Double Pendulum Josh Altic May 15, 2008 Josh Altic Double - PowerPoint PPT Presentation

Double Pendulum Josh Altic May 15, 2008 Josh Altic Double Pendulum

Position x 1 x 2 O θ 1 L 1 x 1 = L 1 sin( θ 1 ) x 2 = L 1 sin( θ 1 ) + L 2 sin( θ 2 ) m 1 y 1 L 2 y 1 = − L 1 cos( θ 1 ) θ 2 m 2 y 2 = − L 1 cos( θ 1 ) − L 2 cos( θ 2 ) y 2 Josh Altic Double Pendulum

Potential Energy: the sum of the potential energy of each mass P = m 1 gy 1 + m 2 gy 2 P = − m 1 gL 1 cos( θ 1 ) − m 2 g ( L 1 cos( θ 1 ) + L 2 cos( θ 2 )) Josh Altic Double Pendulum

Kinetic Energy in General We know that K = 1 / 2 mv 2 . Which brings us to x 2 + ˙ y 2 ) . K = 1 / 2 m (˙ Josh Altic Double Pendulum

Kinetic Energy in the double pendulum system x 2 y 2 x 2 y 2 K = 1 / 2 m 1 (˙ 1 + ˙ 1 ) + 1 / 2 m 2 (˙ 2 + ˙ 2 ) . differentiating: position: x 1 = L 1 cos( θ 1 ) ˙ ˙ θ 1 x 1 = L 1 sin( θ 1 ) x 2 = L 1 cos( θ 1 ) ˙ θ 1 + L 2 cos( θ 2 ) ˙ ˙ θ 2 x 2 = L 1 sin( θ 1 ) + L 2 sin( θ 2 ) y 1 = L 1 sin( θ 1 ) ˙ y 1 = − L 1 cos( θ 1 ) ˙ θ 1 y 2 = − L 1 cos( θ 1 ) − L 2 cos( θ 2 ) y 2 = L 1 sin( θ 1 ) ˙ θ 1 + L 2 sin( θ 2 ) ˙ ˙ θ 2 K = 1 / 2 m 1 ˙ 1 + 1 / 2 m 2 [ ˙ 1 + ˙ 2 + 2 ˙ θ 1 L 1 ˙ θ 2 1 L 2 θ 2 1 L 2 θ 2 2 L 2 θ 1 L 2 cos( θ 1 − θ 2 )] . Josh Altic Double Pendulum

Lagrangian in General The Lagrangian( L ) of a system is defined to be the difference of the kinetic energy and the potential energy. L = K − P . For the Lagrangian of a system this Euler-Lagrange differential equation must be true: � ∂ L � d − ∂ L ∂θ = 0 ∂ ˙ dt θ Josh Altic Double Pendulum

the Lagrangian of our double pendulum system K = 1 / 2 m 1 ˙ 1 + 1 / 2 m 2 [ ˙ 1 + ˙ 2 + 2 ˙ θ 1 L 1 ˙ θ 2 1 L 2 θ 2 1 L 2 θ 2 2 L 2 θ 2 L 2 cos( θ 1 − θ 2 )] . P = − ( m 1 + m 2 ) gL 1 cos( θ 1 ) − m 2 L 2 g cos( θ 2 ) In our case the Lagrangian is 1 ˙ 2 ˙ 2 + m 2 L 1 L 2 ˙ θ 1 ˙ L = 1 / 2( m 1 + m 2 ) L 2 θ 2 1 + 1 / 2 m 2 L 2 θ 2 θ 2 cos( θ 1 + θ 2 )+ ( m 1 + m 2 ) gL 1 cos( θ 1 ) + m 2 L 2 g cos( θ 2 ) . Josh Altic Double Pendulum

