraits the the β mechaical iteractio betwee ad is equilibrated. β β χ, q χq χq χ β β q β x C = = I = ( ) da( x) =, = 1,..., q q τ β q q S t 83

84 We itroduce the iteractio I β def β β χq χq χq χ β β q β I = ( ) t da( x) = ( ) da( x) + β t t τ β t t S t S β t β χq χq β β ( ) N da( x) t t S t Scleroomic coordiate system: I β = Coordiate system compatible with impeetrability coditio: β χq χq β β ( ), N t t I β β χq χq β ( ) τ da( x) β t t S t 84

85 The equatios of motio for the multibody 1 qm Q Q I Q Q N T cif i β ec b = β, = 1 β χq χ β β q β I = I = ( ) t da( x), = 1,..., β q q S t If the q-coordiate system is compatible with impeetrability coditio: β χq χ β q β I = ( ) τ da( x), = 1,..., β q q S t 85

86 The Power 86

87 The Power Propositio 9.1 β β β et = et = = β = P P I q I q I = I I β β β 1 ( + 1) where ( ) Corollary 9.9 If the mechaical iteractio betwee β β P = I. et ad β is equilibrated the 87

88 Thermodyamics Material discotiuity surface Temperature: θ, θ β Eergy balace:,, x β t β + h β β =, x β t S Etropy iequality: β h h β ( ), x S β t θ θ β θ = > S θ β, x β t, β β, β β β x t = h, x t S x S, β β β t, x t 88

89 Thermodyamics β β β, Pet = P + P = x β t β β da( x) = I q β = S t Thermodyamic coditio : x t x S, β β β, t β P ( tqq,, ), ( tq, ) Τ Ω, q et Ideal iteractio: β P ( tqq,, ) =, ( tq, ) Τ Ω, q et Dissipative iteractio: β P ( tqq,, ) <, ( tq, ) Τ Ω, q et 89

90 Ideal iteractio Propositio 9. If the mechaical iteractio betwee the ad β β I = I ( tq, ), ( tq, ) Τ Ω, = 1,,..., β I ( tq, ) =, ( tq, ) Τ Ω, = 1,,..., β is ideal ad if Thus the mechaical iteractio betwee ad β is equilibrated. 9

91 Relative velocity & β, β x C, x x β, β β, β, C = xc, + C, Bt xc g β β, β, xc, =, β C, x S β, t & β, xc, P x& β, C 91

92 Coulomb dry frictio β β β β t = + τ β, β β, β, xc = Nx C, + xc, x t = ( x + x )( N + τ ) = x N + x τ, β β,, β β β, β, β C C, C, C, C, & β, β x C, Bt xc g β S β, t & β, xc, P x& β, C 9

93 Coulomb dry frictio If the parts are i mechaical cotact ad the cotact is characterized by Coulomb β dry frictio the the tagetial composat of the cotact tractio τ satisfies x τ, β β β C, = µ s N x β,, β β β C, C, = µ N β, x C, x τ at stic at slip By taig, β C, = x x t x τ, β,,, C β = x β C, N β + β C, β = x β C, N β, x β t β N x, β C, x x x = β β, β, β, C C, C, S, β β, β β xc, = Pet = xc, N da( x) = S β t 93

94 Coulomb dry frictio x τ, β C, = β µ s N β x β,, β β β C, C, = µ N β, x C, x τ x β, C, x t = x N + x τ = N ( x + µ x ), x S, β β, β β, β β β, β, β β C C, C, C, C, t N x + x β β, β, C, µ C, x x µ β β, β, C C, x =, µ > P = N µ x da( x) <, β β β, β C, et C, S β t 94

95 Coulomb dry frictio 95

96 9.7.1 The iteractio betwee rigid parts x = x = χ ( tqx, ; ) = x = χ ( tqx, ; ), X S, X S β β C C q C C q C C t, C t, β β β The velocity of the two material poits is the give by: x = x + ω ( x x ), sys C O C O x = x + ω ( x x ) β β, sys β C O C O β β The relative velocity of the two material poits is the give by: x = x + ω ( x x ) β,, sys β,, sys β, C O C O ω : = ω ω β, β Impeetrability coditio : β St cotact surface at time t x x + ω ( x x ) β, β, sys β, β, sys β, β C O C O 96

