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Moment Of Inertia Help

  1. Sep 15, 2007 #1
    Moment Of Inertia Help!!

    people ,
    i'm a student , and this is a complicated problem i came across ! please noe that i know da simplest of integration and differentiation and the like .

    it might sound stupid , but can anyone tell why , during the calculation of moment of inertia of a ring the limits during integration are taken as R and 0 , instead of R and -R as in other cases like sphere and rod etc.

    i noe a bit of calculus just to do about these things only .

    please help !!!

    got a hard exam !!
     
  2. jcsd
  3. Sep 15, 2007 #2

    arildno

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    Eeh?
    What is R meant to be?
    A radius?
    An angle?
    The length of a straight line segment?

    And:
    What type of ring are you talking about?
    A 2-D ring with finite or infinitesemal thickness?
    Or something else entirely?
     
  4. Sep 15, 2007 #3
    R is supposed to be the radius , and yes da figure is 2d . the answer is supposed 2 be 0.5 M(R^2) .
     
  5. Sep 15, 2007 #4

    arildno

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    Well, is it a RING or a DISK you are trying to find the moment of inertia to?

    The disk has no hole in the middle, but a ring does.
    Furthermore, if it IS a ring, is the thickness of it negligible, or is it an annulus?
     
  6. Sep 15, 2007 #5
    Man , it is a ring , and the thickness is negligible . Now please help!!
     
  7. Sep 15, 2007 #6

    arildno

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    So:
    There is a big hole in the middle?

    In that case, the given answer is just wrong.

    Are you sure it is not a compact disc?
     
  8. Sep 15, 2007 #7
    I'm guessing that your book exploited symmetry, and calculated half the disk, then multiplied by two.
     
  9. Sep 15, 2007 #8
    can u show the integration as u say , because it doesnt say anything like that
     
  10. Sep 15, 2007 #9

    arildno

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    Again:

    Is it a compact 2-D disk, or is it a ring with a hole in the middle?
     
  11. Sep 15, 2007 #10
    sorry , sorry , my bad

    1. it is a compact disc
    2. thickness is negligible

    now , i beg for your help , got an exam tomorrow !!
     
  12. Sep 15, 2007 #11
    please explain the integration part
     
  13. Sep 15, 2007 #12
    R = Radius of the disc
     
  14. Sep 15, 2007 #13

    arildno

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    Okay! :smile:

    Now, for your question:

    You can consider the disk to be composed of concentric infinitely thin rings at varying radii (hence, they have infinitesemal mass dm), and the disk's moment of inertia is the sum 8i.e, integral) of their inertias:

    [tex]I=\int_{0}^{R}r^{2}dm=\int_{0}^{R}r^{2}(\frac{M}{\pi{R}^{2}})2\pi{r}dr[/tex]
    where:
    [tex]\frac{M}{\pi{R}^{2}}[/tex] is the area density of the disk (M its mass!), whereas the "area" of a ring equals its circumference ([itex]2\pi{r}[/itex]) times its width "dr".

    At all r's between 0 and R, we must put such a ring; hence, the integration limits are 0 and R
    Simplifying this expression, we get:
    [tex]I=\frac{2M}{R^{2}}\int_{0}^{R}r^{3}dr=\frac{2M}{R^{2}}\frac{R^{4}}{4}=\frac{1}{2}MR^{2}[/tex]
     
  15. Sep 15, 2007 #14
    aahhhhhhhh!!


    but what about the moment of inertia of a sphere , where limits are +R and-R (R = radius) ??
     
  16. Sep 15, 2007 #15

    arildno

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    Okay!

    First off, the variable lying between -R and R can't possibly be a radius itself, since it is meaningless to assign a negative number to a radius!

    If we WERE to use a radius as our variable, then it would most naturally go from 0 to R, and we would regard the compact ball as composed of infinitely many spherical SHELLS.


    That decomposition of the ball (or compact sphere if you like) is perfectly valid of course, but not the one implied with integration limits -R and R.


    Rather, our variable in this case runs along a diameter of our ball, having its origin coincide with the ball's center. Let us call that variable z; we are essentially decomposing our ball as a set of stacked on disks/cylinders with infinitesemal heights, and with varying radii as we move along our diameter.

    Get the picture?
     
    Last edited: Sep 15, 2007
  17. Sep 15, 2007 #16
    soooooooo ,

    aahhhhhh!!

    all right , it means that in a sphere , we dont take radius as the limit . so
    1. what are +R and -R then ?
    2. or in other words , what are the limits !

    it would be really nice if u explain it like u did above .
    and can give the integration part of the moment of inertia of a sphere ? thanx a lot for all ur help !
     
  18. Sep 15, 2007 #17
    also the same thing happens in a rod (thin , compact ,NO holes ) , where limits are l/2 and -l/2 where , it is given l=length (please excuse if i am wrong ) . please help!!
     
  19. Sep 15, 2007 #18

    arildno

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    We MAY, but not in this particular decomposition!
    In a ball with radius R, the diameter has length 2R.
    Explained above.
    You have a line segment of length 2R; you choose its midpoint to be the origin; hence, its end points lie at -R and R, respectively.

    Okay, but you need to do a bit yourself as well!

    First off, if the ball's mass is M, its radius R, what is then its volume density?
     
  20. Sep 16, 2007 #19
    density should be M/(4/3 * pi * r^3)
     
  21. Sep 16, 2007 #20
    also can u tell me how is area = circumference * width ?
     
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