Partials of the Lagrangian for θ 1 1 ˙ 2 ˙ 2 + m 2 L 1 L 2 ˙ θ 1 ˙ L = 1 / 2( m 1 + m 2 ) L 2 θ 2 1 + 1 / 2 m 2 L 2 θ 2 θ 2 cos( θ 1 − θ 2 )+ ( m 1 + m 2 ) gL 1 cos( θ 1 ) + m 2 L 2 g cos( θ 2 ) Thus: ∂ L = − L 1 g ( m 1 + m 2 ) sin( θ 1 ) − m 2 L 1 L 2 ˙ θ 1 ˙ θ 2 sin( θ 1 − θ 2 ) ∂θ 1 ∂ L 1 ˙ θ 1 + m 2 L 1 L 2 ˙ = ( m 1 + m 2 ) L 2 θ 2 cos( θ 1 − θ 2 ) ∂ ˙ θ 1 � ∂ L � d 1 ¨ θ 1 + m 2 L 1 L 2 ¨ = ( m 1 + m 2 ) L 2 θ 2 cos( θ 1 − θ 2 ) ∂ ˙ dt θ 1 − m 2 L 1 L 2 ˙ θ 2 sin ( θ 1 − θ 2 )( ˙ θ 1 − ˙ θ 2 ) Josh Altic Double Pendulum

Substituting into the Euler-Lagrange Equation � ∂ L � − ∂ L d ∂θ = 0 ∂ ˙ dt θ 1 ¨ θ 1 + m 2 L 1 L 2 ¨ θ 2 cos( θ 1 − θ 2 ) + m 2 L 1 L 2 ˙ ( m 1 + m 2 ) L 2 θ 2 2 sin( θ 1 − θ 2 )+ gL 1 ( m 1 + m 2 ) sin( θ 1 ) = 0 Simplifying and Solving for ¨ θ 1 : θ 1 = − m 2 L 2 ¨ θ 2 cos( θ 1 − θ 2 ) − m 2 L 2 ˙ θ 2 2 sin( θ 1 − θ 2 ) − ( m 1 + m 2 ) g sin( θ 1 ) ¨ ( m 1 + m 2 ) L 1 Josh Altic Double Pendulum

Partials for θ 2 Once again the Lagrangian of the system is 1 ˙ 2 ˙ 2 + m 2 L 1 L 2 ˙ θ 1 ˙ L = 1 / 2( m 1 + m 2 ) L 2 θ 2 1 + 1 / 2 m 2 L 2 θ 2 θ 2 cos( θ 1 − θ 2 )+ ( m 1 + m 2 ) gL 1 cos( θ 1 ) + m 2 L 2 g cos( θ 2 ) ∂ L = m 2 L 1 L 2 ˙ θ 1 ˙ θ 2 sin( θ 1 − θ 2 ) − L 2 m 2 g sin( θ 2 ) ∂θ 2 ∂ L 2 ˙ θ 2 + m 2 L 1 L 2 ˙ = m 2 L 2 θ 1 cos( θ 1 − θ 2 ) ∂ ˙ θ 2 � ∂ L d � 2 ¨ θ 2 + m 2 L 1 L 2 ¨ = m 2 L 2 θ 1 cos( θ 1 − θ 2 ) ∂ ˙ dt θ 2 − m 2 L 1 L 2 ˙ θ 1 sin( θ 1 − θ 2 )( ˙ θ 1 − ˙ θ 2 ) Josh Altic Double Pendulum

Substituting into the Euler-Lagrange equation for θ 2 � ∂ L � − ∂ L d ∂θ = 0 ∂ ˙ dt θ L 2 ¨ θ 2 + L 1 ¨ θ 1 cos( θ 1 − θ 2 ) − L 1 ˙ θ 2 1 sin( θ 1 − θ 2 ) + g sin( θ 2 ) = 0 . θ 2 = − L 1 ¨ θ 1 cos( θ 1 − θ 2 ) + L 1 ˙ θ 2 1 sin( θ 1 − θ 2 ) − g sin( θ 2 ) ¨ . L 2 Josh Altic Double Pendulum

two dependent differential equations We now have two equations that both have ¨ θ 1 and ¨ θ 2 in them. θ 1 = − m 2 L 2 ¨ θ 2 cos( θ 1 − θ 2 ) − m 2 L 2 ˙ θ 2 2 sin( θ 1 − θ 2 ) − ( m 1 + m 2 ) g sin( θ 1 ) ¨ ( m 1 + m 2 ) L 1 θ 2 = − L 1 ¨ θ 1 cos( θ 1 − θ 2 ) + L 1 ˙ θ 2 1 sin( θ 1 − θ 2 ) − g sin( θ 2 ) ¨ . L 2 Josh Altic Double Pendulum

