Lagrangian Dynamics: Examples

Example: Falling Stick (Continued) 1 2.003J/1.053J Dynamics and Control I Professor Thomas Peacock Lecture 16 Lagrangian Dynamics: Examples Example: Falling Stick (Continued) Figure 1: Falling stick. The surface on which the stick rests is frictionless, so the stick slips. Figure by MIT OCW. A stick slides with out friction as it falls. Length: L Mass: M C: Center of Mass Assume uniform mass distribution. Example: Falling Stick (Continued) 2 xc , φ: Generalized Coordinates Figure 2: The falling stick is subject to a normal force, N at the point of contact. Figure by MIT OCW. Holonomic (constraint forces do no work). No tangential forces at P. Normal force is a constraint force. N does no work. Kinematics r c = xc ı̂ + yc ĵ = xc ı̂ + v c = ẋc ı̂ + L sin φĵ 2 L φ̇ cos φĵ 2 Energy 1 1 m|v c |2 + I|ω|2 2 2 1 L2 2 1 2 = m(ẋc + φ̇ cos2 φ) + Ic φ̇2 2 4 2 T = (1) (2) (3) V = mg L sin φ 2 Generalized Forces Ξxc = 0, Ξφ = 0. (Professor Sarma uses Q for generalized forces.) Professor Williams uses Ξ. Example: Falling Stick (Continued) 3 Forces: Conservative [gravity] + Nonconservative [normal]. The constraint force (normal force) does no work. As we change xc and φ, no virtual work (no dis­ placement in direction of force). Lagrangian (L) T −V =L 2 L= 1 L 2 1 L m(ẋ2c + φ̇ cos2 φ) + Ic φ̇2 − mg sin φ 2 4 2 2 Equations of Motion We can derive 1 equation per generalized coordinate For xc : � � ∂L d ∂L − = Ξxc dt ∂ẋc ∂xc ∂L ∂ẋc = mẋc , ∂L ∂xc = 0, Ξxc = 0. d (mẋc ) = mẍc ⇒ mẍc = 0 dt For φ: � � d ∂L ∂L − = Ξφ dt ∂φ̇ ∂φ � 2 � L ∂L m 2 = φ̇ cos φ + Ic φ̇ 2 2 ∂φ̇ � � L2 L ∂L 1 = m − φ̇2 cos φ sin φ − mg φ ∂φ 2 2 2 Ξφ = 0 (4) (5) (6) (7) Substitute (5), (6), and (7) in (4) to obtain (8): d dt �� � � mL2 mL2 2 L 2 cos φ + Ic φ̇ + φ̇ cos φ sin φ + mg cos φ = 0 4 4 2 (8) Differentiating the first term results in (10). Substituting in (8) and com­ bining terms yields the equation of motion for φ. Example: Falling Stick (Continued) d dt � � ∂L ∂φ̇ � �� � � mL2 d cos2 φ + Ic φ˙ dt 4 � � � � 2 mL mL2 2 ¨ = cos φ + Ic φ + − cos φ sin φφ̇ φ̇ 4 2 = 4 (9) (10) � mL2 mL2 mgL mL2 cos2 φ + Ic φ̈− (cos φ sin φ) φ̇2 + (cos φ sin φ) φ̇2 + cos φ = 0 4 2 4 2 (11) � � mgL mL2 mL2 Ic + cos2 φ φ¨ − (cos φ sin φ)φ̇2 + cos φ = 0 4 4 2 This equation is nonlinear. Equation of Motion by Momentum Principles Let us derive the equations of motion using momentum principles as a compar­ ison. Forces Figure 3: Free body diagram of falling stick. Two forces act on the stick, a normal force, N and a gravitational force, mg. Figure by MIT OCW. Apply linear momentum principles: Example: Falling Stick (Continued) Here P = mẋc ı̂ + m L2 ξ˙ cos ξĵ. � F ext = 5 d P dt dP L L = mẍc ı̂ + m( ξ¨ cos ξ − ξ˙2 sin ξ)ĵ dt 2 2 There are no forces in the x-direction, therefore ẍc = 0 In the y-direction: L L N − mg = m ξ¨ cos ξ − ξ˙2 sin ξ 2 2 We need to eliminate N , so use the angular momentum principle. τcext = d d Hc = (Ic ξ˙) dt dt L cos ξ = Ic ξ¨ 2 After combining equations (12) and (13) and algebra: −N (Ic + (12) (13) mL2 L mL2 ˙2 ξ sin ξ cos ξ + mg cos ξ = 0 cos2 ξ)ξ¨ − 4 4 2 Thus, we have derived the same equations of motion. Some comparisons are given in the Table 