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Homework Help: Calculating the Rotational Inertia

  1. Sep 16, 2007 #1
    1. The problem statement, all variables and given/known data
    A rigid structure consisting of a circular hoop of radius R and mass m, and a square made of four thin bars, each of length R and mass m. The rigid structure rotates at a constant speed about a vertical axis, with a period of rotation of 4.0 s. Assume R = 0.90 m and m = 3.0 kg, calculate the structure's rotational inertia about the axis of rotation.


    2. Relevant equations
    I_rods = (1/12)ML^2
    I_hoop = (1/2)MR^2
    I = I_rods + I_hoop


    3. The attempt at a solution
    The third equation is constructed by noticing that the problem states that the structure consists of both the square made of thin rods and the hoop. The square only has two rods that have moments of inertia because two are perpendicular to the axis and the other two are parallel, which have a moment of inertia of 0 (at least that is what I have been told). So...

    I = 2[I_rods] + I_hoop
    I = 2[(1/12)(3 kg)(0.90 m)^2] + (1/2)(3 kg)(0.90 m)^2
    I = 2.835 kg*m^2

    I know this is wrong and it has to do with my equations. Please help. Thanks for your time.
     
  2. jcsd
  3. Sep 16, 2007 #2

    Doc Al

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    These formulas assume a perpendicular axis through their centers. But where's the axis of rotation in this problem?

    A diagram would help.
     
  4. Sep 16, 2007 #3
  5. Sep 16, 2007 #4

    Doc Al

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    OK. Make use of the parallel axis theorem to find the moments of inertia about the axis of rotation for the rods (two of them, at least) and the hoop.
     
    Last edited: Sep 16, 2007
  6. Sep 16, 2007 #5
    The Parallel Axis Theorem states:

    I = I_com + Md^2

    So the rod furthest to the left will be have a moment of inertia of:

    I = (1/12)(3 kg)(0.09 m)^2 + (3 kg)(0.09 m)^2

    And the hoop's moment of inertia will be:

    I = (1/2)(3 kg)(2*0.09 m)^2 + (3 kg)(0.09 m)^2

    Then the solution will be the summation of those equations? Are these right?
     
  7. Sep 16, 2007 #6

    Doc Al

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    OK.
    No. The rod furthest to the left has all its mass equidistant from the axis.

    But what about the two horizontal rods?

    Why is there a 2 in there?
     
  8. Sep 16, 2007 #7
    Ok, I don't why I put a 2*0.90 in the hoop's moment of inertia equation. That was a mistake. As for the horizontal rods, I have no idea and I need some explanation of them and what to do for them because I thought they had no moment of inertia. The rod furthest to the left must be then I = (1/12)(3 kg)(0.09 m)^2, but why not include its distance from the axis?
     
  9. Sep 17, 2007 #8

    Doc Al

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    Realize that the rods are being rotated about a vertical axis. The equation you have, I_rods = (1/12)ML^2, can be used for a horizontal rod rotating about a vertical axis (through its center of mass). (Use the parallel axis theorem to move the axis.)
    Just the opposite: you must include the distance from the axis. Trick question: What's the rotational inertia of a rod about an axis parallel to the rod (and through its center of mass)? Hint: Only three of the rods contribute to the total rotational inertia.
     
  10. Sep 17, 2007 #9
    I = 0, because the distance away from the axis is 0, thus I = Md^2 = 0
     
  11. Sep 17, 2007 #10

    Doc Al

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    Exactly. With the left rod oriented parallel to the axis, the only thing that counts is its distance from the axis. (If its mass were concentrated at a single point, its I would be the same.)
     
  12. Sep 17, 2007 #11
    Ok, so...

    I_parallelRod = (3 kg)(0.09 m)^2

    The parallel rod is the length of the rod away from the axis.

    I_hoop = (1/2)(3 kg)(0.09 m)^2 + (3 kg)(0.09 m)^2

    The hoop is the length of the radius away from the axis.

    So the summation of these moments of inertia is:

    I = I_horizontalRods + I_parallelRod + I_hoop


    But I don't understand what you are getting at by saying the the rod furthest to the left is equidistant to the axis... how does that change the previous formula I posted?


    Why is this still wrong?
     
  13. Sep 17, 2007 #12

    Doc Al

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    All good. But what about the horizontal rods? What did you use for their rotational inertia?
    Realize that the formula for the vertical rod at distance R from the axis is the same as the formula for a point mass a distance R from the axis.
     
  14. Sep 17, 2007 #13
    Oops I meant to say what is wrong with my horizontal equation that I first posted?
     
    Last edited: Sep 17, 2007
  15. Sep 17, 2007 #14

    Doc Al

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    Tell me exactly which one. (Quote it.)
     
  16. Sep 17, 2007 #15
    Sure...

     
  17. Sep 17, 2007 #16

    Doc Al

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    The rod furthest to the left is a vertical rod, right?

    If you mean that to appy to the horizontal rods, what's the distance between their centers and the axis of rotation?
     
  18. Sep 17, 2007 #17
    Yes, it is a vertical rod. The furthest rod has a distance of 0.90 m away from axis
     
  19. Sep 17, 2007 #18

    Doc Al

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    I thought we finished with the vertical rods in post #11? (Vertical = parallel to the axis.)

    You might want to summarize everything that your adding to the mix, so we get it straight once and for all.
     
  20. Sep 18, 2007 #19
    Sorry, I've been confusing myself and you over this problem.

    I_parallelRod = (3 kg)(0.09 m)^2

    I_hoop = (1/2)(3 kg)(0.09 m)^2 + (3 kg)(0.09 m)^2

    I_horizontalRod = (1/12)(3 kg)(0.09 m)^2 - (3 kg)(0.09 m)^2

    The horizontal rods have a moment of inertia of (1/12)(3 kg)(0.09 m)^2, but it is shifted the left of the axis by the length of a rod.

    Thus...

    I = I_parallelRod + I_hoop + I_horizontalRod


    This is everything wrapped up to now, I apologize for running us in circles.
     
  21. Sep 18, 2007 #20

    Doc Al

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    I'm easily confused.

    Good. This applies to the left vertical rod. (What about the right one?)

    Good.

    Careful. (1/12)(3 kg)(0.09 m)^2 is the rotational inertia for a horizontal rod rotating about a vertical axis through its center. What does the parallel axis theorem tell you to add to that to get its rotational inertia about the required axis? (Left and right don't matter--all that matters is the distance from cm to the axis.)

    Be sure to account for all four rods. (One will be trivial.)
     
  22. Sep 18, 2007 #21
    So if left or right of the axis does not matter then according to the Parallel Axis Theorem (I = I_com + Md^2) the (3 kg)((0.09 m)/2)^2 will be added to the moment of inertia, not subtracted. The 0.09 m is divided by two in the equation because the center of mass has only shifted half of the length of the rod, not the full length like I previously had...

    I_horizontalRod = (1/12)(3 kg)(0.09 m)^2 + (3 kg)((0.09 m)/2)^2

    I_parallelRod = (3 kg)(0.09 m)^2

    I_hoop = (1/2)(3 kg)(0.09 m)^2 + (3 kg)(0.09 m)^2


    And the final equation will be represented in the following way:

    I = I_parallelRod + I_hoop + 2[I_horizontalRod]


    I think that may be right...?
     
  23. Sep 18, 2007 #22

    Doc Al

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    I think you've got it now! :approve:
     
  24. Sep 19, 2007 #23
    It is still wrong! My final answer is 0.07695 kg*m^2. http://www.webassign.net/hrw/hrw7_11-45.gif do I have to change the hoop's equation at all because the diameter is 2R... I don't know anymore. I thought I had it. Unless my calculations are incorrect, but I have checked them numerous times. If anyone else could just try to plug and chug and see what they get. I really want to solve this because it doesn't seem that hard now.
     
  25. Sep 19, 2007 #24

    Doc Al

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    I get the same answer. Realize that webassign can be very picky about the number of significant figures you enter. Round it off.
     
  26. Sep 19, 2007 #25
    I can't seem to get it no matter what I try.
     
    Last edited: Sep 19, 2007
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