Double PendulumJosh AlticMay 15, 2008Josh Altic Double PendulumPositionx1= L1sin(θ1)x2= L1sin(θ1) + L2sin(θ2)y1= −L1cos(θ1)y2= −L1cos(θ1) − L2cos(θ2)x1x2y1y2θ1m1L1L2θ2m2OJosh Altic Double PendulumPotential Energy: the sum of the potential energy of eachmassP = m1gy1+ m2gy2P = −m1gL1cos(θ1) − m2g (L1cos(θ1) + L2cos(θ2))Josh Altic Double PendulumKinetic Energy in GeneralWe know thatK = 1/2mv2.Which brings us toK = 1/2m( ˙x2+ ˙y2).Josh Altic Double PendulumKinetic Energy in the double pendulum systemK = 1/2m1( ˙x21+ ˙y21) + 1/2m2( ˙x22+ ˙y22).position:x1= L1sin(θ1)x2= L1sin(θ1) + L2sin(θ2)y1= −L1cos(θ1)y2= −L1cos(θ1) − L2cos(θ2)differentiating:˙x1= L1cos(θ1)˙θ1˙x2= L1cos(θ1)˙θ1+ L2cos(θ2)˙θ2˙y1= L1sin(θ1)˙θ1˙y2= L1sin(θ1)˙θ1+ L2sin(θ2)˙θ2K = 1/2m1˙θ21L21+ 1/2m2[˙θ21L21+˙θ22L22+ 2˙θ1L1˙θ1L2cos(θ1− θ2)].Josh Altic Double PendulumLagrangian in GeneralThe Lagrangian(L) of a system is defined to be the difference ofthe kinetic energy and the potential energy.L = K − P.For the Lagrangian of a system this Euler-Lagrange differentialequation must be true:ddt∂L∂˙θ−∂L∂θ= 0Josh Altic Double Pendulumthe Lagrangian of our double pendulum systemK = 1/2m1˙θ21L21+ 1/2m2[˙θ21L21+˙θ22L22+ 2˙θ1L1˙θ2L2cos(θ1− θ2)].P = −(m1+ m2)gL1cos(θ1) − m2L2g cos(θ2)In our case the Lagrangian isL = 1/2(m1+ m2)L21˙θ21+ 1/2m2L22˙θ22+ m2L1L2˙θ1˙θ2cos(θ1+ θ2)+(m1+ m2)gL1cos(θ1) + m2L2g cos(θ2).Josh Altic Double PendulumPartials of the Lagrangian for θ1L = 1/2(m1+ m2)L21˙θ21+ 1/2m2L22˙θ22+ m2L1L2˙θ1˙θ2cos(θ1− θ2)+(m1+ m2)gL1cos(θ1) + m2L2g cos(θ2)Thus:∂L∂θ1= −L1g(m1+ m2) sin(θ1) − m2L1L2˙θ1˙θ2sin(θ1− θ2)∂L∂˙θ1= (m1+ m2)L21˙θ1+ m2L1L2˙θ2cos(θ1− θ2)ddt∂L∂˙θ1= (m1+ m2)L21¨θ1+ m2L1L2¨θ2cos(θ1− θ2)−m2L1L2˙θ2sin(θ1− θ2)(˙θ1−˙θ2)Josh Altic Double PendulumSubstituting into the Euler-Lagrange Equationddt∂L∂˙θ−∂L∂θ= 0(m1+ m2)L21¨θ1+ m2L1L2¨θ2cos(θ1− θ2) + m2L1L2˙θ22sin(θ1− θ2)+gL1(m1+ m2) sin(θ1) = 0Simplifying and Solving for¨θ1:¨θ1=−m2L2¨θ2cos(θ1− θ2) − m2L2˙θ22sin(θ1− θ2) − (m1+ m2)g sin(θ1)(m1+ m2)L1Josh Altic Double PendulumPartials for θ2Once again the Lagrangian of the system isL = 1/2(m1+ m2)L21˙θ21+ 1/2m2L22˙θ22+ m2L1L2˙θ1˙θ2cos(θ1− θ2)+(m1+ m2)gL1cos(θ1) + m2L2g cos(θ2)∂L∂θ2= m2L1L2˙θ1˙θ2sin(θ1− θ2) − L2m2g sin(θ2)∂L∂˙θ2= m2L22˙θ2+ m2L1L2˙θ1cos(θ1− θ2)ddt∂L∂˙θ2= m2L22¨θ2+ m2L1L2¨θ1cos(θ1− θ2)−m2L1L2˙θ1sin(θ1− θ2)(˙θ1−˙θ2)Josh Altic Double PendulumSubstituting into the Euler-Lagrange equation for θ2ddt∂L∂˙θ−∂L∂θ= 0L2¨θ2+ L1¨θ1cos(θ1− θ2) − L1˙θ21sin(θ1− θ2) + g sin(θ2) = 0.