The mass of each of the rod segments is
m_1=2 \cdot 0.4=0.8 kg \quad m_2=2 \cdot 0.5=1 kg \quad m_3=2 \cdot 0.3=0.6 kg \quad m_4=2 \cdot 0.2=0.4 kg (a)
The total mass of the system is
m=\sum\limits_{i=1}^4 m_i=2.8 kg (b)
The coordinates of each segment’s center of mass are
\begin{array}{lll}x_{G_1}=0 & y_{G_1}=0 & z_{G_1}=0.2 m \\x_{G_2}=0 & y_{G_2}=0.25 m & z_{G_2}=0.2 m \\x_{G_3}=0.15 m & y_{G_3}=0.5 m & z_{G_3}=0 \\x_{G_4}=0.3 m & y_{G_4}=0.4 m & z_{G_4}=0\end{array} (c)
If the slender rod in Figure 9.15 is aligned with, say, the x-axis, then a=\ell \text { and } b=c=0, so that according to Eqn (9.63),
\left[ I _G\right]=\left[\begin{array}{ccc}\frac{1}{12} m\left(b^2+c^2\right) & -\frac{1}{12} 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] (9.63)
\left[ I _G\right]=\left[\begin{array}{ccc}0 & 0 & 0 \\0 & \frac{1}{12} m \ell^2 & 0 \\0 & 0 & \frac{1}{12} m \ell^2\end{array}\right]
That is, the moment of inertia of a slender rod about the axes normal to the rod at its center of mass is \frac{1}{12} m \ell^2, where m and \ell are the mass and length of the rod, respectively. Since the mass of a slender bar is assumed to be concentrated along the axis of the bar (its cross-sectional dimensions are infinitesimal), the moment of inertia about the centerline is zero. By symmetry, the products of inertia about the axes through the center of mass are all zero. Using this information and the parallel axis theorem, we find the moments and products of inertia of each rod segment about the origin O of the xyz system as follows:
Rod 1:
\begin{array}{ll}\left.I_x^{(1)}=I_{G_1}^{(1)}\right)_x+m_1\left(y_{G_1}^2+z_{G_1}^2\right) & =\left(\frac{1}{12} \cdot 0.8 \cdot 0.4^2\right)+\left[0.8\left(0+0.2^2\right)\right]=0.04267 kg – m ^2 \\\left.I_y^{(1)}=I_{G_1}^{(1)}\right)_y+m_1\left(x_{G_1}^2+z_{G_1}^2\right) & =\left(\frac{1}{12} \cdot 0.8 \cdot 0.4^2\right)+\left[0.8\left(0+0.2^2\right)\right]=0.04267 kg – m ^2 \\\left.I_z^{(1)}=I_{G_1}^{(1)}\right)_z+m_1\left(x_{G_1}^2+y_{G_1}^2\right) & =0+[0.8(0+0)]=0 \\\left.I_{x y}^{(1)}=I_{G_1}^{(1)}\right)_{x y}-m_1 x_{G_1} y_{G_1} & =0-[0.8(0)(0)]=0 \\\left.I_{x z}^{(1)}=I_{G_1}^{(1)}\right)_{x z}-m_1 x_{G_1} z_{G_1} & =0-[0.8(0)(0.2)]=0 \\\left.I_{y z}^{(1)}=I_{G_1}^{(1)}\right)_{y z}-m_1 y_{G_1} z_{G_1} & =0-[0.8(0)(0)]=0\end{array}
Rod 2:
\begin{array}{ll}\left.I_x^{(2)}=I_{G_2}^{(2)}\right)_x+m_2\left(y_{G_2}^2+z_{G_2}^2\right) & =\left(\frac{1}{12} \cdot 1.0 \cdot 0.5^2\right)+\left[1.0\left(0+0.25^2\right)\right]=0.08333 kg – m ^2 \\\left.I_y^{(2)}=I_{G_2}^{(2)}\right)_y+m_2\left(x_{G_2}^2+z_{G_2}^2\right) & =0+[1.0(0+0)]=0 \\\left.I_z^{(2)}=I_{G_2}^{(2)}\right)_z+m_2\left(x_{G_2}^2+y_{G_2}^2\right) & =\left(\frac{1}{12} \cdot 1.0 \cdot 0.5^2\right)+\left[1.0\left(0+0.5^2\right)\right]=0.08333 kg – m ^2 \\\left.I_{x y}^{(2)}=I_{G z}^{(2)}\right)_{x y}-m_2 x_{G_2} y_{G_2} & =0-[1.0(0)(0.5)]=0 \\\left.I_{x z}^{(2)}=I_{G_2}^{(2)}\right)_{x z}-m_2 x_{G_2} z_{G_2} & =0-[1.0(0)(0)]=0 \\\left.I_{y z}^{(2)}=I_{G_2}^{(2)}\right)_{y z}-m_2 y_{G_2} z_{G_2} & =0-[1.0(0.5)(0)]=0\end{array}
Rod 3:
