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Example 7-10 Lagrangian Dynamics Marion and Thornton

  1. May 29, 2017 #1
    1. The problem statement, all variables and given/known data
    A particle of mass m is on top of a frictionless hemisphere centered at the origin with radius a"
    Set up the lagrange equatinos determine the constraint force and the point at which the particle detaches from the hemisphere

    2. Relevant equations
    L=T-U

    3. The attempt at a solution
    this is NOT HW
    solution is there in full (example from book) I am just trying to understand
    why is φ not one of the proper generalized coordinates? can't the particle move sideways as well when released? shouldn't the GC be (r,θ,φ) ie all spherical and the constraint equation r-a=0.
    THanks
     
  2. jcsd
  3. May 30, 2017 #2

    BvU

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    What would the Euler-Lagrange equations give you for ##\phi## ?
     
  4. May 30, 2017 #3
    (sin(2φ)/2)*(dθ/dt)-(d^2/dt^2)(φ)=0
    I though of integrating it but θ is also dependent on t
     
    Last edited: May 30, 2017
  5. May 30, 2017 #4
    even if I replace dθ/dt from what I get from euler-L of θ I would still get cosθ in the above equation!
     
  6. Jun 1, 2017 #5

    BvU

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    (Sorry for the late reaction) ##\quad##I find, with $$ \ T = {1\over 2} m\dot r^2+r^2\dot\theta^2+r^2\sin^2\theta\,\dot\phi^2 \ $$ (here, top of p7) and $$V = -mr\cos\theta\ $$ that $$ {\partial {\mathcal L}\over \partial \phi} = 0 $$ so that $${ d\over dt } 2r^2 \sin^2\theta\,\dot\phi = 0 \ ,$$in other words: ##L_z=## constant (##L## being the angular momentum). The particle is let go with ##L_z=0## so it stays at that value.
     
  7. Jun 3, 2017 #6
    yes absolutely simple one line answer
    forgive my stupidity I copied the T term from Wolfram Alpha and there they use θ for φ and vice vera- (I was too lazy to do this trivial math considering I'm a fanatic I solve every single example and problem in every chapter)
    damn mathematicians-lol(kidding love them)
    that explains why my differential equation above gets complicated and doesn't reveal conservation
    I completely get it the angular momentum does not change in that direction
    Lz is explicitly determined and happens to be a constant multiple of the euler-Lagrangian variation and that's why it never moves sideways I get it supports intuition in this case
    Thank you so much for your help and clarification appreciated A+++++++
     
  8. Jun 3, 2017 #7

    BvU

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    OK, on to the next ...
     
  9. Jun 3, 2017 #8
    Sure
    I am on problem 7-20 so far able to solve them all
    actually I do have a question (clarification) on problem 7-10 should I post here or different thread
    THanks
     
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