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Inertia, Second Moment Vectors, Moments and Products of Inertia, Inertia Dyadics 215

7.9 Principal Axes, Principal Moments of Inertia: Discussion
In view of these results, several questions arise:

1. Are principal unit vectors always mutually perpendicular?
2. What if the roots of the cubic equation Eq. (7.7.12) are not all distinct?
3. What if the roots are not real?
4. Do we always need to solve a cubic equation to obtain the principal moments
of inertia?
5. Will the procedures of the example always work?

To answer these questions consider again the definition of a principal unit vector, having
the property expressed in Eq. (7.7.5). That is, if n is a principal unit vector, then:
a
⋅
In = I n ( no sum on ) a (7.9.1)
a aa a
Similarly if n is also a principal unit vector:
b
⋅
In = I n (7.9.2)
b bb b
If we take the scalar product of the terms of Eqs. (7.9.1) and (7.9.2) with n and n and
b a
subtract the results, we obtain:

⋅⋅
⋅
⋅
⋅
⋅
nI n − n I n = I b n − I n n
b a a b aa b a aa a b (7.9.3)
aa (
= I − I bb) nn ⋅ b
a
Because the inertia dyadic I is symmetric, n · I · n and n · I · n are equal, thus the left
a b b a
side of Eq. (7.9.3) is zero. Then, if the principal moments of inertia I and I are not equal,
aa bb
we have:
⋅
nn = 0 (7.9.4)
a b
This shows that principal unit vectors of distinct principal moments of inertia are mutually
perpendicular.
Next, suppose that the principal moments of inertia are not all distinct. Suppose, for
example, that I and I are equal. Are the associated principal unit vectors still perpen-
aa bb
dicular? To address this question, consider again Eq. (7.7.12):

3
2
λ − I λ + I λ − I = 0 (7.9.5)
I II III
Recall from elementary algebra that if I , I , and I are the roots of this equation then the
aa bb cc
equation may be written in the equivalent form:
( λ − )( λ − )( λ − ) = 0 (7.9.6)
I
I
I
bb
cc
aa