Question 9.SP.15: If a = 3c and b = 2c for the rectangular prism of Sample Pro......
If a = 3c and b = 2c for the rectangular prism of Sample Prob. 9.14, determine (a) the principal moments of inertia at the origin O, (b) the principal axes of inertia at O.
STRATEGY: Substituting the data into the results from Sample Prob. 9.14 gives you values you can use with Eq. (9.56) to determine the principal moments of inertia. You can then use these values to set up a system of equations for finding the direction cosines of the principal axes.
\begin{array}{c}K^3-\left(I_x+I_y+I_z\right) K^2+\left(I_x I_y+I_y I_z+I_z I_x-I_{x y}^2-I_{y z}^2-I_{z x}^2\right) K \\-\left(I_x I_y I_z-I_x I_{y z}^2-I_y I_{z x}^2-I_z I_{x y}^2-2 I_{x y} I_{y z} I_{z x}\right)=0\end{array} (9.56)
MODELING and ANALYSIS:
a. Principal Moments of Inertia at the Origin O. Substituting a = 3c and b = 2c into the solution to Sample Prob. 9.14 gives you
I_x=\frac{5}{3} m c^2 \quad I_y=\frac{10}{3} m c^2 \quad I_z=\frac{13}{3} m c^2
I_{x y}=\frac{3}{2} m c^2 \quad I_{y z}=\frac{1}{2} m c^2 \quad I_{z x}=\frac{3}{4} m c^2
Substituting the values of the moments and products of inertia into Eq. (9.56) and collecting terms yields
K^3-\left(\frac{28}{3} m c^2\right) K^2+\left(\frac{3479}{144} m^2 c^4\right) K-\frac{589}{54} m^3 c^6=0
Now solve for the roots of this equation; from the discussion in Sec. 9.6C, it follows that these roots are the principal moments of inertia of the body at the origin.
K_1=0.568867 m c^2 \quad K_2=4.20885 m c^2 \quad K_3=4.55562 m c^2
K_1=0.569 m c^2 \quad K_2=4.21 \mathrm{mc}^2 \quad K_3=4.56 m c^2
b. Principal Axes of Inertia at O. To determine the direction of a principal axis of inertia, first substitute the corresponding value of K into two of the equations (9.54). The resulting equations, together with Eq. (9.57), constitute a system of three equations from which you can determine the direction cosines of the corresponding principal axis. Thus, for the first principal moment of inertia K_1, you have
\begin{aligned}\left(\frac{5}{3}-0.568867\right) m c^2\left(\lambda_x\right)_1-\frac{3}{2} m c^2\left(\lambda_y\right)_1-\frac{3}{4} m c^2\left(\lambda_z\right)_1 & =0 \\-\frac{3}{2} m c^2\left(\lambda_x\right)_1+\left(\frac{10}{3}-0.568867\right) m c^2\left(\lambda_y\right)_1-\frac{1}{2} m c^2\left(\lambda_z\right)_1 & =0 \\\left(\lambda_x\right)_1^2+\left(\lambda_y\right)_1^2+\left(\lambda_z\right)_1^2 & =1\end{aligned}
Solving yields
\left(\lambda_x\right)_1=0.836600 \quad\left(\lambda_y\right)_1=0.496001 \quad\left(\lambda_z\right)_1=0.232557
The angles that the first principal axis of inertia forms with the coordinate axes are then
\left(\theta_x\right)_1=33.2^{\circ} \quad\left(\theta_y\right)_1=60.3^{\circ} \quad\left(\theta_z\right)_1=76.6^{\circ}
Using the same set of equations successively with K_2 \text { and } K_3 \text {,} you can find that the angles associated with the second and third principal moments of inertia at the origin are, respectively,
\left(\theta_x\right)_2=57.8^{\circ} \quad\left(\theta_y\right)_2=146.6^{\circ} \quad\left(\theta_z\right)_2=98.0^{\circ}
and
\left(\theta_x\right)_3=82.8^{\circ} \quad\left(\theta_y\right)_3=76.1^{\circ} \quad\left(\theta_z\right)_3=164.3^{\circ}