HW10_prob4_cube_v2

# HW10_prob4_cube_v2 - Solid Cube Problem(a Inertia tensor in...

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Solid Cube Problem (a) Inertia tensor in COM frame The cube is located with a vertex at the origin of a Cartesian coordinate system whose x , y and z axes lie along edges of the cube. The edges of the cube are of length a . x z y For this coordinate system, the inertia tensor has the structure (given in class notes) I = b 12 8 3 3 38 3 3 , (1) where b = Ma 2 and M is the mass of the cube. The next step is to f nd the principal moments with respect to the center of mass. In the given coordinate system, the center of mass is at ( a/ 2 ,a/ 2 2) . We use the Parallel Axis Theorem in the tensor form I = I 0 + I R (2) and construct the inertia tensor I R for the center of mass vector R in the original coordinate system with unit basis vectors denoted by x , y ,and z . Here, obviously, R is R =( a/ 2)( x + y + z ) . (3) From the de f nition, I R = M [ R 2 1 RR ] , (4) we obtain the explicit form I R =( 2 / 4)[3 1 ( x + y + z )( x + y + z )] , (5) after noting that R 2 =3 a 2 / 4 , and substituting for R with Eq.(3) and for the unit tensor 1 = xx + yy + zz . (6) 1

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We can write Eq.(5) in a matrix form similar to Eq.(1) as I R = b 4 2 1 1 12 1 1 . (7) By the parallel axis theorem, the inertia tensor I 0 we require is I 0 = I I R , (8) and with Eqs.(1) and (7), we see that in the center of mass coordinate system the inertia tensor is automatically diagonalized I 0 = b 6 100 010 001 . (9) This tells us that when the original coordinate axes are translated to the center of mass, while maintaining their original orientation, they serve as principal axes.
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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_prob4_cube_v2 - Solid Cube Problem(a Inertia tensor in...

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