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Moments of Inertia-Big Confusion

  1. Dec 4, 2006 #1
    Moments of Inertia--Big Confusion

    Hey,

    1. Using the conventional method to calculate moments of inertia of rigid objects which says we should divide the object into small pieces each of mass dm and then calculate the integral. is it an exact calculation or an approxximation? so far my understanding has led me to think its STILL an approximation, because whenever e do it, we ASSUME something, like for example in the case of a solid ring, we assume the thickness to bne infinitely small.

    2. in calculating I for a solid disk, here's what i did:

    divide the disck (radially) into an infinite number of pieces each of mass dm and radius R/2 from the centre, and then calculate the integral which ends up being = MR^2 / 4 which is wrong seeing how the real answer should be MR^2 / 2

    what am i doing wrong here? please guide me/!
     
  2. jcsd
  3. Dec 4, 2006 #2

    Hootenanny

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    Provided the assumptions are good, then we arrive at an exact answer. For example, when we integrate a function to find the area under the curve, do we arrive at an approximation or an exact answer? You may wish to look at the definition of a Riemann Integral for clarification.
    I'm not sure what your doing, but I'll start you off on the right foot. From the general definition of the moment of inertia;

    [tex]I = \int^{M}_{0} r^2 dm[/tex]

    We need to change the variables of the integral (from dm to dr). Now, we know that the density of an object (assuming uniform density) is given by the ratio of its total mass to its volume. Therefore we can write;

    [tex]\rho = \frac{m}{V} \Rightarrow \rho=\frac{dm}{dV} \Rightarrow dm = \rho\cdot dV[/tex]

    Now, can you write an expression for the volume of a cylinder?
     
  4. Dec 4, 2006 #3

    radou

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    Use polar coordinates to describe what [tex]dm = \rho dV[/tex] looks like. Then integrate: [tex]I = \int_{V}r^2 dm[/tex]. If you have problems with that, I can draw a sketch.

    Edit: late again.
     
  5. Dec 4, 2006 #4
    what do u mean assumptions are good? and when u try to find the I of a solid ring, u assume the thikness to be infinitely small. and when u take the limit of the riemann sum (ie the integral) the thickness doesnt change in the process... it doesnt approach zero, it is ASSUMED to be CLOSE TO (read approximately) zero.

    secondly, what i did is: divide the circle (radius R) into inifinitely many 'pie' pieces which are very small and the mass of each is dm. centre of mass of each dm is going to be at the middle , that is R/2 from the centre. so the we will integrate with respect to m while R/2 is the same for all pieces so its constant and taken out of the integral (R/2)^2 int dm = R^2M / 4
     
  6. Dec 4, 2006 #5

    radou

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    This should help: http://usera.imagecave.com/polkijuhzu322/disc.JPG. The sketch is ugly, but you should get the point.

    Edit: assume that the density of the disc is [tex]\rho[/tex], and that the thickness is t.
     
  7. Dec 4, 2006 #6
    The whole idea of taking the limit of something approaching zero is that we can treat it as being zero for most intents and purposes. e.g. for y = x + a, y approaches x as a approaches zero, but in the limit y is exactly equal to x. Hope that clears that up somewhat.
     
  8. Dec 4, 2006 #7
    in my method, im the taking the whole piece as dm, and so the radius between the centre of each piece is R/2 as dm approaches 0...im confused........
     
  9. Dec 4, 2006 #8

    radou

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    You mentioned 'conventional methods' yourself. The one I presented is rather conventional. Try to work it out. Start with [tex]dm = \rho dV[/tex]. Find dV, i.e. the volume of the blue piece (assuming, of course, that the thickness equals t, for example), and then plug the dm into the integral and integrate for [tex]0 \leq r \leq R[/tex] and [tex]0 \leq \phi \leq 2 \pi[/tex]. I did it, and it worked out just fine.
     
    Last edited: Dec 4, 2006
  10. Dec 4, 2006 #9
    ok. try splitting the circle into 'pie' pieces and work it out. show me wat u find out...
     
  11. Dec 4, 2006 #10

    Hootenanny

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    You are confusing the matter, you can't just take the centre of mass of each segment and expect this to give you the moment of inertia, what about the fraction of mass that is inside the centre of mass, what about the fraction of mass that is outside the centre of mass? How do you account for the difference? I.e. there is more mass at the edges of your 'pie section' than there is closer to the centre, the mass is not uniformly distributed. In other words, if we take your pie shaped segment, each of the atoms (point masses) are in general a different distance from the centre of the disc, therefore we need to sum up each of these individual radii, which is why we must integrate with respect to dr not dm.

