Chapter B, Problem B.73P

For the component described in the problem indicated,determine (a) the principal mass moments of inertia at the origin (b) the principal axes of inertia at the origin. Sketch the body and show the orientation of the principal axes of inertia relative to the x, y, and z axes.

Prob. B.58

To determine

(a)

The principal mass moments of inertia at origin.

Answer to Problem B.73P

The principal mass moments of inertia at origin are $2.26\left(\frac{\gamma a^{4}t}{g}\right)$, $17.27\left(\frac{\gamma a^{4}t}{g}\right)$ and $19.08\left(\frac{\gamma a^{4}t}{g}\right)$.

Explanation of Solution

Given information:

The thickness of steel sheet is $t$, the specific weight of steel is $\gamma$, the point $O$ is origin, the radius of the semicircle section 2 is $a$, the side dimension of square plate section 2 is $2a$, the centroidal axis for square section 1 is $x^{\prime }$, $y^{\prime }$ and $z^{\prime }$, the centroidal section for semicircular section 2 is $x^{″}$, $y^{″}$ and $z^{″}$, the perpendicular distance between the centroidal point P and $\text{y}$ -axis is $0$, the perpendicular distance between centroidal point Q and $x$ -axis is $0$.

The following figure illustrates different centroidal axis.

Figure-(1).

Write the expression for the mass of section 1.

$\begin{array}{c}{m}_1=\frac{\gamma }{g}\left(2a\times 2a\times t\right)\\ =\frac{4\gamma a^{2}t}{g}\end{array}$   ....(I)

Here, the mass of the square section 1 is $m_1$, the specific weight of the steel sheet is $\gamma$, the acceleration due the gravity is $g$, the length of square plate is $2a$, the thickness of square plate is $t$.

Write the expression for the mass of section 2.

$\begin{array}{c}{m}_2=\frac{\gamma }{g}\left(\frac{\pi r^2}{2}\times t\right)\\ =\frac{\gamma \pi r^{2}t}{2g}\end{array}$   ....(II)

Here, the mass of the square section 2 is $m_2$, the radius of semicircular plate is $r$.

Write the expression for hypotenuse side $AB$ using Pythagoras theorem in right angled triangle $ABC$.

$AB=\sqrt{{\left(AC\right)}^2+{\left(BC\right)}^2}$   ....(III)

Here, the hypotenuse side in right angled triangle $ABC$ is $AB$, the height of the right angled triangle $ABC$ is $AC$, the base length of right angled triangle $ABC$ is $BC$.

Write the expression for hypotenuse side $OA$ using Pythagoras theorem in right angled triangle $OBA$.

$OA=\sqrt{{\left(AB\right)}^2+{\left(OB\right)}^2}$   ....(IV)

Here, the hypotenuse side in right angled triangle $OAB$ is $OA$, the height of the right angled triangle $OBA$ is $OB$, the base length of right angled triangle $OBA$ is $AB$.

Write the expression of position vector of the line joining $OA$.

$\overrightarrow{OA}=x_1\hat{i}+y_1\hat{j}+z_1\hat{k}$   ....(V)

Here, the vector along the line joining the points $O$ and $A$ is $\overrightarrow{OA}$, the unit vector along $x$ direction is $\hat{i}$, the unit vector along $y$ direction is $\hat{j}$, the unit vector along $z$ direction is $\hat{k}$, the coordinate on $x$ -axis is $x_1$, the coordinate on $y$ -axis is $y_1$, the coordinate on $z$ -axis is $z_1$.

Write the expression for unit vector along $\overrightarrow{OA}$.

${\hat{\eta }}_{\text{OA}}=\frac{\overrightarrow{OA}}{OA}$   ....(VI)

Here, the unit vector along $OA$ is ${\hat{\eta }}_{\text{OA}}$.

Write the expression of mass moment of inertia of section 1 about $x$ -axis.

${\left(I_{\text{x}}\right)}_1=\frac{m_1}{12}\left({\left(L_1\right)}^2+{\left(L_1\right)}^2\right)+m_1\left({\left(L_1\right)}^2+{\left(L_1\right)}^2\right)$   ....(VII)

Here, the mass moment of inertia of section 1 about $x$ -axis is ${\left(I_{\text{x}}\right)}_1$, the dimension of square plate section 1 is $L_1$.

Write the expression of mass moment of inertia of section 2 about $x$ -axis.

