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Period of the motion

  1. Mar 15, 2015 #1
    1. The problem statement, all variables and given/known data
    A light rod of length 0.5 m has two point masses of mass 1.6 kg attached at either end. The rod is pivoted about its centre and is attached to a spring which exerts a moment on the rod that is proportional to the angle the rod is displaced from the equilibrium position and which attempts to return the rod to the equilibrium position. The rod is displaced 0.27 rad from the equilibrium position and released from rest. If the spring constant is 5.4 Nm.rad-1 what is the period of the resulting motion?

    2. Relevant equations
    y = Asin(ωt) = Asin(√((k/m)t))

    3. The attempt at a solution
    I'm not sure if that's even the correct equation that I would use in this situation. All the variables are there, so it must be so... How would I derive period t from that?
     
  2. jcsd
  3. Mar 15, 2015 #2

    rude man

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    You need to use the equation relating torque to angular acceleration.
     
  4. Mar 15, 2015 #3
    Greetings,
    I'll answer your question with resources:
    http://mmsphyschem.com/oscSpring.html [Broken].
    SHM.
    Something about it.
    Please note I am not into this spring and motion stuff...I tried my best gathering the most valuable data related on the internet.


     
    Last edited by a moderator: May 7, 2017
  5. Mar 16, 2015 #4
    τ=ml^2α
    Okay, done. Now what do I do with this equation? I've been given the mass and the length for this equation but that's about it. How is torque necessary in this problem?
     
  6. Mar 16, 2015 #5

    SammyS

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    The moment of inertia of the rod with the two point masses is not ##m \ell^2 \ ##.

    That's how torque is related to the angular velocity.

    How does the spring produce that torque?
     
  7. Mar 17, 2015 #6

    rude man

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    Rewrite α in terms of the angle θ the rod makes with the position corresponding to zero torque. Then combine in the equation relating torque to the moment of inertia I, and θ. What is the torque expression? Hint: think spring constant relating torque to θ. BTW you may also want to think about what poster #5 said about your expression for I.
     
  8. Mar 17, 2015 #7
    I'm not sure if this is correct, but could I relate spring constant to torque? τ=-kθ
    And I realize now that ml^2 isn't the inertia for two-point masses? That should be: I = (mM/m+M)x^2
    If I relate it to τ= Iα then: -kθ = Iα
    Hmmm, can I say that α = ω/t where t is the period and just try to find that? Or did I come up with something that makes no sense?
     
  9. Mar 17, 2015 #8

    rude man

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    Right.
    what is your x? m and M are the same, just use m.
    right.
    No. What is the definition of α really?
     
  10. Mar 17, 2015 #9
    x = 0.5, that's the distance between the two m's. Okay, but is that the right equation for two-point masses? I = (m2/2m)x^2 since m and M are the same?
    α = d2θ/dt2 something like that?
     
  11. Mar 17, 2015 #10

    rude man

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    Don't like your choice of x.
    Let's agree the bar has total length L.
    Suppose you had only one mass m at one end of a bar of length L/2. What would I be if the axis of rotation were at the other end?
    EXACTLY like that!
    So what does the torque equation look like? Leave I = I for the moment until you have that figured out in terms of L and m.
     
  12. Mar 17, 2015 #11
    I always have issues with the moment of inertia. But okay the expression should look like: d2θ/dt2*I = -kθ and I should rearrange everything to find dt which happens to be the period?
    I don't quite understand the inertia but perhaps it is the sum of the contributions from the two masses? So m = 3.2kg multiplied by L2? No that doesn't sound right...
     
  13. Mar 17, 2015 #12

    SammyS

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    The moment of inertia is the sum of:
    The moment of inertia of a rod of length, L, about its center.
    The moment of inertia two point masses at opposite ends of a mass-less rod of length, L.​

    Added in Edit:

    It's a light bar, so ignore its contribution to the moment of inertia.
     
    Last edited: Mar 17, 2015
  14. Mar 17, 2015 #13

    rude man

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    No. You need to solve the differential equation. That will give you the period along with everything else.
    If youi haven't had diff. eq's then you need to compare this equation with the equation of a mass on a spring.
    If I is the inertia of a mass at the end of a bar of length L/2 pivoted at the other end, then the I of two identical bars, identically pivoted, is twice the I of one of the bars.
    Don't use numbers until you have figured out I(L,m).
     
  15. Mar 17, 2015 #14
    ml^2 = I? Multiplied by 2 since the inertia for one of the bars is the same as the other?
    How on earth do I solve this differential equation for period T?
     
  16. Mar 17, 2015 #15
    Look at the Formula sheet Doctor Moore provided you with!
     
  17. Mar 18, 2015 #16

    rude man

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    What's the expression for a point mass a distance L/2 away from the pivot point?
    It's a simple 2nd order linear, constant-coefficient ODE. You should wind up with something like θ = θ0cos(ωt) with θ0 the amplitude and ω a function of some of your given parameters. Surely you can deduce the period of a signal cos(ωt) if ω is given.
    I'll have to sign off here.
     
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