1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Science Advisor
    Homework Helper

    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 :)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook