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Homework Help: Period of rod connected to spring

  1. Apr 14, 2012 #1
    1. The problem statement, all variables and given/known data
    A spring has one of its end connected to the ceiling and the other end connected to the end of horizontal rod. The rod is free to rotate in xy-plane about its vertical axis. The mass of the rod is 0.6 kg and the spring constant is 2000 Nm. If the rod is displaced and undergoes small oscillations, find the period of the oscillation

    2. Relevant equations
    Not sure, maybe:
    T = 2π√(I / κ)
    T = 2π√(I/mgh)
    τ = -κθ
    τ = F . d
    For small angle, sin θ = θ
    I = 1/3 ML2
    F = - kx

    3. The attempt at a solution
    My idea is to find the value of κ using τ = -κθ, then put it into T = 2π√(I / κ)

    Let the rod is displaced by angle θ (where θ is small) with respect to horizontal. Is it correct that there are two forces that contribute to produce torques (weight of rod and restoring force of spring) and they act in the same direction of rotation?

    I tried to do it this way and end up having term cos θ as result of finding the perpendicular distance to the pivot. I take the pivot is the end of the rod which does not connected to spring.

    I think my working was wrong from the beginning...
  2. jcsd
  3. Apr 14, 2012 #2
    I couldnt really understand the question. Is this the complete question? There is no figure?
  4. Apr 14, 2012 #3


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    hi songoku! :smile:
    i assume you mean "the other end connected to the centre of horizontal rod"? :confused:

    and the spring is a torsion spring, with a fixed length, so that the displacement is purely angular?
    rather than trying to apply formulas, isn't it easier just to find the equation relating θ'' and θ, and read the period off that?
    i don't understand the set-up here at all :confused:
  5. Apr 14, 2012 #4
    Sorry I can't attach the real figure but the figure below is really similar to the figure given in the question

  6. Apr 14, 2012 #5


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    so the rod is supported on a pivot under its centre, and the spring is attached to one end?

    but then why does the question give a torsion constant (torque per angle, measured in Nm per radian), instead of an ordinary spring constant (force per distance, measured in N/m)? :confused:
  7. Apr 14, 2012 #6
    Ops sorry my bad. It should be 2000 N/m. And for the figure, there is no support at the center of the rod. I crop the figure from internet and the line in the middle actually has downward arrow and is describing the weight of the rod. Sorry for the confusion :redface:
  8. Apr 18, 2012 #7


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    sorry, songoku, but i don't understand the question :confused:
    without being supported anywhere else, it doesn't seem to make any sense :redface:
  9. Apr 18, 2012 #8
    hi tiny-tim :smile:

    I think the rod oscillates with the spring as the pivot; so it is like pendulum where the spring acts like the rope
  10. Apr 24, 2012 #9
    hi tiny - tim :smile:

    I found a very similar problem. Hope this attachment can clarify everything. Sorry :smile:
  11. Apr 24, 2012 #10


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    but that picture does show a pivot through the centre of the rod

    (and there wouldn't be "small oscillations" about the horizontal position if there wasn't!)

    sooo … are we to proceed on the basis of a similar pivot in the original question? :smile:
  12. Apr 24, 2012 #11
    Let proceed with the question in the attachment because I am afraid that I typed the original question wrongly (I don't have the original question, the book has been taken :shy:)

    Actually I still don't understand the movement of the rod. The rod will move up and down or back and front?
  13. Apr 24, 2012 #12


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    up and down :smile:

    (so the spring stays vertical, there's no torsion)
  14. May 7, 2012 #13
    Sorry for taking long time to reply

    I still don't know how to start doing this but let see:
    T = 2π√(I/Mgh) ; I = 1/3 ML2 and h = 1/2 L
    = 2π√(1/3 ML2/Mg 1/2 L)

    Am I correct this far? If no, where is the mistake? If yes, how to continue? What do we use the spring constant for?

  15. May 8, 2012 #14


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    sorry, i'm totally confused

    what exactly is the question now? :confused:
  16. May 8, 2012 #15
    This is the question we are dealing with :smile:
  17. May 8, 2012 #16


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    hmm … i don't like it when people use a formula that may or may not work :redface:

    it's always safest to start from the basic equation

    for shm, you're looking for a formula (valid for small x) x'' = -kx …

    in this case use torque = spring force x distance = Iα …

    what do you get? :smile:

    (btw, your I must always be about the centre of mass or the centre of revolution … in this case, the same :wink:)
  18. May 8, 2012 #17
    hi tiny-tim :biggrin:
    The formula I know for shm is x'' = -ωx. What is "k" is your formula?

    So, spring force x distance = Iα. Where should I take the pivot? At the center of the rod?
    Let that be the case, then:
    F sin (pi/2 - θ) . distance = Iα (I take θ to be the angle between the rod and horizontal)
    kx cos θ . 1/2 L = 1/12 ML2α

    I am not sure I get it right till this part. What to do with α and cos θ?

    I suppose I still don't understand the question clearly. I take the pivot at the end of the rod where it is connected to the spring so I use 1/3 ML2
  19. May 8, 2012 #18


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    hi songoku! :smile:
    same as ω :biggrin:

    (i prefer k in this case, since ω might get confused with dθ/dt)
    as i said, doesn't work, must be centre of mass or centre of rotation
    correct :smile:
    for "small oscillations", you can assume cosθ = 1 :wink:

    eliminate x by using x = (L/2)θ …

    that gives you an equation of the form θ'' = - kθ ! :smile:
  20. May 8, 2012 #19
    hi tiny-tim :smile:
    Oh, I think for small θ, cos θ = 1 - 1/2 θ2 ; so actually we can simplify it further.

    You take x as the vertical distance of the center of mass of rod with respect to horizontal?

    OK so:
    kx cos θ . 1/2 L = 1/12 ML2α
    k . L/2 θ . 1/2 L = 1/12 ML2α
    α = (4k/M) θ

    Then 4k/M = ω2 --> solve for period. Am I correct?
  21. May 9, 2012 #20


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    hi songoku! :smile:
    are we doing the same question?

    i'm following your diagram, which shows the rod pivoting about a stationary pivot at its centre :confused:
    yes, except the euqations should all have a minus, and isn't that 4 a 3 ? :redface:
  22. May 9, 2012 #21
    hi tiny-tim :smile:
    Yes we are :approve:

    How can you get x = (L/2)θ? By looking at triangle formed by the horizontal, the rod, and vertical distance between center of rod and horizontal?

    Yes, it absolutely should be 3 :blushing:

    And about the minus sign, both side have minus sign because spring force is restoring force so it should be minus and the rod rorates clockwise so the angular acceleration should be minus?
  23. May 9, 2012 #22


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    yes :smile:

    it's a standard formula that you should learn anyway …

    x = rθ

    together with

    v = rω

    a = rα​

    (for example, they convert the standard constant acceleration equations to constant rotational acceleration)
    no :confused:

    if it was plus, the motion wouldn't oscillate, it would be exponential! :rolleyes:
  24. May 9, 2012 #23
    I don't get this part...
  25. May 9, 2012 #24


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    θ'' = k2θ

    has the general solution θ = Aekt + Be-kt

    θ'' = -k2θ

    has the general solution θ = Acoskt + Bsinkt​

    the first is exponential, the second is periodic (oscillatory)
  26. May 9, 2012 #25
    So, actually it is related to the solution of second order differential equation. I haven't covered it yet so for now I juts accept it.

    Thanks a lot for the help :smile:
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