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《航空航天动力学》英文版 lecture 9 Virtual Work And the Derivation of Lagrange's Equations

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16.61 Aerospace Dynamics Spring 2003 Derivation of lagrangian equations Basic Concept: Virtual Work Consider system of N particles located at(, x2, x,,.x3N )with 3 forces per particle(f. f, f..fn). each in the positive
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16.61 Aerospace Dynamics Spring 2003 Lecture #9 rtual Work And the Derivation of Lagrange's Equations Massachusetts Institute of Technology C How, Deyst 2003 (Based on Notes by Blair 2002)

16.61 Aerospace Dynamics Spring 2003 Lecture #9 Virtual Work And the Derivation of Lagrangeís Equations Massachusetts Institute of Technology © How, Deyst 2003 (Based on Notes by Blair 2002) 1

16.61 Aerospace Dynamics Spring 2003 Derivation of lagrangian equations Basic Concept: Virtual Work Consider system of N particles located at(, x2, x,,.x3N )with 3 forces per particle(f. f, f..fn). each in the positive direction 5 4 F (x4,x5,x6) 2 N-2 FN. N-2 Assume system given small, arbitrary displacements in all directions Called virtual displacements No passage of time Applied forces remain constant Massachusetts Institute of Technology C How, Deyst 2003 (Based on Notes by Blair 2002)

16.61 Aerospace Dynamics Spring 2003 Derivation of Lagrangian Equations Basic Concept: Virtual Work Consider system of N particles located at ( x x 1 2 , , x3,…x3N ) ) with 3 forces per particle ( 1 2 3 3 , , , F F F …F N , each in the positive direction. F3 F2 F1 (x1, x2, x3) F5 F6 F4 (x4, x5, x6) xi xj xk FN FN-1 FN-2 (xN-2, xN-1, xN) F3 F2 F1 (x1, x2, x3) F5 F6 F4 (x4, x5, x6) xi xj xk FN FN-1 FN-2 (xN-2, xN-1, xN) Assume system given small, arbitrary displacements in all directions. Called virtual displacements - No passage of time - Applied forces remain constant Massachusetts Institute of Technology © How, Deyst 2003 (Based on Notes by Blair 2002) 2

16.61 Aerospace Dynamics Spring 2003 The work done by the forces is termed virtual work W=∑F6x Note use of &x and not dx Note There is no passage of time e The forces remain constant In vector form 6W=∑F·or Virtual displacements MUSt satisfy all constraint relationships d Constraint forces do no work Massachusetts Institute of Technology C How, Deyst 2003 (Based on Notes by Blair 2002)

16.61 Aerospace Dynamics Spring 2003 The work done by the forces is termed Virtual Work. 3 1 N j j j δW F δ x = = ∑ Note use of δx and not dx. Note: • There is no passage of time • The forces remain constant. In vector form: 3 1 i i i δW δ = = ∑F r • Virtual displacements MUST satisfy all constraint relationships, ! Constraint forces do no work. Massachusetts Institute of Technology © How, Deyst 2003 (Based on Notes by Blair 2002) 3

16.61 Aerospace Dynamics Spring 2003 Example: Two masses connected by a rod ○ R Constraint forces R1=-R2=-R2 Now assume virtual displacements Or, and Sr-but the displacement components along the rigid rod must be equal, so there is a constraint equation of the form Yi Ⅴ irtual work: W=R1·or1+R2·r2 R2e e·r1+R2e·r2 (R2-R2)·or So the virtual work done of the constraint forces is zero Massachusetts Institute of Technology C How, Deyst 2003 (Based on Notes by Blair 2002)

16.61 Aerospace Dynamics Spring 2003 Example: Two masses connected by a rod l R1 R2 àr e l R1 R2 àr e Constraint forces: 1 2 à R R = − = −R e2 r Now assume virtual displacements δ 1r , and δ 2 r - but the displacement components along the rigid rod must be equal, so there is a constraint equation of the form 1 2 e r e r r •δ = r •δ Virtual Work: ( ) 1 1 2 2 2 1 2 2 2 1 à à à 0 R r R r r r r r r r W R e R e R R e = • + • = − • + • = − • = 2 δ δ δ δ δ δ So the virtual work done of the constraint forces is zero Massachusetts Institute of Technology © How, Deyst 2003 (Based on Notes by Blair 2002) 4

16.61 Aerospace Dynamics Spring 2003 This analysis extends to rigid body case Rigid body is a collection of masses e Masses held at a fixed distance i Virtual work for the internal constraints of a rigid bod displacement is zero Example: Body sliding on rigid surface without friction Since the surface is rigid and fixed, Or=0, >w=0 For the body, SW=R.Or, but the direction of the virtual displacement that satisfies the constraints is perpendicular to the constraint force. Thus sW=0 Massachusetts Institute of Technology C How, Deyst 2003 (Based on Notes by Blair 2002)

16.61 Aerospace Dynamics Spring 2003 This analysis extends to rigid body case • Rigid body is a collection of masses • Masses held at a fixed distance. ! Virtual work for the internal constraints of a rigid body displacement is zero. Example: Body sliding on rigid surface without friction Since the surface is rigid and fixed, 0, 0 s δ r W = → = δ For the body, δW = • R1 δ 1 W r , but the direction of the virtual displacement that satisfies the constraints is perpendicular to the constraint force. Thus δ = 0. Massachusetts Institute of Technology © How, Deyst 2003 (Based on Notes by Blair 2002) 5

16.61 Aerospace Dynamics Spring 2003 Principle of virtual Work m = mass of particle i R,- Constraint forces acting on the particle F- External forces acting on the particle 2 For static equilibrium (if all particles of the system are motionless in the inertial frame and if the vector sum of all forces acting on each particle is zero) R.+F=0 The virtual work for a system in static equilibrium is δW=∑(R,+F)·or=0 But virtual displacements must be perpendicular to constraint forces. so R.·6r=0 which implies that we have ∑F·br Principle of virtual work: The necessary and sufficient conditions for the static equilibrium of an initially motionless scleronomic system which is subject to workless bilateral constraints is that zero virtual work be done by the applied forces in moving through an arbitrary virtual displacement satisfying the constraints Massachusetts Institute of Technology C How, Deyst 2003 (Based on Notes by Blair 2002)

16.61 Aerospace Dynamics Spring 2003 Principle of Virtual Work mi = mass of particle i Ri = Constraint forces acting on the particle Fi = External forces acting on the particle ! For static equilibrium (if all particles of the system are motionless in the inertial frame and if the vector sum of all forces acting on each particle is zero) 0 R F i i + = The virtual work for a system in static equilibrium is ( ) 1 0 N i i i i δ δ W = = + ∑ R F • r = But virtual displacements must be perpendicular to constraint forces, so 0 i δ i R r • = , which implies that we have 1 0 N i i i δ = ∑F r • = Principle of virtual work: The necessary and sufficient conditions for the static equilibrium of an initially motionless scleronomic system which is subject to workless bilateral constraints is that zero virtual work be done by the applied forces in moving through an arbitrary virtual displacement satisfying the constraints. Massachusetts Institute of Technology © How, Deyst 2003 (Based on Notes by Blair 2002) 6

16.61 Aerospace Dynamics Spring 2003 Example: System shown consists of 2 masses connected by a massless bar. Determine the coefficient of friction on the floor necessary for static equilibrium. (Wall is frictionless. InIA Ing a 4 N Virtual work oW=mgOx-un,dx, Constraints and force balance: 8x=8x2, N2=2mg Substitution mg(-2)6x=0 Results Massachusetts Institute of Technology C How, Deyst 2003 (Based on Notes by Blair 2002)

16.61 Aerospace Dynamics Spring 2003 Example: System shown consists of 2 masses connected by a massless bar. Determine the coefficient of friction on the floor necessary for static equilibrium. (Wall is frictionless.) Virtual Work: W m 1 2 g x N x δ = δ µ − δ 2 Constraints and force balance: 1 2 2 δ x x = δ , 2 N = mg Substitution: mg (1 2 − µ )δ x = 0 Result: 1 2 µ = Massachusetts Institute of Technology © How, Deyst 2003 (Based on Notes by Blair 2002) 7

