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Extreme Period for a Physical Pendulum

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


    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?


    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.

  2. jcsd
  3. radou

    radou 3,217
    Homework Helper

    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. 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. OlderDan

    OlderDan 3,031
    Science Advisor
    Homework Helper

    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?

    Last edited: Nov 29, 2006
  6. 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. OlderDan

    OlderDan 3,031
    Science Advisor
    Homework Helper

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

  8. 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. OlderDan

    OlderDan 3,031
    Science Advisor
    Homework Helper

    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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