Lecture 9 Hamilton’s Equations - PowerPoint Presentation

Slide1

Lecture 9 Hamilton’s Equations

conjugate momentum

cyclic coordinates

Informal derivation

Applications/examples

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Define

the

conjugate momentum

Start with the Euler-Lagrange equations

The Euler-Lagrange equations can be rewritten as

derivationSlide3

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If we are to set up a pair of sets of odes, we need to eliminate

We can write the Lagrangian

M

is positive definite (coming from the kinetic energy), so

derivation

from which Slide4

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I have the pair of sets of odes

where all the

q

dots in the second set have been replaced by their expressions in terms of

p

derivationSlide5

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We can write this out in detail, although it looks pretty awful

derivation

right hand side

substitute

right hand sideSlide6

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We will never do it this way!

derivation

I will give you a recipe as soon as we know what cyclic coordinates areSlide7

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??Slide8

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If

Q

i and the constraints are both zero (free falling brick, say) we have

and if

L

does not depend on

q

i

, then

p

i

is constant!

Conservation of conjugate momentum

q

i

is a

cyclic coordinate

Given

cyclic coordinatesSlide9

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YOU NEED TO REMEMBER THAT

EXTERNAL FORCESAND/OR CONSTRAINTSCAN MAKE CYCLIC COORDINATES

NON-CYCLIC!Slide10

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??Slide11

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What can we say about an unforced single link

in general?

We see that

x and

y

are cyclic (no explicit

x

or

y

in

L

)

Conservation of linear momentum in

x and

y

directions

We see that

f

is also cyclic (no explicit

f

in

L

)Slide12

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Does it mean anything physically?!

The conserved conjugate momentum is

the angular momentum about the

k

axis

as I will now show youSlide13

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The angular momentum in body coordinates is

The body axes are (from Lecture 3)

and the

k

component of the angular momentum isSlide14

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The only force in the problem is gravity, which acts in the k direction

Any gravitational torques will be normal to

k, so the angular momentum in the k direction must be conserved.

some

algebraSlide15

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Summarize our procedure so far

Write the Lagrangian

Apply holonomic constraints, if any

Assign the generalized coordinates

Find the conjugate momenta

Eliminate conserved variables (cyclic coordinates)

Set up numerical methods to integrate what is leftSlide16

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??Slide17

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Let’s take a look at some applications of this method

Falling brick

Flipping a coin

The axisymmetric topSlide18

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Flipping a coinSlide19

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A coin is axisymmetric, and this leads to some simplification

We have

four

cyclic coordinates, adding

y

to the set and simplifying

f

We can recognize the new conserved term as the angular momentum about

K

Flipping a coinSlide20

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We can use this to simplify the equations of motion

The conserved momenta are constant; solve for the derivatives

(There are numerical issues when

q

= 0; all is well if you don’t start there)

Flipping a coinSlide21

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I can flip it introducing spin about

I

with no change in

f

or

y

This makes

p

4

= 0 =

p

6

Then

Flipping a coinSlide22

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from which

or

Flipping a coinSlide23

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I claim it would be harder to figure this out using the Euler-Lagrange equations

Flipping a coin

I’ve gotten the result for a highly nonlinear problem by clever argument augmented by Hamilton’s equationsSlide24

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??Slide25

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Same Lagrangian

but constrained

symmetric top

How about the symmetric top?Slide26

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Treat the top as a cone

and apply the holonomic constraints

symmetric topSlide27

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After some algebra

and we see that

f

and y are both cyclic here.

f

is the rotation rate about the vertical — the precession

y

is the rotation rate about

K

— the spin

symmetric topSlide28

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The conjugate momenta are

There are no external forces and no other constraints,

so the first and third are conserved

symmetric topSlide29

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We are left with two odes

The response depends on the two conserved quantities

and these depend on the initial spin and precession rates

symmetric top

We can go look at this in MathematicaSlide30

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Let’s go back and look at how the fall of a single brick will go in this method

Falling brick

I will do the whole thing in MathematicaSlide31

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That’s All FolksSlide32

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What is this?!

If

A and

B are equal (axisymmetric body) , this is much simpler

Falling brickSlide34

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We know what happens to all the position coordinates

All that’s left is the single equation for the evolution of p

5.

After some algebra we obtain

Falling brickSlide35

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We can restrict our attention to axisymmetric wheels

and we can choose K to be parallel to the axle without loss of generality

mgzSlide36

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If we don’t put in any simple holonomic constraint (which we often can do)Slide37

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We know v and

w in terms of

q any difficulty will arise from

r

This will depend on the surface

flat, horizontal surface — we’ve been doing this

flat surface — we can do this today

general surface:

z

=

f

(

x

,

y

) — this can be done for a rolling sphere

Actually, it’s something of a question as to where the difficulties will arise in generalSlide38

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We have the usual Euler-Lagrange equations

and we can write out the six equationsSlide40

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The angular momentum in body coordinates is

The body axes are (from Lecture 3)

and the

k

component of the angular momentum isSlide41

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We will have a useful generalization fairly soon The idea of separating the second order equations

into first order equations

More generically