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Moment of inertia of a disc

  1. Aug 3, 2009 #1
    Hi, im a little stuck on this question from a PPOD paper im doing all help appreciated =)

    1. The problem statement, all variables and given/known data
    A solid circular disc of mass 6.6kg and radius 0.25m rotates about an axis which coincides with its diamter. Its angular acceleration is found to be 6.7rad/s. The axis of rotation is changed so that it lies 0.15m from the centre of the disc, and is parallel to the diameter. If the applied moment is the same, determine the angular acceleration.


    2. Relevant equations
    I=mR^2/4
    I=rα


    3. The attempt at a solution
    ive worked out the moment of inertia of the disc before the axis is changed and get

    I = 0.103kgm^2

    I was wondering if the next step is simple to use I=rα and use 0.15 as the value for the radius or should I use the parallel axis theorem as the axis has changed from being perpendicular to parallel??
     
  2. jcsd
  3. Aug 3, 2009 #2

    cepheid

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    You should use the parallel axis theorem, but that is not the correct reason why. You should use the parallel axis theorem because the new axis of rotation is parallel to one of the principal axes (the axes that pass through the centre of mass of the disc).

    What you said is wrong because the orientation of the axis has not changed, only its position. Saying that it has gone from "perpendicular to parallel" makes no sense. It's oriented the same way as it was before (parallel to the diameter of the disc). The only difference is that now it is offset from the diameter by 0.15 m.
     
    Last edited: Aug 3, 2009
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