\begin{array}{ll}\left.I_x^{(3)}=I_{G_3}^{(3)}\right)_x+m_3\left(y_{G_3}^2+z_{G_3}^2\right) & =0+\left[0.6\left(0.5^2+0\right)\right]=0.15 kg – m ^2 \\\left.I_y^{(3)}=I_{G_3}^{(3)}\right)_y+m_3\left(x_{G_3}^2+z_{G_3}^2\right) & =\left(\frac{1}{12} \cdot 0.6 \cdot 0.3^2\right)+\left[0.6\left(0.15^2+0\right)\right]=0.018 kg – m ^2 \\\left.I_z^{(3)}=I_{G_3}^{(3)}\right)_z+m_3\left(x_{G_3}^2+y_{G_3}^2\right) & =\left(\frac{1}{12} \cdot 0.6 \cdot 0.3^2\right)+\left[0.6\left(0.15^2+0.5^2\right)\right]=0.1680 kg – m ^2 \\\left.I_{x y}^{(3)}=I_{G_3}^{(3)}\right)_{x y}-m_3 x_{G_3} y_{G_3} & =0-[0.6(0.15)(0.5)]=-0.045 kg – m ^2 \\\left.I_{x z}^{(3)}=I_{G_3}^{(3)}\right)_{x z}-m_3 x_{G_3} z_{G_3} & =0-[0.6(0.15)(0)]=0 \\\left.I_{y z}^{(3)}=I_{G_3}^{(3)}\right)_{y z}-m_3 y_{G_3} z_{G_3} & =0-[0.6(0.5)(0)]=0\end{array}
Rod 4:
\begin{array}{ll}\left.I_x^{(4)}=I_{G_4}^{(4)}\right)_x+m_4\left(y_{G_4}^2+z_{G_4}^2\right) & =\left(\frac{1}{12} \cdot 0.4 \cdot 0.2^2\right)+\left[0.4\left(0.4^2+0\right)\right]=0.06533 kg – m ^2 \\\left.I_y^{(4)}=I_{G_4}^{(4)}\right)_y+m_4\left(x_{G_4}^2+z_{G_4}^2\right) & =0+\left[0.4\left(0.3^2+0\right)\right]=0.0360 kg – m ^2 \\\left.I_z^{(4)}=I_{G_4}^{(4)}\right)_z+m_4\left(x_{G_4}^2+y_{G_4}^2\right) & =\left(\frac{1}{12} \cdot 0.4 \cdot 0.2^2\right)+\left[0.4\left(0.3^2+0.4^2\right)\right]=0.1013 kg – m ^2 \\\left.I_{x y}^{(4)}=I_{G_4}^{(4)}\right)_{x y}-m_4 x_{G_4} y_{G_4} & =0-[0.4(0.3)(0.4)]=-0.0480 kg – m ^2 \\\left.I_{x z}^{(4)}=I_{G_4}^{(4)}\right)_{x z}-m_4 x_{G_4} z_{G_4} & =0-[0.4(0.3)(0)]=0 \\\left.I_{y z}^{(4)}=I_{G_4}^{(4)}\right)_{y z}-m_4 y_{G_4} z_{G_4} & =0-[0.4(0.4)(0)]=0\end{array}
The total moments of inertia for all the four rods about O are
\begin{array}{lll}I_x=\sum\limits_{i=1}^4 I_x^{(i)}=0.3413 kg – m ^2 & I_y=\sum\limits\limits_{i=1}^4 I_y^{(i)}=0.09667 kg – m ^2 & I_z=\sum\limits_{i=1}^4 I_z^{(i)}=0.3527 kg – m ^2 \\\\I_{x y}=\sum\limits_{i=1}^4 I_{x y}^{(i)}=-0.0930 kg – m ^2 & I_{x z}=\sum\limits_{i=1}^4 I_{x z}^{(i)}=0 & I_{y z}=\sum\limits_{i=1}^4 I_{y z}^{(i)}=0\end{array} (d)
The coordinates of the center of mass of the system of four rods are, from Eqns (a)-(c),
\begin{aligned}& x_G=\cfrac{1}{m} \sum\limits_{i=1}^4 m_i x_{G_i}=\cfrac{1}{2.8} \cdot 0.21=0.075 m \\& y_G=\cfrac{1}{m} \sum\limits_{i=1}^4 m_i y_{G_i}=\cfrac{1}{2.8} \cdot 0.71=0.2536 m \\& z_G=\cfrac{1}{m} \sum\limits_{i=1}^4 m_i z_{G_i}=\cfrac{1}{2.8} \cdot 0.16=0.05714 m\end{aligned} (e)
We use the parallel axis theorems to shift the moments of inertia in Eqn (d) to the center of mass G of the system.
\begin{array}{ll}I_{G_x}=I_x-m\left(y_G^2+z_G^2\right) & =0.3413-0.1892=0.1522 kg – m ^2 \\\\I_{G_y}=I_y-m\left(x_G^2+z_G^2\right) & =0.09667-0.02489=0.07177 kg – m ^2 \\\\I_{G_z}=I_z-m\left(x_G^2+y_G^2\right) & =0.3527-0.1958=0.1569 kg – m ^2 \\\\I_{G_{x y}}=I_{x y}+m x_G y_G & =-0.093+0.05325=-0.03975 kg – m ^2 \\\\I_{G_{x z}}=I_{x z}+m x_G z_G & =0+0.012=0.012 kg – m ^2 \\\\I_{G_{y z}}=I_{y z}+m y_G z_G & =0+0.04057=0.04057 kg – m ^2\end{array}
Therefore, the inertia tensor, relative to the center of mass, is
[ I ]=\left[\begin{array}{ccc}I_{G_x} & I_{G_{x y}} & I_{G_{x z}} \\I_{G_{x y}} & I_{G_y} & I_{G_{y z}} \\I_{G_{x z}} & I_{G_{y z}} & I_{G_z}\end{array}\right]=\left[\begin{array}{ccc}0.1522 & -0.03975 & 0.012 \\-0.03975 & 0.07177 & 0.04057 \\0.012 & 0.04057 & 0.1569\end{array}\right] \left( kg \cdot m ^2\right) (f)
