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Moment of inertia of disk, the easy way out?

  1. Nov 2, 2012 #1
    1. The problem statement, all variables and given/known data
    When calculating moment of inertia of a disk there is something that really bothers me. I've googled this a lot and everywhere i look they 'assume' that the Δa = Δr*2∏r, formula for rectangle, not circle: (area of circle r+Δr - area of circle r) Δa = ∏(r+Δr)^2 - ∏r^2 = ∏r^2 + 2∏Δr*r + ∏Δr^2 - ∏r^2 = 2∏Δr*r + ∏Δr^2. One link is the same but you get the extra Δr^2. is it even possible to integrate this?

    the question is why is this allowed? is it because Δr^2 << Δr*r?


    2. Relevant equations

    I = ∫r^2 dm

    3. The attempt at a solution

    I=∫r^2 dm

    disk with inner diameter D/2, outer diameter D, mass M.

    r = D/2 => r1 = D/2, r2 = D/4
    Δm = M * ΔA / A = M(∏(r+Δr)^2 - ∏r^2)/(∏(r1^2-r2^2))
    ...
    Δm = M (2Δr*r + Δr^2) / (r1^2-r2^2)
    ΔI = Δm * r^2 = ... = 2Mr^3 Δr/(r1^2-r2^2) + M*r^2 Δr^2/(r1^2-r2^2)
    I = ∫..... ??
     
  2. jcsd
  3. Nov 2, 2012 #2

    SteamKing

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    Look at it this way. A disk of radius r has a circumference of 2*pi*r. If the radius is increased by a small amount dr, the circumference will then be 2*pi*(r+dr). The additional volume will be approximately the circumference multiplied by this additional radius dr, or
    dV = 2 * pi * (r+dr)*dr

    dV = 2 * pi * r * dr + (dr)^2

    As dr shrinks toward 0, the (dr)^2 term shrinks faster than dr.
     
  4. Nov 2, 2012 #3
    your dV = 2 * pi * r * dr is equivalent with my 2Δr*r + Δr^2

    what i dislike is the 'removal' of dr^2. It shrinks faster, but they both approach 0.

    on the other hand, is it even possible to solve

    \int(dr + dr^2)
     
  5. Nov 2, 2012 #4
    I am not a math major but one thing you may want to do is take a look at how big that error is as Δr→0 (and by error I mean the extra Δr2 term).

    Try calculating the size of the second order term compared to the first order term as Δr→0 (ie use a ratio).

    I am sure this is some fundamental problem solved in mathematical analysis that someone will come along and shed light on.
     
    Last edited: Nov 2, 2012
  6. Nov 2, 2012 #5
    The problem is I don't know how to solve it.

    I agree that it must have been looked upon, but i have a hard time just accepting things like this. I want to know how and why and see a proof. It's a curse really (:
     
  7. Nov 2, 2012 #6

    haruspex

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    Integration and differentiation are both limit processes. I.e. the answer is defined to be the limit as some small quantity tends to zero. When you write the equation out in detail you find that the zeroth-order terms (the ones not dependent on the delta) cancel out, but the first order terms do not. This means that in the limit the second order (delta-squareds) become irrelevant.
     
  8. Nov 2, 2012 #7
    It's not good enough just to say that and that is what the OP is asking about.

    phenalor, you didn't try what I suggested. For this example:

    (r + Δr)2 = r2 + 2rΔr + Δr2

    Take a look at the size of the second order term compared to the first order term as Δr→0 :

    lim Δr2 / (2rΔr) = Δr / (2r) = 0
    Δr→0

    This means the second order term is insignificantly small compared to the first order term so adding it to the first order term adds nothing as Δr→0.

    A rigorous mathematical justification will be along those lines.
     
    Last edited: Nov 2, 2012
  9. Nov 3, 2012 #8
    You could consider it this way:
    [tex]\frac{dA}{dr} = \lim_{\Delta r\to0}\frac{\Delta A}{\Delta r} = \lim_{\Delta r\to0}\frac{\pi (r+\Delta r)^2 - \pi r^2}{\Delta r} = \lim_{\Delta r\to0}\frac{\pi (2r\Delta r +(\Delta r)^2)}{\Delta r} = 2\pi r[/tex]
    Hence,
    [tex]dA = 2\pi r dr[/tex]
     
  10. Nov 3, 2012 #9

    haruspex

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    I read the OP as being concerned that omitting the delta-squareds was losing precision, i.e. it was only an approximation. If so, the relevant point was the one we both made, that integral is defined as a limit, so omitting the delta-squareds gives the exact answer.
     
  11. Nov 5, 2012 #10
    Sorry I have been gone for so long, lots of examns coming up soon.

    Thank you everyone for explaining this to me. It seems i had forgotten the relationship between [itex]\Delta x[/itex] and [itex]dx[/itex], which is really embarrrasing.

    The reason I got stuck upon this is that i thought [itex]I = \int{r^2}{dm} \rightarrow \Delta I = \Delta m r^2[/itex], which would give me the
    [itex]\Delta I = \frac{2Mr^3\Delta r}{r_1^2-r_2^2}+ \frac{Mr^2\Delta r^2}{r_1^2-r_2^2}[/itex]
    meaning i would get stuck with


    [itex]dI=\frac{2Mr^3}{r_1^2-r_2^2}dr+\frac{Mr^2}{r_1^2-r_2^2}\Delta r dr[/itex]

    edit: also thank you for showing me the awesomeness of LaTeX!
     
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