The Chaotic Motion of a Double Pendulum Carl W. Akerlof September 26, 2012 The following notes describe the kinematics of the double pendulum. The starting point is a pendulum consisting of two point masses, m, and m2, suspended by massless wires of length l1 and l2. The treatment of this case can be found at: http://scienceworld.wolfram.com/physics/DoublePendulum.html For a real system, the equations of motion depend in a more complicated way on the distribution of mass that is essential for modeling the physical pendulum used in this experiment. Figure 1. Point mass double pendulum. Figure 2. Extended mass double pendulum. University of Michigan 1 Department of Physics Double pendulum with point masses: xl11 sin 1 yl11 cos 1 xl21sin l 2 sin 2 yl21cos 122 lcos xl11 cos 11 yl sin 11 11 xl21 cos 112 l cos 22 yl21 sin 112 l sin 22 222 rl111 222 22 rl21112 2 ll cos 121222 l Double pendulum with distributed masses: xu11 sin 11 v cos 1 yu11 cos 111 v sin xl21 sin 1 u 2 sin 2 v 2 cos 2 yl2 1 cos 12 u cos 22 v sin 2 xu11 cos 111 v sin 11 yu sin v cos 11 111 11 xl21 cos 112 u cos 222 v sin 22 yl21 sin 112 u sin 222 v cos 22 222 222 rr111 ; ruv 1 11 222 22 2 22 rl21112 2 lu cos 12 v 2 sin 12 122 r 2 ; ruv 2 22 Assume mirror symmetry: vv12 0 KE1 m r22 m l 22 2 m u l cos m r 22 2 11 1 211 2 21 2 1 2 2 2 PE m11 u g cos 1 m 21 l g cos 1 m 2 u 2 g cos 2 University of Michigan 2 Department of Physics Thus, the Lagrangian for the system is: LTV1 mr 22 ml 22 2 mul cos mr 22 2 11 1 211 2 2 1 2 1 12 2 2 2 mu11 g cos 1 mlg 21 cos 1 mu 2 2 g cos 2 This leads directly to the equations of motion which we shall investigate shortly. Dynamics of the physical pendulum The aim of this experiment is to compare the actual dynamical behavior of a real physical pendulum with a mathematical simulation. To this end, we need to characterize the pendulum properties as accurately as possible and incorporate these into the appropriate equations of motion. One approximation will be involved: the motion will be assumed frictionless. This simplification is driven principally by the lack of any very elegant fundamental theory although it would actually be fairly trivial to incorporate velocity-dependent damping in the dynamical modeling. The moving parts for the real pendulum are: 2 primary axis bearings 2 primary arms 2 secondary axis bearings 2 secondary axis spacers 1 secondary axis cap screw 1 secondary axis hex nut 1 secondary arm The vendor for this apparatus, chaoticpendulums.com, has conveniently provided the dimensions for these parts as shown in Figure 3. Note that all dimensions are in millimeters. The required spatial mass moments can be calculated analytically for each item. The contribution that each one provides to the kinetic and potential energies is given below. For future reference, the distance between the primary and secondary axes is denoted l1 and equals 173 mm. Bearing: A bearing facilitates low-friction axial rotation shear while constraining the shaft location. Thus, the inner race of the bearing can be rotating with angular velocity, , while the outer race is rotating with an angular velocity of . To keep the modeling of this component simple, it is assumed that the bearing is homogeneous with a rotation rate at radius, r, that is linearly interpolated by the distances from the inner and outer radii. Primary axis bearing: University of Michigan 3 Department of Physics KE1 m r 22 2 b 1 PE 0 Secondary axis bearing: KE1 ml22 mr 2 22 mr 2 mr 2 2 2 bb11 1 b 12 b 2 PE mb l11 g cos Primary arm: 1 22 KE m11 r 2 PE m11 u g cos Secondary axis components: The cap screw, spacers and hex nut are all constrained to rotate rigidly with the secondary arm. KE1 m r22222 m r m r 2 csn2 PE mcsn 2 m m l11 g cos Secondary arm: 1 22 2 2 KE m21 l 12 m 2 u l 1 1 2 cos 2 1 m 2 r 2 2 PE m21 l g cos 1 m 2 u g cos 2 Thus, the Lagrangian can be represented by: La1122 ab cos a c cos c cos 2211 1 12 12 2 1 1 2 22 2 1 1 2 2 where: amrmrmmmmml2422 22 2 11 1bbcsn 2 1 amr2 2 12 b 22 222 amrmrmrmrmr22 2 22bcsn bmul12 