97 Defiitio: e = e( t), e e = 1 x = ex, x,, β, sys β β O O O Prismatic joit β, ω = β e x = x + ω ( x x ) = ex, C S β,, sys β,, sys β, β β C O C O O t Impeetrability coditio : x + ω ( x x ) β, β, sys β, β O C O x = ex, x β, β, sys β β β O C C β e = β The cotact surface ormal has to be perpedicular to the directio β of relative traslatio e, St is a prismatic surface. I cylidrical coordiates (, r θ, z) with e = e z the cotact surface is defied by r = r( θ ). 97

98 Revolute joit Defiitio: ( O, e), e = e( t), e e = 1 β,, sys β O = x,, ω β = e ω β, ω β x = ω e ( x x ), x S, β, sys β β C C O C t e Impeetrability coditio : ω e ( x x ), ω, x S β β β β C O C t β e ( x x ) =, x S β C O C t It may be show that i this case S ( O, e) i.e. r= rz ( ). See exercise 5:1! β t is a surface of revolutio with axis 98

99 Simple joits β e β e β e β a) Cylidrical b) Prismatic c) Screw β β e = β e d) Revolute joit e) Spherical f) Plaar 99

100 Simple joits c = ( c c c ), c = e Rotatio axis: ( O, e) x : = x x β,, sys, sys β, sys O O O Defiitio of compoets: β c O c1 c = e 3 1

101 Simple joits β e β e β e β a) Cylidrical b) Prismatic c) Screw β β e = β e 11 d) Revolute joit e) Spherical f) Plaar

102 No-simple joits β Poit cotact Lie cotact β Aother type of cotacts or joits, ot based o a cotact surface, are poit or lie cotacts. This secod category of joits is called o-simple. Examples of poit cotacts ca be foud i ball bearigs ad helical gears o oparallel shafts. Lie cotact is characteristic of cams, roller bearigs ad most gears. The rigid wheel o a rigid surface is also a example of a lie cotact. 1

103 Uiversal (Hoo) joit e γ O e 1 β Permits a trasmissio of rotatioal motio betwee itersectig but o-parallel axes. This is accomplished by usig two revolute joits i successio betwee orthogoal axes. The uiversal joit ivolves a itermediary cruciform rigid part γ. 13

104 The equatios of motio 1 qm Q Q I Q Q N T cif i β ec b = β, = 1 The iteractio: β χq χ β β q β I = I = ( ) t da( x), = 1,..., β q q S t 14

105 Rigid parts i cotact Joit O β β M O f β β O O β f β M O f β = f β M β O = M β O 15

106 Rigid part iteractio β, sys β β Q = x O ; f + ω; MO, = 1,...,, sys x x O ; = q, sys O ω ω = q ;, sys, sys O = xo ; q = x ω = ω = ; q β,, sys, sys β, sys, sys β, sys β,, sys O = O O = ( O ; O ; ) q = O ; q = = x x x x x x x β,, sys O ; β, β β, = = ( ; ; ) q = ; = = ω ω ω ω ω ω ω β, ; q 16

107 Rigid part iteractio The mechaical iteractio: I = Q + Q β β β 17

108 Costrait o a relative velocity compoet Uit vector: c = c () t Costrait: β,, sys β,, sys O O ; = c x = c x q = β O c β = g (, tqq ) =, β g (, tq) = c() t x (, tq), β, sys O ; 18

109 Costrait o a relative agular velocity compoet Uit vector: c = c () t Costrait: β,, β ; = c ω = c ω q = O β = h (, tqq ) =, h tq t tq β, β (, ) = c() ω; (, ) 19

110 Costrait o the relative velocity x = x = x β,, sys, sys β, sys O O O c() t = ( c () t c () t c ()) t 1 3 β i = g ( tqq, ) =, i= 13,,, g (, tq) = c () t x (, tq) β, β i i O; O 11

111 Costrait o the relative velocity x = Τ Ω C = Τ Ω β,, sys O ( tqq,, ), t, q, q ( tq, ), t, q Kiematical costrait Geometrical costrait (, tq) β C = χ (, tqx ; ) χ (, tqx ; ) a, t Τ, q Ω q O q O O 111

112 Rollig ad pivotig ω = ω + ω, β β β β p r ω β β r = β Pivotig compoet: ω, Rollig composat: ω = c ω, + c ω, p β β β r 1 r1 r Π β ω B t c C c = 3 β β B t x = x + ω ( x x ) β, β, sys β, β, sys β β C O r C O 11