creating two second order differential equations − m 2 L 1 ˙ θ 2 1 sin( θ 1 − θ 2 ) cos( θ 1 − θ 2 ) + gm 2 sin( θ 2 ) cos( θ 1 − θ 2 ) − m 2 L 2 ˙ θ 2 2 sin( θ 1 − θ 2 ) − ( m 1 + m 2 ) g sin( θ 1 ) ¨ θ 1 = L 1 ( m 1 + m 2 ) − m 2 L 1 cos 2 ( θ 1 − θ 2 ) m 2 L 2 ˙ θ 2 2 sin( θ 1 − θ 2 ) cos( θ 1 − θ 2 ) + g sin( θ 1 ) cos( θ 1 − θ 2 )( m 1 + m 2 ) + L 1 ˙ θ 2 1 sin( θ 1 − θ 2 )( m 1 + m 2 ) − g sin( θ 2 )( m 1 + m 2 ) ¨ θ 2 = L 2 ( m 1 + m 2 ) − m 2 L 2 cos 2 ( θ 1 − θ 2 ) Josh Altic Double Pendulum

converting to a system of first order differential equations If I define new variables for θ 1 , ˙ θ 1 , θ 2 and ˙ θ 2 I can construct a system of four first order differential equations that I can then solve numerically. differentiating I get This gives me z 1 = ˙ ˙ θ 1 z 1 = θ 1 z 2 = ˙ z 2 = θ 2 ˙ θ 2 z 3 = ˙ z 3 = ¨ θ 1 ˙ θ 1 z 4 = ˙ z 4 = ¨ θ 2 . ˙ θ 2 . Josh Altic Double Pendulum

A system of four first order differential equations z 1 = ˙ ˙ θ 1 z 2 = ˙ ˙ θ 2 − m 2 L 1 z 2 4 sin( z 1 − z 2 ) cos( z 1 − z 2 ) + gm 2 sin ( z 2 ) cos( z 1 − z 2 ) − m 2 L 2 z 2 4 sin( z 1 − z 2 ) − ( m 1 + m 2 ) g sin( z 1 ) z 3 = ˙ . L 1 ( m 1 + m 2 ) − m 2 L 1 cos 2 ( z 1 − z 2 ) m 2 L 2 z 2 4 sin( z 1 − z 2 ) cos( z 1 − z 2 ) + g sin( z 1 ) cos( z 1 − z 2 )( m 1 + m 2 ) + L 1 z 2 4 sin( z 1 − z 2 )( m 1 + m 2 ) − g sin( z 2 )( m 1 + m 2 ) z 4 = ˙ . L 2 ( m 1 + m 2 ) − m 2 L 2 cos 2 ( z 1 − z 2 ) Josh Altic Double Pendulum

example of cyclical behavior of the system m1 m2 1.5 1 y1 and y2 0.5 0 −0.5 −1.5 −1 −0.5 0 0.5 1 1.5 x1 and x2 Josh Altic Double Pendulum

example of cyclical behavior of the system m1 1.5 m2 1 0.5 y1 and y2 0 −0.5 −1 −1.5 −2 −2 −1 0 1 2 x1 and x2 Josh Altic Double Pendulum

example of nearly cyclical behavior of the system 2 m1 1.5 m2 1 0.5 y1 and y2 0 −0.5 −1 −1.5 −2 −1 0 1 2 x1 and x2 Josh Altic Double Pendulum

example of nearly cyclical behavior of the system 2 m1 1.5 m2 1 0.5 y1 and y2 0 −0.5 −1 −1.5 −2 −1 0 1 2 x1 and x2 Josh Altic Double Pendulum

Example of Chaotic behavior of the system m1 1 m2 0.5 y1 and y2 0 −0.5 −1 −1.5 −2 −1 0 1 2 x1 and x2 Josh Altic Double Pendulum