1 / 9

# Double Pendulum - PowerPoint PPT Presentation

Double Pendulum. Double Pendulum. The double pendulum is a conservative system. Two degrees of freedom The exact Lagrangian can be written without approximation. l. q. m. l. f. m. Make substitutions: Divide by mgl t  t ( g / l ) 1/2. Find conjugate momenta as angular momenta.

Related searches for Double Pendulum

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about 'Double Pendulum' - sadah

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

### Double Pendulum

• The double pendulum is a conservative system.

• Two degrees of freedom

• The exact Lagrangian can be written without approximation.

l

q

m

l

f

m

Divide by mgl

tt(g/l)1/2

Find conjugate momenta as angular momenta.

Dimensionless Form

Divide by mgl

tt(g/l)1/2

Find conjugate momenta as angular momenta.

Hamilton’s Equations

• For small angles the Lagrangian simplifies.

• The energy is E = -3.

• The mode frequencies can be found from the matrix form.

• The winding number W is irrational.

• The cotangent manifold T*Q is 4-dimensional.

• Q is a torus T2.

• Energy conservation constrains T*Q to an n-torus

• Take a Poincare section.

• Hyperplane q= 0

• Select dq/dt > 0

q

f

1

2

Jf

• The greatest motion in f-space occurs when there is no energy in the q-dimension

• Points must lie within a boundary curve.

Jf

2

1

f

• For small angle deflections there should be two fixed points.

• Correspond to normal modes

Jf

2

1

f

• An orbit on the Poincare section corresponds to a torus.

• The motion does not leave the torus.

• Motion is “invariant”

• Orbits correspond to different energies.

• Mixture of normal modes

Jf

f

next