16.61 Aerospace Dynamics Spring 2003 So far we have approached this as a statics problem, but this is a dynamics course!! Recall d' alembert who made dynamics a special case of statics 6W=∑(R+F-mi)oor=0 →∑(F m1)·Or=0 2 So we can apply all of the previous results to the dynamics problem as well ommen Virtual work and virtual displacements play an important role in analytical dynamics, but fade from the picture in the application of the methods However, this is why we can ignore the calculation of the constraint forces Massachusetts Institute of Technology C How, Deyst 2003 (Based on Notes by Blair 2002)

16.61 Aerospace Dynamics Spring 2003 So far we have approached this as a statics problem, but this is a dynamics course!! Recall díAlembert who made dynamics a special case of statics: ( ) ( ) i 1 1 0 0 R F r r F r r N i i i i i N i i i i i W m m = = = + − • − • = ∑ ∑ "" "" δ δ δ = ⇒ ! So we can apply all of the previous results to the dynamics problem as well. Comments: " Virtual work and virtual displacements play an important role in analytical dynamics, but fade from the picture in the application of the methods. " However, this is why we can ignore the calculation of the constraint forces. Massachusetts Institute of Technology © How, Deyst 2003 (Based on Notes by Blair 2002) 8

16.61 Aerospace Dynamics Spring 2003 Generalized Forces Since we have defined generalized coordinates, we need generalized forces to work in the same"space Consider the 2 particle problem X, X X X 1:2. Coordinates: Xi,x2,x3,x4 Constraint: (Gx3)+(x2-x4)=l DOF 4-1=3 Select n-3 generalized coordinates x1+x3 (x,+x q2 tan x3-x1 Can also write the inverse mapping 41,q2,q3,…q Massachusetts Institute of Technology C How, Deyst 2003 (Based on Notes by Blair 2002)

16.61 Aerospace Dynamics Spring 2003 Generalized Forces Since we have defined generalized coordinates, we need generalized forces to work in the same ìspace.î Consider the 2 particle problem: (x1, x2) (x3, x4) xi xj l θ (x1, x2) (x3, x4) xi xj l θ ( ) ( ) 1 234 2 2 2 1 3 2 4 Coordinates: , , , Constraint: DOF: 4 1 3 x x x x x x − + − x x = l − = • Select n=3 generalized coordinates: q x x q x x q x x x x 1 1 3 2 2 4 3 1 4 2 3 1 2 2 = + = + = − − − ( ) ( ) tan ( ) ( ) • Can also write the inverse mapping: x f i i = (q1 2 , , q q3,…qn , t) Massachusetts Institute of Technology © How, Deyst 2003 (Based on Notes by Blair 2002) 9

16.61 Aerospace Dynamics Spring 2003 3N Virtual work W ∑ FSx Constraint relations x=∑ oxj Sg Substitution GW=∑∑F1n Define generalized force Q=∑ 7 Work done for unit displacement of qi by forces acting on the system when all other generalized coordinates remain constant o=∑gn If qi is an angle, Q is a torque If gi is a length, @ i is a force hav q,'s are independent, then for static equilibrium must Q Massachusetts Institute of Technology C How, Deyst 2003 (Based on Notes by Blair 2002)

16.61 Aerospace Dynamics Spring 2003 Virtual Work: 3 1 N j j j δW F δ x = = ∑ Constraint relations: 1 n j j i i i x x q q =   ∂ δ = δ   ∂   ∑ Substitution: 3 1 1 N n j j i j i i x W F q = =   ∂ δ = δ   q ∂   ∑∑ Define Generalized Force: 3 1 N j i j j i x = q Q F   ∂ =   ∂  ∑ ! Work done for unit displacement of qi by forces acting on the system when all other generalized coordinates remain constant. ⇒ = = δ δ W Q ∑ i i i n 1 q n • If is an angle, is a torque i q Qi • If is a length, is a force i q Qi • If the ís are independent, then for static equilibrium must have: i q 0, 1, 2, Q i i = = … Massachusetts Institute of Technology © How, Deyst 2003 (Based on Notes by Blair 2002) 10

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