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Finding the Euler-Lagrange equation for a special pendulum

  1. Nov 3, 2015 #1
    1. The problem statement, all variables and given/known data

    Qn3 problem sheet 1 dynamics.PNG
    Qn3 problem sheet 1 dynamics#2.PNG


    2. Relevant equations
    Euler - Lagrange equation:
    ##\frac{d}{dt}(\frac{\partial L}{\partial \dot\theta})=\frac{\partial L }{\partial \theta}##

    Hamilton's equations:

    ##\frac{\partial H}{\partial \theta}=-p_{\theta}\text{ and }\frac{\partial H}{\partial p_{\theta}}=\dot\theta##

    3. The attempt at a solution


    Part a and b was basically finding the lagrangian of the system, for which i got (and is correct):

    ##L=\frac{m}{2}[l^2\dot\theta^2+\dot{\tilde{y}}^2+2l\dot{\tilde{y}}\dot{\theta}\sin{\theta}]-mg[\tilde{y}-l\cos{\theta}]##

    All of the derivatives rules are getting me confused for part c and d.

    ##\sin{\theta}\approx \theta \text{ and } \cos{\theta}\approx 1-\frac{1}{2}\theta^2##

    New lagrangian with approximation:

    ##L\approx\frac{m}{2}[l^2\dot\theta^2+\dot{\tilde{y}}^2+2l\dot{\tilde{y}}\dot{\theta}\theta]-mg[\tilde{y}-l+\frac{1}{2}l\theta^2]##

    ##\textbf{Part C}##

    ##\frac{\partial L}{\partial \dot\theta}=\frac{m}{2}[2l^2\dot\theta+2l\dot{\tilde{y}}\theta]=ml^2\dot\theta+ml\dot{\tilde{y}}\theta##
    ##\frac{d}{dt}(\frac{\partial L}{\partial \dot\theta})=ml^2\ddot\theta+ml\ddot{\tilde{y}}\theta+ml\dot{\tilde{y}}\dot\theta##
    ##\frac{\partial L}{\partial \theta}=ml\dot{\tilde{y}}\dot\theta-mg\theta##

    Therefore, Legrange's equation reads:

    ##ml^2\ddot\theta+ml\ddot{\tilde{y}}\theta+ml\dot{\tilde{y}}\dot\theta=ml\dot{\tilde{y}}\dot\theta-mlg\theta##

    Does this look correct?

    The last part of question 3 asks us to discuss the physical meaning for arbitrary frequency and for ##\omega\ll 1##. I think this implies that I should put ##\tilde{y}=y_0\sin{\omega t}## into the Euler-Lagrange equation... but then I have a problem of computing ##\dot{\tilde{y}}\text{ and }\ddot{\tilde{y}}##.

    Do these look correct?

    ##\tilde{y}\approx y_o\dot{\theta}t##
    ##\dot{\tilde{y}}\approx y_0\ddot{\theta}t+y_0\dot\theta##
    ##\ddot{\tilde{y}} \approx y_0\dddot{\theta}t+y_0\ddot{\theta}+y_0\ddot{\theta}=y_0\dddot{\theta}t+2y_0\ddot{\theta}##

    If this is correct, after subbing these into the Euler-Lagrange equation, i'm still not sure of the physical significance of arbitrary omega and when omega is far less than one.

    I will not do part D in this same post. Thanks for any help!
     
    Last edited: Nov 3, 2015
  2. jcsd
  3. Nov 3, 2015 #2

    TSny

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    Check the last term on the right. It doesn't have the same dimensions as the other terms.

    Check last term on the right.

    Why the dot over the y here?
    Since you are given ##y## as a function of time, you can just take derivatives of this function to get expressions for ##\dot{y}## and ##\ddot{y}##.

    I don't understand these expressions.
     
