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Applied optimization to SHM problem? Hard?

  1. Sep 10, 2014 #1
    [Note: Post restored from archive - gneill, PF Mentor]

    1. The problem statement, all variables and given/known data

    A pendulum consists of a uniform circular disk of radius r which is free to turn about a horizontal axis perpendicular to it's plane. Find the position of the axis for which the periodic time ( for small amplitude oscillations) is a minimum.


    τ: torque
    Ι (i): moment of inertia
    m: mass
    g:gravity
    θ: angle between force mg and the radius length
    r: radius
    ω: angular frequency
    T: period
    Sinθ=θ for small amplitudes.
    L: length (offset for parallel axis theorem.) (unknown?)


    2. Relevant equations

    Equation (1): τ= -mgL θ

    Equation (2): θ'' + (mgL/ I ) θ = 0

    Equation (3): ω = (gL / (.5 r^2 + L^2)) ^.5

    3. The attempt at a solution
    So I solved for ω and my reasoning was that in order to have the minimum periodic time ω has to be at it's maximum. So I used applied optimization and set dω/dL = 0. However, I am not sure if this was the right way to do it since I squared both sides of the equation 3 to get rid of the square root and then used implicit differentiation. But I think this might be wrong because it's like if I differentiates the inside of Eq. 1 and that differentiation does not take into account the square root. So I am not sure if I did it right. I differentiated directly but the derivative was a monster and I couldn't factor out the L to solve for it. Is this a right way to do it correctly?
     

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    Last edited by a moderator: Sep 10, 2014
  2. jcsd
  3. Sep 10, 2014 #2

    BvU

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    You removed the post ? I thought you did just fine!
     
  4. Sep 10, 2014 #3
    It was correct?
     
  5. Sep 10, 2014 #4
    Null
     
    Last edited: Sep 10, 2014
  6. Sep 10, 2014 #5

    It was correct?
     
  7. Sep 10, 2014 #6

    BvU

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    Can't find no mistakes :smile: (PF isn't really meant for stamp approving, but I think you're doing fine).

    Don't know of a different path to the same answer. Your squaring is justified: ##\omega## is known to be > 0 , so no problem.
     
  8. Sep 10, 2014 #7
    Oh alright. Got it. Thanks for your guidance then ;p
     
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