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Rotational Dynamics-Moment of inertia

  1. Aug 31, 2009 #1
    1. The problem statement, all variables and given/known data
    A meter stick of mass 0.44kg rotates in the horizontal plane about a vertical axis passing through the 30cm mark. What is the moment of intertia of the stick? (Treat it as a long uniform rod).


    2. Relevant equations
    Moment of inertia for a long uniform rod (axis through center): (1/12)ML2
    Moment of inertia for a long uniform rod (axis through end): (1/3)ML2
    I=mr2

    3. The attempt at a solution

    LA=1.0m;
    LB=0.70m (I'm not sure how to determine the length/axis for LB???)
    I=1/3LA2 + 1/3LB2
    I=1/3(1.0)2 + 1/3(0.70)2
    I= 0.50 m x N
     
  2. jcsd
  3. Aug 31, 2009 #2
    [tex]L_{B} = 0.2m [/tex] (distance between center of mass and axis if you use moment of inertia on center)

    [tex]I = \frac{1}{12}ML^2 + M(0.2)^2 [/tex]
     
  4. Aug 31, 2009 #3
    Ok thanks.

    However, wouldn't the formula for both terms on the right side of the equation be 1/12ML2 (I.e., LA and LB, respectively)

    Also, if the L is determined by the distance from the centre of gravity, does that mean LA=0.5? or does is it still equal to 1.0?

    Sorry for so many questions, I just want to make sure I understand this.
     
  5. Aug 31, 2009 #4

    rock.freak667

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    The parallel axis theorem: I=Ic+mr2

    r= distance from the centre of rotation

    Ic= moment of inertia about the centre.
     
  6. Aug 31, 2009 #5
    I'm not sure what the parallel axis theorem is, as we haven't covered it in any of the lessons for this course...Is there a way to solve the problem without that theorem?
    Thanks :)
     
  7. Aug 31, 2009 #6
    of course, you can always use its definition:

    [tex] I = \int_{M} r^2dm[/tex]
     
  8. Aug 31, 2009 #7
    The course I am in is a very basic course, and as such, calculus is outside the scope of this course. The only formulas we have been given are for moment of inertia (i.e. I=1/12ML2, etc.) Can the problem be solved using those formulas?

    Thanks very much for your help though :)
     
  9. Aug 31, 2009 #8

    rock.freak667

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    It can. You know the moment of inertia about the centre as being 1/12 ML2.


    You want to move the axis to 0.3m mark. At the centre, the axis is at 0.5m. So the distance from the centre to the 0.3m mark is 0.2m.

    So the moment of interia about the 0.3m mark = 1/12 ML2+ M(0.2)2

    Which is how kyiydnlm got the answer.
     
  10. Aug 31, 2009 #9
    Oh okay - thanks I understand what you're saying now.

    For the eq'n: 1/12 ML2+ M(0.2)2

    Will the L be equal to the full distance of 1.0m, or will it equal 0.30 (the distance from the end to the axis)?
     
  11. Aug 31, 2009 #10

    rock.freak667

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    L is the total length of the rod...or metre stick as it may be
     
  12. Aug 31, 2009 #11
    Okay, thanks for all your help!!
     
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