    In addition, dm is your variable of integration, which in this context is assumed to [itex]\to0[/itex] anyway. So saying that [itex]dm\to0[/itex] is meaningless. The integral

    [tex]\int r^2 dm[/tex]

    Means, integrate r2 with respect to m, therefore it is meaningless to say;

    [tex]\int r^2[/tex]

    as you have not identified your variable of integration.

    I hope this is clear.
     
    Last edited: Dec 4, 2006
  12. Dec 4, 2006 #11
    BUT, when i divide it, remember we are taking the limit as dm->0 which means there are infinite no. of pie pieces (the more the pie pieces, the straighter they get which makes my assumption for their centre of mass to be R/2 from the centre valid). Thats the method used by my book in calculating I for a thin ring.
     
  13. Dec 4, 2006 #12

    Doc Al

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    Two problems here:
    (1) Even in the limit, the pie pieces are still triangles, not straight segments. The center of mass of these triangles is not at R/2. (This should make intuitive sense, since there's more mass between R/2 and R than between 0 and R/2.)
    (2) Regardless of the location of the center of mass, what you need to be adding is the rotational inertia of each pie piece. What's the rotational inertia of each pie piece?

    A thin ring has all its mass at the same distance from the axis--very different from your pie slices.

    Bottom line: Dividing the disk into pie pieces is not a good plan for caclulating the rotational inertia.
     
  14. Dec 4, 2006 #13
    But, even with the thin ring, ur mass elements (dm) will hav more mass on one side than on the other, and this difference diminishes as we divide the ring into more and more mass pieces
    (ie limit of the sum to infinity). im following the same procedure with the circle, (the mass diff between 0-R/2 and R/2 - R diminishes as we add more and more pies.) :(:(:( i still am unsure. can someone ENLIGHTEN ME? it all seems the same for me:(:(
     
  15. Dec 4, 2006 #14
    sorry for double post. responding to point 1. the more pies u add the straighter they get right? so if u add infinite pies they become infinitely straight (ie rectangles).
     
  16. Dec 4, 2006 #15

    Doc Al

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    No, the mass of a section of thin ring is all at the same distance from the center in the limit. (And as you make the ring thinner and thinner, that gets truer and truer.) Not so with the pie slice! It's always a triangle shape, and the mass is always distributed from 0 to R.

    The mass difference gets smaller, but not the ratio, which is what determines the location of the center of mass. Why don't you figure it out and see? Take a thin triangle and figure out where its center of mass is. Does the location of its center of mass depend on just how thin it is?

    In any case, what will you do with those pie slices once you find their correct center of mass?

    Much better would be to divide the disk into thin rings, which have an easy-to-calculate rotational inertia--unlike the pie slices.

    They get thinner, but remain triangular.
     
    Last edited: Dec 4, 2006
  17. Dec 4, 2006 #16
    i know im being [retty stubborn and dumb but pls bear with me:
    the mass sections in the ring: arent they just like triangles with a big portion (from the pointy end of a trangle to some point in the middle) cut from them? if so, how come it's valid to consider their centre of masses the same from the centre? i mean for a big portion the centre of mass will b different from that of a small portion... wouldnt it?
     
  18. Dec 4, 2006 #17

    Doc Al

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    Right. But as soon as you start talking about thin sections of rings, the mass is concentrated between X and X + dx, where dx goes to zero. (X is the distance from the inner ring radius to the center of the disk.) So in the limit of a thin ring, all the mass is at X.

    But when you slice it up not in rings, but in pie slices, the mass is always spread out from X = 0 to X = R. That remains no matter how thin you slice it sideways.
     
  19. Dec 4, 2006 #18
    wait up, in my textbook, the RING is sliced SIDEWAYS...so 'your' dx doesnt go to zero
     
  20. Dec 4, 2006 #19

    Doc Al

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    Why in the world would you slice a thin ring sideways? But even if you did, it doesn't matter, since the distance to the center is still the same for all the mass. So the rotational inertia is easy to calculate.

    Why don't you scan in a picture of what your book is doing?
     
  21. Dec 4, 2006 #20
    [​IMG]

    this is the best i could do. each slice is referred to as 'dm'
     
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