Question 11.9: Find the principal moments of inertia and the principal axes...

We seek the nontrivial solutions of the system [I]{e} = λ{e}. 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 Eq. (11.54),

J_1 = 100 + 300 + 500 = 900

J_2=\left|\begin{array}{rr}100 & -20 \\-20 & 300\end{array}\right|+\left|\begin{array}{rr}100 & -100 \\-100 & 500\end{array}\right|+\left|\begin{array}{rr}300 & -50 \\-50 & 500\end{array}\right|=217,100                         (b)

J_3=\left|\begin{array}{rrr}100 & -20 & -100 \\-20 & 300 & -50 \\-100 & -50 & 500\end{array}\right|=11,350,000

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

\lambda_1=532.052 \quad \lambda_2=295.840 \quad \lambda_3=72.1083  \left( kg \cdot m ^2\right)               (d)

We substitute each of these, in turn, back into Eq. (a) to find its corresponding principal direction.
Substituting \lambda_1=532.052  kg \cdot m ^2 into Eq. (a) we obtain

Since the determinant of the coefficient matrix is zero, at most two of the three equations in Eq. (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)} and e_z^{(1)} using any two of the independent equations in Eq. (e). With e_x^{(1)} = 1, the first two of Eq. (e) become

-20.0000 e_y^{(1)}-100.000 e_z^{(1)}=432.052

-232.052 e_y^{(1)}-50.000 e_z^{(1)}=20.0000                         (f)

Solving these two equations for e_y^{(1)} 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

\hat{ e }_1=\frac{ e ^{(1)}}{\left\| e ^{(1)}\right\|}=\frac{1.00000 \hat{ i }+0.882793 \hat{ j }-4.49708 \hat{ k }}{\sqrt{1.00000^2+0.882793^2+(-4.49708)^2}}

or

\hat{ e }_1=0.213186 \hat{ i }+0.188199 \hat{ j }-0.958714 \hat{ k }  \left(\lambda_1=532.052  kg \cdot m ^2\right)                    (h)

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

\hat{ e }_2=0.17632 \hat{ i }-0.972512 \hat{ j }-0.151609 \hat{ k }  \left(\lambda_2=295.840  kg \cdot m ^2\right)                        (i)

The two unit vectors \hat{ e }_1 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 Eq. (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 Eqs. (h) and (i) into this cross product, we find that

\hat{ e }_3=-0.960894 \hat{ i }-0.137114 \hat{ j }-0.240587 \hat{ k }  \left(\lambda_3=72.1083  kg \cdot m ^2\right)                            (j)

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

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

[ Q ]=\left[\begin{array}{rrr}0.213186 & 0.188199 & -0.958714 \\0.176732 & -0.972512 & -0.151609 \\-0.960894 & -0.137114 & -0.240587\end{array}\right]                      (l)

Indeed, if we apply the transformation in Eq. (11.49), \left[ I ^{\prime}\right]=[ Q ][ I ][ Q ]^T we find

=\left[\begin{array}{ccc}532.052 & 0 & 0 \\0 & 295.840 & 0 \\0 & 0 & 72.1083\end{array}\right]  \left( kg \cdot m ^2\right)