Chapter 9.6, Problem 9.181P

(a)

To determine

Find the principal mass moment of inertia at the origin.

Answer to Problem 9.181P

The principal mass moment of inertia at the origin is $\underset{\_}{K_1=414\times {10}^{-3}\text{kg}\cdot {\text{m}}^2,K_2=29.8\times {10}^{-3}\text{kg}\cdot {\text{m}}^2,\text{and}{K}_3=32.3\times {10}^{-3}\text{kg}\cdot {\text{m}}^2}$.

Explanation of Solution

Calculation:

Refer to problem 9.145 and 9.149.

$\begin{array}{l}{I}_x=26.4325\times {10}^{-3}\text{kg}\cdot {\text{m}}^2\\ I_y=31.1726\times {10}^{-3}\text{kg}\cdot {\text{m}}^2\\ I_z=8.5773\times {10}^{-3}\text{kg}\cdot {\text{m}}^2\end{array}$

$\begin{array}{l}{I}_{xy}=2.5002\times {10}^{-3}\text{kg}\cdot {\text{m}}^2\\ I_{yz}=4.0627\times {10}^{-3}\text{kg}\cdot {\text{m}}^2\\ I_{zx}=8.8062\times {10}^{-3}\text{kg}\cdot {\text{m}}^2\end{array}$

Substitute the values of $26.4325\times {10}^{-3}\text{kg}\cdot {\text{m}}^2$ for $I_x$, $31.1726\times {10}^{-3}\text{kg}\cdot {\text{m}}^2$ for $I_y$, $8.5773\times {10}^{-3}\text{kg}\cdot {\text{m}}^2$ for $I_z$, $2.5002\times {10}^{-3}\text{kg}\cdot {\text{m}}^2$ for $I_{xy}$, $4.0627\times {10}^{-3}\text{kg}\cdot {\text{m}}^2$ for $I_{yz}$ and $8.8062\times {10}^{-3}\text{kg}\cdot {\text{m}}^2$ for $I_{zx}$ into Equation 9.56.

$\begin{array}{l}\left\{\begin{array}{l}{K}^3-\left[\left(26.4325+31.1726+8.5773\right){10}^{-3}\right]K^2+\\ \left[26.4325\left(31.1726\right)+\left(31.1726\right)\left(8.5773\right)+\left(8.5773\right)26.4325\right]\\ -{\left(2.5002\right)}^2-{\left(4.0627\right)}^2-{\left(8.8062\right)}^2\left({10}^{-6}\right)K-\\ \left[\begin{array}{l}\left(26.4325\right)\left(31.1726\right)\left(8.5773\right)-\left(26.4325\right){\left(4.0627\right)}^2-\\ 31.1726{\left(8.8062\right)}^2-\left(8.5773\right){\left(2.5002\right)}^2-2\left(2.5002\right)\left(4.0627\right)\left(8.8062\right){10}^{-9}\end{array}\right]\end{array}\right\}=0\\ K^3-\left(66.1824\times {10}^{-3}\right)K^2+\left(1217.76\times {10}^{-6}\right)K-\left(3981.23\times {10}^{-9}\right)=0\end{array}$

Solve the above Equation.

$\begin{array}{l}{K}_1=4.1443\times {10}^{-3}\text{kg}\cdot {\text{m}}^2\\ K_2=29.7840\times {10}^{-3}\text{kg}\cdot {\text{m}}^2\\ K_3=32.2541\times {10}^{-3}\text{kg}\cdot {\text{m}}^2\end{array}$

Thus, the principal mass moment of inertia at the origin is $\underset{\_}{K_1=414\times {10}^{-3}\text{kg}\cdot {\text{m}}^2,K_2=29.8\times {10}^{-3}\text{kg}\cdot {\text{m}}^2,\text{and}{K}_3=32.3\times {10}^{-3}\text{kg}\cdot {\text{m}}^2}$.

(b)

To determine

Find the principal axis of inertia at the origin.

Answer to Problem 9.181P

The principal axis of inertia at the origin for $K_1$ is $\underset{\_}{{\left({\theta }_x\right)}_1=67.8°,{\left({\theta }_y\right)}_1=80.1°,\text{and}{\left({\theta }_z\right)}_1=24.6°}$.

The principal axis of inertia at the origin for $K_2$ is $\underset{\_}{{\left({\theta }_x\right)}_2=31.6°,{\left({\theta }_y\right)}_2=71.4°,\text{and}{\left({\theta }_z\right)}_2=114.5°}$.

