Solutions To Tensor Algebra Problem

 In a certain rectangular coordinate system, the directions of whose axes are given by the unit vectors i, j and k, the inertia tensor of an object is given by:

I = K x= 

(1….0…..0)

(0….1…..1)

(0….1… ..1)

a) What are the principal moments of inertia of the object (the moments of inertia along the principal axis) relative to the origin of the above coordinate system?

b) What is the direction of the principal axis corresponding to the principal moment of inertia and equal to K?

c)If the origin of the above rectangular coordinate system is at the center of mass of the object and the total mass of the object is M, what is the change in the inertial tensor of the object if the rectangular coordinate system is displaced parallel to itself a distance r_o in the direction

 

 (1/ Ö2)j +  (1/Ö2)k?

Solutions:

a) We have:  I = K x 

(1….0…..0)

(0….1…..1)

(0….1… ..1)

 We write out the determinant with eigenvalue  l:

(1 - l….0…..0)

(0….1 - l   ..1)

(0….1… ..1 - l)

Leading to the characteristic equation:

(1 - l)³ – (1 - l) = 0

Factoring:

(1 - l) [ ((1 - l)² – 1] = 0 

Or:

(1 - l) (l² –  2l) = 0

Yielding eigenvalues: l= 0, l = 2

Then:

T = Kl,  so:  T1 = 0, T2 = K, and T3 =2K

Or: (0, K, 2K)  for the principal moments of inertia.

b) We have to take:  (I – T1)C

So that:

=

 (0….0…..0) (x)

  (0….0…..1) (y)

  (0….1… ..0) (z)   =   0

 Finally:  0  =

(0)

(z)

(y)  

 Where ‘0’ for the x  element in column matrix implies the direction is i.

c) By the analog of the parallel axis theorem:

I _o = I_G -   M(R² I – RR) 

D I =  I _o -  I_G   =    M(R² I – RR)

RR =   

And:

D I =   

Finally:  D I =