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Question about the inertia matrix of a bar

  1. Mar 30, 2017 #1
    1. The problem statement, all variables and given/known data
    756c9e.png

    2. Relevant equations
    ml2/12

    3. The attempt at a solution
    5bf3a7.png


    So according to my databook:

    3267d3.png

    The axis x'-x' in the question corresponds to axis ZZ in the databook image above. That means in terms of radius, the moment of inertia about axis x'-x' is mr2/2. So in light of this, why is the constant outside the matrix ml2/12? Why not mr2/2?
     
    Last edited: Mar 30, 2017
  2. jcsd
  3. Mar 30, 2017 #2

    Orodruin

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    The moment of inertia for a bar around any axis perpendicular to it is ##m\ell^2/12##.
     
  4. Mar 30, 2017 #3
    I understand that but the x'-x' axis in the question is parallel to the bar and not perpendicular?

    Thanks
     
  5. Mar 30, 2017 #4

    Orodruin

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    Exactly, and the moment of inertia relative to that axis is zero, as shown in the solution.
     
  6. Apr 1, 2017 #5
    If the moment of inertia relative to the parallel axis is zero, then why is the moment of inertia about axis ZZ (in my data book) not 0? This axis is parallel to the cylinder?
     
  7. Apr 1, 2017 #6

    Orodruin

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    Because the problem you are solving is idealising the rod as infinitely thin - i.e., having negligible diameter.
     
  8. Apr 3, 2017 #7
    Ah that makes sense! This question is from an exam paper set by my university's engineering department a few years ago, is there any way to determine that this rod is infinitely thin from the wording of the question? Does the ''uniform slender'' part suggest this?

    Also, I was wondering why (according to the mark scheme) the angle is -β? I mean the diagram shows a counter clock wise rotation so shouldn't the angle be β?
     
  9. May 31, 2017 #8
    I
    I was wondering whether anyone could help me out with this?
     
  10. May 31, 2017 #9

    Orodruin

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    So you need a negative angle to counter that offset. The y-axis is pointing into the screen so a positive rotation would typically be clockwise.
     
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