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Rotational Dynamics

  1. Jan 11, 2006 #1
    Wanted to check this question also:

    A meter stick of mass 0.44 kg rotates, in the horizontal plane, about a vertical
    axis passing through the 30 cm mark. What is the moment of inertia of the stick?
    (Treat it as a long uniform rod)

    Answer: I = 0.1628 kg.m^2

    Correct??
     
  2. jcsd
  3. Jan 11, 2006 #2

    Doc Al

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    Staff: Mentor

    Rather than just give your answer, always show how you got your answer. (At least give the formulas that you used.) I assume you meant to calculate the moment of inertia about that axis.
     
  4. Jan 11, 2006 #3
    My Work

    So I was a little confused but I calculated using the formula I = mr^2 for both sides of the stick and then added them:

    I = (0.132 kg)(0.3m)^2 + (0.308 kg)(0.7m)^2
    = 0.1628 kg.m^2
     
  5. Jan 11, 2006 #4

    Doc Al

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    What's the rotational inertia of a rod about one end? It's not [tex]I = m L^2[/tex] (where L is the length of the rod).
     
  6. Jan 12, 2006 #5
    Inertia through end of uniform rod

    That formula is I = 1/3ML^2

    So the answer is 0.147 kg.m^2 ??
     
  7. Jan 12, 2006 #6

    Doc Al

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    You have the correct formula for the rotational inertia, but recheck your calculation.
     
  8. Jan 12, 2006 #7
    Is the correct answer 0.0132 kg.m^2

    Using 0.3m for the Length?
     
  9. Jan 13, 2006 #8

    Doc Al

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    In post #3 you showed your work using the incorrect formula for rotational inertia. How would you modify that calculation using the correct formula?
     
  10. Jan 13, 2006 #9
    Ok hopefully I've got it this time

    I = 1/3 ML1^2 + 1/3 ML2^2
    = 1/3 (.132)(.3)^2 + 1/3 (.308)(.7)^2
    = 0.064 kg.m^2
     
  11. Jan 13, 2006 #10

    Doc Al

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    Your method is correct, but I suspect a typo in your final answer. (I get 0.054)
     
  12. Jan 13, 2006 #11
    Thank You!!!

    Thanks so much for your help!
     
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