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Pendulum problem using Lagrangian

  1. Feb 18, 2016 #1
    1. The problem statement, all variables and given/known data
    I am studying a question in Marion's classical mechanics:
    Screen Shot 2016-02-18 at 10.44.12 PM.png
    I am successful in obtain the equation of motion, which is Screen Shot 2016-02-18 at 10.46.42 PM.png where theta is the theta shown in Screen Shot 2016-02-18 at 10.47.38 PM.png . However, in the second part of the solution, Screen Shot 2016-02-18 at 10.48.21 PM.png , it puts derivative of theta to be zero and I can't understand this. Screen Shot 2016-02-18 at 10.43.57 PM.png Also, the solution concludes theta=pi/2 is the solution. It is very confusing.
    2. Relevant equations
    3. attempt
    In my mind, I can image that derivative of theta is zero at equilibrium as the mass stops moving but i don't know why such criteria is still valid when expanding around a certain angle. pi/2 is the equilibrium angle which I can derive by finding minimum of the potential energy but the result does't seem to be related to the "half-angle" asked.
     

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  3. Feb 18, 2016 #2

    TSny

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    I believe I follow the solution, but I'm not clear on what you are asking. You mention "derivative of theta is zero", but I don't see anywhere in the solution where there is any mention of the derivative of theta being zero. So, I'm not sure what you are referring to here.

    The pendulum will swing between some minimum value of theta, ##\theta_{min}##, and some maximum value of theta, ##\theta_{max}##. For small swings, ##\theta_{min}## is close to ##\theta_{max}##. Let ##\theta_0## be any fixed angle that has a value within or near the range of swing of the pendululm. ##\theta_0## is not initially assumed to be the midpoint of the swing. The only assumption initially is that ##\theta## for the pendulum will never deviate very much from ##\theta_0## during the small oscillations. So, if you define ##\varepsilon = \theta - \theta_0##, then ##\varepsilon## will always be small. Equation (9) is the solution for ##\varepsilon## corresponding to a particular value of ##\theta_0##. Inspection of equation (9) shows how ##\theta_0## should be chosen if you want ##\theta_0## to be the midpoint of the swing.
     
  4. Feb 18, 2016 #3
    I mean to derive Eq. (8), I have to expand cosθ about ##\theta_0## (I understand this part) as well as taking -R(##\theta_0##')^2 to be zero. (this is where I lost it)
     
  5. Feb 18, 2016 #4

    TSny

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    As you noted, for small angles, you can neglect the term ##R \dot{\theta}^2## in the equation of motion because this term is of "second order" in small quantities.

    Likewise, ##\theta \ddot{\theta}## may be replaced by ##\theta \ddot{\varepsilon} \approx \theta_0 \ddot{\varepsilon}##
     
  6. Feb 18, 2016 #5
    I do not understand why ##R \dot{\theta}^2## can be neglected. Although, the angle deviation is small, does it imply that its time derivative is small too?
     
  7. Feb 18, 2016 #6
    Also, if "second order" involving theta is neglected, why woudln'd θ¨θ be set to zero too?
     
  8. Feb 18, 2016 #7

    TSny

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    Yes, the time derivative of ##\theta## will also be small. For small swings, the motion is approximately SHM. For SHM, ##\dot{\theta}_{max} = \omega \theta_{max}##.
     
  9. Feb 18, 2016 #8
    But I can only use the SHM corresponds to a equilibrium point but in this question, the approximation is for a certain unknown position that might not be a equilibrium point?
     
  10. Feb 18, 2016 #9

    TSny

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    ##\ddot{\theta}## is a small "first-order" quantity. But ##\theta## is not necessarily small. ##\theta \approx \theta_0##.
     
  11. Feb 18, 2016 #10
    But I can only use the SHM corresponds to a equilibrium point? but in this question, the approximation is for a certain unknown position that might not be a equilibrium point?
     
  12. Feb 18, 2016 #11

    haruspex

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    If we are taking an equation that's supposed to be valid over time and applying a small perturbation approximation then necessarily we are discussing the neighbourhood of an equilibrium point.
     
  13. Feb 18, 2016 #12

    TSny

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    We know ##\dot{\theta} = \dot{\varepsilon}##.

    We can estimate the order of magnitude of ##\dot{\varepsilon}## by considering an average value of ##\dot{\varepsilon}## to be ##\overline {\dot{\varepsilon} } \approx \varepsilon_{max}/T## where ##T## is the period of the motion. You could argue that we should use something like ##T/4## rather than ##T## in estimating the average, but we are just trying to get an order of magnitude.

    So, by making the oscillations small enough, we can make ##\varepsilon_{max}## small enough that ##\overline {\dot{\varepsilon}}## becomes a small first order quantity. (##T## should not change much as ##\varepsilon_{max}## is made very small.) The maximum value of ##\dot{\varepsilon}## should be of the same order of magnitude as the average value of ##\dot{\varepsilon}##. So, ##\dot{\theta} = \dot{\varepsilon}## should be a small first-order quantity throughout the motion.

    I'm not sure if this type of argument is rigorous enough to be satisfying. It seems to me that the physical setup is such that it is clear that small oscillations will be at least approximately SHM with small amplitude. Thus, the angular velocity ##\dot{\theta}## will also be small.
     
  14. Feb 19, 2016 #13

    haruspex

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    I find this question and solution unsatisfactory in several ways.
    It is simplest to avoid Cartesian coordinates in the first place. We can immediately write down the energy equation
    ##\frac 12m(l-R\theta)^2\dot\theta^2=mg(H-R\cos(\theta)+(l-R\theta)\sin(\theta))##
    for some constant H.
    Since we know we are looking for solutions around pi/2, it will be more convenient to work with the complement. I'll write L=l-πR/2.
    ##(L+R\phi)^2\dot\phi^2=2g(H-R\sin(\phi)+(L+R\phi)\cos(\phi))## (1)
    At extreme values of φ, KE is zero
    ##H-R\sin(\phi)+(L+R\phi)\cos(\phi)=0##
    The part of the question where it asks for the midpoint of the swing angle does not make it clear whether this is just for small angles. But that equation will be nasty in general, so let's assume that it is. Out of interest, we can find the first order departure from symmetry. Suppose it swings between -φ and φ+ε, small ε. I my algebra is right, we can show the range to be -φ to tan(φ).
    What about SHM?
    On the RHS of (1), the first order phi terms cancel, leaving 2g(H+L-Lφ2/2), which is what we need for SHM. But on the left we have a coefficient (L2+2LRφ). Now, that does not look like SHM to me. Instead, we end up with the first order equation ##\dot\phi=\frac{\sqrt{2g(H+L)}}{L+R\phi}##.
    Have I gone wrong somewhere?
     
  15. Feb 19, 2016 #14

    TSny

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    This is multiplied by ##\dot{\phi}^2## which is already second order.

    So, the overall expression for the left side can be written to second order as ##L^2\dot{\phi}^2##
     
  16. Feb 19, 2016 #15

    haruspex

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    I don't buy that.
    We have to get a non-constant expression for ##\dot\phi##. Taking the leading terms on both sides we have ##\dot\phi=\frac{\sqrt{H+L-\frac 12L\phi^2}}{L+R\phi}##. The φ2 term in the numerator disappears before the φ term in the denominator.
     
  17. Feb 19, 2016 #16

    TSny

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    I get ##\dot{\phi}^2 = \frac{2g}{L^2}\left(H+L \right) - \frac{g}{L} \phi^2## which is SHM and shows that ##\dot\phi## is not constant.

    If you take a simple pendulum oscillating in small angles, you get the similar form

    ##\dot{\phi}^2 = \frac{2E}{mL^2} - \frac{g}{L} \phi^2##, where ##E## is the total energy (a second-order quantity).

    I don't think the ##\phi^2## terms disappears first. Isn't ##H+L## a second order quantity like ##\phi^2##?
     
  18. Feb 19, 2016 #17

    haruspex

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    Not sure how you get that. Maybe I made a mistake. Do you agree with my equation (1)?
    No, H+L is constant.
    By the way, just to be clear, I didn't mean to imply that the first order equation I wrote at the end of post #13 can be an adequate description of small oscillations. The φ2 term on the right cannot be discarded since it has the essential role of ensuring that the angular velocity eventually becomes zero.
     
  19. Feb 19, 2016 #18

    TSny

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    Yes, I agree with (1). When I keep terms accurate to second order, I find that (1) reduces to the first equation I posted in #16.

    H+L is a constant during the oscillations. But, for small oscillations, H+L is a small, second order quantity which is proportional to the total energy. When the pendulum is just hanging at rest, you can check that H+L = 0 according to (1). When the pendulum is swinging with small oscillations, then H is just a little bit larger than -L.

    So, in your expression ##\dot\phi=\frac{\sqrt{H+L-\frac 12L\phi^2}}{L+R\phi}## both terms in the square root are of the same order and you cannot neglect the ##\phi^2## term relative to the ##H+L## term. However, you can neglect the ##R\phi## term in the denominator compared to L.
     
  20. Feb 20, 2016 #19

    haruspex

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    OK,I think I see it.
    The awkward thing is that what constitutes a sufficiently small amplitude to approximate SHM depends on the length ratios. We need R2(H+L)<<L3, so the smaller L is in relation to R, the smaller the amplitude needs to be. That's what makes it non-obvious.
    In fact, we are told R<l/π. It would be more of an issue if we only had R<2l/π.
    Interestingly, (1) has an analytic solution.
    Thanks.
     
  21. Feb 20, 2016 #20
    Thanks so much for helping me out! But in fact I already can't follow the discussion here :frown: as I have only learnt the Lagrangian equation but nothing about the hamilton that you guys mentioned yet. Anyway, I think the question has some problems itself as thanks to https://www.physicsforums.com/threads/pendulum-problem-using-lagrangian.858008/members/haruspex.334404/ [Broken], it was already pointed out that "a small perturbation approximation then necessarily we are discussing the neighbourhood of an equilibrium point" but the question attended for an arbitrary angle but not angles near the equilibrium. And if seeing the question as if it was asking about oscillation near the equilibrium, I have no problems about it.
     
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