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Finding Moment of Inertia of Infinitely Thin Hoop using a Double Integral

  1. Nov 10, 2006 #1
    Here is the problem:


    Is it possible to determine this moment of inertia in this problem using double integrals of the form:


    I could do this problem using double integrals if the region R was a solid circle using [tex]x^2+y^2=R^2[/tex] and then simplifying the integration by switching to polar coordinates. However, the region is a hoop whose width is infinitely thin.

    How do I modify my region R in order to take into account the "uniform thin hoop" that has no width? Is there a way I could evaluate a hoop of width [tex]\Delta R[/tex] and find moment of inertia in terms of [tex]\Delta R[/tex] then let find the limiting value as [tex]\Delta R[/tex] approaches zero?

    If I used that approach, what would I use for the density of the infinitely thin hoop?
    Last edited: Nov 10, 2006
  2. jcsd
  3. Nov 10, 2006 #2
    I think you are talking about comparing two disks of radius R and R+dr... and look as the limit of dr goes to 0.

    When I first read this tho, I went about it a slightly different way...
    Note how their example has you integrating over little pieces of mass (along the whole loop). In this case since it's infinitely thin, in some sense you are really doing a line integrals, not area integrals. If you want to use the pendendicular axis theorem the line intergrals are just more difficult because they have constraints that are more difficult, but you still use density as mass per unit length... not mass per unit volume.... and what would that value be? (Hint -- what is the circumference of the loop?)

    Actually, if you are familiar with "delta functions", the constraints and integrals are pretty easy.

    Then - once you get an answer for one of the perdendicular axes, can you use symmetry arguments to check your answer? (Hint: How should the magnitudes of Ix, Iy compare?)
  4. Nov 10, 2006 #3
    I think I'm almost there if I use a line integral. I am off by a factor of [tex]R[/tex]. Why?

    The density will mass per unit length

    [tex]\rho = \frac{M}{L}[/tex]

    The length of the curve is [tex]2\pi R[/tex] which gives a density of

    [tex]\rho = \frac{M}{{2\pi R}}[/tex]

    The equation of the circle is

    [tex]x^2 + y^2 = R^2[/tex]

    Parameterizing the curve gives

    [tex]\vec r(t) = x\left( t \right)\hat i + y\left( t \right)\hat j[/tex]

    [tex]\vec r(t) = R\cos \left( t \right)\hat i + R\sin \left( t \right)\hat j[/tex]

    Setting up the line integral we have

    [tex]\int_C^{} {\left( {\frac{M}{{2\pi R}}} \right)} ds[/tex]

    [tex]ds = \sqrt {x'(t)^2 + y'(t)^2 }dt [/tex]

    [tex]ds = \sqrt {\left( { - R\sin t} \right)^2 + \left( {R\cos t} \right)^2 }dt[/tex]

    [tex]ds = \sqrt {R^2 \sin ^2 t + R^2 \cos ^2 t}dt[/tex]

    [tex]ds = \sqrt {R^2 \left( {\sin ^2 t + \cos ^2 t} \right)}dt[/tex]

    [tex]ds = \sqrt {R^2 }dt [/tex]

    [tex]ds = Rdt[/tex]

    The integral for the moment of inertia of the line

    \frac{M}{{2\pi R}}\int_0^{2\pi R} {Rdt} = \frac{M}{{2\pi }}\int_0^{2\pi R} {dt = } \frac{M}{{2\pi }}\left[ t \right]_{t = 0}^{t = 2\pi R} = MR

    Where does the other factor of R need to come from?

    What are delta functions?
    Last edited: Nov 10, 2006
  5. Nov 11, 2006 #4


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    It comes from the definitiuon of moment of inertia, and using the correct limits of integration. t is not a length. It is a parameter that ranges from zero to 2*pi

    [tex]\frac{M}{{2\pi R}}\int_0^{2\pi} {R^2 Rdt} [/tex]
  6. Jan 11, 2009 #5
    As you said it yourself, you have simplified the integration by switching to polar coordinates. The additional R comes from polar coordinates. When switching the variables of integration, you have to include the Jacobian determinant in the integral. The Jacobian determinant is the determinant of the matrix of all first-order partial derivatives of a vector-valued function.

    In our case, we have to find the partial derivatives of [tex]x=R\cdot \cos(\phi)[/tex] and [tex]y=R\cdot \sin(\phi)[/tex] (x and y expressed in polar coordinates) using R and [tex]\phi[/tex] as derivative parameters:
    [tex]\left|\frac{dx}{dR} \ \frac{dx}{d\phi}\right|[/tex]
    [tex]\left|\frac{dy}{dR} \ \frac{dy}{d\phi}\right|[/tex]

    For the integration parameters:
    [tex]\phi[\tex] then ranges from 0 to 2[tex]\pi[\tex] and R ranges from 0 to R.
  7. Jun 24, 2009 #6
    I still got some doubts about this.. I've tried many ways to find the moments of inertia [tex]I_{x} = I_{y} [/tex] but i always get some constants wrong.
    If i parametrize [tex] x(t) = R Cos(t) [/tex], what's wrong with taking the integral [tex] \int _{0} ^{2 \pi} \rho x^{2} dt [/tex]? If i do that, i get the correct factor of [tex] 1/2 [/tex] but R instead of [tex] R^{2} [/tex] :
    [tex] \int _{0} ^{2 \pi} \rho x^{2} dt = M/{2 \pi R} \int _{0} ^{2 \pi} {R^{2}cos^{2} (t) dt = {M R}/{4 \pi} [/tex] .. i don't even get the 1/2 not to mention that [tex] \pi [/tex]
    If i try to set x as a function of y, then i get some [tex] 1/3 [/tex] around from taking the integral of [tex] y^{2} [/tex]
    this calculations frustate me...
    Last edited: Jun 24, 2009
  8. Jun 24, 2009 #7
  9. Jun 25, 2009 #8
    I've found it! i'm gonna post it just ftw.
    I had some mistakes in my first post, for a start, the integral [tex] \int _{0} ^{2 \pi} x^2 \rho dt [/tex] should be done with ds instead (so the [tex] \rho ds [/tex] would give the mass of that bit, and all makes physical sense). Since [tex] ds = R dt [/tex] we have:
    [tex] I_y = \int_{0} ^{2 \pi} R^2 cos^2 (t) \rho R dt =
    {M R^2}/{2 \pi} \int_{0} ^{2 \pi} cos^2 (t) dt [/tex]
    i thought that the integral was just 1/2, but i was wrong. applying integration by parts twice i get that the integral is [tex] \pi [/tex] and it all comes out nice...
    got to be careful with those integrals and differentials!
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