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Minimum Period of Oscillation Disk

  1. Feb 9, 2016 #1
    1. The problem statement, all variables and given/known data
    how far from the rim of a disk of Radius R must the pivot point be located in order for its period of oscillation to be a minmum where R is the distance from the point to the centre of mass?

    I'm stuck at the derivative because I saw a similar problem where the answer is T=R/sqrt(2) but if you look at where I am right now the derivative of the whole thing is a bit too big. So I think I might have made a mistake or I have no idea how to do a derivative at all (also I'm wondering what to do with R^2 since it is a constant) and I would like to have some input from other people as to if I'm on the right track, because I feel lost.

    Any help would be greatly appreciated!! :)
    2. Relevant equations
    2+d^{2}}{dg}}.gif
    3. The attempt at a solution


    IMG_20160209_160016.jpg
     

    Attached Files:

  2. jcsd
  3. Feb 9, 2016 #2

    collinsmark

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    I think you mean to say d is the distance from the point to the centre of mass (not R; that's the radius).
    I think you mean to say [itex] d = \frac{R}{\sqrt{2}} [/itex].
    You seem to be on the right track so far, except somewhere along the way you lost a value of "2" in one of your terms.

    Try that again starting with
    [tex] T = 2 \pi \sqrt{\frac{d}{g} + \frac{R^2}{2gd}} [/tex]

    and keep plugging away with the derivative of that.

    There's no need to simplify too much along the way. Since you are setting the derivative equal to zero, some of the more scary parts go way quite quickly. What's left over is pretty manageable.

    [Edit: Oh, and one last thing: don't forget that the problem isn't asking for d itself, but rather R - d, since it says "how far from the rim." But you can make that as your final step.]

    [Another edit: I corrected an unintentional omission of [itex] 2 \pi [/itex] in my equation.]
     
    Last edited: Feb 9, 2016
  4. Feb 9, 2016 #3
    Thank You!!
     
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