Solutions_HW10

Solutions_HW10 - Theoretical Dynamics November 19, 2010...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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,...
View Full Document

This note was uploaded on 12/29/2011 for the course PHYSICS 601 taught by Professor Hassam during the Fall '11 term at Maryland.

Page1 / 15

Solutions_HW10 - Theoretical Dynamics November 19, 2010...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online