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Double Pendulum (One pendulum hanging from the other)

  1. Mar 19, 2013 #1

    CAF123

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    1. The problem statement, all variables and given/known data
    A body of mass m hangs by a light inelastic string of length a from a body of mass 3m which in turn hangs from a fixed point O by a string of length a. The masses are confined to motion in a vertical plane through O. Denote the horizontal displacements of the two particles from the vertical by ##x_1## (for upper mass) and ##x_2##(for lower mass). Assume that they are both small compared to a.

    1) Determine the kinetic and potential energy in terms of ##x_1##, ##x_2##.
    2) Rescale the coordinates ##x_i = \mu_i z_i## where ##\mu_i## are constants such that that the kinetic energy would take the form ##T = \frac{1}{2}(\dot{z_1}^2 + \dot{z_2}^2)##. Rewrite the total energy in terms of these new rescaled coordinates.

    3. The attempt at a solution
    I think I have made a good attempt at the above but I want to be sure of my answer before I start the next part of the question.

    Define coord system with +ve y down and +ve x right. Then x1 = x1 , y1 = acosθ1 = ##\sqrt{a^2 - x_1^2}.## For the other mass: x2 = x2, y2 = ##\sqrt{a^2 - x_1^2} + \sqrt{a^2 - x_2^2}.##
    By Taylor expansion/simplification I get to $$\frac{m_1gx_1^2}{2a} + \frac{m_2gx_2^2}{2a} + \frac{m_1}{2}[\dot{x_1}^2] + \frac{m_2}{2}[\dot{x_2}^2],$$ m1 = 3m and m2 = m.

    To get the rescaled version, I just subbed in what they give (after taking the derivative etc..). To get the required T, I think the condition ##\mu_1^2 m_1 = \mu_2^2m_2 = 1## must hold. Provided this is right, $$E = \underline{z}^T G \underline{z} + \frac{1}{2} \underline{\dot{z}}^T \dot{z}.$$

    I can then rearrange this into the form ##\underline{\ddot{z}} + n^2 G \underline{z} = 0, ##with ##n = \pm \sqrt{2}## and ##G## a diagonal matrix being ##\frac{g}{2a} I_2##, ##I_2## 2x2 identity.

    Many thanks.
     
  2. jcsd
  3. Mar 19, 2013 #2
    Note the expression for energy you got is the sum of energies of two independent simple pendula. Can this be correct?
     
  4. Mar 19, 2013 #3

    CAF123

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    I would agree, this does not make sense since they are clearly coupled. However, is it not the case that by rescaling the coordinates the system becomes decoupled? (Is that not the purpose of the rescaling)?

    (The physical reasoning behind this is not clear, although)
     
  5. Mar 19, 2013 #4
    The original energy equation is already essentially decoupled, rescaling has nothing to do with that. So you must have made a mistake composing it.
     
  6. Mar 19, 2013 #5

    CAF123

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    Hmm..I'll look over it again and get back to you

    EDIT: Yes, I see the error. I'll fix it and then post the new expression.
     
  7. Mar 19, 2013 #6

    CAF123

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    If I didn't make any further mistakes, then I have:$$E = \frac{m_1g x_1^2}{2a} + \frac{m_2g x_2^2}{2a} + \frac{m_1}{2}[\dot{x_1}^2] + \frac{m_2}{2}[\dot{x_2}^2 + \frac{x_1^2}{a^2} + \frac{2x_1 x_2}{a^2} + \frac{x_2^2}{a^2}]$$
     
  8. Mar 19, 2013 #7
    I see that you have tried to account for the fact that the velocity of the lower mass depends on the velocity of the upper mass. But I don't understand why it is so complex. It should really be along the lines (v + u)^2.

    Then, how about the potential energy of the lower mass?
     
  9. Mar 19, 2013 #8

    CAF123

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    I got something of the form (u + v)^2 but then I expanded this. Perhaps I neglected too many terms. What I get is:$$\frac{m}{2}\left[\underline{\dot{x_2}^2} + \left(\frac{-x_1}{\sqrt{a^2 - x_1^2}} - \frac{x_2}{\sqrt{a^2 - x_2^2}}\right)^2\right]$$ When I expand I get: $$ \frac{m}{2}\left[\underline{\dot{x_2}}^2 + \frac{x_1^2}{a^2 - x_1^2} + \frac{2x_1 x_2}{a^2} + \frac{x_2^2}{a^2 - x_2^2}\right]$$

    This is for the last kinetic energy term (due to the lower mass).
    Is there something wrong with it?
     
  10. Mar 19, 2013 #9
    Before you go expanding: what is the velocity of the lower mass?

    For potential energy: what is the vertical position of the lower mass?
     
