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Rotational Inertia

  1. May 5, 2007 #1
    1. The problem statement, all variables and given/known data
    I need to rank these masses from greatest to least rotational inertia. We have not been taught integration, so I know we won't be using it. I have a chart of formulas for certain shapes though. I am just not sure if I even need them.
    [​IMG]



    2. Relevant equations I=Mr^2. <-- I dont think I will use this. I know that one thin rod about the perpendicular is I=1/12ML^2. . . but the book does not give one for about the length. . . this leads me to believe that the formulas are not necessary. I think it is sort of a trick question, but I am not sure why.



    3. The attempt at a solution
    Any hints would be great.
     
    Last edited: May 5, 2007
  2. jcsd
  3. May 5, 2007 #2

    Hootenanny

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    Try considering where the centre of mass might lie on each of the individual rods (i.e. consider the vertical and horizontal rods seperately); then apply your formula for a point mass.
     
  4. May 5, 2007 #3
    I am not sure that I follow.
     
  5. May 5, 2007 #4
    Anyone? I am not sure what Hootenanny is saying. Use 1/12ML^2 for each individual rod? Then what?
     
  6. May 5, 2007 #5

    Doc Al

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    1/12ML^2 is the rotational inertia of a rod about a perpendicular axis through the center of the rod. What if the rod were rotated about one end? What if the axis was parallel to the rod?

    Consider how rotational inertia depends on how the mass is distributed with respect to the axis. If the mass is closer to the axis of rotation, does that increase or decrease the rotational inertia?
     
  7. May 5, 2007 #6

    mjsd

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    i think Hootenanny's idea is (as you mentioned....doing this without calculus and just by definition of I) that if you know where the centre of mass of the system is located, then you can then work out Inertia by just applying I=MR^2 where M is the mass of the entire structure (the same for all your "T" shape structures, I believe), and R would just be the perpendicular distance from CM to the axis of rotation... so clearly, bigger R shall mean bigger I. But Hoontenanny has suggested to break the "T" shape down to two rods and then find the CM for each and then you can sum up the two contribution that make up [tex]I = \sum_{i} m_i R^2_i[/tex]

    edit: I personally would think that using the rod result is much more intuitive.
     
    Last edited: May 5, 2007
  8. May 5, 2007 #7
    If the center of mass is closer to the axis of rotation, its rotational Inertia is less than that of one further. I see Doc Al is hinting at some parellel-axis thereom action. . . maybe I have just lost it, but I am having some difficulty trying to find the center of mass of these T's (quantitatively). . . I will refer to my text.
     
  9. May 5, 2007 #8
    Okay. So I can see intuitively that A>D>B>C. . . But I need help finding out how to calculate I about an axis that passes through the center of mass. . . than I can apply Parellel-axis Theorem, right?
    Again, my intuition tells me that since these are uniform objects, their centers of mass are somewhere on the vertical line of symmetry (that divides the T into two "L"s). . . my guess is somewhere closer to the horizontal end than the top. . . but how can I calculate this quantitatively. . .

    Thanks for your patience. . . I went retarded in this chapter for some reason:tongue:
    BTW. . . that picture is a clickable thumbnail if you didn't notice.
     
  10. May 6, 2007 #9

    Doc Al

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    Think of the object as the sum of two rods, not as a T. For each case, find the rotational inertia of each rod about the given axis and add them up. There are only three cases:

    (1) Rod rotating about a perpendicular axis through its center (you have the formula)
    (2) Rod rotating about a perpendicular axis through one end (you should have the formula)
    (3) Rod rotating about a parallel axis (this one is easier than you think; hint: all of the mass is at the same distance from the axis)
     
  11. May 6, 2007 #10
    That is the problem. . . I do not have a formula for this one. I have one for a solid cylinder about the end and center, but not a thin rod. This is why I am so flabbergasted. Plus, the formula for the rods involves a radius, which was not given.
     
  12. May 6, 2007 #11

    Doc Al

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    It takes less than 10 seconds to find it on the web!

    You quoted the formula for a thin rod about its center:
    No mention of radius here!

    Here ya go: Common Moments of Inertia
     
  13. May 6, 2007 #12
    I am sure it does, but I know my professor, and that is not how she operates things. For this kind of assignment, we would only use information that has been given either in our text or lecture notes.

    As for the formulas that I have: I need to find I about the perp. center and the end for a thin rod. I only have one for about the center not the end.

    I am stuck.
     
  14. May 6, 2007 #13

    Doc Al

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    I have no idea what you've covered in your lectures or what's in your text, but I'd be surprised if such formulas are not given. (What text are you using?) In any case, given one formula you can find the other using the parallel axis theorem (if you've covered that).

    You can also reason conceptually based upon your understanding of what contributes to rotational inertia. For example: Which has the greater rotational inertia: (1) A rod rotating about its center, or (2) A rod rotating about one end?

    Another one. Which has the greater rotational inertia: (1) A rod rotating about a parallel axis right through the rod, or (2) A rod rotating about a parallel axis some distance from the rod?
     
  15. May 6, 2007 #14
    also, you can apply the parallel axis theorum by adding an imaginary additional segment equal to the radius of the one rod to the one that abuts it. That make sense?
     
  16. May 6, 2007 #15
    I am not sure.
     
  17. May 6, 2007 #16
    If you figured out the I's for the case of the thin rod about its center and about its end and are still worried about the 1/2radius offset in the one case where it abuts, the same parallel axis theorum can be used to compute the small difference. Not sure where you're at with this problem, focus on what Doc Al and get a conceptual feel for the effect of lever arm, think about the ice skater as she brings in her arms as she twirls on the ice.
     
  18. May 6, 2007 #17
    It is not a conceptual misunderstanding, it is a lack of numerical proof. But here is what I am going with:
    The center of mass is 3/4 of the way down the vertical rod in the Ts. Thus, using the parellel-axis theorem I get:
    I_a=I_cm+M(.75L)^2
    I_b=I_cm+M(.25L)^2
    I_c=I_cm+M(0L)^2
    I_d=I_cm+M(.5L)^2

    I am quite sure that this will suffice. The only thing that is troubling me is this, I know that the center of mass is 3/4 of the way down the vertical rod, but how can I state that mathematically? What formula can I use? I know each rods com is 1/2L being that they are uniform bodies. And I know that by definition the com is the point where the T will be balanced in a uniform gravitational field. But how can I show that it is where I say it is? The hard part is over, now I am just curious:tongue2:
     
  19. May 7, 2007 #18

    Doc Al

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    Careful here. Realize that the axes used in a & b are perpendicular to those used in c & d. I_cm depends on the orientation of the axis; it's not the same throughout.

    Since the T is composed of two indentical rods, the center of mass of the T will be right in the middle between the centers of each rod.
     
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