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Min Period of oscillation of a rod

  1. May 31, 2014 #1
    1. The problem statement, all variables and given/known data

    An iron rod of length L is hung at a common point with threads of length 'l' which are attached to the two ends of the rod. The rod is displaced a bit in the plane of the threads. What is the length of the threads if the period of the swinging of the rod is the least, and what is this period?


    2. Relevant equations



    3. The attempt at a solution

    Moment of Inertia of the rod about O is $$ I = \frac{ML^2}{12}+My^2 $$

    Write torque equation , ## I\ddotθ = -Mgyθ ##

    $$ \ddotθ = -\frac{Mgy}{I}θ $$

    $$ ω^2 = \frac{Mgy}{I} $$

    $$ ω = \sqrt{\frac{Mgy}{I}} $$

    $$ T = 2\pi \sqrt{\frac{I}{Mgy}} $$

    $$ T = 2\pi \frac{(l^2-\frac{L^2}{6})^\frac{1}{2}}{g(l^2-\frac{L^2}{4})^\frac{1}{4}} $$

    Should I minimize this ? This looks a little more complicated than usual .

    Edit : Fixed the errors
     

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    Last edited: Jun 1, 2014
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  3. Jun 1, 2014 #2

    TSny

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    Check this expression. What is the moment of inertia of a rod about its center of mass?

    Otherwise, your work looks good to me.

    You might find it easier to express the period in terms of the variable y (instead of ##l##) and minimize the square of the period with respect to y.
     
  4. Jun 1, 2014 #3
    OK...I have fixed the errors. Does post#1 look alright now?

    $$ T = 2\pi \frac{(y^2+\frac{L^2}{12})^\frac{1}{2}}{g(y)^\frac{1}{4}} $$

    $$ T^2 = 4\pi^2 \frac{(y^2+\frac{L^2}{12})}{g^2(y)^\frac{1}{2}} $$

    Now I should minimize T2 . Right ?
     
  5. Jun 1, 2014 #4

    TSny

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    In the denominator, the 1/4 power on y is not correct.

    Right.
     
  6. Jun 1, 2014 #5
    The minimum value occurs at y=L/6 or at ## l= \frac{\sqrt{10}}{6}L ##

    Is it correct ?
     
    Last edited: Jun 1, 2014
  7. Jun 1, 2014 #6

    TSny

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    That's not what I get.
     
  8. Jun 1, 2014 #7
    The minimum value occurs at ##y=\frac{L}{\sqrt{6}}## or at ##l= \sqrt{\frac{5}{12}}L##

    Does this make sense ?
     
  9. Jun 1, 2014 #8

    TSny

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    Still not what I'm getting. What is your expression for T2?
     
  10. Jun 1, 2014 #9
    Another attempt

    The minimum value occurs at ##y=\frac{L}{\sqrt{12}}## or at ##l= \frac{L}{\sqrt{3}}##
     
  11. Jun 1, 2014 #10

    haruspex

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    You should get into the habit of checking equations for dimensional consistency. The exponents of both g and y are wrong in the denominator.
    I agree with TSny that you are still getting the wrong answer. Please post all your steps from the above.
    Clearly the constant factor in the equation isn't going to affect the value of y, so you can dispense with that.
     
  12. Jun 1, 2014 #11

    TSny

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    Bingo.
     
  13. Jun 1, 2014 #12
    Thanks.

    This is something I would like to understand .

    This technique is used often where we need to max/min a function involving square root and instead we max/min the square of the function .How do we know that the maximum and minimum of both the function and its square occur at the same point ?
     
    Last edited: Jun 1, 2014
  14. Jun 1, 2014 #13

    TSny

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    The square root function increases whenever its argument increases. It's a "monatonically increasing" function. So, the square root function will take on its maximum (minimum) value where its argument is maximum (minimum).
     
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