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MI of athin uniform rod

  1. Apr 20, 2007 #1
    If i take a long rod and calculate the MI about an axis running through the rod the MI when i take the origin as the center of the rod as origin is 1/6mr^2(i.e.1/12+1/12 with the held of perpendicular axis theorem). If i take the MI about the axis keeping the origin at the end of the rod the value becomes 1/3+1/3 i.e. 2/3.
    Isn't the MI about the axis through the cylinder supposed to be the same because i cannot imagine any difference in the motion of the rod.
    But there is supposed to be some difference? Where is it?
     
  2. jcsd
  3. Apr 20, 2007 #2

    Doc Al

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    Staff: Mentor

    I don't understand how you are applying the perpendicular axis theorem to a thin rod (I assume you mean a cylinder). I assume that r is the radius of the cylinder's cross-section, not its length.

    If you apply the perpendicular axis theorem to a disk (I = 0.5mr^2, perpendicular to its center), then you can deduce that I = 0.25mr^2 about a diameter. To apply that to a rod you'll have to integrate along the length (and also use the parallel axis theorem).
     
  4. Apr 21, 2007 #3
    I think there is a misinterpretation of the letter r. I meant 1/12ml^2 etc. So the r of the previous note is my way of writing l
     
  5. Apr 21, 2007 #4

    Doc Al

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    I still don't understand how you are applying the perpendicular axis theorem to the rod. What makes you think you can do that? (For one thing, the perpendicular axis theorem only directly applies to planar objects.)

    Also, what do you think the rotational inertia of a thin rod is about an axis running through the rod?
     
  6. Apr 21, 2007 #5
    I am sorry i could not present my problem very well. But is this statement correct. The Perpendicular axis theorem is not applicable in unsymmetrical situations.
    And secondly is MI a vector scalar or what???
     
  7. Apr 21, 2007 #6
    What do you mean by planar objects i have not understood?
     
  8. Apr 21, 2007 #7

    Doc Al

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    I would say that is not true: the theorem can be applied to nonsymmetric mass distributions--but they need to be planar (flat).
    In general, an object is characterized by a moment of inertia tensor, but if you rotate it about a principal axis then the moment of inertia can be described by a scalar. (Check your mechanics text for more on this.)

    A planar object is one that is flat--like a piece of sheet metal. (planar = like a plane) A thin disk would be an example of a planar object.

    I think you need a better understanding of the perpendicular axis theorem. Here's a good start: Perpendicular Axis Theorem
     
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