¨θ2=−L1¨θ1cos(θ1− θ2) + L1˙θ21sin(θ1− θ2) − g sin(θ2)L2.Josh Altic Double Pendulumtwo dependent differential equationsWe now have two equations that both have¨θ1and¨θ2in them.¨θ1=−m2L2¨θ2cos(θ1− θ2) − m2L2˙θ22sin(θ1− θ2) − (m1+ m2)g sin(θ1)(m1+ m2)L1¨θ2=−L1¨θ1cos(θ1− θ2) + L1˙θ21sin(θ1− θ2) − g sin(θ2)L2.Josh Altic Double Pendulumcreating two second order differential equations¨θ1=−m2L1˙θ21sin(θ1− θ2) cos(θ1− θ2) + gm2sin(θ2) cos(θ1− θ2)−m2L2˙θ22sin(θ1− θ2) − (m1+ m2)g sin(θ1)L1(m1+ m2) − m2L1cos2(θ1− θ2)¨θ2=m2L2˙θ22sin(θ1− θ2) cos(θ1− θ2) + g sin(θ1) cos(θ1− θ2)(m1+ m2)+ L1˙θ21sin(θ1− θ2)(m1+ m2) − g sin(θ2)(m1+ m2)L2(m1+ m2) − m2L2cos2(θ1− θ2)Josh Altic Double Pendulumconverting to a system of first order differential equationsIf I define new variables for θ1,˙θ1,θ2and˙θ2I can construct asystem of four first order differential equations that I can thensolve numerically.This gives mez1= θ1z2= θ2z3=˙θ1z4=˙θ2.differentiating I get˙z1=˙θ1˙z2=˙θ2˙z3=¨θ1˙z4=¨θ2.Josh Altic Double PendulumA system of four first order differential equations˙z1=˙θ1˙z2=˙θ2˙z3=−m2L1z24sin(z1− z2) cos(z1− z2) + gm2sin(z2) cos(z1− z2)−m2L2z24sin(z1− z2) − (m1+ m2)g sin(z1)L1(m1+ m2) − m2L1cos2(z1− z2).˙z4=m2L2z24sin(z1− z2) cos(z1− z2) + g sin(z1) cos(z1− z2)(m1+ m2)+L1z24sin(z1− z2)(m1+ m2) − g sin(z2)(m1+ m2)L2(m1+ m2) − m2L2cos2(z1− z2).Josh Altic Double Pendulumexample of cyclical behavior of the system−1.5 −1 −0.5 0 0.5 1 1.5−0.500.511.5x1 and x2y1 and y2 m1m2Josh Altic Double Pendulumexample of cyclical behavior of the system−2 −1 0 1 2−2−1.5−1−0.500.511.5x1 and x2y1 and y2 m1m2Josh Altic Double Pendulumexample of nearly cyclical behavior of the system−2 −1 0 1 2−1.5−1−0.500.511.52x1 and x2y1 and y2 m1m2Josh Altic Double Pendulumexample of nearly cyclical behavior of the system−2 −1 0 1 2−1.5−1−0.500.511.52x1 and x2y1 and y2 m1m2Josh Altic Double PendulumExample of Chaotic behavior of the system−2 −1 0 1 2−1.5−1−0.500.51x1 and x2y1 and y2 m1m2Josh Altic Double