Continued from earlier instalment:
3.Diagonalizing
tensors.
This is analogous to obtaining the eigenvalues for a matrix in linear algebra.
Consider the object below for which we want the principal axis.
_{}^{}
Fig. 3. We wish to find
the principal axis, a _{i j}.
And
we have a _{i j} =
With T’
= A × I × A^{t}
Where
t denotes the transpose. Then we obtain,
T’ =
(15 …..0……..0)
(0…….11 ....3Ö2)
(0……..3Ö2…..8 )
Which
is to be diagonalized. Writing this out:
(15  l…...0…….....0)
(0……..11 l ......3Ö2)
(0……..3Ö2…....8 l)
This leads to a cubic  i.e. with triple roots  which are:
l1 = 15, l2
= 5, and l3 = 14
Substituting l1 in the matrix we get:
(0 ….....0……......0) ( c _{x} )
(0……....4 ......(3Ö2)) ( c _{y} )
(0……..(3Ö2)…..(7) ) ( c _{z} )
For
which the separate equations, e.g. in c
_{x},
c
_{y}
and c _{z} can be solved. For example,
4 c _{y}  3Ö2 c _{z} = 0
 3Ö2 c _{y }7 c _{z } = 0
After
working through all the solutions, we obtain:
C =  Ö (2/3) e_{2}^ + 1/
Ö3 (e_{3}^)
Example 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) (l^{2} – 2l) = 0
Yielding
eigenvalues: l= 0, l = 2
Then:
T = Kl, so:
T1 = 0, T2 = K, and T3 =2K
Or: (0, K, 2K)
b) We have to take: (I –
T1)C
So
that:
K [(1….0…..0)
[(0….1…..1)
[(0….1… ..1)

K (1….0…..0)](x)
(0….1….. 0)] (y)
(0….0… ..1)] (z)
=
(0….0…..0)
(x)
(0….0…..1) (y)
(0….1… ..0) (z) = 0
So
that: 0
=
(0)
(z)
(y)
Which 0 for the x component implies the answer is i.
c)
By the analog of the parallel axis theorem:
I
_{o }= I_{G}  M(R^{2} I – RR)
D I = I _{o } I_{G} =
M(R^{2} I – RR)
RR
= r _{o} ^{2 }x
(0….0………0)
(0….1/2…..1/2)
(0….1/2…
..1/2)
D I = M r _{o}
^{2 }x
[(1….0…..0)
[(0….1…..0)
[(0….0… ..1)

(0….0……..0)]
(0….1/2.. 1/2)]
(0….1/2… ..1/2)]
=
M
r _{o} ^{2 } x
(1….0………..0)
(0….1/2…..1/2)
(0….1/2… ..1/2)
Problems:
1) For the example problem given, use the remaining two eigenvalues, l2 and l3,
to diagonaliize the matrix and obtain solutions in c _{x}, c
_{y}
and c _{z}.
2)
Thereby obtain the principal axes in terms of: e^’_{x}, e ^’_{y} and e^’
_{z}
.
3) Show
in particular that:
e^’ _{z} =  1/ Ö3 e1^  1/ Ö3 e2^
+ 1/ Ö3 e3^
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