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Inertia, Second Moment Vectors, Moments and Products of Inertia, Inertia Dyadics 211
Interestingly, these coefficients may be shown (see Reference 7.4) to be independent of the
choice of unit vectors n in which I is expressed. As a consequence, I , I , and I are
II
i
III
I
sometimes called invariants of I.
Equation (7.7.12) is a cubic equation for λ which in general has three distinct roots: λ ,
1
λ , and λ . This implies that there are at least three solutions for each at the a , a , and a 3
2
2
1
3
of Eq. (7.7.10). To obtain these solutions, we can select one of the roots (say, λ ) and then
1
substitute λ for λ in Eq. (7.7.10). This produces three equations that are dependent because
1
λ is a solution of Eq. (7.7.11). Hence, only two of the three equations are independent.
1
However, we can obtain a third independent equation by recalling that n is a unit vector
a
and thus a , a , and a also satisfy the equation:
1
3
2
a + a + a = 1 (7.7.16)
2
2
2
1 2 3
Then, by selecting any two equations from Eq. (7.7.10) together with Eq. (7.7.16), we can
solve the resulting system of three equations and determine the values of a , a , and a .
2
3
1
Next, by letting λ be λ , we can repeat the process and find a second set of components
2
(a , a , and a ) of a principal unit vector n . Similarly, for λ = λ , we obtain a third principal
3
2
a
1
3
unit vector.
By following this procedure we obtain the components of three principal unit vectors
defining three principal directions of inertia of the system. The values λ , λ , and λ are
2
3
1
then the corresponding principal moments of inertia. This procedure and these concepts
are illustrated in the example in the following section.
7.8 Principal Axes, Principal Moments of Inertia: Example
Consider a system whose central inertia matrix relative to mutually perpendicular unit
vectors n , n , and n is:
1 2 3
I I I 92 −3 4 3 3 4
11 12 13
I = I I I = −3 4 39 8 3 8 (7.8.1)
ij [] 21 22 23
I I I 33 4 3 8 37 8
31 32 33
To determine the principal values, the principal unit vectors, and the principal directions,
we need to form Eq. (7.7.10). That is,
( 92 − ) +− ( 3 4)a 2 + 3 3 4a 3 = 0
λ a
1
− ( 34) +( 39 8 − ) ( 3 8) a = 0 (7.8.2)
+
a
λ a
1 2 3
( 33 4) +( 3 8) a 2 ( 37 8 − ) λ a 3 = 0
+
a
1
Because these equations are linear and homogeneous (right sides are zero), there is a
nontrivial solution for the a , a , and a only if the determinant of the coefficients is zero
1 2 3
(Eq. (7.7.11)). Hence,