${\left(I_{\text{x}}\right)}_2=\frac{m_2}{4}\left(r^2\right)+m_2\left({\left(L_1\right)}^2+{\left(r\right)}^2\right)$   ....(VIII)

Here, the mass moment of inertia of section 2 about $x$ -axis is ${\left(I_{\text{x}}\right)}_2$.

Write the expression of total mass moment of inertia about $x$ -axis.

$I_{\text{x}}={\left(I_{\text{x}}\right)}_1+{\left(I_{\text{x}}\right)}_2$   ....(IX)

Here, the total mass moment of inertia about $x$ -axis is $I_{\text{x}}$.

Write the expression of mass moment of inertia of section 1 about $y$ -axis.

${\left(I_{\text{y}}\right)}_1=\frac{m_1}{12}\left(L_1^2\right)+m_1\left(r^2\right)$   ....(X)

Here, the mass moment of inertia of section 1 about $y$ -axis is ${\left(I_{\text{y}}\right)}_1$.

Write the expression of mass moment of inertia of section 2 about $y$ -axis.

${\left(I_{\text{y}}\right)}_2=m_2{\left(\frac{L_1}{2}\right)}^2\left(\frac{1}{2}-\frac{16}{9{\pi }^2}\right)+m_2\left({\left(\frac{4r}{3\pi }\right)}^2+\frac{L_1^2}{2}\right)$   ....(XI)

Here, the mass moment of inertia of section 2 about $y$ -axis is ${\left(I_{\text{y}}\right)}_2$.

Write the expression of total mass moment of inertia about $y$ -axis.

$I_{\text{y}}={\left(I_{\text{y}}\right)}_1+{\left(I_{\text{y}}\right)}_2$   ....(XII)

Write the expression of mass moment of inertia of section 1 about $z$ -axis.

${\left(I_{\text{z}}\right)}_1=\frac{m_1{L}_1^2}{12}+m_1{\left(\frac{L_1}{2}\right)}^2$   ....(XIII)

Here, the mass moment of inertia of section 1 about $z$ -axis is ${\left(I_{\text{z}}\right)}_1$.

Write the expression of mass moment of inertia of section 2 about $z$ -axis.

${\left(I_{\text{z}}\right)}_2=m_2{\left(\frac{L_1}{2}\right)}^2\left(\frac{1}{4}-\frac{16}{9{\pi }^2}\right)+m_2\left({\left(\frac{4r}{3\pi }\right)}^2+{\left(2a\right)}^2\right)$   ....(XIV)

Here, the mass moment of inertia of section 2 about $z$ -axis is ${\left(I_{\text{z}}\right)}_2$.

Write the expression of total mass moment of inertia about $z$ -axis.

$I_{\text{z}}={\left(I_{\text{z}}\right)}_1+{\left(I_{\text{z}}\right)}_2$   ....(XV)

Here, the total mass moment of inertia about $z$ -axis is $I_{\text{z}}$.

From, the symmetry in above figure about $x^{\prime }$, $y^{\prime }$ and $z^{\prime }$.

$\begin{array}{c}{\left(I_{\text{<msup><mo> x</mo><mo>′</mo></msup><msup><mo> y</mo><mo>′</mo></msup>}}\right)}_1={\left(I_{\text{<msup><mo> y</mo><mo>′</mo></msup><msup><mo> z</mo><mo>′</mo></msup>}}\right)}_1\\ ={\left(I_{\text{<msup><mo> z</mo><mo>′</mo></msup><msup><mo> x</mo><mo>′</mo></msup>}}\right)}_1\\ =0\end{array}$   ....(XVI)

Write the expression for product of mass moment of inertia in $x$ - $y$ plane.

$\begin{array}{c}\left(I_{\text{xy}}\right)=\sum _{\text{i}=1}^{\text{i}=2}\left({\left(I_{x^{\prime }{y}^{\prime }}\right)}_{\text{i}}+m_{\text{i}}\overline{x_{\text{i}}}\overline{y_{\text{i}}}\right)\\ \left(I_{\text{xy}}\right)=\left({\left(I_{\text{<msup><mo> x</mo><mo>′</mo></msup><msup><mo> y</mo><mo>′</mo></msup>}}\right)}_1+m_1\overline{x_1}\overline{y_1}\right)+\left({\left(I_{x^{\prime }{y}^{\prime }}\right)}_2+m_2\overline{x_2}\overline{y_2}\right)\end{array}$   ....(XVII)

Here, the product mass moment of inertia is $\left(I_{\text{xy}}\right)$, the subscript limit is $\text{i}$, the perpendicular distance between $x^{\prime }$ -axis and $x$ axis is $\overline{x_1}$, the perpendicular distance between $y^{\prime }$ axis and $y$ -axis is is $\overline{y_1}$, the perpendicular distance between $x^{″}$ -axis and $x$ axis is $\overline{x_2}$, the perpendicular distance between $y^{″}$ axis and $y$ -axis is is $\overline{y_2}$.