The principal axis of inertia at the origin for $K_3$ is $\underset{\_}{{\left({\theta }_x\right)}_3=111.2°,{\left({\theta }_y\right)}_3=21.2°,\text{and}{\left({\theta }_z\right)}_3=91.5°}$.

Explanation of Solution

Calculation:

Use Equation 9.54 and 9.57 to find the direction cosines ${\lambda }_x,{\lambda }_y,{\lambda }_z$ of each principal axis.

For $K_1$:

Use Equation 9.54a and 9.54b.

$\begin{array}{l}\left(I_x-K_1\right){\left({\lambda }_x\right)}_1-I_{xy}{\left(\lambda \right)}_1-I_{zx}{\left({\lambda }_z\right)}_1=0\\ I_{xy}{\left({\lambda }_x\right)}_1+\left(I_y-K_1\right){\left(\lambda \right)}_1-I_{yz}{\left({\lambda }_z\right)}_1=0\end{array}$

Substitute $4.1443\times {10}^{-3}\text{kg}\cdot {\text{m}}^2$ for $K_1$, $29.7840\times {10}^{-3}\text{kg}\cdot {\text{m}}^2$ for $K_2$, $32.2541\times {10}^{-3}\text{kg}\cdot {\text{m}}^2$ for $K_3$, $26.4325\times {10}^{-3}\text{kg}\cdot {\text{m}}^2$ for $I_x$, $31.1726\times {10}^{-3}\text{kg}\cdot {\text{m}}^2$ for $I_y$, $2.5002\times {10}^{-3}\text{kg}\cdot {\text{m}}^2$ for $I_{xy}$, $4.0627\times {10}^{-3}\text{kg}\cdot {\text{m}}^2$ for $I_{yz}$, and $8.8062\times {10}^{-3}\text{kg}\cdot {\text{m}}^2$ for $I_{zx}$.

$\begin{array}{l}\left[\left(26.4325-4.1443\right)\left({10}^{-3}\right)\right]{\left({\lambda }_x\right)}_1-\left(2.5002\times {10}^{-3}\right){\left({\lambda }_y\right)}_1-\left(8.8062\times {10}^{-3}\right){\left({\lambda }_z\right)}_1=0\\ -\left(2.5002\times {10}^{-3}\right){\left({\lambda }_x\right)}_1+\left[31.1726-4.1443\right]\left({10}^{-3}\right)\left[{\left({\lambda }_y\right)}_1-\left(4.0627\times {10}^{-3}\right)\right]{\left({\lambda }_z\right)}_1=0\end{array}$

Simplifying,

$8.9146{\left({\lambda }_x\right)}_1-{\left({\lambda }_y\right)}_1-3.5222{\left({\lambda }_z\right)}_1=0$ (1)

$-0.0925{\left({\lambda }_x\right)}_1-{\left({\lambda }_y\right)}_1-0.1503{\left({\lambda }_z\right)}_1=0$ (2)

Solving the Equations (1) and (2) for ${\left({\lambda }_z\right)}_1$.

${\left({\lambda }_z\right)}_1=2.4022{\left({\lambda }_x\right)}_1$

Substitute into $2.4022{\left({\lambda }_x\right)}_1$ for ${\left({\lambda }_z\right)}_1$ in Equation (1).

$\begin{array}{c}{\left({\lambda }_y\right)}_1=\left[8.9146-3.5222\left(2.4022\right){\left({\lambda }_x\right)}_1\right]\\ =0.45357{\left({\lambda }_x\right)}_1\end{array}$

Substitute into Equation 9.57:

$\begin{array}{l}{\left({\lambda }_x\right)}_1^2+{\left[0.45357{\left({\lambda }_x\right)}_1\right]}^2+{\left[2.4022{\left({\lambda }_x\right)}_1\right]}^2=1\\ {\left({\lambda }_x\right)}_1=0.37861\\ {\left({\lambda }_z\right)}_1=0.90950\\ {\left({\lambda }_x\right)}_1=0.17173\end{array}$

$\begin{array}{l}{\left({\theta }_x\right)}_1=67.8°\\ {\left({\theta }_y\right)}_1=80.1°\\ {\left({\theta }_z\right)}_1=24.6°\end{array}$

Thus, the principal axis of inertia at the origin for $K_1$ is $\underset{\_}{{\left({\theta }_x\right)}_1=67.8°,{\left({\theta }_y\right)}_1=80.1°,\text{and}{\left({\theta }_z\right)}_1=24.6°}$.

