Question 9.9: Find the principal moments of inertia and the principal axes......

We seek the nontrivial solutions of the system [ I ]\{ e \}=\lambda\{ e \} \text {, that is, }

\left[\begin{array}{ccc}100-\lambda & -20 & -100 \\-20 & 300-\lambda & -50 \\-100 & -50 & 500-\lambda\end{array}\right]\left\{\begin{array}{l}e_x \\e_y \\e_z\end{array}\right\}=\left\{\begin{array}{l}0 \\0 \\0\end{array}\right\}               (a)

From Eqn (9.54),

Thus, the characteristic equation is

\lambda^3-900 \lambda^2+217,100 \lambda-11,350,000=0                              (c)

The three roots are the principal moments of inertia, which are found to be

\boxed{\lambda_1=532.052 \quad \lambda_2=295.840 \quad \lambda_3=72.1083}                         (d)

Each of these is substituted, in turn, back into Eqn (a) to find its corresponding principal direction.
Substituting \lambda_1=532.052  kg \cdot m ^2 into Eqn (a) we obtain

Since the determinant of the coefficient matrix is zero, at most two of the three equations in Eqn (e) are independent. Thus, at most, two of the three components of the vector e ^{(1)} can be found in terms of the third. We can therefore arbitrarily set e_x^{(1)}=1 and solve for e_y^{(1)} \text { and } e_z^{(1)} using any two of the independent equations in Eqn (e).With e_x^{(1)}=1, the first two of Eqn (e) become

Solving these two equations for e_y^{(1)} \text { and } e_z^{(1)} yields, together with the assumption that e_x^{(1)}=1,

e_x^{(1)}=1.00000 \quad e_y^{(1)}=0.882793 \quad e_z^{(1)}=-4.49708                              (g)

The unit vector in the direction of e^{(1)} is

or

Substituting \lambda_2=295.840  kg \cdot m ^2 into Eqn (a) and proceeding as above we find that

The two unit vectors \hat{ e }_1 \text { and } \hat{ e }_2 define two of the three principal directions of the inertia tensor. Observe that \hat{ e }_1 \cdot \hat{ e }_2=0, as must be the case for symmetric matrices.
To obtain the third principal direction \hat{ e }_3, we can substitute \lambda_3=72.1083  kg \cdot m ^2 into Eqn (a) and proceed as above. However, since the inertia tensor is symmetric, we know that the three principal directions are mutually orthogonal, which means \hat{ e }_3=\hat{ e }_1 \times \hat{ e }_2. Substituting Eqns (h) and (i) into the crossproduct, we find that

We can check our work by substituting \lambda_3 \text { and } \hat{ e }_3 into Eqn (a) and verify that it is indeed satisfied:

The components of the vectors \hat{ e }_1, \hat{ e }_2, \text { and } \hat{ e }_3 define the three rows of the orthogonal transformation [Q] from the xyz system into the \text{x}^{\prime} \text{y}^{\prime} \text{z}^{\prime} system aligned along the three principal directions:

Indeed, if we apply the transformation in Eqn (9.49), \left[ I ^{\prime}\right]=[ Q ][ I ][ Q ]^T, we find