Write the expression for product mass moment of inertia in $y$ - $z$ plane.

$\begin{array}{c}\left(I_{\text{yz}}\right)=\sum _{\text{i}=1}^{\text{i}=2}\left({\left(I_{\text{<msup><mo> y</mo><mo>′</mo></msup><mo> z'</mo>}}\right)}_{\text{i}}+m_{\text{i}}\overline{y_{\text{i}}}\overline{z_{\text{i}}}\right)\\ \left(I_{\text{yz}}\right)=\left({\left(I_{\text{<msup><mo> y</mo><mo>′</mo></msup><msup><mo> z</mo><mo>′</mo></msup>}}\right)}_1+m_1\overline{y_1}\overline{z_1}\right)+\left({\left(I_{\text{<msup><mo> y</mo><mo>′</mo></msup><msup><mo> z</mo><mo>′</mo></msup>}}\right)}_2+m_2\overline{y_2}\overline{z_2}\right)\end{array}$   ....(XVIII)

Here, the product mass moment of inertia is $\left(I_{\text{yz}}\right)$, the subscript limit is $\text{i}$, the perpendicular distance between $y^{\prime }$ axis and $y$ -axis is $\overline{y_1}$, the perpendicular distance between $z^{\prime }$ -axis and $z$ -axis is $\overline{z_1}$, the perpendicular distance between $y^{″}$ axis and $y$ -axis is $\overline{y_2}$, the perpendicular distance between $z^{″}$ -axis and $z$ -axis is $\overline{z_2}$.

Write the expression for product mass moment of inertia in $z$ - $x$ plane.

$\begin{array}{c}\left(I_{\text{zx}}\right)=\sum _{\text{i}=1}^{\text{i}=2}\left({\left(I_{\text{z'<msup><mo> x</mo><mo>′</mo></msup>}}\right)}_{\text{i}}+m_{\text{i}}\overline{x_{\text{i}}}\overline{z_{\text{i}}}\right)\\ \left(I_{\text{zx}}\right)=\left({\left(I_{\text{<msup><mo> z</mo><mo>′</mo></msup><msup><mo> x</mo><mo>′</mo></msup>}}\right)}_1+m_1\overline{x_1}\overline{z_1}\right)+\left({\left(I_{\text{<msup><mo> z</mo><mo>′</mo></msup><msup><mo> x</mo><mo>′</mo></msup>}}\right)}_2+m_2\overline{x_2}\overline{z_2}\right)\end{array}$   ....(XIX)

Here, the product mass moment of inertia in $z$ - $x$ plane is $\left(I_{\text{zx}}\right)$, the perpendicular distance between $z^{\prime }$ -axis and $z$ -axis is $\overline{z_1}$, the perpendicular distance between $z^{″}$ -axis and $z$ -axis is $\overline{z_2}$, the perpendicular distance between $x^{\prime }$ -axis and $x$ axis $\overline{x_1}$, the perpendicular distance between $x^{″}$ -axis and $x$ axis $\overline{x_2}$.

Write the expression of mass moments of inertia at origin.

$\left[\begin{array}{l}{K}^3-\left(I_x+I_y+I_z\right)K^2+\left(I_x{I}_y+I_y{I}_z+I_z{I}_x-I_{xy}^2-I_{yz}^2-I_{zx}^2\right)K-\\ \left(I_x{I}_y{I}_z-I_x{I}_{yz}^2-I_y{I}_{zx}^2-I_z{I}_{xy}^2-2I_{xy}{I}_{yz}{I}_{zx}\right)=0\end{array}\right]$   ....(XX)

Here, the principle mass moment of inertia at origin is $K$.

Calculation:

Substitute $a$ for $x_1$, $2a$ for $y_1$, a for $z_1$ in Equation (V).

$\overrightarrow{OA}=a\hat{i}+2a\hat{j}+a\hat{k}$

Substitute $a$ for $AC$, $a$ for $BC$ in Equation (III).

