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Using Calculus To Derive Forumlas

  1. Oct 15, 2013 #1
    Can someone explain this proof of the moment of inertia of a sphere? And how you can use the same principals to solve for other formulas?

    https://www.physicsforums.com/attachment.php?attachmentid=62987&stc=1&d=1381895479http://

    I understand each individual variable. But I don't understand how they were able to get the moment of inertia by the orginal integrand, and where did that integrand come from?
     
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  3. Oct 16, 2013 #2

    Simon Bridge

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    The image didn't come out.

    The general principle is to divide the object into a lot of small masses dm.
    Work out the contribution to the moment of inertia for each dm and add them up.
    Since it is a continuous sum, you use an integration sign instead of a sigma.

    The approach is general - however, the math is easier if you can exploit some aspect of the symmetry of the system. i.e. for the sphere, they probably exploited spherical symmetry by using spherical-polar coordinates.

    There are many examples of this sort of calculation online - have you had a look?
    Mainly you just have to understand how integration works as a sum of small bits rather than as a collection of fancy tricks and formulas.
     
  4. Oct 16, 2013 #3
    I followed along with a proof of the moment of inertia of a solid cylinder, but I suppose the only thing i don't understand, is what formulas they start with. I get that you always start with I=∫r2dm is it all just a matter of logic, and how you can substitute dm with ρdV?
     
  5. Oct 16, 2013 #4

    arildno

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    "how you can substitute dm with ρdV? "

    Mass equals density times volume, does it not?
     
  6. Oct 16, 2013 #5

    HallsofIvy

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    It does if density is a constant. More generally, "mass equals the integral of density over the volume" which immediately leads to the differential formula "[itex]dm= \rho dV[/itex]".
     
  7. Oct 16, 2013 #6
    It does, i think you just answered my ultimate question. How does substitution work in any integral/derivative? so you can substitute mass with ρdV, simply because mass = ρv? why do you say its with respect to volume and not density? Same with velocity as a derivative of position. d/dx s(t) = v. why is that?
     
  8. Oct 16, 2013 #7
    @Hallsoflvy why are you saying mass equals the integral of density over volume? ρ = m/V. rearranged, m = ρV.
     
  9. Oct 16, 2013 #8

    arildno

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    Remember that we always have M/V=density (when we regard density as the average density over V)

    When we in maths shrink, in the limiting process, the volume element, the right hand side (i.e, the density) goes towards a non-zero, constant value we call the LOCAL density. The LHS has then become dm/dV if you like.
     
  10. Oct 16, 2013 #9
    ohhh okay, so we take the differential of Volume? because we are splitting the sphere up into tiny spheres with near-zero volume? And since density stays constant, its simply with respect to volume?
     
  11. Oct 16, 2013 #10

    arildno

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    It "becomes" a constant, if we make the volume small enough, within our mathematical modelling.
    --------------------------------------
    If we look at the product:
    [tex]\rho(x)(\bigtriangleup{x}|_{x=x_{0}}) (*)[/tex]
    where the volume [itex]\bigtriangleup{x}|_{x=x_{0}}[/itex] is CENTERED about [itex]x=x_{0}[/itex], while we let the density be allowed to vary across the volume, we have, approximately:
    [tex]\rho(x)\approx\rho(x_{0})+\frac{d\rho}{dx}|_{x=x_{0}}(x-x_{0})[/tex]
    and (*) can then be written as:
    [tex]\rho(x)(\bigtriangleup{x}|_{x=x_{0}})=\rho(x_{0})(\bigtriangleup{x}|_{x=x_{0}})+\frac{d\rho}{dx}|_{x=x_{0}}(x-x_{0})(\bigtriangleup{x}|_{x=x_{0}})[/tex]

    Note that the second term is, basically, of magnitude volume SQUARED, while the first is of magnitude "volume".

    Thus, as you let "volume" shrink in magnitude, the second term gets negligible, relative to the first, sincy tiny*tiny=even tinier.

