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Lecture 9 Hamilton’s Equations

Lecture 9 Hamilton’s Equations. Informal derivation. conjugate momentum. cyclic coordinates. Applications/examples. derivation. Start with the Euler-Lagrange equations. Define the conjugate momentum. The Euler-Lagrange equations can be rewritten as. derivation.

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Lecture 9 Hamilton’s Equations

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  1. Lecture 9 Hamilton’s Equations Informal derivation conjugate momentum cyclic coordinates Applications/examples

  2. derivation Start with the Euler-Lagrange equations Define the conjugate momentum The Euler-Lagrange equations can be rewritten as

  3. derivation If we are to set up a pair of sets of odes, we need to eliminate We can write the Lagrangian from which M is positive definite (coming from the kinetic energy), so

  4. derivation I have the pair of sets of odes where all the q dots in the second set have been replaced by their expressions in terms of p

  5. derivation We can write this out in detail, although it looks pretty awful right hand side substitute right hand side

  6. derivation We will never do it this way! I will give you a recipe as soon as we know what cyclic coordinates are

  7. ??

  8. cyclic coordinates Given If Qi and the constraints are both zero (free falling brick, say) we have and if L does not depend on qi, then pi is constant! Conservation of conjugate momentum qi is a cyclic coordinate

  9. YOU NEED TO REMEMBER THAT EXTERNAL FORCES AND/OR CONSTRAINTS CAN MAKE CYCLIC COORDINATES NON-CYCLIC!

  10. ??

  11. What can we say about an unforced single linkin general? We see that xand 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)

  12. The conserved conjugate momentum is Does it mean anything physically?! the angular momentum about the k axis as I will now show you

  13. The angular momentum in body coordinates is The body axes are (from Lecture 3) and the k component of the angular momentum is

  14. some algebra 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.

  15. 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 left

  16. ??

  17. Let’s take a look at some applications of this method Flipping a coin The axisymmetric top Falling brick

  18. Flipping a coin

  19. Flipping a coin 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

  20. Flipping a coin 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)

  21. Flipping a coin I can flip it introducing spin about I with no change in f or y This makes p4 = 0 = p6 Then

  22. Flipping a coin from which or

  23. Flipping a coin I claim it would be harder to figure this out using the Euler-Lagrange equations I’ve gotten the result for a highly nonlinear problem by clever argument augmented by Hamilton’s equations

  24. ??

  25. symmetric top How about the symmetric top? Same Lagrangian but constrained

  26. symmetric top Treat the top as a cone and apply the holonomic constraints

  27. symmetric top 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

  28. symmetric top The conjugate momenta are There are no external forces and no other constraints, so the first and third are conserved

  29. symmetric top We are left with two odes The response depends on the two conserved quantities and these depend on the initial spin and precession rates We can go look at this in Mathematica

  30. Falling brick Let’s go back and look at how the fall of a single brick will go in this method I will do the whole thing in Mathematica

  31. That’s All Folks

  32. Falling brick What is this?! If A and B are equal (axisymmetric body) , this is much simpler

  33. Falling brick We know what happens to all the position coordinates All that’s left is the single equation for the evolution of p5. After some algebra we obtain

  34. We can restrict our attention to axisymmetric wheels and we can choose K to be parallel to the axle without loss of generality mgz

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

  36. We know v and w in terms ofq any difficulty will arise from r Actually, it’s something of a question as to where the difficulties will arise in general 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

  37. We have the usual Euler-Lagrange equations and we can write out the six equations

  38. The angular momentum in body coordinates is The body axes are (from Lecture 3) and the k component of the angular momentum is

  39. We will have a useful generalization fairly soon The idea of separating the second order equations into first order equations More generically

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