$\begin{array}{c}AB=\sqrt{{\left(a\right)}^2+{\left(a\right)}^2}\\ =\sqrt{2a^2}\\ =\sqrt{2}a\end{array}$

Substitute $\sqrt{2}a$ for $AB$, $2a$ for $OB$ in Equation (IV).

$\begin{array}{c}OA=\sqrt{{\left(\sqrt{2}a\right)}^2+{\left(2a\right)}^2}\\ =\sqrt{2a^2+4a^2}\\ =\sqrt{6}a\end{array}$

Substitute $a\hat{i}+2a\hat{j}+a\hat{k}$ for $\overrightarrow{OA}$, $\sqrt{6}a$ for $OA$ in Equation (VI).

$\begin{array}{c}{\hat{\eta }}_{\text{OA}}=\frac{a\hat{i}+2a\hat{j}+a\hat{k}}{\sqrt{6}a}\\ =\frac{a\left(\hat{i}+2\hat{j}+\hat{k}\right)}{\sqrt{6}a}\\ =\frac{1}{\sqrt{6}}\hat{i}+\frac{2}{\sqrt{6}}\hat{j}+\frac{1}{\sqrt{6}}\hat{k}\end{array}$

Substitute $2a$ for $L_1$ in Equation (VII).

$\begin{array}{c}{\left(I_{\text{x}}\right)}_1=\frac{m_1}{12}\left({\left(2a\right)}^2+{\left(2a\right)}^2\right)+m_1\left({\left(a\right)}^2+{\left(a\right)}^2\right)\\ =\frac{2m_1{a}^2}{3}+2m_1{a}^2\\ =\frac{8m_1{a}^2}{3}\end{array}$

Substitute $2a$ for $L_1$, $a$ for $r$ in Equation (VIII).

$\begin{array}{l}{\left(I_{\text{x}}\right)}_2=\frac{m_2}{4}\left(a^2\right)+m_2\left({\left(2a\right)}^2+{\left(a\right)}^2\right)\\ =\frac{m_2{a}^2}{4}+5m_2{a}^2\\ =\frac{21m_2{a}^2}{4}\end{array}$

Substitute $\frac{8m_1{a}^2}{3}$ for ${\left(I_{\text{x}}\right)}_1$, $\frac{21m_2{a}^2}{4}$ for ${\left(I_{\text{x}}\right)}_2$, $\frac{4\gamma a^{2}t}{g}$ for $m_1$, $\frac{\gamma \pi r^{2}t}{2g}$ for $m_2$, $a$ for $r$ in Equation (IX).

$\begin{array}{c}{I}_{\text{x}}=\left(\frac{8a^2}{3}\right)\left(\frac{4\gamma a^{2}t}{g}\right)+\left(\frac{21a^2}{4}\right)\left(\frac{\gamma \pi r^{2}t}{2g}\right)\\ =\left(\frac{32\gamma t{a}^4}{3g}\right)+\left(\frac{21\pi \gamma a^{4}t}{8g}\right)\\ =\frac{256\gamma t{a}^4+63\pi \gamma t{a}^4}{24g}\\ =18.91335\left(\frac{\gamma t{a}^4}{g}\right)\end{array}$

Substitute $2a$ for $L_1$, $a$ for $r$ in Equation (X).

$\begin{array}{c}{\left(I_{\text{y}}\right)}_1=\frac{m_1}{12}{\left(2a\right)}^2+m_1\left(a^2\right)\\ =\frac{m_1{a}^2}{3}+m_1{a}^2\\ =\frac{4m_1{a}^2}{3}\end{array}$

Substitute $2a$ for $L_1$, $a$ for $r$ in Equation (XI).

$\begin{array}{c}{\left(I_{\text{y}}\right)}_2=m_2{\left(\frac{2a}{2}\right)}^2\left(\frac{1}{2}-\frac{16}{9{\pi }^2}\right)+m_2\left({\left(\frac{4a}{3\pi }\right)}^2+\frac{{\left(2a\right)}^2}{2}\right)\\ =0.31987\left(m_2{a}^2\right)+1.18012\left(m_2{a}^2\right)\\ =1.499999m_2{a}^2\end{array}$

Substitute $\frac{4m_1{a}^2}{3}$ for ${\left(I_{\text{y}}\right)}_1$, $1.4999999m_2{a}^2$ for ${\left(I_{\text{y}}\right)}_2$, $\frac{4\gamma a^{2}t}{g}$ for $m_1$, $\frac{\gamma \pi r^{2}t}{2g}$ for $m_2$, $a$ for $r$ in Equation (XII).

