Example: Spinning Hoop with Sliding Mass (Continued) 1 2.003J/1.053J Dynamics and Control I, Spring 2007 Professor Thomas Peacock 4/25/2007 Lecture 19 Lagrangian Dynamics: Spinning Hoop with Sliding Mass, Linearization of Equations of Motion, and Bifurcations Example: Spinning Hoop with Sliding Mass (Continued) Lagrangian 1 L = m(a 2 sin2 θΩ2 + a 2˙θ2) + (−mga cos θ)2 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]. Figure 1: Spinning hoop with sliding mass. Figure by MIT OCW.Example: Spinning Hoop with Sliding Mass (Continued) 2 Lagrange’s Equation for θ 2g¨ θ − sin θ cos θΩ + sin θ = 0 (1) a Equilibrium Points θ = 0, π, arccos g aΩ2The third point only exists if gaΩ2 ≤ 1. Stability Analysis Stability around θe = arccos(g/aΩ2) θ = arccos ge 2 ⇒aΩstable. θ = θe + � ⇒ �¨+ Ω2sin2θe� = 0 Oscillatory Behavior. Stability around θe = 0 ˙ ¨θe = 0 ⇒ consider small changes θ = θe + �, θ = �˙, θ = �¨ g�¨ �Ω2− g + � = 0 ⇒ �¨ + ( Ω2− )� = 0 (2) a a Ω2: Controlled parameter. If Ω2 is small, behavior is stable. If Ω2 > ga, behav-ior is unstable. Stable: Ω2 < ga Unstable: Ω2 > ga If we look for a solution to Equation 1 of the form � = Aeλt, we have: λ2 2 Aλt ge + ( − Ω )Aeλt= 0 a g λ = ± (Ω2− ) a If Ω2 < g�a , λ is imaginary ⇒ oscillation. 2 g If Ω >a, λ is real ⇒ exponential growth. 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: Spinning Hoop with Sliding Mass (Continued) Stability around θ e = π θe = π θ = θe + � = π + � From Equation (1) ¨� − sin(π + �) cos(π + �)Ω2 + g sin(π + �) = 0 a sin(π + �) = sin π cos � + cos π sin � ≈ −� cos(π + �) = cos π cos � − sin π sin � ≈ −1 �¨− �Ω2 − g� = 0 a �¨−(Ω2+ g )� = 0 ⇒ Unstable, because of the negative sign in front of the � term. a Regime Diagram Plot a regime diagram. Figure 2: Regime diagram for modeled system. Depending on the the angle and angular velocity, the system may be in a stable or an unstable regime. Figure by MIT OCW. The solutions are symmetrical around the x-axis: the mass can rise on either side. The solutions are symmetrical around the y-axis (Ω and −Ω), because the hoop can spin clockwise or counterclockwise with the same behavior. For θe = π, −π, the equilibrium point is unstable for all Ω. For θe = 0, the equilibrium point is stable until Ω2 = ag . Beyond that Ω the point is unstable. 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].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]. Alternative Method to Derive Linearized Perturbation Equations 4 For θe = arccos ( g ), the equilibrium point exists for g = Ω2 and then θ grows aΩ2 a as Ω2 increases. The equilibrium point is stable. This system exhibits stability behavior known as a pitchfork bifurcation. At first, the mass oscillates about θ = 0. As Ω is increased towards Ω2 = g/a those oscillations continue until Ω2 > g/a. At that point, the mass rises to arccos g/aΩ2 on one side of the hoop. θ continues to increase but it will not reach π or −π. Because perturbation can cause the mass to rise either at θ or −θ, the behavior is called a bifurcation. Alternative Method to Derive Linearized Perturbation Equations This method makes approximations at the level of the Lagrangian. Then one differentiates to obtain linearized equations of motion. Do not use this method if there is any uncertainty about the equilibrium points. This method does not give the nonlinear equations of motion needed to find the equilibrium points. Cart with Pendulum and Spring Example Revisited Figure 3: Cart with Pendulum and Spring. Figure by MIT OCW. This example is explained in full in Lectures 16, 17, and 18. 1 2 1 2 1 k(s−l)2L = (M + m) ˙x + m( ˙s + s 2θ˙2 +2 ˙x( ˙s sin θ +sθ˙ cos θ))+ mgs cos θ − 2 2 2 Alternative Method: 1. Assume equilibrium solution is known. 2. Consider small changes to the variables. 3. Expand L up to its quadratic terms. 4. Use the approximate L in Lagrange’s Equations to obtain linearized quations for stability analysis.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]. � � � � � � � � � � � � Alternative Method to Derive Linearized Perturbation Equations 5 Assume Equilibrium Solution is Known Equilibrium Solution: θ = 0, s0 = l + mg , s = s0 + �, x =any value. Use 0 for k this example. This method helps if you already know all the equilibria. Consider Small Changes to the Variables � � ���� L =1 (M + m) ˙2 +1 m �˙2 + (s0 + �)2 x �θ + (s0 + �) θ˙ θ˙2 x θ˙2 + 2 ˙ ˙ 1 − 2 2 2 θ2 1 2 + M g (s0 + �) 1 − − k (s0 + � − l)2 2 Sine and Cosine Approximations sin θ ≈ θ − θ3 3! cos θ ≈ 1 − θ2 2 Expand L up to its quadratic terms L = 1(M +m) ˙x 2 + 1(˙�2 +s02θ˙2 +2 ˙xs0θ˙)+mgs0 +mg�−mgs0 θ2 − 1 k(s0 +�−l)2 2 2 2 2 1 1 1 − k(s0 + � − l)2 = − k(s0 − l)2 − k(s0 − l)� − k�2 2 2 2 Retained only quadratic nonlinearity in Lagrangian. Keep only quadratic terms because when one differentiates, the quadratic terms will become linear terms. Use Lagrange’s Equations to Obtain Linearized Equations of Motion x: d ∂L ∂L − = 0 dt ∂x˙ ∂x ∂L = (M + m) ˙x + ms0θ˙ ∂x˙∂L = 0 ∂x d ∂L ¨ ¨ = (M + m)¨x + ms0θ = 0 ⇒ (M + m)¨x + ms0θ = 0 dt ∂x˙Cite as: