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Pendulum with time dependent length

  1. Sep 11, 2011 #1
    1. The problem statement, all variables and given/known data
    A mass m is suspended by a massless string of varying length l = l0 - vt, where v is constant. The mass is released at angle [\theta]0 from rest.

    (a) Write down the Lagrangian and find the equation of motion
    (b) Show that these equations reduce to those of a simple pendulum for the case v -> 0
    (c) (Now this is the hard part) Solve for the approximate motion for small amplitude. The amplitude at time t = 0 is [itex]\theta[/itex]0. Qualitatively describe the motion.


    2. Relevant equations

    U = -mgy = -mg(l0-vt)cos([itex]\theta[/itex])
    T = 1/2mV2 = (1/2)mv2 + (1/2)m[itex]\ddot{\theta}[/itex]2(l0 - vt)2


    3. The attempt at a solution
    (a) Using L=T-U, the Lagrangian is given by:
    L = 1/2mV2 = (1/2)mv2 + (1/2)m[itex]\ddot{\theta}[/itex]2(l0 - vt)2 + mg(l0-vt)cos([itex]\theta[/itex])​
    The equation of motion is found by using the Euler-Lagrange equation. I found that
    [itex]\ddot{\theta}[/itex] = -gsin[itex]\theta[/itex] / (l0 - vt)​

    (b) This one is just painfully obvious. I think it was only asked as a check for the students.

    (c) This is the part I can't seem to get. I know that in the small angle approximation sin[itex]\theta[/itex] [itex]\rightarrow[/itex] [itex]\theta[/itex]. My intuition tells me that I should end up with a periodic function with an amplitude that decreases with time and some sort of time dependence such that the frequency of oscillation increases with time . As the string becomes shorter the mass cannot swing as far. The period of a simple pendulum in proportional to [itex]\sqrt{l}[/itex]. Since l is decreasing at a constant rate, so does the period. So I expect to have a cos((l0 - vt)[itex]\theta[/itex]) term. Or something similar with an exponential. It's just solving the differential equation that has got me stuck.

    Any suggestions will be appreciated. Thank you for your help!
     
    Last edited: Sep 11, 2011
  2. jcsd
  3. Sep 11, 2011 #2
    I haven't thought about this for more than a few minutes, but one thing that jumps out is that the kinetic energy term should be more complicated than what you have there. The angular motion and the vertical motion caused by the pendulum string being pulled upward are not independent, because the angular motion has a vertical component. So for example, the kinetic energy is smaller when the mass is swinging downward in its arc than when it is swinging upward. You need to think about how to describe this in your definition of kinetic energy.
     
  4. Sep 12, 2011 #3
    I'm almost certain the kinetic energy is correct.

    T = [itex] \frac{1}{2}m v'^2[/itex] where v2 is defined as [itex] \dot{x}^2 + \dot{y}^2 [/itex]

    x = l(t)sin[itex]\theta[/itex] = (l - vt)sin[itex]\theta[/itex]
    so [itex]\frac{dx}{dt}[/itex] = [itex]\frac{\partial}{\partial t}dt + \frac{\partial}{\partial \theta}d\theta[/itex]
    giving [itex] \dot{x} = -vsin\theta + (l-vt)\dot{\theta}cos\theta [/itex]

    You get something similar for [itex]\dot{y}[/itex] but with some convenient minus signs. Squaring the two and adding them gives
    v'2 = [itex] l^2\dot{\theta}^2 + 2vtl\dot{\theta} +v^2t^2\dot{\theta}^2 [/itex]
    Gathering the terms gives
    v'2 = [itex]v^2 + \dot{\theta}^2(l + vt)^2 [/itex]​
    as in my previous post.

    Please let me know if you see any error in my logic. Thanks for your time!
     
  5. Sep 13, 2011 #4
    You are correct. So much for my intuition! The Lagrangian is right.

    However, your equation of motion is off, because the time derivative in the Euler-Lagrange equation of motion applies to everything. I end up getting, with the small-angle approximation:

    [itex]
    \ddot{\theta}(\ell - vt) + 2v \dot{\theta} - g \theta = 0
    [/itex]

    I messed with this for a while, but couldn't solve it. The best I could do was set:

    [itex]
    \theta = e^{\phi}
    [/itex]

    And then obtain a first-order but (unfortunately) nonlinear equation:

    [itex]
    (\ell - vt) (\dot{\omega} + \omega^2) + 2v\omega - g = 0
    [/itex]

    Where

    [itex]
    \omega = \dot{\phi}
    [/itex]

    This equation was very difficult to solve, so I'm not sure I was on the right track in going that way.

    I'm sorry I couldn't be more help. If there's an easy way to do this, we're both missing it.

    Edit: Okay, I think I found a way to move forward. Take a look at the "Exact Solutions" section on this page: http://en.wikipedia.org/wiki/Differential_equation

    #7 looks like what we have here. You need to multiply through by an integrating factor like it says there, but after that you can just integrate, and you obtain an (implicit, unfortunately) solution.

    Let me know if you have any questions about how to do any of this.
     
    Last edited: Sep 13, 2011
  6. Sep 13, 2011 #5
    I spoke with the professor. He advised me to try using successive approximations. I haven't gotten back to the problem yet, but I'll give both suggestions a go.
     
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