  4. Nov 3, 2015 #3
    Hi TSny and thank you for your reply.

    My mistake, there should be an ##l## in front of the ##\frac{1}{2}\theta^2##

    So, $$L\approx\frac{m}{2}[l^2\dot\theta^2+\dot{\tilde{y}}^2+2l\dot{\tilde{y}}\dot{\theta}\theta]-mg[\tilde{y}-l+\frac{1}{2}l\theta^2]$$

    Oops, it should be:

    $$\frac{d}{dt}(\frac{\partial L}{\partial \dot\theta})=ml^2\ddot\theta+ml\ddot{\tilde{y}}\theta+ml\dot{\tilde{y}}\dot\theta$$

    There shouldn't be a dot, that was a typo. I will amend all of these mistakes in the original post.

    ##\tilde{y}=y_0\sin{\omega t} \approx y_0\omega t =y_0\dot{\theta}t ##

    I then took consecutive time derivatives although i am a little confused as to whether it is a total or partial derivative.
     
  5. Nov 3, 2015 #4

    TSny

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    ##\omega## does not equal ##\dot{\theta}##.

    ##\omega## is a constant representing the angular frequency of the point of support.
    ##\dot{\theta}## is not a constant and it represents the rate of change of the angle of the pendulum.

    Also, I don't think you can assume ##\omega t## is small here.
     
  6. Nov 3, 2015 #5

    TSny

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    Where did the negative sign come from for the first term on the right?

    (The last term on the left still needs a correction.)
     
  7. Nov 3, 2015 #6
    Oh of course, how silly of me. Thanks for clearing that up.

    That negative should not have been there. I think all the corrections have been made now..

    so after plugging in ##\ddot{\tilde{y}}=-y_0\omega^2\sin{\omega t}## and cancelling the m's and a factor of l i getfor the equation of motion:

    $$l\ddot\theta+(g-y_0\omega^2\sin{\omega t})\theta=0$$

    Although i'm still not sure of the physical significance of arbitrary omega and small omega.. except that when omega is small the equation would be simpler to solve since we can approximate ##\ddot{\tilde{y}} \approx 0## and the non linearity disappears?

    Also, this is slightly irrelevant, but is that differential equation one of bessels ?
     
  8. Nov 3, 2015 #7
    Wait a second.... In the Lagrange equations of motion the dynamics are independent of the mass of the bob (at least in my attempt). But when I find Hamilton's equations they are dependent upon mass.. that can't be right can it?
     
  9. Nov 3, 2015 #8

    TSny

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    OK. I think that's right. I believe this would be classified as a linear differential equation with variable coefficients. Both ##\ddot{\theta}## and ##\theta## occur to just the first power. It does not have the form of a Bessel differential equation (at least not that I'm familiar with).

    I'm not sure what they want for a physical interpretation. Could you interpret the expression in parentheses as an "effective, time-dependent gravitational acceleration (or field)"?

    You could take the ##y_0 \omega^2 sin(\omega t) \theta## to the other side and interpret it as a harmonic "driving" term for the pendulum. But it does not have the typical form of a driving term since it is proportional to ##\theta##; whereas, a typical driving term would not depend on ##\theta##.
     
  10. Nov 3, 2015 #9

    TSny

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    ...
    The mass will appear in the individual Hamilton's equations. But if you combine the equations to get a second order differential equation for ##\ddot{\theta}## you will find that the mass cancels out. Note that the momentum conjugate to ##\theta## is proportional to ##m## when expressed in terms of ##\theta## and ##\dot{\theta}##.
     
  11. Nov 4, 2015 #10
    Ok yeah that makes sense.

    To get the Hamiltonian (and thus Hamilton's equations of motion for this system), do I only need to sub in ##\dot{\theta}=\frac{p_{\theta}}{ml^2}## ?
     
  12. Nov 4, 2015 #11

    TSny

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    I don't believe that's the correct relation between ##\dot{\theta}## and ##p_{\theta}##.

    Yes, you will need to make a substitution for ##\dot{\theta}## in order to express the Hamiltonian in terms of ##p_{\theta}## and ##\theta##.
     
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