2 1 cmummmmmlg1122 2 bc2 sn 1 cmug22 The canonical momenta can be calculated from the Lagrangian: University of Michigan 4 Department of Physics L paab11111212212 cos 1 L pab cos a 21212211222 2 The Hamiltonian is given by: Hp11 p 2 2 L 11aab22 cos ac cos c cos 2211 1 12 12 2 1 1 2 22 2 1 1 2 2 By replacing i by pi , one can convert the Hamiltonian into a form that depends only on i and pi . ap22 1 bp 2 1 2 aa11 22 b bp1112 a p 2 2 aa11 22 b and ba12 b 12 cos 2 1 Thus: 22 1 ap22 12 a 12 b 12 cos 2 1 ppap 1 2 11 2 Hcccos cos 2 2 1122 aa11 22 a 12 b 12 cos 2 1 The coupled equations of motion follow: University of Michigan 5 Department of Physics H ap22 1 a 12 b 12 cos 2 1 p 2 1 p 2 1 aa11 22 a 12 b 12 cos 2 1 H ab12 12 cos 2 1 pap 1 11 2 2 p 2 2 aa11 22 a 12 b 12 cos 2 1 H pcqpp1111212 sin , , , 1 H pcqpp 2221212 sin , , , 2 ab12 12 cos 2 1 papapab 1 11 2 22 1 12 12 cos 2 1 pb 2 12 sin 2 1 qp121, , , p2 2 2 aa a b cos 11 22 12 12 2 1 Figure 3. Double pendulum dimensions. University of Michigan 6 Department of Physics Dynamical stability and Lyapunov exponents In this experiment, we would like to explore how sensitively the dynamical evolution of a system depends on the initial conditions. Do small perturbations lead to small variations or huge ones? As a starting point, consider a one-dimensional system whose evolution is governed by: qfqt Suppose we want to find out how q will evolve for slightly different initial conditions in the neighborhood of qo: f qfqtfqt o q q Thus, in the limit, q 0, qf qt() et q 0 qq qo The value, λ, is the Lyapunov exponent at qo. If it is greater than zero, the evolution of the system diverges exponentially with time, preventing all attempts to compute the motion at arbitrary times. We can take this example to the next level of complexity by considering the simple pendulum. The Lagrangian for this system is: LTV1 mr 22 mug cos 2 The canonical momentum is: L pmr2 and thus the Hamiltonian is: University of Michigan 7 Department of Physics p2 Hp L mugcos 2 mr 2 Hp p mr 2 H pmug sin We would like to find out how the motion evolves for small perturbations from the initial condition specified by θ0 and p0. 22HH p p pp 2 mr 2 22HH ppmug cos 2 p This can be represented by the matrix equation below: 1 0 2 mr p p mug cos 0 We seek solutions of the form: aaa mm11 21 bbbmm12 22 where qabp(0) is an eigenvector for the matrix with an eigenvalue, λ, and thus qt() et q (0). To find λ, the following equation must be solved: mm11 21 Det 0 mm12 22 For this particular case, the resulting equation is: mu cos 2 g mr 2 University of Michigan 8 Department of Physics For , the values of λ are imaginary. However, for , one of the values is positive and 2 2 real, leading to divergent growth. The eigenvalues and eigenvectors (normal modes) are thus: mug cos : i 2 mr 2 1 e imumrg2 cos mug cos : 2 2 mr 1 e mu mr 2 gcos To understand the long term behavior of the system, one must average the Lyapunov exponents over the dynamical path in the phase space. It is worth remarking about a generic property of Lyapunov exponents stemming from the Hamiltonian description. The Jacobian matrix that transforms the phase space volume, 11pp 2 2 must have the form: 22HH 2 H 2 H pp 2 p pp 11 1 1 2 12 22HH 2 H 2 H J 2 1112212pp The diagonal elements of this matrix cancel in pairs so that the trace of J is zero. This guarantees that the sum of eigenvalues, ie. the Lyapunov exponents, is also zero. The physical implication of this is the invariance of phase space volume for a conservative system. For a dissipative system, this would not be true. University of Michigan 9 Department of Physics This analysis can be extended to the double pendulum although the expressions become algebraically messy. In this case, we will only be interested in obtaining numerical solutions. The Jacobian matrix for this problem is a 4 × 4 array: j11jjj 12 13 14 j jjj J 21 22 23 24 j31jjj 32 33 34 j41jjj 42 43 44 By defining the following intermediate quantities, the matrix elements can be computed fairly easily.