Advanced Dynamics. - Lecture 1 Constraints. Paolo Tiso Spring Semester 2017 ETH Zürich

Transcription

1 Advanced Dnamics - Lecture 1 Constraints Paolo Tiso Spring Semester 2017 ETH Zürich

2 LECTURE OBJECTIVES 1. Introduce generalized coordinates 2. Formall introduce constraints 3. Distinguish between holonomic and non-holonomic constraints 1. Mathematical form 2. Phsical meaning 4. Pfaffian form and integrabilit: is a constraint reall nonholonomic? Mechanical sstems are subject to constraints. In some cases, the constraints limit the possible configurations; in others, the dictate how the sstem might move from one configuration to the subsequent one. This lecture mathematicall formalizes constraints, categorizes them and provides methods to cast them into convenient forms and check integrabilit.

3 1.1 GENERALIZED COORDINATES Generalized Coordinates (GCs): variables required to define the position of a bod uniquel (tpicall, positions and angles) pitch aw r Particle, 2D: 2 CGs r P q =[, ] P Rigid bod, 2D: 3 CGs z q =[ P, P, ] r q =[,, z] Particle, 3D: 3 CGs Choice of GCs is not unique. But not all choices lead to unique determination of position, e.g.: r P P Q z r P q =[ P, P, Q ] position not uniquel determined! P roll Rigid bod, 3D: 6 CGs q =[ P, P, z P,,, ]

4 1.2 CONSTRAINTS Scalar function of GCs, gen. velocities (GVs), and time. f(q, q, t) =0 Form (1) is the most general and is called non-holonomic. When GVs dependenc is not present, the constraint is called holonomic (*). Further distinguish two cases: Scleronomic (**) f(q) =0 Rheonomic (***) f(q, t) =0 Holonomic constraints govern how a sstem arrives at a certain configuration. EXAMPLE: rigid link L B A r B - r A = L No velocit, no eplicit time dependenc: scleronomic constr. (1) (*) from Greek, holos =entirel, nomos =law, rules the sstem entirel. (**) sclero = rigid + nomos =law. (***) rheo = flow + nomos =law: relates to something changing with time.

5 EXAMPLE 1.1: PENDULUM WITH MOVING SUPPORT (*) B L A A = 0 A = X sin(!t) scleronomic ( B - A ) 2 +( B - A ) 2 - L 2 = 0 scleronomic A = X sin(!t) Rheonomic (eplicit time dependenc) Given prescribed motion (*) Note: A, B are assumed as point masses EXAMPLE 1.2: THIN WHEEL ROLLING WITHOUT SLIPPING z precession C e spin (on the disk) B P(on the ground) Disk is a 3D rigid bod: 6 GCs Spin angle Precession angle nutation angle r C = q = C C z C 2 4 C C z C 3 5

6 EXAMPLE: THIN WHEEL ROLLING WITHOUT SLIPPING (cont d) z R e C B P v C = ẋ C e + ẏ C e Constraints: 1. Disk plane stas vertical = 0 (1) 2. Disk does not leave or penetrate the ground z C - R = 0 (2) 3. Rolling without slipping Working out (3): ẋ C e + ẏ C e - R cos e - R sin e = 0 Velocit transfer formula v B = 0! v C +! r CB = 0 (3)! = e z - e r CB = -Re z = e z - [cos e - sin e ] ẋ C - R sin = 0 ẏ C - R cos = 0 (4) (5) Contain velocities: non-holonomic constraint (?) (1) and (2) eliminate 2 GCs (3) and (4) can eliminate GSs onl if the can be integrated -> Holonomic constraints eliminate GCs

7 Degrees of Freedom (DoFs): minimum number of GCs required to determine the position of a bod p = n - r p n r : # DoFs : # GCs : # hol. constraints r P n = 3; r = 2; p = 3-2 = 1 EXAMPLE 1.3: PLANAR MULTIBODY SYSTEM k Spring is not a constraint! n = 3 9 = 27; r = = 25; p = = 2 Each hinge takes awa two GCs. The roller acts on one GC.

8 If non-holonomic constraints can be integrated, the will become holonomic and also eliminate DoFs. Ever holonomic constraint can be given in Pfaffian form (eact differential): f(q, t) =0! df = Or in velocit form b dividing b : df n dt = X In practice, man constraints are given in a linear velocit form: nx a i (q, t) q i + b(q, t) =0 i=1 i=1 nx i=1 i q i i dq i + B relating velocities, the establish a restriction on the manner in which the sstem can move, given its current state. This is formall a non-holonomic constraint. However, if we can show that the coefficients of (o) are those of a Pfaffian form, then the constraint can be integrated, and it is therefore (o) dt = = 0

9 In other words, we would like to show that equivalent to If so, then the velocit form is an eact differential We must show that Recall that: df n dt = X nx a i (q, t) q i + b(q, t) =0 i=1 i = Ca i q i + df n dt = X a i (q, t) q i + b(q, 2 f(q, i f(q, k k Strateg: check the = 0 i ; Eact velocit form Non-holonomic-like constraint = Cb where C = k

10 EXAMPLE 1.4: ROLLING WITHOUT SLIPPING IN 3D f 1 (q, q) =ẋ C - R sin = 0 f 2 (q, q) =ẏ C - R cos = 0 In general form, we can write q =[ C, C,, ] a 11 ẋ C + a 12 ẏ C + a 13 + a14 + b 1 = 0 a 11 ẋ C + a 12 ẏ C + a 13 + a14 + b 1 = 0 where the coefficients are given b a 11 = 1; a 12 = 0; a 13 = -R sin ; a 14 = 0; b 1 = 0 a 11 = 0; a 12 = 1; a 13 = -R cos ; a 14 = 0; b 1 = 0 Let s look at f 1. The dependenc on C is dropped: C( C, C,, )! C( C,, ) Likewise: a 12 = C a 14 = = = 0 z = C = 0 C( C,, )! C( C, ) C

11 EXAMPLE 1.4: rolling without slipping in 3D (cont d) Now C( C, )a 13 = -C( C, )R sin and therefore a 14 C, )R sin ) = (@C( C, )a 14 ) @(C( C, )R sin ) but = C( C, )R cos 6= So onl possible if C=0. The integration constant does not eist, therefore the constraint is non-holonomic. Non-holonomic constraints do not eliminate GCs, but onl prescribe the wa a motion has to take place from a certain configuration. Note: if prescribed (track), then the constraint is holonomic! z C q =[ C, C,, ] Show proof

12 EXAMPLE 1.5: VEHICLE/SKATE e 2 e v G e 1 Constraint: velocit of point G alwas aligned with ais e 1. Holonomic or non-holonomic? G q =[ G, G, ] e v G e 2 = 0 (1) B epressing both terms in fied coordinates v G = ẋ G e + ẏ G e e 2 = - sin e + cos e (1) becomes - sin ẋ G + cos ẏ G = 0 non-holonomic constraint