For $K_2$:

Use Equation 9.54a and 9.54b.

$\begin{array}{l}\left(I_x-K_2\right){\left({\lambda }_x\right)}_2-I_{xy}{\left({\lambda }_y\right)}_2-I_{zx}{\left({\lambda }_z\right)}_2=0\\ -I_{xy}{\left({\lambda }_x\right)}_2+\left(I_y-K_2\right){\left({\lambda }_y\right)}_2-I_{yz}{\left({\lambda }_z\right)}_2=0\end{array}$

Substitute $4.1443\times {10}^{-3}\text{kg}\cdot {\text{m}}^2$ for $K_1$, $29.7840\times {10}^{-3}\text{kg}\cdot {\text{m}}^2$ for $K_2$, $32.2541\times {10}^{-3}\text{kg}\cdot {\text{m}}^2$ for $K_3$, $26.4325\times {10}^{-3}\text{kg}\cdot {\text{m}}^2$ for $I_x$, $31.1726\times {10}^{-3}\text{kg}\cdot {\text{m}}^2$ for $I_y$, $2.5002\times {10}^{-3}\text{kg}\cdot {\text{m}}^2$ for $I_{xy}$, $4.0627\times {10}^{-3}\text{kg}\cdot {\text{m}}^2$ for $I_{yz}$, and $8.8062\times {10}^{-3}\text{kg}\cdot {\text{m}}^2$ for $I_{zx}$.

$\begin{array}{l}\left[26.4325-29.7840\times {10}^{-3}\right]\left[{\left({\lambda }_x\right)}_2-\left(2.5002\times {10}^{-3}\right){\left({\lambda }_y\right)}_2-\left(8.8062\times {10}^{-3}\right){\left({\lambda }_z\right)}_2\right]=0\\ -\left(2.5002\times {10}^{-3}\right){\left({\lambda }_x\right)}_2+\left[\left(31.1726-29.7840\right)\left({10}^{-3}\right)\right]{\left({\lambda }_y\right)}_2-\left(4.0627\times {10}^{-3}\right){\left({\lambda }_z\right)}_2=0\end{array}$

Simplifying

$-1.3405{\left({\lambda }_x\right)}_2-{\left({\lambda }_y\right)}_2-3.5222{\left({\lambda }_z\right)}_2=0$ (3)

$-1.8005{\left({\lambda }_x\right)}_2+{\left({\lambda }_y\right)}_2-2.9258{\left({\lambda }_z\right)}_2=0$ (4)

Solving Equations (3) and (4) for ${\left({\lambda }_z\right)}_2$:

${\left({\lambda }_z\right)}_2=-0.48713{\left({\lambda }_x\right)}_2$

Substitute $-0.48713{\left({\lambda }_x\right)}_2$ for ${\left({\lambda }_z\right)}_2$ in Equation (3).

$\begin{array}{c}{\left({\lambda }_y\right)}_2=\left[-1.3405-3.5222\left(-0.48713\right)\right]{\left({\lambda }_x\right)}_2\\ =0.37527{\left({\lambda }_x\right)}_2\end{array}$

Substitute into Equation 9.57:

$\begin{array}{l}{\left({\lambda }_x\right)}_2^2+{\left[0.37527{\left({\lambda }_x\right)}_2\right]}^2+{\left[-0.48713{\left({\lambda }_x\right)}_2\right]}^2=1\\ {\left({\lambda }_x\right)}_2=0.85184\\ {\left({\lambda }_z\right)}_2=-0.41496\\ {\left({\lambda }_y\right)}_2=0.31967\end{array}$

$\begin{array}{l}{\left({\theta }_x\right)}_2=31.6°\\ {\left({\theta }_y\right)}_2=71.4°\\ {\left({\theta }_z\right)}_2=114.5°\end{array}$

Thus, the principal axis of inertia at the origin for $K_2$ is $\underset{\_}{{\left({\theta }_x\right)}_2=31.6°,{\left({\theta }_y\right)}_2=71.4°,\text{and}{\left({\theta }_z\right)}_2=114.5°}$.

For $K_3$:

Use Equation 9.54a and 9.54b.

