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Help with moment of inertia

  1. Feb 24, 2004 #1
    I don't understand how to do the follow problem so I was hoping that someone could possibly explain it to me with a possible example. thanks


    Find the position of the center of mass of a right circular cone which has a base of radius a, a height h, a mass M, and uniform density. Find the moment of inertia of the cone for rotations about an axis through its tip and the center of the circle making up its base.
  2. jcsd
  3. Feb 25, 2004 #2
    help please

    Will someone please help me?
  4. Feb 25, 2004 #3

    matt grime

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    You know what the volume of the cone is?

    Imagine a plane chopping off the top of the cone at a distance d from the tip and parallel to the base. you know its volume in terms of d - its base can be found from simple geometry: imagine a cross section going through th vertical axis the ratio of d to the raduis of the base of the smaller cone is the same as the ratio of the height to the radius of the base of the large cone.

    now find the d that makes that volume half of the volume of the larger cone.

    as for the moment of inertia, well if you couldn't do the previous one, i need to know that you 1. know the formula for moment on inertia 2. can do triple integrals, and 3. know about cylindrical polar coordinates (to make life easier).
  5. Feb 25, 2004 #4


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    Are you telling us that you have never ever seen any problems like this? There are absolutely no formulas or examples in your textbook?

    What your teacher and textbook are hiding from you is that the center of mass has the property that each of its coordinates, x for example, times the mass is equal to the integral of that coordinate, x in this case, times density, over the volume.

    Likewise, the moment of inertia, of rotation about an axis, is the integral of "distance from axis", squared" times density integrated over the volume.

    Example: a cylinder of radius R, height h, uniform density.
    (Since the density is uniform, it is a constant, δ.)

    The mass is the integral of δ over volume which, since δ is a constant is just δtimes the volume: δπR2h.
    To find the x-coordinate of the center of mass integrate xδ over that same volume: [tex]\int\int\int \delta x dV [/tex]

    Since this is a cylinder, I would recommend cylindrical coordinates:
    x= r cosθ y= r sinθ and dV= rdrdθdz.

    The integral is [tex]\int_{\theta=0}^{2\pi} \int_{r=0}^R\int_{z=0}^h \delta r^2 cos\theta drd\thetadz [/tex]. That turns out to be easy because
    [tex]\int_0^{2\pi} cos\theta d\theta= 0[/tex]. We have δπr2h x= 0 so x= 0.

    Similarly the y coordinate of the center of mass is 0 (that should not be a surprise).

    To find the z coordinate do almost exactly the same thing: find
    [tex]\int_\theta=0^{2\pi} \int_r=0^R\int_z=0^h \delta r z drd\thetadz [/tex] (x= r cosθ has been replaced by z)
    [tex]\int_0^R r dr= \frac{1}{2}R^2 [/tex]
    [tex]\int_0^{2\pi} d\theta= 2\pi [/tex] and
    [tex]\int_0^h zdz= \frac{1}{2}h^2[/tex].

    The integral is (1/2)πδR2h2
    dividing by the "mass",δπR2h, we get the z-coordinate of the center of mass (1/2)h (again, no great surprise).

    To find the moment of inertia, with the axis of rotation the "main" axis (through the centers of the two ends) we note that the distance of any point to that axis is just r itself so we need to integrate r2: the integral is
    [tex]\int_{\theta=0}^{2\pi} \int_{r=0}^R\int_{z=0}^h \delta r^3 dr d\theta dz [/tex]
    Last edited: Feb 25, 2004
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