1. Advantages of Lagrange Disadvantages of Lagrange Less Algebra Scalar quantities No accelerations No dealing with workless constant forces No consideration of normal forces Less feel for the problem Table 1: Comparison of Newton and Lagrange Methods Example: Simple Pendulum 6 Example: Simple Pendulum Figure 4: Simple pendulum. The length of the pendulum is l. Figure by MIT OCW. Kinematics θ is the generalized coordinate. Holonomic system (nomral force at P does not move as θ changes. Does no work). Energy 1 m(lθ̇)2 2 V = −mgl cos θ T = Lagrangian L =T −V = 1 m(lθ̇)2 + mgl cos θ 2 Generalized Forces Ξθ = 0 (No Generalized Forces) Example: Simple Pendulum 7 Equations of Motion ∂L ∂θ̇ = ml2 θ̇, ∂L ∂θ � � d ∂L ∂L − = Ξθ dt ∂θ̇ ∂θ � � d ∂L = −mgL sin θ, Ξθ = 0, dt = ml2 θ¨ ∂θ̇ ml2 θ¨ + mgL sin θ = 0 Alternative View What if we did not realize that gravity is a conservative force? Figure 5: Moving pendulum. When the pendulum rotates by δθ, the distance traversed is lδθ. Figure by MIT OCW. What happens to Lagrange’s Equations? Lagrangian 1 m(lθ̇)2 2 V =0 1 L = T − V = m(lθ̇)2 2 No potential forces, because gravity is not conservative for the argument. T = Example: Cart with Pendulum and Spring 8 Virtual Work and Generalized Forces δwN C = −mgl sin θδθ Ξθ = −mgl sin θ When you displace by δθ the displacement will have a vertical component. It is in the opposite direction of the force. Equations of Motion ∂L ∂θ̇ = ml2 θ̇, ∂L ∂θ = 0,Ξθ = −mgl sin θ � � ∂L d ∂L − = Ξθ dt ∂θ̇ ∂θ mL2 θ¨ − 0 = −mgl sin θ mL2 θ¨ + mgl sin θ = 0 Same equation. It does not matter if you recognize a force as being conservative, just do not account for the same force in both V and Ξ. Example: Cart with Pendulum and Spring 3 degrees of freedom Figure 6: Cart with pendulum and spring. Figure by MIT OCW. Example: Cart with Pendulum and Spring 9 Kinematics Three generalized coordinates x, θ, s Cart: xm = x ym = 0 Pendulum: xm = x + s sin θ ẋm = ẋ + ṡ sin θ + s cos θθ̇ ym = −s cos θ ẏm = −ṡ cos θ + s sin θθ̇ Energy Kinetic Energy (T): 1 1 M ẋ2 + [(ẋ + ṡ sin θ + s cos θθ̇)2 + (s sin θθ̇ − ṡ cos θ)2 ] 2 2 Potential Energy (V): 2 conservative forces - Spring and Gravity 1 V = −mgs cos θ + k(s − l)2 2 Generalized Forces in System No work from normal forces because cart rolls without slipping. Ξx = Ξθ = Ξs = 0 Lagrangian L = T −V = 1 1 1 (M +m)ẋ2 + m(ṡ2 +s2 θ̇2 +2ẋ(ṡ sin θ+sθ̇ cos θ))+mgs cos θ− k(s−l)2 2 2 2 Example: Cart with Pendulum and Spring 10 Equations of Motion For Generalized Coordinate x x: � � ∂L d ∂L − = Ξx dt ∂ẋ ∂x ∂L = (M + m)ẋ + m(ṡ sin θ + sθ̇ cos θ) ∂ẋ ∂L ∂x = 0, Ξx = 0 d [(M + m)ẋ + m(ṡ sin θ + sθ̇ cos θ)] = 0 dt ẋ(t), ṡ(t), s(t), θ(t) are all functions of time. (M + m)ẍ + ms̈ sin θ + mṡ(cos θ)θ̇ + mṡθ̇ cos θ + msθ¨ cos θ − msθ̇(sin θ)θ̇ = 0 For Generalized Coordinate θ θ: � � d ∂L ∂L − = Ξθ dt ∂θ̇ ∂θ ∂L = ms2 θ̇ + mxs ˙ cos θ ∂θ̇ ∂L = mẋs˙ cos θ − mxs ˙ θ̇ sin θ − mgs sin θ, Ξθ = 0 ∂θ d [ms2 θ̇ + mxs ˙ cos θ] − mx˙ s˙ cos θ + mxs ˙ θ̇ sin θ + mgs sin θ = 0 dt 2msṡθ̇+ms2 θ̈+m¨ xs cos θ+mx˙ s˙ cos θ−mxs ˙ sin θθ̇−mx˙ s˙ cos θ+mxs ˙ θ̇ sin θ+mgs sin θ = 0 sθ¨ + 2ṡθ̇ + ẍ cos θ + g sin θ = 0 For Generalized Coordinate s ms̈ + mẍ sin θ − msθ̇2 − mg cos θ + k(s − l) = 0