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Moment of Inertia Tensor Cylinder.

  1. Dec 29, 2013 #1
    I am computing the [itex]\hat{I}[/itex] - moment of inertia tensor - of a cylinder with height 2h and radius R, about its axis of symmetry at the point of its centre of mass.

    I am working in cartesian coordinaes and am not sure where I am going wrong. (I can see the cylindirical coordiates would be the best option here and have computed it correctly in these coordinates, but would like to know where I am going wrong please...)

    I have defined the z axis to be in line with the symmetry axis of the cylinder . I am then working in the principal axes s.t non-diagonal elements are 0.

    So by symmetry , I can see that Ixx = Iyy.

    Computing Izz:

    Moment of Inertia tensor formula: [itex]_{vol}[/itex][itex]\int[/itex] dv[itex]\rho[/itex] (r[itex]^{2}[/itex]δ[itex]_{\alpha\beta}[/itex]-r[itex]_{k,\alpha}[/itex]r[itex]_{k,\beta}[/itex])

    x ranges from R to -R, as does y.
    z ranges from h to -h.

    So Izz= [itex]_{vol}[/itex][itex]\int[/itex] ( x^2 and y^2) [itex]\rho[/itex] dV
    where dV = dx dy dx

    This yields: 8R^3M/3∏

    So a PI is present, so I can clearly see I have gone wrong. I think this might be due to my ranges .

    If someone could point me in the right direction, that would be greatly appeacted :).
  2. jcsd
  3. Dec 29, 2013 #2


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    I think you are getting tripped up on the notation you are using.

    Izz = [itex]\int_{vol}ρ(x^{2} + y^{2}) dV[/itex]

    Assuming the lengthwise axis of the cylinder is parallel to the z-axis, then re-writing the integral above:

    Izz = [itex]\int[\int_{dA}ρ(x^{2} + y^{2})dA] dz[/itex]

    If the density ρ is constant, then the area integral above is constant w.r.t. z. The area integral also represents the polar moment of inertia of the circular cross-section of the cylinder.
  4. Dec 29, 2013 #3


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    Hello bingagsss

    Surely you mistyped this.

    If you let x range from -R to R, then for a given x the range of y will not be -R to R.

    Wouldn't that make the height of the cylinder 2h? (Maybe you're letting h be half the height of the cylinder?)
    Last edited: Dec 29, 2013
  5. Dec 30, 2013 #4
    Sorry I meant p =M/[itex]\pi[/itex]R[itex]^{2}[/itex]2h.

    Post 1 - it is a cylinder of height 2h

    Thanks I think I see - taking a circle cross-section of the cylinder to lie in the xy plane, it has equation: x[itex]^{2}[/itex]+y[itex]^{2}[/itex]=a[itex]^{2}[/itex].

    If I then let x range from R to -R, y must be a function of x , i.e (a[itex]^{2}[/itex]-x[itex]^{2}[/itex])^1/2.
    You then need to split the integral into two pieces , one where y ranges from +(a[itex]^{2}[/itex]-x[itex]^{2}[/itex])^1/2. and the other -(a[itex]^{2}[/itex]-x[itex]^{2}[/itex])^1/2(both cases x ranging from R to -R).

    So for one of these (+(a[itex]^{2}[/itex]-x[itex]^{2}[/itex])^1/2.) this yields:

    [itex]^{h}_{-h}[/itex][itex]\int[/itex][itex]^{r}_{-r}[/itex][itex]\int[/itex][itex]^{(a^{2}-x^{2})}_{0}[/itex][itex]\int[/itex] x[itex]^{2}[/itex]+y[itex]^{2}[/itex] dydxdz
    = [itex]^{h}_{-h}[/itex][itex]\int[/itex][itex]^{r}_{-r}[/itex][itex]\int[/itex]x[itex]^{2}[/itex](a[itex]^{2}[/itex]-x[itex]^{2}[/itex])[itex]^{1/2}[/itex] + (a[itex]^{2}[/itex]-x[itex]^{2}[/itex])[itex]^{3/2}[/itex]/3 dx dz

    Is this correct? (I'm guessing a trig substition would then be needed to solve this?)
  6. Dec 30, 2013 #5


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    Goodness, you have made your problem more complicated.

    Starting with:

    Izz = [itex]\int[\int_{A}ρ(x^{2} + y^{2})dA] dz[/itex]

    Then, if we want to evaluate the inner area integral to determine the polar moment of inertia of the
    circular cross section, a change to polar coordinates would make things easier:

    x = r cos [itex]\theta[/itex]
    y = r sin [itex]\theta[/itex]

    Using these substitutions, evaluating the following integral

    [itex]\int_{A}(x^{2} + y^{2}) dA[/itex]

    is much easier.
  7. Dec 30, 2013 #6


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    OK. (Sorry I missed that.)

    You don't need to split it up into two separate integrations over y, but it's ok if you do. By symmetry, each of the separate integrations will be equal.

    It looks correct to me. Note, your trig substitution is probably going to be equivalent to going over to polar coordinates at this point.
  8. Dec 30, 2013 #7


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    Right. In the first post binbagsss stated that he had already worked it out in polar coordinates but that he wanted to also do it in Cartesian coordinates as an exercise.
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