  11. Mar 19, 2013 #10

    CAF123

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    All of the expression in the squared brackets in the previous post.
    Edit: I forgot the chain rule - the expression i wrote intially is dimensionally ridiculous.

    ##\sqrt{a^2 - x_1^2} + \sqrt{a^2-x_2^2}##
     
  12. Mar 19, 2013 #11
    The square of the velocity of the upper mass is ##\dot{x}_1^2 + \dot{y}_1^2 ##. The lower mass: ## (\dot{x}_1 + \dot{x}_2)^2 + (\dot{y}_1 + \dot{y}_2)^2##. Are these what you have been using before expanding and linearizng?

    Correct. But then I don't understand why the potential energy term for the lower mass does not have ##x_1##. Or was that combined with its kinetic energy?
     
  13. Mar 19, 2013 #12

    CAF123

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    Yes for the upper mass but no to the lower mass. I was doing for the lower mass: $$\frac{m}{2}(\dot{x_2}^2 + \dot{y_2}^2)$$ and expressing y2 in terms of x1 and it's derivative.


    Because it is yet another silly error. Full potential term: $$V = V_1 + V_2 = -m_1gy_1 - m_2gy_2 = -m_1g\sqrt{a^2 - x_1^2} - m_2g(\sqrt{a^2 - x_1^2} + \sqrt{a^2 - x_2^2})$$
     
  14. Mar 19, 2013 #13

    CAF123

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    My final expression for T: $$\frac{m_1}{2}[\underline{\dot{x_1}}^2 + \frac{x_1^2 \dot{x_1}^2}{a^2 - x_1^2}] + \frac{m_2}{2}[\underline{\dot{x_2}^2} + (\frac{-x_1 \dot{x_1}}{\sqrt{a^2 - x_1^2}} - \frac{x_2 \dot{x_2}}{\sqrt{a^2 - x_2^2}})^2 ]$$

    before I expand.
     
  15. Mar 19, 2013 #14
    I just noticed that you defined ## y_2 ## as the sum of ## y_1 ## and second mass's vertical displacement relative the first. So my expression is indeed incorrect in this notation. Yours is too, anyway, because the horizontal component of the velocity is ## \dot{x}_1 + \dot{x}_2 ##.

    I would still suggest that you redefine ## y_2 ## to be the second mass's vertical displacement relative the first, that should simplify the expression.

    This I agree with.
     
  16. Mar 20, 2013 #15

    CAF123

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    Yes, because I defined my coordinates wrt ceiling. So my kinetic term is: $$\frac{m_1}{2}\left[\frac{\dot{x_1}^2a^2}{a^2 - x_1^2}\right] + \frac{m_2}{2}\left[(\dot{x_1}+ \dot{x_2})^2 + \left(\frac{-2x_1\dot{x_1}}{\sqrt{a^2 - x_1^2}} + \frac{x_2\dot{x_2}}{\sqrt{a^2 - x_2^2}}\right)^2\right]$$,
    where $$\dot{y_1} + \dot{y_2} = \frac{-2x_1\dot{x_1}}{\sqrt{a^2 - x_1^2}} + \frac{x_2\dot{x_2}}{\sqrt{a^2 - x_2^2}}$$
     
  17. Mar 20, 2013 #16
    Why -2?
     
  18. Mar 20, 2013 #17

    CAF123

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    I just skipped a step and brought two terms together:
    $$\dot{y_1} + \dot{y_2} = \frac{-x_1 \dot{x_1}}{\sqrt{a^2 - x_1^2}} -\frac{x_1\dot{x_1}}{\sqrt{a^2 - x_1^2}} - \frac{x_2 \dot{x_2}}{\sqrt{a^2 - x_2^2}}$$
     
  19. Mar 20, 2013 #18
    I do not understand why the vertical velocity of the lower mass has two x1-based terms. Physically, it does not seem to make sense.
     
  20. Mar 20, 2013 #19

    CAF123

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    For the upper mass, ##y_1 = \sqrt{a^2 - x_1^2}##. For the lower mass, ##y_2 = \sqrt{a^2 - x_1^2} + \sqrt{a^2 - x_2^2}##

    Then ##\dot{y_1} + \dot{y_2} ## gives the result, no?
     
  21. Mar 20, 2013 #20
    If you insist that ## y_2 ## is defined this way, let me denote ##y_2' = y_2 - y_1##, which is the vertical velocity of the lower mass relative the upper mass. The total vertical velocity of the lower mass is either ## \dot{y}_1 + \dot{y}_2' ## or simply ## \dot{y}_2 ## in your notation, but not ##\dot{y_1} + \dot{y_2} ##
     
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