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Rotational inertia axis

  1. Apr 26, 2005 #1
    I would like to further understand rotational inertia. I understand that for a point mass, I = MR^2 and for a continuous object it is basically the sum of all "little" MR^2 for each element of that object. I get a little fuzzy when actually solving for I for an object. For example, if we have a cone, the mass M of the cone is the density * volume or rho*(1/3)pi*r^2. So to find inertia we take the integral of the product of mass and R^2 but over what interval? My book says to take it over the volume but I am not 100% sure what that means. Would it be the triple integral dx dy dz (or more easily dr d(theta) dz)? If so, what would be the integrand?

    TIA for any help
  2. jcsd
  3. Apr 26, 2005 #2


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    First, always be clear about where the axis of rotation is, relative to the position of the object.
    If we let the z-axis denote the rotation axis, then the moment of inertia of the object is given by [tex]I=\int_{V}\rho(x^{2}+y^{2})dV[/tex]
    where x,y are orthogonal coordinates in a plane defined by the rotation axis.
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