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.

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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.

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

[latex]\begin{aligned}&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}\end{aligned}[/latex]

Substituting the values of the moments and products of inertia into Eq. (9.56) and collecting terms yields

[latex]\begin{gathered}K^{3}-\left(I_{x}+I_{y}+I_{z} ight) 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} ight) 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} ight)=0\end{gathered}[/latex] (9.56)

[latex]K^{3}-\left(\frac{28}{3} m c^{2} ight) K^{2}+\left(\frac{3479}{144} m^{2} c^{4} ight) K-\frac{589}{54} m^{3} c^{6}=0[/latex]

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.

[latex]\begin{array}{lll}K_{1}=0.568867 m c^{2} & K_{2}=4.20885 m c^{2} & K_{3}=4.55562 m c^{2} \K_{1}=0.569 m c^{2} & K_{2}=4.21 m c^{2} & K_{3}=4.56 m c^{2}\end{array}[/latex]

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 [latex]K_{1}[/latex], you have

[latex]\begin{array}{r}\left(I_{x}-K ight) \lambda_{x}-I_{x y} \lambda_{y}-I_{z x} \lambda_{z}=0 \-I_{x y} \lambda_{x}+\left(I_{y}-K ight) \lambda_{y}-I_{y z} \lambda_{z}=0 \-I_{z x} \lambda_{x}-I_{y z} \lambda_{y}+\left(I_{z}-K ight) \lambda_{z}=0\end{array}[/latex] (9.54)

[latex]\lambda_{x}^{2}+\lambda_{y}^{2}+\lambda_{z}^{2}=1[/latex] (9.57)

[latex]\begin{array}{r}\left(\frac{5}{3}-0.568867 ight) m c^{2}\left(\lambda_{x} ight)_{1}-\frac{3}{2} m c^{2}\left(\lambda_{y} ight)_{1}-\frac{3}{4} m c^{2}\left(\lambda_{z} ight)_{1}=0 \-\frac{3}{2} m c^{2}\left(\lambda_{x} ight)_{1}+\left(\frac{10}{3}-0.568867 ight) m c^{2}\left(\lambda_{y} ight)_{1}-\frac{1}{2} m c^{2}\left(\lambda_{z} ight)_{1}=0 \\left(\lambda_{x} ight)_{1}^{2}+\left(\lambda_{y} ight)_{1}^{2}+\left(\lambda_{z} ight)_{1}^{2}=1\end{array}[/latex]

Solving yields

[latex]\left(\lambda_{x} ight)_{1}=0.836600\quad\left(\lambda_{y} ight)_{1}=0.496001 \quad\left(\lambda_{z} ight)_{1}=0.232557[/latex]

The angles that the first principal axis of inertia forms with the coordinate axes are then

[latex]\left( heta_{x} ight)_{1}=33.2^{\circ}\quad\left( heta_{y} ight)_{1}=60.3^{\circ} \quad\left( heta_{z} ight)_{1}=76.6^{\circ}[/latex]

Using the same set of equations successively with [latex]K_{2}[/latex] and [latex]K_{3}[/latex], you can find that the angles associated with the second and third principal moments of inertia at the origin are, respectively,

[latex]\left( heta_{x} ight)_{2}=57.8^{\circ}\quad\left( heta_{y} ight)_{2}=146.6^{\circ} \quad\left( heta_{z} ight)_{2}=98.0^{\circ}[/latex]

and

[latex]\left( heta_{x} ight)_{3}=82.8^{\circ}\quad\left( heta_{y} ight)_{3}=76.1^{\circ} \quad\left( heta_{z} ight)_{3}=164.3^{\circ}[/latex]

Question 9.SP.15: If a = 3c and b = 2c for the rectangular prism of Sample Pro...

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