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Lagrangian of a sliding ladder

  1. Apr 17, 2014 #1
    1. The problem statement, all variables and given/known data
    A ladder of length 2l and mass m leans against a smooth wall and rests on a smooth floor. The ladder initially makes an angle θ0 to the vertical. It slides downwards maintaining contact with both the wall and the floor. Calcula the the Lagrangian and the conjugate momentum, and find the equation of motion. (The moment of inertia of a rod of length a and mass M about an axis through its centre perpendicular to the rod is Ma2/12.)

    3. The attempt at a solution
    I have L=(2/3)ml2(dθ/dt)2-mglcosθ with θ the time varying angle to the vertical. I'm confident this is correct as a similar result is obtained here for a slightly different definition of θ http://mathhelpforum.com/calculus/129710-lagrangian-function-ladder-smooth-wall.html.

    Therefore why is the question asking for a conjugate momentum?
     
  2. jcsd
  3. Apr 17, 2014 #2

    BvU

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    Once you have the Lagrangian, Euler-Lagrange gives you the equation of motion. Conjugate momentum is the momentum associated with the generalized coordinates, in this case only ##\theta##. It occurs in the EOM.
     
  4. Apr 17, 2014 #3
    Ah, so the conjugate momentum for theta is the partial derivative of L wrt theta dot. I got confused thinking a conjugate momentum had to be conserved, but I realise now it doesn't.

    So this means the angular momentum of the ladder is varying due to the external torques?
     
  5. Apr 17, 2014 #4

    BiGyElLoWhAt

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    Is that the correct Lagrangian? I feel like there should be a translational portion to the KE terms. Yea, the ladder is rotating, but it's center of mass is moving down, and (potentially neglegibly, depending on your prof) out away from the wall it's leaning on. I say potentially neglegibly because in my intermediate mechanics class we had a system of 2 rods fixed at one end and a spring connecting the other 2 ends with some masses in between the springs, and we were told to neglect the translational motion of the blocks as the springs expanded and contracted (as the springs expand, the blocks move closer to the fixed pivot, and vice versa)... But of course we were told this after the test... lol.
     
  6. Apr 17, 2014 #5

    CAF123

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    There is a translational component. If ##\theta## is the angle from the vertical, then the C.O.M can be described by ##\ell \cos \theta \hat{y} + \ell \sin \theta \hat{x}##. From this you can extract the x and y components and take their time derivative to give the translational kinetic energy.
     
  7. Apr 17, 2014 #6

    BiGyElLoWhAt

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    Ok that's what I thought, just wanted to make sure before OP left here with the wrong L.

    Only 1 question, is the ##\ell \cos (\theta) \hat{y}## a typo? seems as though the ##\hat{x}## & ##\hat{y}## should be switched. Also I am almost certain it should be ##\frac{\ell}{2}## vs. ##\ell## (assuming that's the description of it's position, of course)
     
  8. Apr 17, 2014 #7

    CAF123

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    Hi BiGyElLoWhAt,
    No, I don't think so, provided I interpreted the question correctly. See sketch.

    It is given that the length is 2l, so from the C.O.M to either end, we have to consider a length l.
     

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  9. Apr 17, 2014 #8

    BiGyElLoWhAt

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    OOOOO
    AAAAA

    And that reading thing strikes again...
     
  10. Apr 17, 2014 #9

    BvU

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    Big: Physik has the correct kinetic energy of translation and rotation in the Lagrangian. That's how he/she got the 2/3 factor.

    no typo, no error. ##\theta## is wrt vertical, length of ladder is 2 ##\ell##. It is completely as described in post #1.

    Right: there has to be a symmetry to get a conserved quantity. See Noether's[/PLAIN] [Broken] theorem -- extremely important in physics.

    And what happens to L is what happens to ##\dot \theta##: Its time derivative is definitely not zero, as you can see from the EOM.

    [edit]caf beat me to it
     
    Last edited by a moderator: May 6, 2017
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