Lagrangian Dynamics: Derivations of Lagrange s Equations

Transcription

1 Constraints and Degrees of Freedom 1.003J/1.053J Dynamics and Control I, Spring 007 Professor Thomas Peacock 4/9/007 Lecture 15 Lagrangian Dynamics: Derivations of Lagrange s Equations Constraints and Degrees of Freedom Constraints can be prescribed motion Figure 1: Two masses, m 1 and m connected by a spring and dashpot in parallel. Figure by MIT OCW. degrees of freedom If we prescribe the motion of m 1, the system will have only 1 degree of freedom, only x. For example, x 1 (t) = x 0 cos ωt x 1 = x 1 (t) is a constraint. The constraint implies that δx 1 = 0. The admissible variation is zero because position of x 1 is determined. For this system, the equation of motion (use Linear Momentum Principle) is mẍ = k(x x 1 (t)) c(ẋ ẋ 1 (t)) mẍ + cẋ + kx = cẋ 1 (t) + kx 1 (t) Cite as: Thomas Peacock and icolas Hadjiconstantinou, course materials for.003j/1.053j Dynamics and Control I, Spring 007. MIT OpenCourseWare ( Massachusetts Institute of Technology.

2 Lagrange s Equations cẋ 1 (t) + kx 1 (t): known forcing term differential equation for x (t): ODE, second order, inhomogeneous Lagrange s Equations For a system of n particles with ideal constraints Linear Momentum D Alembert s Principle ṗ = f ext + f constraint (1) i i i (f ext + f constraint ṗ ) = 0 () i i i f constraint = 0 i (f ext ṗ ) δr i = 0 (3) i i Choose ṗ i = 0 at equilibrium. We have the principle of virtual work. Hamilton s Principle ow ṗ = m i r i, so we can write: i (m i r i f ext ) δr i = 0 (4) i δw = f ext δr i, (5) i which is the virtual work of all active forces, conservative and nonconservative. [ ] d m i r i δr i = m i (ṙ i δr i ) ṙ i δṙ i (6) dt d (6) is obtained by using dt (ṙ δr) = rδr + ṙδṙ Cite as: Thomas Peacock and icolas Hadjiconstantinou, course materials for.003j/1.053j Dynamics and Control I, Spring 007. MIT OpenCourseWare ( Massachusetts Institute of Technology.

3 Lagrange s Equations 3 ṙ i δṙ i can be rewritten as 1 δ(ṙ r) by using δ(ṙ ṙ) = ṙδṙ. Substituting this in (6), we can write: d 1 m i r i δr i = m i (ṙ i δr i ) δ m(ṙ i ṙ i ) (7) dt The second term on the right is a kinetic energy term. δ 1 ow we rewrite (4) as: m(ṙ i ṙ i ) = δ(kinetic Energy) = δt Substitute (5) and (7 into (8) to obtain: m i r i δr i f ext δr i = 0 (8) i Rearrange to give d m i (ṙ i δr i ) δt δw = 0 dt d δt + δw = m i (ṙ i δr i ) (9) dt Cite as: Thomas Peacock and icolas Hadjiconstantinou, course materials for.003j/1.053j Dynamics and Control I, Spring 007. MIT OpenCourseWare ( Massachusetts Institute of Technology.

4 Lagrange s Equations 4 Integrate (9) between two definite states in time r(t 1 ) and r(t ) Figure : Between t 1 and t, there are admissible variations δx and δy. We are integrating over theoretically admissible states between t 1 and t that satisfy all constraints. Figure by MIT OCW. t t d (δw + δt )dt = m i (ṙ i δr i )dt (10) dt t 1 t 1 = m i ṙ i δr i (11) t The right hand side, m iṙ i δr i = 0. t 1 t Why? ṙ i δr i = 0, because at a particular time, δr i (t i ) = 0. Also, we know t 1 the initial and final states. It is the behavior in between that we want to know. t t 1 The result is the extended Hamilton Principle. t (δw + δt )dt = 0 (1) t 1 Cite as: Thomas Peacock and icolas Hadjiconstantinou, course materials for.003j/1.053j Dynamics and Control I, Spring 007. MIT OpenCourseWare ( Massachusetts Institute of Technology.

5 Lagrange s Equations 5 Generalized Fores and the Lagrangian m δw = δw conservative + δw nonconservative = δv + Q j δq j Conservative δw : j=1 onconservative δw : δw = f cons δr i i V f cons = i r i V δw = δr i = δv r i Q j δq j m Q j δq j j=1 m: Total number of generalized coordinates Q j = Ξ j : Generalized force for nonconservative work done q j = ξ j : Generalized coordinate Substitute for δw in (1) to obtain: Define Lagrangian t m (δt δv + Q j δq j )dt = 0 (13) t 1 j=1 L = T V The Lagrangian is a function of all the generalized coordinates, the generalized velocities, and time: (13) can now be written as L = L(q j, q j, t) where j = 1,, 3..., m t [ m ] δl(q j, q j, t) + Q j δq j dt = 0 (14) t 1 j=1 Cite as: Thomas Peacock and icolas Hadjiconstantinou, course materials for.003j/1.053j Dynamics and Control I, Spring 007. MIT OpenCourseWare ( Massachusetts Institute of Technology.

