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Vibrations - Modeling system, equation of motion

  1. Nov 21, 2017 #1
    Vibrations - Modelling system, equation of motion
    Hi,

    In the first question (question 4) in the attached file, how would you go about modelling the system and finding the equation of motion? All those masses are confusing me, I don't even know where to start.

    I don't know whether the angle theta has a significant effect on the displacement of the spring or not, it the masses are fixed or not.

    Any ideas??
     

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  3. Nov 21, 2017 #2

    BvU

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    Hello Pmm, :welcome:

    PF Guidelines require a bit more from you before we're allowed to assist.

    But I can say a few things:
    More the other way around: if the spring contracts or expands, ##\theta## changes.
    Masses are fixed on their rods. m1 m2 m3 is a fixed triangle that can rotate around ##O##
    And you may assume small oscillations -- so the spring stays vertical
     
  4. Nov 21, 2017 #3
    Thank you!

    I drew a FBD, and for the spring side I'm getting Fspring = kx = N. For the angle part I'm getting ω = θt. I tried finding the moments about 0, but don't know the angles to use. I feel like it's something to do with Asin(ωt) but I'm not sure how to get there.
     
  5. Nov 21, 2017 #4

    BvU

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    Please post your work in detail. You don't explain what t in ω = θt is, but it sure looks wrong.
    What do you think of figure 5 ? Any conclusions/reasonable assumptions ?
    And if that doesn't light the light :smile:, what about the wording of parts 4, 5 and 6 ?
     
  6. Nov 21, 2017 #5
    My bad, t is time, ω is angular velocity in rad/s.

    My assumptions from the figure:
    1. Damping due to air is negligible.
    2. There are no external forces acting on the system.
    3. The mass m3 is displaced in the vertical direction only.
    4. It is a 1 DOF system (θ is enough to fully describe the position of all parts).

    I'm still stuck on where to go from there. In 4,5,6 they talk about angular velocity as a function of time. I know vertical displacement from a rotating object is given by Asinωt, velocity is it's derivative wrt t so, ωAcosωt, acceleration is -ω2Asinωt......I have no clue what A is though. Isn't the amplitude just the length of the circle's radius? So l3 in this case?

    And the graph underneath...I don't get it, it's a plot of θ vs θ isn't it? ωnt = θ, and so is angular displacement right??
     
    Last edited: Nov 21, 2017
  7. Nov 21, 2017 #6

    BvU

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    4 is the important one. ##\theta## is your main character in this whole play. The whole thing is at equilibrium when ##\theta = 0 ## as you can see from fig. 5. Out of equilibrium there is a restoring force (magnitude ?) and there is some inertia, so there will be oscillations, just like with a mass on a spring. Since we are talking angles, we should speak the language: torque, angular acceleration, and so on.
    It's time for some equations, and sure enough the exercise starts off with that. Any suggestion ?
     
  8. Nov 21, 2017 #7
    I know that ∑M0= I0θ'' ... but I don't know how to find I. I tried using I = ml2 and got ΣM0 = -θ'' (m1l12 + m2l22 + m3l32) + kl3sinθ but I can't see where that gets me.

    Can you tell me, what am I looking to even arrive at with these types of problems? I know it's the equation of motion, but how do I know when I've found it, what am I looking to express?

    And my lecture notes keep using ωn and fn both for the natural frequency interchangeably. Which is correct? I thought ωn was angular velocity?
     
  9. Nov 21, 2017 #8

    BvU

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    Yep, ##\omega = 2\pi f##.
    You are working towards somehing that (see the figure) looks like a sine function of time. Like with a mass on a spring. The restoring force is opposite to and proportional with the distance from equilibrium: ma = -kx


    Here you have mass(es) and a spring, but this time it is not an ##x## or ##y## but something with an angle. So you have to translate F = ma to angular lingo.
     
  10. Nov 21, 2017 #9
    That's what I did in my last post with " ΣM0 = -θ'' (m1l12 + m2l22 + m3l32) + kl3sinθ". Is that the correct answer?
     
  11. Nov 22, 2017 #10

    BvU

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    Sorry I overlooked that.
    Sort of, yes: There is only one torque: ##kl_3\sin\theta ## ( except I that miss a sign) .
    So you get ##\ I\alpha = \sum\tau\ ## .

    And for small ##\theta## you may approach ##\sin\theta \approx \theta##
     
  12. Nov 22, 2017 #11
    Thank you!

    What do you mean by only one torque though?
    And what is α in Iα= Σtorques?
     
