homework8sol.20100402.4bb602e61df795.39866753

# homework8sol.20100402.4bb602e61df795.39866753 - TAM 412...

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TAM 412, Homework 8, Selected answers Harry Dankowicz Mechanical Science and Engineering University of Illinois at Urbana-Champaign [email protected] 1. By symmetry, the moment of inertia matrix J about the center of the cube with respect to axes parallel to the edges must be a multiple of the identity matrix. From the relationship J = H T JH the same follows for the moment of inertia matrix about the center of the cube with respect to any other orthogonal axes. For the moment of inertia matrix about a vertex with respect to axes parallel to the edges the principal moments of inertia equal 11 ma 2 / 3, 11 ma 2 / 3, and 2 ma 2 / 3 with corresponding principal axes - e x + e z , - e x + e y , and e x + e y + e z , respectively. 2. The moment of inertia matrix is obtained from the Mathematica command 3. For a rod, the moment of inertia matrix about the center of mass with respect to axes with the
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Unformatted text preview: z-axis along the symmetry axis of the rod equals J = ml 2 3 ml 2 3 ⇒ J = ml 2 3 ( ω 1 e x + ω 2 e y ) Furthermore, c = l 2 e z , v = u + ω × l e z . 4. A line passing through the center of mass is a principal axis if it is parallel to a vector v = v 1 e x + v 2 e y + v 3 e z such that J v 1 v 2 v 3 = λ v 1 v 2 v 3 for some λ . The moment of inertia matrix about a point shifted by a vector v from the center of mass is then given by J + m v 2 2 + v 2 3-v 1 v 2-v 1 v 3-v 1 v 2 v 2 1 + v 2 3-v 2 v 3-v 1 v 3-v 2 v 3 v 2 2 + v 2 3 and v 2 2 + v 2 3-v 1 v 2-v 1 v 3-v 1 v 2 v 2 1 + v 2 3-v 2 v 3-v 1 v 3-v 2 v 3 v 2 2 + v 2 3 v 1 v 2 v 3 = . 1...
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