    For EACH number x_0, then, each with an infinitesemal volumedV attached to it, dM=rho*dV, where the density function is evaluated at x_0 (and we drop the subscript _0 in the following integration)
     
  12. Oct 16, 2013 #11

    Simon Bridge

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    Almost - splitting the sphere into tiny spheres is unlikely to help because spheres don't tesselate well.

    The exact shape of the volume element is something you choose according to the coordinate system (which you choose according to symmetry).

    In cartesian coordinates, you divide the sphere into lots of little cubes.

    A rectangular volume would be given by ΔV=ΔxΔyΔz [with one corner at position (x,y,z)]
    A smaller volume would be represented by a small delta: δV=δxδyδz ... it's a notation meaning "small change in..."

    An infinitesimal volume would be dV=dx.dy.dz

    The amount of mass in the infinitesimal volume dV at position (x,y,z) is dm=ρ(x,y,z)dxdydz

    In your example the density is uniform so you just get dm=ρdxdydz


    In cylindrical coordinates, a volume element is a kind-of curved brick: dV=dr(r.dθ)dz
    Because rdθ is the arc-length at radius r inside angle dθ... dr is the width and dz is the height.


    It can take a while to get used to - but it is key to understanding how to set-up integrals.

    i.e. an area would be given by: ##A=\int dA## ... "doh!" right?
    But lets say it's the area of a rectangle with sides b and w... then dA=dxdy and:
    $$A=\int_0^w\int_0^b dxdy$$
    How about the area of a circle radius R ... then dA=r.dθdr
    $$A=\int_0^{2\pi}\int_0^R r.drd\theta$$
     
  13. Oct 16, 2013 #12
    I think i would understand this alot better if I had a much better understanding of integrals. In highschool AP calculus, we are just learning the product and quotient rule of derivatives, I just read ahead and try and scrounge the internet for more information
     
  14. Oct 16, 2013 #13

    Simon Bridge

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    Ah right ... so right now integration is a spooky magic black-box process where you apply a bunch of rules.

    You know yu can find the area of an irregular polygon by dividing it up into rectangles and triangles, finding the areas of each of them then adding them all up? Integration is a way of doing that in a more systematic way.

    If you want the area under f(x) inside a<x<b ... then you can divide the regeon into N smaller ones with width Δx = (b-a)/N and use those to make a bunch of rectangles.

    The 1st rectangle has height f(a+Δx), the third has height f(a+2Δx) and the third f(a+3Δx) and so on ... the nth one has height f(a+nΔx) and the Nth one f(a+NΔx)=f(b) ... all these rectangles sort-of follow the curve.

    So the nth rectangle has area: f(a+nΔx)Δx ... and the area of the entire curve is approximately $$\sum_{n=1}^N f(x_n)\Delta x:x_n=a+nΔx$$ If we make N bigger and Δx smaller and smaller in such a way that NΔx = (b-a) doesn't change, then the approximation gets closer and closer to the actual area you want to find.

    When Δx is infinitesimally small ... the sum becomes an integral sign and the Δx becomes a dx and the xn's are just values on the x axis and you write: $$A=\int_a^bf(x)\; dx$$... which should be familiar to you.

    When you study sums like this, you discover there are lots of shortcuts that mean you don't have to go back to dividing a function into strips every single time. That's what you are learning.


    Dividing into strips was just a handy shortcut also. You can also divide the y axis up into M sections and you get a lot of small boxes - then you count up the boxes and mutiply by the area.
    ... that will require two sums, which leads to two integrals as you make the boxes very small.


    See also:
    http://tutorial.math.lamar.edu/Classes/CalcI/CalcI.aspx
     
  15. Oct 17, 2013 #14
    okkayy, i totally understand that! thank you so much!
     
  16. Oct 17, 2013 #15

    Simon Bridge

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    It takes a while to get used to it.
    Enjoy.
     
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