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Overdamping limit

  1. Sep 16, 2009 #1
    1. The problem statement, all variables and given/known data

    Find the conditions under which it is valid to approximate the equation

    [tex]mL^2\ddot{\theta}+b\dot{\theta}+mgLsin\theta=\Gamma[/tex]

    by its overdamped limit
    [tex]b\dot{\theta}+mgLsin\theta=\Gamma[/tex]

    2. Relevant equations

    The formula is for an overdamped pendulum where b is the damping coefficient, and Gamma is a torque on the pendulum.

    3. The attempt at a solution

    I know from past experience that the conditions are [tex]b^2>>m^2gL^3[/tex], but I'm not sure how to obtain that from the equation.

    Thanks,
    Jeans
     
    Last edited: Sep 17, 2009
  2. jcsd
  3. Sep 17, 2009 #2

    Mark44

    Staff: Mentor

    If [itex]\theta[/itex] is "small," [itex]sin(\theta)~\approx~\theta[/itex]. Under this condition you can replace [itex]sin(\theta)[/itex] in your differential equation to make it a linear d.e with constant coefficients. Is that enough of a start?
     
  4. Sep 17, 2009 #3
    I had considered doing that and then equalizing the equations, but the problem is implying that it is done for all theta.

    Here's an email from my teacher regarding it. I don't really know how much of it I understand (I just got the email this morning):

    "No you don't need to solve for any ODE. Rescale the problem with time scale T, the coefficient preceding theta' should be 1, the coefficient preceding theta'' should be much less than 1. Check the overdamped bead on the hoop problem for similar procedure."
     
  5. Sep 17, 2009 #4
    Ok! I think I get it! Essentially I had to non-dimensionalize the equation by getting rid of the double-dot term.

    [tex]mL^2\ddot{\theta}+b\dot{\theta}+mgLsin\theta=\Gamma[/tex]

    [tex]mL^2\frac{d^2\theta}{dt^2}+b\frac{d\theta}{dt}+mgLsin\theta=\Gamma[/tex]

    [tex]\frac{L}{g}\frac{d^2\theta}{dt^2}+\frac{b}{mgL}\frac{d\theta}{dt}+sin\theta=\frac{\Gamma}{mgL} [/tex]

    Define: [tex]\tau=\frac{mgL}{b}t --> t=\frac{b}{mgL}\tau[/tex]

    [tex]\frac{d\theta}{dt}=\frac{d\theta}{\frac{b}{mgL}d\tau}=\theta'[/tex]
    [tex]\frac{d^2\theta}{dt^2}=\frac{d^2\theta}{\frac{b^2}{m^2g^2L^2}d\tau}=\theta''[/tex]

    Apply the new definition:

    [tex]\frac{L*m^2g^2L^2}{gb^2}\theta''+\theta'+sin\theta=\frac{\Gamma}{mgL} --> \frac{m^2gL^3}{b^2}\theta''+\theta'+sin\theta=\frac{\Gamma}{mgL}[/tex]

    And there is my [tex]\frac{m^2gL^3}{b^2}[/tex] which agrees with what I thought was the answer.

    Then if we multiply out the function under Gamma on the right side of the equation and substitute t back in, we are left with the original equation minus the double-dot term. Yay!

    QATC?

    thanks a bunch,
    Jeans
     
    Last edited: Sep 17, 2009
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