Advanced Dynamics. - Lecture 4 Lagrange Equations. Paolo Tiso Spring Semester 2017 ETH Zürich

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1 Advanced Dynamics - Lecture 4 Lagrange Equations Paolo Tiso Spring Semester 2017 ETH Zürich

2 LECTURE OBJECTIVES 1. Derive the Lagrange equations of a system of particles; 2. Show that the equation of motion of the 2D rigid body could be obtained by d Alembert s principle by projecting onto virtual displacements; 3. Derive the equation of motion of the 2D rigid body via Lagrange equations as well. Lagrange equations are obtained by rewriting of the d Alembert s principle by using energies. As a result, they are algebraically less involved and can be automatized in a symbolic code.

3 4.1 LAGRANGE EQUATIONS Motivation: d Alembert s principle still requires accelerations in terms of GCs, which are cumbersome to compute. The idea is to start from energies, rather that positions. Consider a system of N particles, ad define the kinetic energy as: Note that: T = d dt 1 2 m iṙ i ṙ i = m i r i 1 2 m i ṙ i 2 and therefore, we can write the d Alembert s principle as apple px apple - F i r i = 0 ) - F i i q k = 0 i If the GCs are kinematically admissible (i.e. identically satisfy the constraints), then: apple - F i = 0 k = 1,..., p i (4) r i q (2) (1) (3)

4 apple i - F i = 0 k = 1,..., p (5) Rewrite this differently Note i = i i and therefore: i Substitute (7) into (5): i d i i d + d i d + @r i i - i = 0 k = 1,..., p k (8) We would like to work only with q k, and eliminate r i. Consider the red-boxed i = i (9)

5 Note also that, by taking the partial derivative with the velocity w.r.t. ṙ i = i q k @ṙ q q k i and therefore, the blue-boxed term could be simply written as q k (11) (10) - Q k = 0, k = 1,..., p (12) The set of equations (12) is called Lagrange equations (LEs)., we get q k Giuseppe Ludovico Lagrangia ( ) Ø LEs are nothing else than d Alembert s principle, written in terms of kinetic energies. Ø As such, the chosen generalized coordinates are supposed to identically satisfy the constraints (i.e. they are unconstrained coordinates) Ø By working with energy, the inertial terms involve less algebra than the d Alembert s principle.

6 In some cases, (some of) the active forces are potential, i.e.: F i = + F nc i Therefore, the generalized forces become Q k = F i = i + F nc i = + Q nc k And the LEs could be written as: q k - Q nc k = 0, k = 1,..., p

7 EXAMPLE 4.1 x y R s m 1 Find the equation of motion using Lagrange Equations. Use y and as generalized coordinates. Write kinematics to satisfy constraints identically: L g -R x b 1 = 0 ẋ 1 = 0 y m 1 = y ẏ 1 = ẏ 2 (1) x 2 = L cos ẋ 2 = -L sin (2) R b We need the kinetic and potential energy: y 2 = y + L sin ẏ 2 = ẏ + L cos T = 1 2 m 1( x y 2 1 )+ 1 2 m 2( x y 2 2 ) V = -m 2 gx 2 By using (1) and (2): T = 1 2 m 1ẏ m 2[ẏ 2 + L Lẏ cos ] V = -m 2 gl cos Lagrange Equations = @ = 0 (m 1 + m 2 )ÿ + m 2 L cos - m 2 L 2 sin = 0 m 2 L 2 + m2 Lÿ cos + m 2 gl sin = 0

8 EQUATION OF MOTION OF 2D RIGID BODY USING D ALEMBERT S PRINCIPLE e y r k r P k r Pk P B The position of the body is fully specified by the position of an arbitrary point P belonging to and one angle. B Therefore, the position of a generic point k is given by r k = r P + r Pk ( ) (1) B O e x The corresponding virtual displacement is r k = r P + r Pk ( ) (2) In order to obtain the virtual displacement the rigid body constraint: r Pk r Pk r Pk = const ) 2 r Pk r Pk = 0 r Pk, we can take the variation of Which yields, as expected, that has to be orthogonal to. (see rigid link example). This can be expressed by posing (1) r Pk = e z r Pk (4) (3) r Pk (1) Alternatively, we could use the velocity transfer formula and use the kinematical method v k = v P + e z r Pk! r k = r P + e z r Pk

