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Simple Harmonic Motion of a Rigid Body

  1. Dec 2, 2009 #1
    1. The problem statement, all variables and given/known data

    A very light, rigid rod with a length of 0.570 m extends straight out from one end of a meter stick. The stick is suspended from a pivot at the far end of the rod and is set into oscillation. Determine the period of oscillation.


    2. Relevant equations

    omega = (Mgd/I)^1/2
    omega = 2pi/period.

    3. The attempt at a solution

    I solved this question, but I was just looking for clarification as to why it was right. The moment of inertia of the rod would be 1/3 ML^2 and using the parallel axis theorem I would have to add Md^2 on top of that.

    The masses cancel in the first equation, but I was wondering about the "d" in the numerator. What does it represent? I originally thought it was the distance from the centre of mass to the pivot but it wasn't the case for this problem. Any insight is appreciated.

    I when i subbed d=0.57 into the equation i got the correct answer of 2.15s, but I don't understand why.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 2, 2009 #2

    kuruman

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    Homework Helper
    Gold Member

    The parallel axes theorem says that the moment of inertia about distance d from the center of mass is given by

    I = ICM + md2, where ICM = (1/12)mL2 for a uniform rod not (1/3)mL2. If you got the right answer, you must have made two errors that cancelled each other.
     
  4. Dec 2, 2009 #3
    True. I was using the moment of inertia about one end of the stick, which is indeed (1/3)mL2. In my case i guess the D was still the distance from the old axis to the new axis. In using the moment of inertia about the CM, and making d the distance between the pivot and the CM, I was able to again get the correct answer.

    Thanks for your help
     
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