6 Lagrange s Equations 6 Lagrange s Equations We would like to express δl(q j, q j, t) as (a function) δq j, so we take the total derivative of L. ote δt is 0, because admissible variation in space occurs at a fixed time. m [( ) ( ) ( ) ] L L L δl = δq j + δq j + δt q j q j t j=1 t m [( ) ( ) ] L L (δl)dt = δq j + δq j dt (15) t 1 t 1 j=1 q j q j t To remove the δq j in (15), integrate the second term by parts with the following substitutions: ( ) L u = q j ( ) d L du = dt q j y = δq j dy = δq j t m ( ) m t ( ) L L δq j dt = δq j dt t 1 j=1 q j j=1 t 1 q j { } m ( ) [ ( ) ] L t t d L = δq j δq j dt q j t 1 t 1 dt q j j=1 ( ) L t δq j = 0 q j t 1 t m ( ) t m ( ) L d L δq j dt = δq j dt (16) t 1 j=1 q j t1 j=1 dt q j Combine (14), (15), and (16) to get: t m [( ) ( ) ] L d L δq j δq j + Q j δq j dt = 0 q j dt q j t1 j=1 m [ ( ) ( ) ] t d L L + +Q j δq j dt = 0 t 1 j=1 dt q j q j Cite as: Thomas Peacock and icolas Hadjiconstantinou, course materials for.003j/1.053j Dynamics and Control I, Spring 007. MIT OpenCourseWare ( Massachusetts Institute of Technology.

7 Example 1: -D Particle, Horizontal Plane 7 dt has finite values. δq j are independent and arbitrarily variable in a holonomic system. They are finite quantities. Thus, for the integral to be equal to 0, ( ) ( ) d L L + +Q j = 0 dt q j q j Equations of Motion (Lagrange): or: ( ) ( ) d L L Q j = dt q j q j ( ) ( ) d L L Ξ j = dt ξ j ξ j Where Q j = Ξ j = generalized force, q j = ξ j = generalized coordinate, j = index for the m total generalized coordinates, and L is the Lagrangian of the system. Although these equations were formally derived for a system of particles, the same is true for rigid bodies. Example 1: -D Particle, Horizontal Plane Figure 3: -D Particle on a horizontal plane subject to a force F. Figure by MIT OCW. Cite as: Thomas Peacock and icolas Hadjiconstantinou, course materials for.003j/1.053j Dynamics and Control I, Spring 007. MIT OpenCourseWare ( Massachusetts Institute of Technology.

8 Example 1: -D Particle, Horizontal Plane 8 Cartesian Coordinates q 1 = x q = y r = xî + yĵ ṙ = ẋî + ẏĵ v = ṙ ṙ = ẋ + ẏ = q 1 + q Q 1 = F cos θ Q = F sin θ V = 0 For q 1 or (x) L = T V 1 T = m(ṙ ṙ) 1 = m( q 1 + q ) (in horizontal plane, position with respect to gravity same at all locations) L = 0 q 1 L = mq 1 q 1 d L = mq 1 dt q 1 mq 1 0 = F cos θ mq = F sin θ Cite as: Thomas Peacock and icolas Hadjiconstantinou, course materials for.003j/1.053j Dynamics and Control I, Spring 007. MIT OpenCourseWare ( Massachusetts Institute of Technology.

9 Example 1: -D Particle, Horizontal Plane 9 Polar Coordinates Figure 4: -D Particle subject to a force F described by polar coordinates. Figure by MIT OCW. q 1 = r q = φ F = F r ê r + F φ ê φ r = r(t)ê r ṙ = ṙê r + rφ ê φ v = ṙ + r φ q 1 : q : 1 L = T V = m( q 1 + q 1 q ) + 0 L = mq 1 q q 1 ( ) d L = mq 1 dt q 1 L = 0 q Cite as: Thomas Peacock and icolas Hadjiconstantinou, course materials for.003j/1.053j Dynamics and Control I, Spring 007. MIT OpenCourseWare ( Massachusetts Institute of Technology.

10 Example: Falling Stick 10 ( ) d L d = (mq 1 q ) = m(q 1 q 1q + q 1 q ) dt q dt q (r): Q 1 1 = F r Q = F φ r: moment. m( q 1 q 1 q + q 1 q ) = F φ q 1 Example: Falling Stick m( q 1 q + q 1 q ) = F φ mq 1 mq 1 q = F r Figure 5: Falling stick. The stick is subject to a gravitational force, mg. The frictionless surface causes the stick to slip. Figure by MIT OCW. G: Center of Mass l: length Constraint: 1 point touching the ground. degrees of freedom q 1 = x G q = φ Must find L and Q j. Look for external nonconservative forces that do work. one. ormal does no work. Gravity is conservaitve. Q 1 = Q = 0 Cite as: Thomas Peacock and icolas Hadjiconstantinou, course materials for.003j/1.053j Dynamics and Control I, Spring 007. MIT OpenCourseWare ( Massachusetts Institute of Technology.

11 Example: Falling Stick 11 Lagrangian L = T V Rigid bodies: Kinetic energy of translation and rotation 1 1 T = m(ṙ G ṙ G ) + I G (ω ω) l y G = sin φ ẏ G = l cos φφ ω = φ kˆ ṙ G = ẋ G î + ẏ G ĵ = ẋ G î + l cos φφ ĵ ṙ G ṙ G = ẋg + cos φφ 4 [ ] ( ) 1 l 1 1 ml T = q 1 + cos q q + q 4 1 See Lecture 16 for the rest of the example. l Cite as: Thomas Peacock and icolas Hadjiconstantinou, course materials for.003j/1.053j Dynamics and Control I, Spring 007. MIT OpenCourseWare ( Massachusetts Institute of Technology.