9 The linear momentum principle of each particle of the rigid body is given by m k r k - F k - R k = 0 (5) by projecting onto the virtual displacement and summing over all the particles, we obtain r k (m k r k - F k - R k )=0 (6) The constraint forces do not work over virtual displacements r k R k = 0 (7) By substituting (7) and (4) into (6), one obtains ( r P + e z r Pk ) (m k r k - F k )=0 The arbitrariness of the virtual displacements allows to write 2 independent eqns: r P (m k r k - F k )=0 (9) ( e z r Pk ) (m k r k - F k )=0 (10) (8)

10 Equation (10) could be rewritten 2 as r Pk (m k r k - F k ) e z = 0 (11) The resulting equations are m k r k = m k r Pk r k = F k r Pk F k We still need to express (12) and (13) in terms of r P and ; we need to find the accelerations, by using the velocity transfer formula and differentiate in time: (12) (13) ṙ k = ṙ P + e z r Pk (14) r k = r P + e z r Pk + e z ṙ Pk = r P + e z r Pk + e z ( e z r Pk ) = r P + e z r Pk - 2 r Pk (15) (2) Recall the identity a (b c) =b (c a) =c (a b)

11 Recall the definition of the the center of mass r PC = 1 m tot m k r Pk, m tot = Substituting (15) and (16) in (12) and (13): m tot ( r P + e z r PC - 2 r PC )= m tot r PC r P + Note that: m k F k = F m k r Pk (e z r Pk ) = m k r Pk (e z r Pk )= (16) (17) O e y Center of mass r k e x r P r Pk F k = M P m k r Pk 2 e z = I P e z (19) Finally, the equation of motion for the 2D rigid body become: m tot ( r P + e z r PC - 2 r PC )=F (20) m tot r PC r P + I P ez = M P (21) k r Pk P (18) C r PC B

12 (20) (21) m tot ( r P + e z r PC - 2 r PC )=F m tot r PC r P + I P ez = M P Resulting force For a general choice of P, the equations are coupled. However, note that: 1. If P is fixed, than (21) becomes r P = 0 I P ez = M P (22) Resulting torque w.r.t reference point P and (2) vanishes, as. We ill later see that (20) can be used to compute the reaction force at the hinge at P. 2. If P=C (center of mass), then r PC =0 and thus the equations become uncoupled: m tot r C = F I C ez = M C (23) (24)

13 EQUATION OF MOTION OF 2D RIGID BODY LAGRANGE EQUATIONS We first set up the kinetic energy for the body according to its definition and the kinematic relations across arbitrary points of the body: T = 1 2 = 1 2 = 1 2 m k ṙ k ṙ k m k (ṙ P + e z r Pk ) (ṙ P + e z r Pk ) m k ṙ P ṙ P m k ( e z r Pk ) ( e z r Pk )+ m k ṙ P ( e z r Pk ) = 1 2 m totṙ P ṙ P I P 2 + m tot ṙp (e z r PC ) Coupling term for P not being the center of mass where we used the definition of the moment of inertia and the center of mass introduced in the previous section. We now just need to evaluate the different terms of the Lagrange equations.

14 For the translation: d = m tot P + m tot ez r PC = m tot P + m tot ez r PC + m 2 tot e z (e z r PC ) = P For the rotation: = I P + m tot ṙ P (e z r = I P + mtot r P (e z r PC )+m tot ṙp (e z (e z r = m ṙ tot P e PC = m tot ṙp z (e z r PC ))

15 The generalized forces are, for the translation DoFs: and for rotation: Q = yields, as expected, the same equation of motion obtained by using the d Alembert s principle. It is convenient to work with the center of mass, as the kinetic energy simplifies: T = 1 2 Q rp = F = N X m k ṙ P ṙ P I P 2 + m tot ṙp (e z r PC ) r PC = 0 ) T = 1 2 F p = F k (e z r Pk )= By carrying out the Lagrange equations: d - Q rp = 0 P dt m k ṙ C ṙ C I C 2 F k e z (r Pk F k )=e z M - Q = 0

16 EXAMPLE 4.2 k O C r m, I C g A disk of radius r, mass m and centroidal moment of inertia I C rolls without slipping on a circular profile of radius R. The center of the disk is connected to the vertical axis by a spring of stiffness k and zero unstretched length. The spring force always acts in the horizontal direction, while gravity acts in the vertical direction, as indicated. Denote with ψ the rotation angle of the disk about its center C, and with θ the angular position of the center of the disk with respect to the center of the circular profile. Find the equation of motion using Lagrange Equations. Use θ as generalized coordinate. e y R Rolling without slipping, integrable: R = r Spring is unstretched when θ=0. e x We need to write the kinetic and potential energies of the body. The kinetic energy w.r.t. the center of mass C writes: T = 1 2 mṙ C ṙ C I C! 2

