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Pendulum and spring

  1. May 29, 2016 #1
    1. The problem statement, all variables and given/known data

    I have to derive equation of motion for this system. I want to use a moment of force, but i have a problem with moment of force spring.
    2. Relevant equations

    3. The attempt at a solution
    What i've done is:
    M(Fg)=-mgLsinα
    M(N)=0
    M(Fb)=mω^2 Lsinα*Lcosα
    mL^2*α''=ΣM
    M(Fs)=?
     

    Attached Files:

  2. jcsd
  3. May 29, 2016 #2

    BvU

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    You still haven't provided the full problem statement.
    Nor any equations.
    Are you supposed to do this classically, or with a Lagrangian ?
    And, as I stated, ##F= - kx## is usual for a spring.

    Did you read through the guidelines ?
    Did a mentor delete your post (and my reply) ?
     
  4. May 30, 2016 #3
    So, i have to find equation of motion for this system. We know length of the line (L) and L1(picture). Initially the angle between spring and line(L) is 90, so initial length of the spring is √(L1^2-L^2). What's more the the pendulum rotates around the main rod (
    angular velocity ω).
    Yes my post was deleted.
     
  5. May 30, 2016 #4

    BvU

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    That all you have to describe the system (the full problem statement) ?
    If so, are you allowed to make a small-angle approximation (##x = x_0 + L(\theta-\theta_0)## ) ?
    If not, you have some trig to work out.
     
  6. May 30, 2016 #5
    I'm supposed to this classically, but i'm not allowed to make a small-angle approximation. There is also dissipation F=-cl'. I can't even imagine how this system works. I worked out length of the spring (dependent on angle) using law of cosines, but i don't know if it's a good idea.
     
    Last edited: May 30, 2016
  7. May 30, 2016 #6

    BvU

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    Pretty awkward indeed. Don't know how to make it simpler, I'm afraid ... :frown:

    Apparently the full problem statement is still more involved ? Although -cL' is a constant (?)
     
  8. May 30, 2016 #7
    No -cl' isn't constatnt, l(small L) is the length of the spring(dependent on the angle)
     
  9. May 30, 2016 #8

    BvU

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    That's not dissipation ! That's the spring (what I called ##
    F= - kx ## ). Very nifty to use l' and L' for different quantities :wink: . Confusion assured !

    The whole thing looks a bit like a steam engine governor with an extra spring: ##\omega## pushes the weight outward, gravity + spring pull it back.

    And if it doesn't say what the equilibrium length is, you're stuck ! (unless your problem statement says somehow that it is zero).
     
  10. May 30, 2016 #9
    In my exercise is:
    Assume forces:
    1. Spring elasticity: F=-kΔl
    2. Dissipation/damping: F=-cl'
    So I don't think it's the same.
     
  11. May 30, 2016 #10

    TSny

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    So l' is rate of change of l?

    I still don't see a complete statement of the problem.
     
  12. May 30, 2016 #11
    Yes l' is a derivative. It is all I have in my assignment
     
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