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Moment of Inertia and MR2

  1. Dec 4, 2014 #1
    How do we know that moment of inertia cannot exceed MR2 for a solid body like solid cylinder, solid sphere, hollow sphere etc? R is the radius of the object in consideration.

    I know it has something to do with- that for solid bodies (in some cases hollow also) there are a lot of point masses in the body which are not at a distance of R from the axis of rotation. They are closer (less than R) to the axis of rotation than the point masses on the boundary of the body (at a distance of R from the body). Still, how does it prove that in such cases the moment of inertia would be less than MR2?

    ∑miri =
    img2730.png
    or

    img2731.png
    So how do the above two equations show that if there are masses which are located at a distance of less than R from the axis of rotation then the moment of inertia would be less than MR2?
     
  2. jcsd
  3. Dec 4, 2014 #2

    ShayanJ

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    The moment of inertial of a cylindrical shell around its axis is just [itex] M R^2 [/itex]. Now I can place anything I want at the centre of the shell(actually it can be anywhere!) and somehow connect it to the shell. Then the moment of inertial of the whole thing around the shell's axis is more that [itex] M R^2 [/itex]. So what you say is not true and so doesn't have a proof!
     
  4. Dec 4, 2014 #3

    robphy

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    I think a part of your question deals with what M and R represent.
    Is M the total mass of the object? Is R some maximum distance that a piece of the object is from the axis of rotation?

    I teach moment of inertia by saying it has the form kMR^2, where M is the total mass and R is some characteristic length (like the radius).
    The goal is to compute k.

    If I have a cylindrical shell of mass M and radius R and a mass m is added at radius r (say aR where 0<=a<=1),
    then the moment of inertia is
    [tex]MR^2+m(aR)^2=(M+ma^2)R^2=\frac{M+ma^2}{M+m} (M+m)R^2=k M_{total}R^2.[/tex]
    Note: for nonzero m, the coefficient k equals 1 only when a=1, otherwise k<1.
     
  5. Dec 4, 2014 #4

    ShayanJ

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    So take a>1. Then k>1 !

    EDIT:Sorry, I know this is wrong now!
     
    Last edited: Dec 4, 2014
  6. Dec 4, 2014 #5
    I am pretty sure on this part that moment of inertia can never cross MR2. My main question was that why in most of the cases of solid bodies, it is always less than MR2 and never equal to it?

    Have a look at this link (its a PDF file): http://www.colorado.edu/physics/phys1110/phys1110_fa08/lectures/Lec31.pdf [Broken]
    Look at page 4 and 5. And look at the line which says
    "Not all of the mass is at a distance R from the axis. So we end up with something less than MR2."

    So my main question was that why do we end up with something less than MR2? I don't find "Not all of the mass is at a distance R from the axis" to be a sufficient enough reason for Moment of Inertia to be less than MR2.
     
    Last edited by a moderator: May 7, 2017
  7. Dec 4, 2014 #6

    ShayanJ

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    Sorry for post #4. It was a misunderstanding!
    I think we should think more carefully about the calculation presented by robphy!
     
  8. Dec 4, 2014 #7
    Yeah Shyan. Now when I look at the calculations of robphy closely I am able to understand how MOI would always be less than MR2. Thanks for that robphy!
     
  9. Dec 4, 2014 #8
    But then what happens if we put a mass at a position which is greater than R? That is, what happens when a>1? Can this happen? Why or why not?
     
  10. Dec 5, 2014 #9

    ShayanJ

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    That's exactly my misunderstanding in post #4. If you have a mass with some length scale R and then add another mass somewhere farther, you're just extending the the object and so you'll just have a new length scale!
     
  11. Dec 5, 2014 #10
    So you mean wherever we place the new object that becomes the new R?
     
  12. Dec 5, 2014 #11

    ShayanJ

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    Yes, that's the point. Because actually by R, we mean a length scale which shows the extent of the object we're considering. I think its good to look at the moments of inertia of some shapes that don't involve circles!
     
  13. Dec 5, 2014 #12
    Hmm...yeah I got that now. :)
    And, I am talking about basic figures like cylinder, disc, ring, sphere etc. because that's what my curriculum asks for. But I think it would be the same explanation for other simple geometrical figures like square plate, cube, cuboid, rod etc. Won't it be?
     
  14. Dec 5, 2014 #13

    ShayanJ

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    Take a look at here! You'll notice the difference.
     
  15. Dec 5, 2014 #14
    what is a?
     
  16. Dec 5, 2014 #15

    robphy

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    a is a number between 0 and 1 describing the position r as a fraction of R, presumed to be the outer radius of the object.
     
  17. Dec 5, 2014 #16
    The difference is that R changes to L in case of rods or l2+b2 in case of squares, cubes and cuboids. The explanation that robphy gave still remains true even after different representations of R. Don't you agree?
     
  18. Dec 5, 2014 #17

    ShayanJ

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    Yeah but about infinite disk and polygons things seem to be different. Maybe I'm confused again!
     
  19. Dec 5, 2014 #18

    sophiecentaur

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    If you look at what you wrote in the initial post, you will see an Integral. That, in words, says that the Moment of Inertia is the sum (for all elemental parts of the object) of the mass of each particular element of mass times the distance from the CM squared. the maximum value that it can have (for a given diameter / width / length of object) is when all the elemental masses are distance R from the CM. If any masses are less than R from the CM, the MI will, of course be less than that Integral. That's just a bit of simple logic.
     
  20. Dec 7, 2014 #19
    I can't see how this is simple logic. Firstly, how can we add all the small moment of inertia to get the overall moment of inertia?
    I (Moment of Inertia) = dm1r12+dm2r22....+ dmnrn2 = kMR2
    Now if in some of the above terms , ri is less than R and some it is equal than R then how can we know that the sum would be less than R?
    Secondly, I don't understand how the above equation that I wrote can be ever mathematically proven to RHS. I mean if you take all the dm's common (I am assuming that all the differential point masses are of same mass) from the summation (in this case an integral) then you are just left with : dm(r12 + r22 + r32....rn2)
    It would eventually be something like- dm x (Some distance much greater than R2 ), which is not even close to how the actual MOI is.

    All the sum of squares in the parenthesis above would, by all means, be greater than R2. Which is simply not possible! That's the reason I can't understand the way you put it in integral/summation. But what robphy wrote, it makes sense and also doesn't create any ambiguity.[/SUP]
     
    Last edited: Dec 7, 2014
  21. Dec 7, 2014 #20

    A.T.

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    How can any ri be greater than the radius of the object? R = max(ri) per definition.
     
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