113 Rollig ad pivotig without slippig,,sys β C = x,, sys i C i = c x β = g β q β, β, =, i= 13,, g (, tq ) = c x i i C; sys β 3 3 = g β i q =, i= 1, = g q = C (, tq) = Holoomic costrait No-holoomic costrait Ci (, tq) = 113

114 Holoomic costraits Holoomic costrait: C (, tq) =, C (, tq) q 1 C ( tq, ) =, q Ω, C (, tq) l q, for some 1 l l 1 l 1 l+ 1 q = f( tq,, q,..., q, q,..., q) 114

115 Holoomic ad iematical costraits Holoomic costrait: C (, tq) = Kiematical costrait: d C (, tq) C ( tqt, ( )) = gq =, g = g( tq, ) =, = 1,,..., dt q = Kiematical costrait: β = g (, tqq ) = / Holoomic costrait: 115

116 Holoomic ad o-holoomic costraits Kiematical costrait: β = g (, tqq ) = If there exists γ = γ(, tq), G=G(, tq) such that G(, tq) g ( tq, ) = γ ( tq, ), = 1,,..., q The, sice γ, G gq = q = G= G = C = G c= def γ γ = = q The fuctio γ = γ(, tq) is called a itegratig factor to the iematical costrait. If it is ot possible to fid fuctios γ ad G, the the iematical coditio is said to be oholoomic. 116

117 ω = & ϕ O g j x C i 117

118 118

119 Kiematical costraits c = c( t), c c = 1 c x β = g β, (, tqq ) = c ω β = h β (, tqq ) =,, sys O = = Propositio 9.7 The q - coordiate system is compatible with the iematical costraits β β above if ad oly if g (, tq) =, h (, tq) =, (, tq) Τ Ω, = 1,,...,. 119

120 The iteractio betwee rigid parts β, β sys β, β β I = x O ; f + ω; MO, = 1,..., β M O f β β O O f β β M O 1

121 Kiematical costraits f β = cf β + f β, MO = cm O + MO, β β β f =, M = I =, = 1,,...,, The iteractio is the equilibrated! β β β O, 11

122 Iteractio ad et power c = ( c c c ), c = c ( ) 1 3 i i t g (, tq) = c () t x (, tq), h(, tq) = c () t ω (, tq), i= 13,,, = 1,..., β,, sys β, i i O; i i ; O Propositio 9.9 The iteractio may be writte 3 I β = ( g β f β + h β M β ), = 1,,..., i i i i i= 1 where f β i = c β i f ad M β i = c β i M O, i= 13,, are compoets of the costrait forces ad momets relative to the basis c. 1

123 Net power β Pet = I q = β 3 3 β β β β β β β β β Pet = ( ( gi fi + hi Mi )) q = (( gi q ) fi + ( hi q ) Mi ) = i= 1 i= 1 = = I β 3 P β = (( g β q ) f β + ( h β q ) M β ) et i i i i i= 1 = = 13

124 Ideal revolute joit Revolute axis: ( O, c 3 ) β Costraits: x =, c ω =, c ω = β,, sys β, β, O 1 g q =, i= 13,,, h q =, i= 1, i = = i c 3 Ideal joit: β P = ( h q ) M = et 3 3 = 3, 3 = h3 q M3 = I = gi fi + h1 M1 + h M = i= 1 c ω β β A priori uow reactio compoets: f1, f, f3, M1, M 14

125 Ideal revolute joit: The plaar pedulum, a homogeeous sleder rod Cosider a pedulum cosistig of, apart from groud with legth L. Assume that these parts are coected at a fixed poit O by a ideal revolute with axis ( O, ). 1 j O O i θ 1 c:( xy, ) 15

126 Ideal revolute joit: plaar pedulum Groud: x, O = ω = Pedulum: x = x + ω p O c co, ω = ω 1 1 Geeralized coordiates: xyθ,,, p = ( isi θ + j( cos θ)) Oc L 1 x c = ix+ jy, ω 1 = θ 1 L L x O = ix+ jy + θ pco = i( x θcos θ) + j( y θsi θ) = L ix+ jy ( icosθ + jsi θθ ) 16

127 Ideal revolute joit: plaar pedulum 1, 1 L L xo = xo xo = i( x θcos θ) + j( y θsi θ) Kiematical costraits: L x θ cosθ = 1, x O = L y θ siθ = 17

