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Time of oscillation of a pendulum

  1. Jun 23, 2016 #1
    1. The problem statement, all variables and given/known data
    A rigib poll of length 2L is made into a V shape so that each leg has length L. What is the period of oscilation for small angle. The angle between the legs is 120 degrees
    2. Relevant equations
    3. The attempt at a solution

    I tried to calculate the period by imagining a rigid poll that would take the originals place. Since the lenght of one leg is L, the poll which would take the place is located in the middle and of lenght ##L/2## since its 30,60,90 triangle, Moment of inertia is ##I=\frac{M(L/2)^2}{3}## and that nakes the period ##T=6.28*\sqrt{\frac{I}{Mg(L/2)}}## and that just doesnt seem right? What do you think?
     
  2. jcsd
  3. Jun 23, 2016 #2
    You have to calculate the moment of inertia I of the V shape body around the point at the vertex of V (which I suppose is the pivot point). I believe it will be ##I=2M\frac{L^2}{3}##.Also you need to find the center of mass of the V shape body. I believe it will be at distance ##R=\frac{L}{4}## from the pivot point. Then the period will be

    ##T=2\pi\sqrt\frac{I}{MgR}## where R is the distance of the center of mass from the pivot point (the vertex of V).
     
  4. Jun 23, 2016 #3

    haruspex

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    That's rather too much assistance straight up. Better would have been to leave it as
    "You have to calculate the moment of inertia I of the V shape body around the point at the vertex of V (which I suppose is the pivot point).. Also you need to find the center of mass of the V shape body. "

    Anyway, your expression for I is wrong.
     
  5. Jun 24, 2016 #4
    Sorry where I am wrong, the body is originally 2L and each V side has L length. Moment of intertia is aroung the peak of V not around the c.o.m.

    Ok I see now the factor of 2 should be omitted.
     
    Last edited: Jun 24, 2016
  6. Jun 24, 2016 #5

    haruspex

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    Right.
     
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