  13. Nov 22, 2017 #12

    BvU

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    ##\alpha = \dot \omega = \ddot \theta##, the angular acceleration.
    Very good observation !
    And I realize I am making a mistake: the center of mass of m1, m2, m3 is not at ##O##.
    But from fig 5 equilibrium is at ##\theta = 0 ## and the motion is a sine, so:
    My best estimate is that the exercise composer intended you to ignore the torque contribution due to gravity working on the center of mass: the angle of l2 is not given and you would need it.

    Would @haruspex agree in this ?
     
  14. Nov 22, 2017 #13
    I'm confused now :P I thought each mass contributes to the torque about point 0? That lecturer is a sneaky one so i wouldn't be surprised if he's trying to confused the class.
     
  15. Nov 22, 2017 #14

    BvU

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    Very good observation (again! There is something to be said for sneaky lecturers :smile: ...) .

    True. I took into account that l1 and l3 are almost horizontal. So their torques are constant (to within ##\theta^2## ) and the contribution goes into the equilibrium position. l2 looks pretty vertical, so the contribution from m2 varies proportional to ##\theta## and that is the same order of magnitude as that of the spring. The orientation of l2 is not given. With a vertical orientation you would get a torque of 10 kg x 10 m/s2 0.27 m x ##\sin \theta## in a direction opposite to that of the spring -- effectively reducing the 120 N/m to 93 N/m.

    If I were you I'd ignore it (if you were supposed to do something with that the exercise would become rather complicated and the orientation of l2 would have been given ) and only consider the contributions of the masses to the moment of inertia.

    I don't mean to confuse you. Had overlooked the ##m_i\, g\; l_i \cos\phi_i## torques -- but I suspect the exercise composer had too !
     
  16. Nov 22, 2017 #15
    Ok...I think i kind of get it....

    For part b though, it asks for the natural frequency of the system (not frequencies), if its just the usual sqrt(k/m)*(1/2π), what do I use for the mass value if it's the natural frequency of the entire system?

    Is the natural frequency of this system even sqrt(k/m)*(1/2π)?
     
    Last edited: Nov 22, 2017
  17. Nov 22, 2017 #16

    haruspex

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    Without seeing the data table it is hard to be sure, but the lack of angle constants implies to me that l2 is supposed to be at right angles to the others. And I gather the equilibrium position is with the main rod horizontal. Unclear whether that is the relaxed spring position, but I don't think that matters.

    That all being so, deltas to the gravitational torques of m1 and m3 are second order quantities and can be ignored. That from m2 is first order.

    So we get ##\ddot\theta\Sigma m_il_i^2+(-gm_2l_2+kl_3)\theta=0##, yes?
     
  18. Nov 22, 2017 #17
    I get the ##\ddot\theta\Sigma m_il_i^2## part and the kl3 part, but why does m2 effect the moment about 0 and not m1 and m3? I'd have thought if anything it would be the opposite, m1 and m3 effect it but m2 doesn't because it's virtually vertical....

    Why not θ''Σmili2+(-Σgmili+kl3)θ=0??

    Also, is the natural frequency just √(k/m), and the value of m to use just m1+m2+m3?
     
    Last edited: Nov 22, 2017
  19. Nov 22, 2017 #18

    haruspex

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    We are concerned with the small change to the moment that results from the small angular displacement. If a mass m is on rod length l at angle α+θ measured anticlockwise from positive x axis, then the gravitational moment is mgl cos(α+θ). The difference from the equilibrium position is mgl(cos(α+θ)-cos(α)). For small θ that approximates -mgl sin(α)θ. For m1 and m3, sin(α)=0; for m2 sin(α)=1.

    You can think of m2 as an upside-down pendulum.
     
  20. Nov 22, 2017 #19
    thanks I get it now!!!

    To find the formula for natural frequency; from the equation of motion, I see that there is a θ'' and a θ but no θ'. I tried solving the characteristic equation by letting the θ' coefficient = 0 but that didn't work. Am I on the right track? And for the natural frequency value do you use (m1+m2+m3) for the mass?
     
  21. Nov 22, 2017 #20

    haruspex

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    Right, but that is how it should be for undamped SHM.
    Use that for the mass where? You cannot plug this into an off-the-shelf equation for a mass on a spring. Work with the differential equation.
     
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