Question 11.10: Find the moments of inertia of the rod in Example 11.5 (Fig....
Find the moments of inertia of the rod in Example 11.5 (Fig. 11.15) about its center of mass G.
From Example 11.5,
\left[ I _A\right]=\left[\begin{array}{ccc}\frac{1}{3} m\left(b^2+c^2\right) & -\frac{1}{3} m a b & -\frac{1}{3} m a c \\-\frac{1}{3} m a b & \frac{1}{3} m\left(a^2+c^2\right) & -\frac{1}{3} m b c \\-\frac{1}{3} m a c & -\frac{1}{3} m b c & \frac{1}{3} m\left(a^2+b^2\right)\end{array}\right]
Using Eq. (11.62)_1 and noting the coordinates of the center of mass in Fig. 11.15,
I_{G_x}=I_{A_x}-m\left[\left(y_G-0\right)^2+\left(z_G-0\right)^2\right]=\frac{1}{3} m\left(b^2+c^2\right)-m\left[\left(\frac{b}{2}\right)^2+\left(\frac{c}{2}\right)^2\right]=\frac{1}{12} m\left(b^2+c^2\right)
I_{P_x}=I_{G_x}+m\left(y_{G / P}^2+z_{G / P}^2\right) (11.62)_1
Eq. (11.62)_4 yields
I_{G_{x y}}=I_{A_{x y}}+m\left(x_G-0\right)\left(y_G-0\right)=-\frac{1}{3} m a b+m \cdot \frac{a}{2} \cdot \frac{b}{2}=-\frac{1}{12} m a b
I_{P_{x y}}=I_{G_{y y}}-m x_{G / P} y_{G / P} (11.62)_4
The remaining four moments of inertia are found in a similar fashion, so that
\left[ I _G\right]=\left[\begin{array}{ccc}\frac{1}{12} m\left(b^2+c^2\right) & -\frac{1}{2} m a b & -\frac{1}{12} m a c \\-\frac{1}{12} m a b & \frac{1}{12} m\left(a^2+c^2\right) & -\frac{1}{12} m b c \\-\frac{1}{12} m a c & -\frac{1}{12} m b c & \frac{1}{12} m\left(a^2+b^2\right)\end{array}\right] (11.63)