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Unformatted text preview: Theoretical Dynamics November 19, 2010 Homework 10 Instructor: Dr. Thomas Cohen Submitted by: Vivek Saxena Note : Goldstein problems 5 . 8 and 5 . 14 are solved at the end. Problem 1 (a) The coefficients of the moment of inertia tensor in the usual 3 3 matrix representation are given by I jk = Z V ( r )( r 2 jk x j x k ) dV (1) Therefore, for the given mass density, we have to evaluate the integral, I jk = Z  Z  Z  dxdy dz e x 2 + y 2 + z 2 + xy 2 l 2 ( r 2 jk x j x k ) (2) The argument of the exponential is a quadratic form which can be written as x 2 + y 2 + z 2 + xy 2 l 2 = 1 2 x T A x (3) where x = x y z , A = 1 l 2 1 2 l 2 1 2 l 2 1 l 2 1 l 2 (4) so, A is real and symmetric. Define F jk = Z  Z  Z  dxdy dz e 1 2 x T A x x j x k (5) and G = Z  Z  Z  dxdy dz e 1 2 x T A x (6) In terms of F jk , the moment of inertia coefficient I jk is I jk = [( F 11 + F 22 + F 33 ) jk F jk ] = [ tr ( F ) jk F jk ] (7) To find the coefficient, we use Wicks Theorem, stated below. 10 1 Wicks Theorem : If A is a real symmetric N N matrix, and x is an N 1 vector, then h x i x j x k x l i = Z  Z  Z   {z } N integrations dx 1 dx 2 ...dx N e 1 2 x T A x x i x j x k x l Z  Z  Z   {z } N integrations dx 1 dx 2 ...dx N e 1 2 x T A x = X ( A 1 ) ab ( A 1 ) cd (8) where the sum runs over all permutations { a,b, ,c,d } of the set of indices { i,j, ,k,l } . In particular, for N = 3, we have h x j x k i = F jk G = ( A 1 ) jk (9) To evaluate G , we use the following result on Gaussian integration. Result : If A is a real symmetric N N matrix, and x is an N 1 vector, then Z  Z  Z   {z } N integrations dx 1 dx 2 ...dx N e 1 2 x T A x = (2 ) N det[ A ] 1 / 2 (10) For the matrix A defined in Eqn. (4), det( A ) = 3 4 l 6 . For N = 3, G = s (2 ) 3 3 / 4 l 6 = 4 r 2 3 3 / 2 l 3 (11) From Eqns. (9) and (10), we get F jk = 4 r 2 3 3 / 2 l 3 ( A 1 ) jk (12) The inverse of A is A 1 = 4 l 2 3 2 l 2 3 2 l 2 3 4 l 2 3 l 2 (13) So, using Eqn. (12), the matrix F is found to be F = 16 3 q 2 3 3 / 2 l 5 8 3 q 2 3 3 / 2 l 5 8 3 q 2 3 3 / 2 l 5 16 3 q 2 3 3 / 2 l 5 4 q 2 3 3 / 2 l 5 (14) 10 2 and finally, using Eqn. (7), the matrix representation of the moment of inertia tensor is I = 28 3 q 2 3 3 / 2 l 5 8 3 q 2 3 3 / 2 l 5 8 3 q 2 3 3 / 2 l 5 28 3 q 2 3 3 / 2 l 5 32 3 q 2 3 3 / 2 l 5 (15) (b) The moment of inertia matrix obtained above is of the form I = a b b a 0 0 c (16) where a = 28 3 q 2 3 3 / 2 l 5 , b = 8 3 q 2 3 3 / 2 l 5 and c = 32 3 q 2 3 3 / 2 l 5 . If is an eigenvalue, then it satisfies the secular equation,...
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This note was uploaded on 12/29/2011 for the course PHYSICS 601 taught by Professor Hassam during the Fall '11 term at Maryland.
 Fall '11
 Hassam
 mechanics, Inertia, Work

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