17 EXAMPLE 4.2 cont d Use the velocity transfer formula where: ṙ C = ṙ 0 +! r OC ṙ 0 = 0 The velocity of the center of mass is therefore! = -( + )e z = r OC = r sin e x + r cos e y ṙ C = (r + R)(- sin e x + cos e y ) 1 + R e r z and the kinetic energy results in T = 1 m k ṙ 2 C ṙ C I C! 2 = 1 2 m 2 (r + R) I (r + R) 2 C r 2 2 The potential energy is given by the contributions on the gravity and the spring: V = V g + V s = 1 2 k[(r + r) sin ]2 + mg(r + r) cos The Lagrange equation is given by (no non-conservative forces are present) = 0 resulting in applem(r + R) 2 (r + R) 2 + I C + k(r + r) 2 r 2 sin cos - mg(r + r) sin = 0

18 4.2 LAGRANGE EQUATIONS FOR CONSTRAINED COORDINATES So far, we used GCs that were compatible with constraints, i.e. holonomic constraint equations were identically satisfied. This allows to construct virtual displacements always normal to constraint forces, which can then be eliminated using d Alembert s principle/lagrange equations. However, we might face the following situations: Ø Constraint equations are complicated, so it is hard to use them to eliminate coordinates: EXAMPLE 4.3 n= 4 x3 =12 GCs; r=10 holonomic constraints; p=12-10=2 DoFs Use θ 1 and θ 3 as independent GCs, and try to eliminate θ 2 : 2 = f( 1, 3 ) tan 3 = L 1 sin 1 + L 2 sin 2 L 1 cos 1 + L 2 cos 2 - H sin 2 = tan 3 cos 2 + tan 3 L1 L 2 cos 1 - H L 2 difficult! - L 1 L 2 sin 1

19 Ø Some active forces depend on constraint forces: Ø Constraints are non-holonomic: z y e C x B P Ø We simply want to find some constraint forces right away. R In the system on the left, the friction force f on the collar C depends on the constraint force N C : f = -sgn(ẋ C )µn C The equation of motion must include the reaction force N C in order to determine the friction force Some GCs cannot be eliminated since they are related by non-integrable differential equations ẋ C - R sin = 0 ẏ C - R cos = 0 This roller coaster is a complex multibody system. For the design of the rail structure, the reactions between the train and the rail might be the only ones of immediate interest.

20 Idea: choose generalized coordinates that do not satisfy (some of the) constraints to let constraint forces appear in the EoM ê n, q n. Virtual displacement not compatible with the constraint n ˆr f(q, t) =0 trajectory n = rf rf ê 1, q 1 ê 2, q 2 In this case, the virtual displacement does not lay on the tangent to the constraint surface! ˆr n 6= 0 In other words, the mapping between GCs and positions does not identically satisfy the m constraints applied to the system. Constraints have to be enforced f i (q, t) =0, r = r(q, t) i = 1,..., m

21 Consider again a system of N particles. The linear momentum principle for each particle is given by m i r k - F k - R k = 0 The virtual work done on the virtual displacement is r k (m k r k - F k - R k )=0 Unlike than the case of unconstrained 1 coordinates, the virtual work of the reaction forces does not vanish r k R k 6= 0 as the virtual displacements are not compatible with the constraints. What is known is the direction of the reaction force, which is aligned with the gradient of the constraint R k = mx irf i ˆr

22 The constraint forces are directed as the gradient of the constraint equation. Let us consider a system of N particles subject to m holonomic constraints f i (q, t) =0, r k = r k (q, t), i = 1,..., m k = 1,..., N Then, the constraint force on particle k due to constraint i is given by R (i) k i k Where the scalar λ i is proportional to the magnitude of the scalar force*. The total constraint force on the particle k due to all m constraints is given by R k = (*) Note that λ i is not the magnitude of the constraint force, as, in k 6= 1. mx R (i) k = m X k Consider now the virtual work of the linear momentum principle onto virtual displacements not compatible with the constraints: r k (m k r k - F k - R k )=0

23 Clearly, the virtual work of the constraint forces does not vanish: r k R k 6= 0 And the constraint forces cannot be eliminated. R k r k = R = mx i q = N X q = n X j=1 mx j Therefore, the LEs gain an additional term, representing the generalized constraint forces q j + Q nc j - mx Which must be complemented by m constraints f i (q, t) =0, i = 1,..., m j = 0, q q j j = 1,..., n Note that the unknowns are now the n GCs q j and the Lagrange multipliers λ i.

