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Extreme Period for a Physical Pendulum

  1. Nov 29, 2006 #1
    1. The problem statement, all variables and given/known data

    [​IMG]


    solid, uniform disk of mass M and radius a may be rotated about any axis parallel to the disk axis, at variable distances from the center of the disk

    If you use this disk as a pendulum bob, what is T(d), the period of the pendulum, if the axis is a distance d from the center of mass of the disk?

    and

    The period of the pendulum has an extremum (a local maximum or a local minimum) for some value of d between zero and infinity. Is it a local maximum or a local minimum?

    2. Relevant equations

    From the picture, I come up with the moment of inertia of the solid disk around its center of mass
    I = 1/2Ma^2
    From the question, we are asked to find the period of the pendulum if the axis distance d from the center of mass.

    The period T for this is P= 2pi (sqrt L/g) where g is the gravitation force
    and L is the lenght.
    From my understanding is that because of the new lenght, we need to use the Parallel Theorem to find the new lenght

    I am not sure about this, so hope someone can help

    Iend = Icm + Md^2
    Iend = 1/2Ma^2 + Md^2

    So the period is P = 2pi (sqrt(( a^2 +d^2)/g))
    But this is not correct.

    Thank
     
  2. jcsd
  3. Nov 29, 2006 #2

    radou

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    Good thoughts, but your expression for the period doesn't seem correct. You may want to look at this: http://hyperphysics.phy-astr.gsu.edu/hbase/pendp.html. Your period expression has to include the moment of inertia.
     
  4. Nov 29, 2006 #3
    So is this right :

    I = 1/2Ma^2

    P = P= 2pi (sqrt 1/2Ma^2/(1/2Mgd^2)) == then we can cancel out the M to get

    P = 2pi (sqrt 1/2a^2/(1/2gD^2)) === cancel 1/2 ===

    P = sqrt(a^2/gd^2)
     
  5. Nov 29, 2006 #4

    OlderDan

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    You need to use the parallel axis theorem to find the moment of inertia of the disk about the pivot point. I don't see why you have a 1/2 in the denominator of the fraction. What became of the 2pi?

    http://hyperphysics.phy-astr.gsu.edu/hbase/parax.html#pax
     
    Last edited: Nov 29, 2006
  6. Nov 30, 2006 #5
    Thank for the useful info.

    Is this the correct moment of inertia of the disk about the pivot point.

    I= 1/2Ma^2 + Md^2 = M ( 1/2a^2 + d^2)

    So the period of the disk is

    P = 2pi (sqrt (M(1/2a^2 + d^2))
     
  7. Nov 30, 2006 #6

    OlderDan

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    No. Check the formula for the period of a physical pendulum

    http://hyperphysics.phy-astr.gsu.edu/hbase/pendp.html
     
  8. Nov 30, 2006 #7
    I think I've got it.

    P = 2pi(sqrt(a^2/2gd + d/g))

    If you look at this, how can you determine whether it has a local maximum or local minimum for some value of d?
     
  9. Nov 30, 2006 #8

    OlderDan

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    Take a derivative wrt d and set it to zero to see if there is an extremum. If there is one, determine if it is a max or min. If you have not taken calculus, graph it.
     
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