$\begin{array}{l}\left(I_x-K_3\right){\left({\lambda }_x\right)}_3-I_{xy}{\left({\lambda }_y\right)}_3-I_{zx}{\left({\lambda }_z\right)}_3=0\\ -I_{xy}{\left({\lambda }_x\right)}_3+\left(I_y-K_3\right){\left({\lambda }_y\right)}_3-I_{yz}{\left({\lambda }_z\right)}_3=0\end{array}$

Substitute $4.1443\times {10}^{-3}\text{kg}\cdot {\text{m}}^2$ for $K_1$, $29.7840\times {10}^{-3}\text{kg}\cdot {\text{m}}^2$ for $K_2$, $32.2541\times {10}^{-3}\text{kg}\cdot {\text{m}}^2$ for $K_3$, $26.4325\times {10}^{-3}\text{kg}\cdot {\text{m}}^2$ for $I_x$, $31.1726\times {10}^{-3}\text{kg}\cdot {\text{m}}^2$ for $I_y$, $2.5002\times {10}^{-3}\text{kg}\cdot {\text{m}}^2$ for $I_{xy}$, $4.0627\times {10}^{-3}\text{kg}\cdot {\text{m}}^2$ for $I_{yz}$, and $8.8062\times {10}^{-3}\text{kg}\cdot {\text{m}}^2$ for $I_{zx}$.

$\begin{array}{l}\left[\left(26.4325-32.2541\right){10}^{-3}\right]{\left({\lambda }_x\right)}_3-\left(2.5002\times {10}^{-3}\right){\left({\lambda }_y\right)}_3-\left(8.8062\times {10}^{-3}\right){\left({\lambda }_z\right)}_3=0\\ \left[\begin{array}{l}-\left(2.5002\times {10}^{-3}\right){\left({\lambda }_x\right)}_3+\left[31.1726-32.2541\left({10}^{-3}\right)\right]\left({\lambda }_3\right)+\\ \left[31.1726-32.2541\left({10}^{-3}\right)\right]\left[{\left({\lambda }_y\right)}_3-\left(4.0627\times {10}^{-3}\right){\left({\lambda }_z\right)}_3\right]=0\end{array}\right]\end{array}$

Simplifying

$-2.3285{\left({\lambda }_x\right)}_3-{\left({\lambda }_y\right)}_3-3.5222{\left({\lambda }_z\right)}_3=0$ (5)

$-2.3118{\left({\lambda }_x\right)}_3-{\left({\lambda }_y\right)}_3-3.7565{\left({\lambda }_z\right)}_3=0$ (6)

Solving Equations (5) and (6) for ${\left({\lambda }_z\right)}_3$.

${\left({\lambda }_z\right)}_3=0.071276{\left({\lambda }_x\right)}_3$

Substitute $0.071276{\left({\lambda }_x\right)}_3$ for ${\left({\lambda }_z\right)}_3$ in Equation (5).

$\begin{array}{c}{\left({\lambda }_y\right)}_3=\left[-2.3285-3.5222\left(0.071276\right){\left({\lambda }_x\right)}_3\right]\\ =-2.5795{\left({\lambda }_x\right)}_3\end{array}$

Substitute into Equation 9.57:

${\left({\lambda }_x\right)}_3^2+{\left[-2.5795{\left({\lambda }_x\right)}_3\right]}^2+{\left[0.07127{\left({\lambda }_x\right)}_3\right]}^2=1$

$\begin{array}{l}{\left({\lambda }_x\right)}_3=0.36134\\ {\left({\lambda }_z\right)}_3=-0.025755\\ {\left({\lambda }_y\right)}_3=0.93208\end{array}$

$\begin{array}{l}{\left({\theta }_x\right)}_3=68.8°\\ {\left({\theta }_y\right)}_3=158.8°\\ {\left({\theta }_z\right)}_3=88.5°\end{array}$

Find the direction cosines corresponding to the labelled axis, take the negative root of ${\left({\lambda }_x\right)}_3$.

${\left({\lambda }_x\right)}_3=-0.36134$

$\begin{array}{l}{\left({\theta }_x\right)}_3=111.2°\\ {\left({\theta }_y\right)}_3=21.2°\\ {\left({\theta }_z\right)}_3=91.5°\end{array}$

Thus, the principal axis of inertia at the origin is $\underset{\_}{{\left({\theta }_x\right)}_3=111.2°,{\left({\theta }_y\right)}_3=21.2°,\text{and}{\left({\theta }_z\right)}_3=91.5°}$.

Sketch the orientation of the principal axis to the $x,y,z$ axis as shown in Figure 1.

Refer to Figure 1.

Principal axis 3 has been labelled so that the principle axes form a right handed set.