HW10_prob1_cone inertia ellipsoid

# HW10_prob1_cone inertia ellipsoid - Problem - Inertia...

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Problem - Inertia ellipsoid of a right circular cone The height of the cone is h and its base diameter is 2 a . The cone is oriented so that its symmetry axis lies along the z axis of a Cartesian frame whose origin is at the vertex of the cone. In this orientation, there are only three non-zero elements of the inertia tensor, the diagonal elements. All of the o f -diagonal elements are zero. In other words, these xyz axes are the principal axes. We will verify this below. For the inertia ellipsoid of the right circular cone to be that of a sphere, all of the principal moments must be equal. This means that our task becomes one simply of f nding general expressions for the three principal moments in terms of h and a andtheneva luat ingtherat io a/h that makes them all equal. Start with the general expression for the αβ component of the inertia tensor ( α , β = x , y , z ) I αβ = Z ρ ( r )( r 2 δ αβ x α x β ) d 3 r. (1)

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## This note was uploaded on 12/19/2010 for the course PHYSICS ph 409 taught by Professor Wilemski during the Fall '10 term at Missouri S&T.

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HW10_prob1_cone inertia ellipsoid - Problem - Inertia...

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