Double Pendulum - Classical Mechanics - Lecture Slides, Slides of Classical Mechanics

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Double

Pendulum

Double

Pendulum

-^

The

double

pendulum

is

a

conservative

system.

-^

Two

degrees

of

freedom

-^

The

exact

Lagrangian

can

be

written

without

approximation.

^

2

sin ( ) cos ( 2

^

^

l

l m T^





l

l

m

m

^

2

sin()

(

sin ( ))

cos()

(

cos ( 2

^

^

^

l l l l m T

^

2 2

cos)

( 2

2 2

^

^

^

ml T

^

^

^

)

cos(

cos 2

cos)

( 2

2 2

2

2 2

^

^

mgl

ml L^

Hamilton’s

Equations

-^

Make

substitutions:

-^

Divide

by

mgl

-^

t^ 

t (

g /

1/2 l )

-^

Find

conjugate

momenta

as

angular

momenta.

cos(

cos 2

cos 3

cos 2 (^3) (

cos (^1) ( 2

2

2



 

^

^

J J J J H E





^

cos 3

cos (^1) ( 2

2

^

J

J

H J

^

sin(

sin 2

^

^

H

 J





^

cos 3

cos 2 (^3) ( 2

) cos (^1) ( 2

^

J

J

H J

^

^

sin(

cos 3

cos 2 (^3) (

cos (^1) ( 2

sin 2

2 cos 3 sin 2

2

2

2

2

      

^

^

J J J J J J J H

 J

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Small

Angle

Approximation

-^

For

small

angles

the

Lagrangian

simplifies.^ –

The

energy

is

E

-^

The

mode

frequencies

can

be

found

from

the

matrix

form.

-^

The

winding

number

is

irrational.

0

1

2 1

2 1

5 3

2

2

2

2



















2

2 2 1

  

2

2

 

   ^

^

     T

2

2

 

^

V

2

2 2 2

  

(^2 )

 

Boundaries

-^

The

greatest

motion

in

‐

space

occurs

when

there

is

no

energy

in

the

‐

dimension

-^

Points

must

lie

within

a

boundary

curve.





^

J

J^

cos

1 (^



J^ 

0

2 cos 3

)

cos (^1) ( 2

2



















^

J

J













 cos 2 2

) , , )

cos (^1) ( , 0 (

2

 







J

E

J

J

H E bound

Fixed

Points

-^

For

small

angle

deflections

there

should

be

two

fixed

points.^ –

Correspond

to

normal

modes 

J^ 

0

1

2 1

2 1

5 3

2

2

2

2

  



  









i i

i

i

i

i

^ 

2

2 2 1

  

2

2 2 2

  

  

         

1

2

1

^1 ^1

  

         

1

2

1

^1 ^1

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