128 Ideal revolute joit: plaar pedulum 1, 1 L L xo = xo xo = i( x θcos θ) + j( y θsi θ) 1, 1, 1, L xox ; = i, xoy ; = j, xo ; θ = ( icosθ + jsi θ) 1, ω = θ = θ ω =, ω =, ω = 1, 1, 1, ; x ; y ; θ Iteractio: I = x f + ω M = i f = 1 1, 1 1, 1 1 x Ox ; ; x O x f I = x f + ω M = j f = 1 1, 1 1, 1 1 y Oy ; ; y O y f 1 1, 1 1, 1 L 1 1 I = x θ O; θ f + ω; θ MO = ( icosθ + jsi θ) f + MO = L L f cosθ f siθ + M x y Oz, 18

129 Ideal revolute joit: plaar pedulum O f y O f x j i θ 1 c:( xy, ) 1 Ix = fx 1 ideal M Oz, = I y = f y 1 L Iθ = ( fxcosθ + fysi θ) 19

130 Ideal revolute joit: plaar pedulum Geeralized coordiates: θ = O x 1,, θθ, x 1, O ; θ = 1, ω = θ θ ω 1, ; θ = I = x f + ω M = M = M = 1 1, 1 1, 1 1 θ O; θ ; θ O O z 13

131 Ideal screw joit Screw axis: ( O, c 3 ) Costraits: β, β,, sys ω β, β, β, x O = c3 L, ω = c3ω, ω, π g q =, i= 1,, h q =, i= 1, i = = i screw lead: L β, ω g3q = L, h3 q = ω π = = β, β = L ( g3 h3) q = π c 3 131

132 Ideal screw joit Ideal joit: ω P = g q f + h q M = Lf + M = β, β, β et ( 3 ) 3 ( 3 ) 3 3 ω 3 = = π ω β, L f3 M3 π + = ( L β I = g f + g f + g h ) f h M h M π A priori uow reactio compoets: f1, f, f3, M1, M 13

133 Revolute joit with frictio β No relative motio: β,, sys β, β, x O =, c1 ω =, c ω =, 3 β c ω = M3 M3c, = M3c, ( f1, f, f3, M1, M) c 3 P β et = 3 β I = ( g f+ hm), = 1,..., i i i i i= 1 A priori uow reactio compoets: f1, f, f3, M1, M, M3 133

134 Revolute joit with frictio Relative motio: β,, sys β, β, x O =, c1 ω =, c ω = M3 > M3c, M = M, ( ω ; f, f, f, M, M ), ω = c ω β, β, β, 3 3r β, β et ( 3 ) 3 ω3 3 = P = h q M = M < if ω > = > < β, β, 3 3r, ( ω3 ; 1,, 3, 1, ) β, if ω3 M f f f M M 3 β = i , r i= 1 I g f h M h M h M A priori uow reactio compoets: f1, f, f3, M1, M 134

135 Exteral forces Q = x ; f + ω; M, = 1,..., e, sys e e O O, sys x x O ; = q, sys O, sys, sys O = xo ; q = x ω ω ω = q ; = ω = ; q M e O f e 135

136 Exteral forces M e, OO f e, Electric motor O ψ ω = e ψ&,sys O = x ω = eψ x, sys O = ω = Revolute joit ω = ; ψ e β I,, = x O; f + ω; MO = ω; MO = e MO = sys ψ ψ ψ ψ = M 136

137 The Lagragia ad the Power theorem qm = Q + Q+ Q + Q + Q T cif i ic ec b Q ic N = 1 β=, 1 I β, i ec b co o Q + Q + Q = Q + Q Coservative geeralized force: Q co V = q Potetial: V= Vtq (, ) ( V = V + V ) e b No-coservative geeralized force: o Q The Lagragia: L= Ltqq (,, ): = Ttqq (,, ) Vtq (, ) The Mechaical eergy: E= pq T L, p = q = 1 137

138 The Lagragia ad the Power theorem Sum of et powers: P ic et N = β, 1 < = β P β et, P (, tqq, ), ic et P = I q β et = β Power expeded by iteral cotact forces: Q ic N = β, 1 < β = I β def ic ic ic = = = 1 P Q q Q q Propositio 9.31 P = P + I where ic ic ic et I N ic = I β β, = 1 < β Corollary 9.11 For a multibody with equilibrated iteractios we have cosequetly ic P =. P = I ad ic ic et 138

139 The Power theorem Propositio 9.3 E P P I L t o ic ic = + et 139

140 The Power theorem Corollary 9.1 For a multibody with equilibrated iteractios E L t o = P Corollary 9.13 o ic L E P I t with the equality sig preset i equatio if the iteral iteractios are ideal. 14

