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Pendulum Problem

  1. Dec 9, 2013 #1
    1. The problem statement, all variables and given/known data

    A pendulum is composed of a thin rod of length L = 50 cm, a moment of inertia of I = 0.2kgm2 with respect to its center and mass M = 400g. Attached to the bottom of the rod is a point mass (m = 200g). The pendulum swings in a vertical plane, attached by the top end to the ceiling.

    2. Relevant equations

    T = 2∏√(I/Mg x d)
    *I think*

    3. The attempt at a solution

    T = 2∏√(.2/(.4)(10) x .25)

    No radius was given for the point mass, so I assumed the moment of inertia of center of mass is .2kgm2 and therefore the distance of center of mass (d) would be L/2 or .25m

    Did I go about solving this correctly?

    Also, the second problem asks: How long will it take for the pendulum to perform 100 oscillations?

    Would I simply multiple the period I found from the first part by 100?

    Thank you all for your time and consideration.
     
    Last edited: Dec 9, 2013
  2. jcsd
  3. Dec 10, 2013 #2

    Simon Bridge

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    Don't guess.
    It helps to sketch the situation out:

    - Which part of the rod is attached to the ceiling?
    - Which part of the rod has the point mass attached to it?

    You are given the moment of inertia for a rod pivoted at the center of mass - but that is not where the rod is pivoted - so you need to adjust this (hint: parallel axis theorem).

    ... yes.
    The question is basically checking if you understand what "one period" means.

    It's good to see you thinking beyond what's in the question - just remember that everything you want to assert has to be justified by something said inside the problem statement.
     
  4. Dec 10, 2013 #3
    Hmm, I see.

    The equation for moment of inertia at end of rod is Iend = mL2/3

    Therefore, Iend = 1.33 kgm2

    From my understanding of using the parallel axis theorem to calculate moment of inertia of a system, one would simply add up the inertias.

    Isystem = Icom + Iend of rod + Ipoint mass

    Isystem = .2kgm2 + 1.33kgm2 + 2.00kgm2

    Isystem = 3.53kgm2

    Does this sound more reasonable?
     
  5. Dec 10, 2013 #4

    Simon Bridge

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    That's not how the parallel axis theorem works.

    You should only have two terms to calculate inertia: ##I_{tot}=I_{rod}+I_{point}##

    You can just look up the inertia for the rod-pivoted-about-one-end.
    However, since the question tells you the com inertia, they are expecting you to use that and the parallel axis theorem. Either way is correct but both together is not.

    Aside: interesting... the PF spell-checker does not accept either "inertias" or "inertiæ" (accepted plurals for "inertia").
     
    Last edited: Dec 10, 2013
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