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Parallel axis theorem

  1. Oct 13, 2016 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations
    Parallel axis theorem: Ip = Icm + Md^2
    Icm = I = ML²/12 + 2 * mr²

    3. The attempt
    Ip = Icm + Md^2 ==> wrong
    I = Md^2 ==> right

    Why don't I need to add "Icm"?
    Thanks.
     
    Last edited: Oct 13, 2016
  2. jcsd
  3. Oct 13, 2016 #2

    PeroK

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    Why do you think you need the parallel axis theorem?
     
  4. Oct 13, 2016 #3
     
  5. Oct 13, 2016 #4

    PeroK

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    You mean because the question has the word "parallel" in it?
     
  6. Oct 13, 2016 #5
    My teacher told me that when ever the axis isn't past though center of mass, we can use it...
    Isn't it?
     
  7. Oct 13, 2016 #6

    PeroK

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    You can use it. But, because you haven't analysed the question properly, you haven't thought about the moment of inertia of the bar in this question.
     
  8. Oct 13, 2016 #7
    "Part A
    Find the moment of inertia of this combination about an axis perpendicular to the bar through its center.
    Express your answer with the appropriate units.

    I = 2.27 kg⋅m2

    Correct"


    I have calculated the Icm, and it is right...
     
  9. Oct 13, 2016 #8

    PeroK

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    For which direction of rotation have you calculated the moment of inertia?
     
  10. Oct 13, 2016 #9
    Do you mean the axis of rotation (dimension) is different this time?
    But why don't we need to calculate Icm in the required direction and then add Md^2?
    Thanks
     
  11. Oct 13, 2016 #10

    PeroK

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    There isn't only one moment of inertia for a rigid body for rotation about its centre of mass. It depends on the direction you rotate it. Can you see how to rotate a bar so that its moment of inertia is 0? Hint: you are assuming the bar is one-dimensional in this problem.
     
  12. Oct 13, 2016 #11
    meaning no rotation?
    I understand this, that means answer in part a can't apply on this question. But, why don't we need to calculate Icm in the required direction for this question? (Let Icm pass through centre of mass and parallel to the bar)
    Huhh, no radius of the rod is given, do you mean it is assuming the radius of the rod is zero?
     
  13. Oct 13, 2016 #12

    PeroK

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    Yes, unless the rod is given a radius, you treat it as a one-dimensional body. So, you must take it to have 0 moment of inertia if it is rotated about its axis. The masses at each end are points, so they have 0 moment of inertia about any axis through them.

    Perhaps it was something of a trick question. But, it did catch you out, because you applied the parallel axis theorem without thinking about the problem carefully enough. Don't let them catch you out again!
     
  14. Oct 13, 2016 #13
    thanks
     
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