24 EXAMPLE 4.3 m 1 M m 2 P Disk rolls without slipping and if is pulled by the joke by the force F. The force P is always normal to the radius of the disk. Find the Lagrange equations using x and θ as generalized coordinates. (a) Find the constraint. We have to express the fact that the horizontal velocity component of the pin B due to the rotation of the disk must match the velocity of the joke. The velocity of the pin B can be expressed using the velocity transfer formula where v A = R e x and therefore! = - e z v B =(R + cos )e x - sin e y The constraint can be expressed as (o) v B = v A +! r AB r AB = (sin e x + cos e y ) v B e x = ẋ ) ẋ - (R + cos ) = 0 It can be shown that the constraint is integrable, and therefore holonomic. Indeed, the differential form of (o) is an exact differential: dx - (R + cos )d = @x = d = -(R + cos ) ) f = x - R - sin - C =

25 EXAMPLE 4.3 cont d m 2 (b) Find the Lagrange Equations. The kinetic energy writes T = 1 2 m 1 v A I m 2ẋ 2 = 3 4 m 1R m 2ẋ 2 m 1 M P where we used the moment of inertia of a solid disk w.r.t. its center of mass I = 1 2 m 1R 2 Note that the potential function is constant. The center of mass of both bodies stays on the same line, so no gravity potential is present. Also, no springs or other forms of conservative forces are present. The generalized forces associated to the constraint are given by where Q c =[Q c x; Q c ]= r q f r q f =[-1; cos + R] while the generalized forces associated associated to the forces P and F can be calculated by computing their virtual work 1. W = F xe x + P r B + M 1. Note that the same procedure can be followed to obtain the generalized constraint forces. Here, we just used the result developed in the previous theoretical treatment.

26 EXAMPLE 4.3 cont d m 2 P Use the kinematical method to find the virtual displacement v B =(R + cos )e x - sin e y! r B =[(R + cos )e x - sin e y ] The force P is given by m 1 M The Lagrange equations become @x - Q = 0 @ - = 0 f(x, ) =0 P = P cos e x - P sin e y and therefore, the virtual work becomes W = F x +(P + M) = Q x x + Q (1) (2) (3) m 2 ẍ - F + = m 1R 2 - (P + M) - ( cos + R) =0 ẋ - (R + cos ) = 0 m 2 m 2 P P m 1 M m 1 M (1) Is the linear momentum principle (2) Is the angular momentum principle w.r.t. O

27 EXAMPLE 4.3 cont d m 2 Suppose now the presence of friction F f and the vertical slot. We assume that between the pin B P ü The friction force is always opposite to the motion; ü Is proportional to the normal force through the friction coefficient; m 1 M F f = -µ sgn(v B e y )e y magnitude of normal force Velocity component along the guide Note that the friction force F f is an active force, that depends on a constraint force. This is a case in which the constrained Lagrange equations come very handy. We need to include the contribution F f to the generalized forces. For this, we compute the virtual work done by F f on the virtual displacements. W f = F f r B = -µ sgn(v B e y )e y [(R + cos )e x - sin e y ] = -µ sgn( sin ) sin Generalized friction force

28 EXAMPLE 4.3 cont d The Lagrange equations become (4) (5) (6) m 2 ẍ - F + = m 1R 2 - (P + M) - ( cos + R)+µ sgn( sin ) sin = 0 ẋ - (R + cos ) = 0 Generalized constraint forces Generalized friction force A single equation in θ can be always retrieved: from (4): = -m 2 ẍ + F from (6): ẍ =(R + cos ) + sin 2 Then substitute in (5)

29 The same framework is valid also for non-holonomic constraints, which are given in the form No surface exists, nx only tangent plane a i (q, t) q i + b(q, t) =0 n Holding time: nx a i (q, t) q i = 0 Compactly, we can write (o) as: Therefore, the reaction force is defined as q a i q i + b i = @q + Qnc - a q = 0 with a =[a 1,...,a n ] i = 1,..., m mx R = a ia i = 0, ˆr jq 2 R n dˆr