Cite as: Thomas Peacock and Nicolas Hadjiconstantinou, course materials for 2.003J/1.053J Dynamics and Control I, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. � � Example: Cart with Pendulum and Spring (continued) 1 2.003J/1.053J Dynamics and Control I, Spring 2007 Professor Peacock 4/18/2007 Lecture 17 Lagrangian Dynamics: Examples and Equilibrium Analysis Example: Cart with Pendulum and Spring (continued) Figure 1: Cart with pendulum and spring. Figure by MIT OCW. x, θ, s: Generalized Coordinates Lagrangian L = T − V = 1(M + m) ˙x 2 +1 m[ ˙s 2 + s 2θ˙2 + 2 ˙x( ˙s sin θ + sθ˙ cos θ)] + mgs cos θ − 1 k(s − l)2 2 2 2 Equations of Motion x: d ∂L ∂L − = Ξxdt ∂x˙ ∂xCite as: Thomas Peacock and Nicolas Hadjiconstantinou, course materials for 2.003J/1.053J Dynamics and Control I, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. � � � � � � � � 2 Example: Cart with Pendulum and Spring (continued) (M + m)¨x + ms¨sin θ + 2ms˙θ˙ cos θ + ms θ ¨cos θ − msθ˙2 sin θ = 0 (1) θ: d ∂L ∂L − = Ξθdt ∂θ˙ ∂θ sθ ¨+ 2 ˙sθ˙+ ¨x cos θ + g sin θ = 0 (2) s: d ∂L ∂L − = Ξsdt ∂s˙ ∂s ∂L = ms˙ + mx˙ sin θ ∂s˙ ∂L = msθ˙2 + mx˙θ˙ cos θ + mg cos θ − k(s − l)∂s d ∂L = ms¨+mx¨ sin θ+mx˙ θ˙ cos θ−msθ˙2 −mx˙θ˙ cos θ−mg cos θ −k(s−l) = 0 dt ∂s˙ms¨ + mx¨ sin θ − msθ˙2 − mg cos θ + k(s − l) = 0 (3) These equations are highly non-linear. In general, the equations of a real-world system will be highly nonlinear, what can we do analytically? → Identify equilibrium points and analyze their stability (2.003) → Nonlinear dynamics (2.05J) Equilibrium Analysis Let us look for equilibrium points.Cite as: Thomas Peacock and Nicolas Hadjiconstantinou, course materials for 2.003J/1.053J Dynamics and Control I, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. 3 Example: Cart with Pendulum and Spring (continued) Finding Equilibrium Points Set velocities and acceleration to zero (Remaining generalized coordinates have values constant in time). x˙ = x¨ = θ˙= θ ¨= s˙ = ¨s = 0 For (1), 0 = 0 For (2), g sin θ = 0 For (3), −mg cos θ + k(s − l) = 0 In general, the equations can be much more complex: 3 nonlinear coupled equa-tions to be solved simultaneously (can try fsolve in MATLAB) Solve: g sin θ = 0 ⇒ θ0 = 0, π mg −mg cos θ + k(s − l) = 0 ⇒ s = l + cos θ k mg θ0 = 0 ⇒ s0 = l + k mg θ0 = π ⇒ s0 = l − k x is any value = �.θ0 = 0 θ0 = π mg mg s0 = l + s0 = l − k k Stable. x0 = � Unstable. x0 = � Cite as: Thomas Peacock and Nicolas Hadjiconstantinou, course materials for 2.003J/1.053J Dynamics and Control I, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. 4 Example: Cart with Pendulum and Spring (continued) Explaining their physical meaning θ0 = 0: Spring and pendulum hanging vertically down 0 = 0 means cart can take any x value as long as the cart is at rest. θ0 = π: Spring and pendulum standing vertically up. Analyzing Their Stability Close to their equilibrium points, nonlinear systems behave like linear systems. Thus, if we linearize the equations of motion about those static equilibrum points and analyze the linearized system mathematically, we can draw conclu-sions about the nonlinear system’s behavior and stability. mg x = 0, θ0 = 0, s0 = l + k If we make a small disturbance, what happens? Imagine x << 1, θ0 << 1, s = s0 + � where � << 1. � is a small perturbation away from s0. ( ˙s = �, ˙ s¨ = ¨�) Linearize by keeping only linear terms.Cite as: Thomas Peacock and Nicolas Hadjiconstantinou, course materials for 2.003J/1.053J Dynamics and Control I, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. � � 5 Example: Cart with Pendulum and Spring (continued) Approximations: For small angles: sin θ ≈ θ, cos θ ≈ 1 1. (M + m)¨x + m(¨�θ + ˙�θ˙+ ˙�θ˙+ (s0 + �)θ ¨ − (s0 + �)θ˙2θ) ¨ ≈ (M + m)¨x + ms0θ = 0 (1L) 2. (s0 + �)θ ¨+ 2 ˙�θ˙+ ¨x + gθ = 0 ¨ s0θ + ¨x + gθ = 0 (2L) 3: m�¨+ m¨xθ − m(s0 + �)θ˙2 − mg + k(s0 + � − l) = 0 mg m�¨− mg + k(l + + � − l) = 0 ⇒ m�¨+ k� = 0 (3L) k (3L) is uncoupled from x, θ. Consequently, the spring oscillations are indepen-dent of x and θ in the linearized system. k(3L) is decoupled ⇒ describes oscillatory motion with frequency m . (1L) and (2L) are coupled. From equation (1L) we have: ¨ (M + m)¨x = −ms0θ Substitute in (2L) ¨ ms0 ¨ s0θ − θ + gθ = 0 (M + m) s0M ¨ θ + gθ = 0 (M + m) Describes oscillations with frequency g(MM +m) . To solve, assume a solution A sin ωt + B cos ωt. If we perturb pendulum, an oscillation will result.Cite as: Thomas Peacock and Nicolas Hadjiconstantinou, course materials for 2.003J/1.053J Dynamics and Control I, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. 6 Example: Cart with Pendulum and Spring (continued) ¨ From (1L), x¨(M + m) = −ms0θ. Integrate twice with respect to time to get x and θ relationship. (M + m) ¨ x¨ = θ ⇒ We know that x must be oscillatory. −ms0 We know that: x0 = θ0 = 0, s = l + mg k Small disturbances give oscillations. is stable. Let us do the same for x0 = 0, θ0 = π, s0 = l − mg k Approximations: Note cos θ ≈ −1 for θ ≈ π sin(θ0 + δθ) = sin θ0 + d dθ sin θ|θ0 δθ = sin θ0 + cos θ0δθ = sin π + cos πδθ = 0 − δθ so sin(θ + δθ)|θ0 ≈ −δθ = −φ (4) (5) (6) (7) x << 1, θ = π + φ, s = s0 + �(φ, � << 1) θ˙ = ˙φ, ˙s = ˙� ¨ θ = ¨ φ, ¨s = ¨� φ = δθ −φ = −δθ Linearize equations about this equilibrium point 1: (M + m)¨x + m¨s sin θ + 2m ˙sθ˙ cos θ + ms ¨ θ cos θ − msθ˙2 sin θ = 0 (M +m)¨x+m�¨sin(π+φ)+2m�˙φ˙(−1)+m(s0+�)φ¨(−1)−m(s0+�)φ˙2 sin(π+φ) = 0Cite as: Thomas Peacock and Nicolas Hadjiconstantinou, course materials for 2.003J/1.053J Dynamics and Control I, Spring 2007. MIT OpenCourseWare