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Rotational Motion Problem

  1. Nov 12, 2012 #1
    1. The problem statement, all variables and given/known data

    Hey all - I'm doing a physics lab on rotational motion, and I can not make sense of the last part, the additional analysis. The question is:

    Since our masses are not actually point-masses, the theoretical model I = MR^2 is not very accurate, so for a better theoretical comparison we
    should assume that each of the slotted masses is a disk of uniform density (neglect the effect of the slot) instead of a point mass. The moment of inertia for a single disk about an axis through the center of mass and parallel to the face of the disk is shown in the figure at right, where M is the mass of the disk, R is the radius of the disk and L is the thickness of the disk. You will need to use the parallel-axis theorem, check your textbook for details. Is this theoretical value in better agreement with the experimental data? Can you derive an equation to determine the moment of inertia of the masses including the cut-out slots? How does this value compare to the measured moment of inertia?

    This is due in the next hour and a half, and I've been trying to make sense of it for over an hour. How can we equate the parallel-axis theorem to what they give us in the problem?

    2. Relevant equations

    τ = r x F
    τ = I α
    I = I (cm) + Mh^2
    I = I(0) + Ii
    I (cm) = (1/4)Mr^2 + (1/3)ML^2

    3. The attempt at a solution

    Because it is more of a conceptual problem, I am not sure where to even start. If someone could help me out, it would mean a lot! Thanks!!! If you need more information about the lab, let me know.
     
  2. jcsd
  3. Nov 12, 2012 #2

    tiny-tim

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    hi wizzpowa! :smile:

    hmm … half an hour to go …

    i'm not sure what your question is :confused:

    if the mass is a disc of mass m and diameter r at distance R from the centre of rotation, O,

    then the moment of inertia IO about O is mR2 plus IC, the moment of inertia about the centre of the disc, C,

    and τC = ICα :wink:
     
  4. Nov 12, 2012 #3

    tiny-tim

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    hi wizzpowa! :smile:

    thanks for the pm

    as always on this forum, you need to show us your work!

    (or at least the bit that's bothering you)
     
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