141 Coordiate chage Regular coordiate chage: q q q= qtq ˆ(, ) qˆ E = E + p, t P = P + Q ˆ, t o o o q ic ic ic qˆ P = P + Q t qˆ qˆ L(, tq, q ) Ltqtq (, ˆ = (, ), + q ) t q L E P P E P P t o ic o ic = + = + L t The Power theorem is ivariat uder a regular coordiate chage! 141

142 Rigid part iteractio β, β, sys β, β β I = x O ; f + ω; MO, = 1,...,, x x x ω = ω ω β,, sys, sys, sys O ; = O ; β O ;, β β ; ; ; x x = q, sys, sys O O ;, ω ω = q ;, sys, sys O = xo ; q, = x ω = ω = ; q, e, sys, e, e Q = x O ; f + ω; MO, = 1,..., 14

143 Equatios of motio ad the ideal revolute joit qm = Q + Q+ Q + Q + Q N ic 1 Q = I β β, = 1 T cif i ic ec b β c 3 Revolute joit costraits: giq =, i= 13,, h,, i q = i= 1 = = Revolute joit iteractio: 3 β = i i i= 1 I g f h M h M A priori uow reactios : f1, f, f3, M1, M 143

144 Equatios of motio ad the ideal spherical joit qm = Q + Q+ Q + Q + Q N ic 1 Q = I β β, = 1 T cif i ic ec b Revolute joit costraits: = g q =, i= 13,, i Revolute joit iteractio: I 3 β = gi fi i= 1 A priori uow reactios : f1, f, f3 144

145 Exercise 5:5 Solutio Cofiguratio coordiates: β, 1 RON-basis fixed to bet bar : f = ( f1 f f3) RON-basis fixed to the wheel : e = ( e e e )

146 Exercise 5:5 Solutio e = f ( si θ) + f cosθ Agular velocity of bet bar : ω = e, 3 ω 1 = e 1 ; 3, = ; β ω 1 Agular velocity of the wheel : ω = ω + e ( β) = e + f ( β) = f ( β si θ) + f cos θ, 1 3 ω; = f ( si θ) + f cos θ = e, ω = f ; β

147 Exercise 5:5 Solutio 1 Kietic eergy of bet bar : Kietic eergy of the wheel : T = ω I ω = e I e = e I e = I C 3 C 3 3 C 3 T = 1 v v m+ 1 ω I ω O O C = I 147

148 Exercise 5:5 Solutio A e = f ( si θ) + f cosθ Cotact poit betwee ad : A, v = v + ω p A O OA vo = v B + ω pco = ω pbo = ω ( pba + pao ) = e3 ( f1rcosθ + f3rsi θ + f3r) = = ( f ( si θ) + f cos θ ) ( f Rcos θ+ f ( Rsi θ+ r)) = f ( R+ rsi θ )

149 Exercise 5:5 Solutio A v = v + ω p = f ( R+ rsi θ ) + ( f ( β si θ) + f cos θ) f ( r) = f ( R r β) A O OA R Rollig without slippig: va = f( R r β) = R r β = β = r R β = + cost. r 149

150 Exercise 5:5 Solutio A Kietic eergy of the wheel : T = 1 v v m+ 1 ω I ω = O O C 1 1 ( R+ rsi θ) m+ ( f1( β si θ) + f3cos θ) IC( f1( β si θ) + f 3cos θ) = 1 1 ( R+ rsi θ) m+ (( β + si θ) f1 IC f1+ cos θ f3 IC f3) = m( ( R+ rsi θ) + r cos θ ) = mr = mr 4 15

151 Exercise 5:5 Solutio A 1 Kietic eergy of the multibody : T = T( ) = T + T = I m( ( R rsi ) r cos ) + + θ + θ = I + m( ( R+ rsi ) + r cos ) 4 θ θ 4 T 3 1 = I + m( ( R+ rsi θ) + r cos θ), 4 T = 151

152 Exercise 5:5 Solutio A V b g Potetial eergy i the gravitatioal field: Vg = Vg( ) = cost. Q = = 3 The iteractio betwee ad : I = x f + ω M = ω M = ω M = e M = 3 3,, sys 3 3, B; ; B ; B ; B 3 B M Lagrage s equatios: d T T 3 1 = I m( ( R rsi ) r cos ) M dt + + θ + θ = 4 ω 15