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Bead sliding on a rigid straight wire

  1. Apr 14, 2012 #1
    1. The problem statement, all variables and given/known data
    A bead is free to slide along a rigid, straight wire, whilst the wire is rotating at angular velocity ω about the z-axis and is tilted away from the z-axis at angle α. I have the equation of motion (EOM) and need to find an explicit solution for the distance of the bead along the wire (q) as a function of time.


    2. Relevant equations
    EOM: [itex]\ddot{q} - ω^2sin^2(α)q = -gcos(α)[/itex]


    3. The attempt at a solution
    I assume the ansatz

    [itex]q = Ae^{Bt} [/itex]

    which differentiates twice to give

    [itex]\ddot{q} = B^2q[/itex]

    and substitute the second expression into the EOM, which gives (after a little rearrangement):

    [itex]q = -\frac{gcos(α)}{B^2 - ω^2sin^2(α)}[/itex]

    I don't know what I've done wrong, but I'm sure I must have made a mistake somewhere since the result is an expression for q which has no dependence on time, despite my ansatz that q is a function of time. Can anyone explain what I should do differently here?
     
  2. jcsd
  3. Apr 14, 2012 #2

    tiny-tim

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    Hi Favicon! :smile:
    no, it's AeBt = that

    but wouldn't it be simpler to write [itex]\ddot{q} - ω^2sin^2(α)(q + gcos(α)/ω^2sin^2(α))[/itex], and make the obvious linear substitution? :wink:
     
  4. Apr 14, 2012 #3
    Please bear with me as I'm quite rusty on my differential equation solving. By "the obvious linear substitution" I take it you mean [itex]p(t) = q(t) + \frac{gcos(\alpha)}{\omega^2sin^2(\alpha)}[/itex]

    Then, since [itex]\alpha[/itex] and [itex]\omega[/itex] are both constant, we have [itex]\ddot{p} = \ddot{q}[/itex]

    Substituting those into the EOM (in the form you suggested) gives

    [itex]\ddot{p} - \omega^2sin^2(\alpha)p = 0[/itex]

    No I take the ansatz [itex]p = Ae^{Bt}[/itex] and substitute into the previous equation to get

    [itex]B^2Ae^{Bt} - \omega^2sin^2(\alpha)Ae^{Bt} = 0[/itex]

    [itex]B = ±\omega sin(\alpha)[/itex]

    If I remember rightly, this means a general solution takes the form

    [itex]p(t) = A_1e^{\omega sin(\alpha)t} + A_2e^{-\omega sin(\alpha)t}[/itex]

    Substituting for q then yields

    [itex]q(t) = A_1e^{\omega sin(\alpha)t} + A_2e^{-\omega sin(\alpha)t} - \frac{gcos(\alpha)}{\omega^2sin^2(\alpha)}[/itex]

    Am I right so far? I can see that to go any further would require knowledge of q(t) at some specific time, e.g q(0). However, the textbook from which I'm working suggests that to determine a specific solution for q(t) would require knowledge of [itex]\dot{q}(0)[/itex] at some specific time also. Where does this requirement come from?
     
  5. Apr 14, 2012 #4

    tiny-tim

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    yes :smile:

    now you have two constants of integration, A1 and A2, so you need two initial conditions to find both (which will usually be a condition on q and a condition on q')
     
  6. Apr 14, 2012 #5
    Ah yes, that's a good point. Thanks very much for your help Tiny :)
     
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