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Rotational Dynamics of a meter stick

  1. Apr 27, 2008 #1
    1. The problem statement, all variables and given/known data
    A meter stick of mass .44 kg rotates, in the horizontal plane, about a vertical axis passing thru the 30 cm mark. What is the moment of inertia of the stick? Treat it as a long uniform rod.

    2. Relevant equations

    3. The attempt at a solution

    Once I calculate this, is the the problem complete? It seems too simple.
  2. jcsd
  3. Apr 27, 2008 #2


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    Hi Lma12684,

    The formula I=(1/2)MR^2 is not the moment of inertia for a meter stick. Different shapes have different moment of inertia formulas (although these formulas all are derived from the same integral definition).

    You would probably want to use the formula for moment of inertia of a rod here (this assumes that the width of the meterstick is much less than its length, which is probably reasonable). You can probably find a formula in a table for a rod with a rotational axis through the center of the rod and another formula with the axis through one end.

    Once you have that, how do you use that expression to find [itex]I[/itex] for an axis at 30cm?
  4. Apr 27, 2008 #3
    Should I assume that 30 cm is the middle of the rod? Thanks.
  5. Apr 27, 2008 #4


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    No, you can't do that since it's a meter stick. But I would first find the formula that applies when it is rotating about the center of the rod. What do you get for that case?

    Once you have that, you can use that formula with the parallel axis theorem to find the moment of inertia for the case of a rotation axis that is not at the center.
  6. Apr 27, 2008 #5
    Relavent formula: I=(sum of t)/angular acceleration
  7. Apr 27, 2008 #6


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    Do you have your textbook available? If so, I think you should be able to find a table that has some shapes and the formulas for their moments of inertia. (There will probably be some for spheres, rods, disks, etc.) You'll probably be using that table quite a bit so it's important to locate it.

    If you don't have your book available, you can look here:


    (The formula you quoted I=(sum of t)/angular acceleration is a general relationship that is the angular analogue of Newton's law.)

    Once you have that moment of inertia for a rotation about the 50cm mark, then the parallel axis theorem will allow you to find it at the 30cm mark.
  8. Apr 28, 2008 #7
    My mistake, I misunderstood you. The formula I found for a long uniform rod was I=1/12(M)(L^2).

    For L, so I consider it as 1 meter? Or do I use .5 m?

    I am not sure what you mean by parellel axis theorem.

  9. Apr 28, 2008 #8


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    In that formula the L is the total length of the rod.

    I would think the parallel axis theorem is in your textbook. Once you find the moment of inertia through the center of mass, it allows you to find the moment of inertia about any axis that is parallel to that; all you need is the mass and the distance between the old and new axes. What do you get?
  10. Apr 28, 2008 #9
    I found that I=.0366 kg/m.

    Now, I used I=I(cm) + m(d^2).
    .0366=I + (.44)(.3^2)

    I know I did something wrong. Did I use the wrong distance?
  11. Apr 28, 2008 #10


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    The value you found (0.0366) was the moment of inertia through the center of mass [itex]I_{\rm cm}[/itex], and so goes on the right side of the equation. You are solving for the new I in this equation.

    Also, the d is the distance between the two axes. How far is the center of mass and the new axis? It's not 30cm.
  12. Apr 28, 2008 #11
    I think I figured it out. I had my I(cm) in the wrong place of the equation. I now found that I=.0762. THanks!!!!
  13. Apr 28, 2008 #12
    Sorry, I just saw your post. Let me check it again. Thanks.
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