$\begin{array}{c}{I}_{\text{y}}=\left(\frac{4a^2}{3}\right)\left(\frac{4\gamma a^{2}t}{g}\right)+\left(1.499999a^2\right)\left(\frac{\gamma \pi r^{2}t}{2g}\right)\\ =\left(\frac{4a^2}{3}\right)\left(\frac{4\gamma a^{2}t}{g}\right)+\left(1.49999a^2\right)\left(\frac{\gamma \pi a^{2}t}{2g}\right)\\ =\left(\frac{\gamma a^{4}t}{g}\right)\left(\left(\frac{16}{3}\right)+\frac{1.49999\pi }{2}\right)\\ =7.68937\left(\frac{\gamma a^{4}t}{g}\right)\end{array}$

Substitute $2a$ for $L_1$ in Equation (XIII).

$\begin{array}{c}{\left(I_z\right)}_1=\frac{4m_1{a}^2}{12}+m_1{\left(\frac{2a}{2}\right)}^2\\ =\frac{m_1{a}^2}{3}+\frac{4m_1{a}^2}{4}\\ =\frac{4m_1{a}^2+12m_1{a}^2}{12}\\ =\frac{4m_1{a}^2}{3}\end{array}$

Substitute $2a$ for $L_1$, $a$ for $r$ in Equation (XIV).

$\begin{array}{c}{\left(I_{\text{z}}\right)}_2=m_2{\left(\frac{2a}{2}\right)}^2\left(\frac{1}{4}-\frac{16}{9{\pi }^2}\right)+m_2\left({\left(\frac{4a}{3\pi }\right)}^2+{\left(2a\right)}^2\right)\\ =0.069873m_2{a}^2+4.18012m_2{a}^2\\ =4.2499m_2{a}^2\end{array}$

Substitute $\frac{m_1{a}^2}{3}$ for ${\left(I_{\text{z}}\right)}_1$, $4.21506m_2{a}^2$ for ${\left(I_{\text{z}}\right)}_2$, $\frac{4\gamma a^{2}t}{g}$ for $m_1$, $\frac{\gamma \pi r^{2}t}{2g}$ for $m_2$ in Equation (XV).

$\begin{array}{l}{I}_{\text{z}}=\left(\frac{4a^2}{3}\right)\left(\frac{4\gamma a^{2}t}{g}\right)+4.2499\left(\frac{\gamma \pi r^{2}t}{2g}\right)a^2\\ =\left(\frac{4a^2}{3}\right)\left(\frac{4\gamma a^{2}t}{g}\right)+4.2499\left(\frac{\gamma \pi a^{2}t}{2g}\right)a^2\\ =\left(\frac{\gamma a^{4}t}{g}\right)\left(\left(\frac{16}{3}\right)+\frac{13.35145}{2}\right)\\ =12.00922\left(\frac{\gamma a^{4}t}{g}\right)\end{array}$

Substitute $0$ for ${\left(I_{\text{<msup><mo> x</mo><mo>′</mo></msup><msup><mo> y</mo><mo>′</mo></msup>}}\right)}_1$, $0$ for ${\left(I_{\text{<msup><mo> x</mo><mo>′</mo></msup><msup><mo> y</mo><mo>′</mo></msup>}}\right)}_2$, $\frac{4\gamma a^{2}t}{g}$ for $m_1$, $\frac{\gamma \pi r^{2}t}{2g}$ for $m_2$, $a$ for $\overline{y_1}$, $\frac{4a}{3\pi }$ for $\overline{x_2}$, $0$ for $\overline{x_1}$, $2a$ for $\overline{y_2}$ in Equation (XVII).

$\begin{array}{c}\left(I_{\text{xy}}\right)=\left(0+0\left(\frac{4\gamma a^{2}t}{g}\left(a\right)\right)\right)+\left(0+\frac{\gamma \pi r^{2}t}{2g}\left(\frac{4a}{3\pi }\right)2a\right)\\ =\left(\frac{\gamma \pi a^{2}t}{2g}\left(\frac{4a}{3\pi }\right)2a\right)\\ =1.3333\left(\frac{\gamma a^{2}t}{g}\right)\end{array}$

Substitute $0$ for ${\left(I_{\text{<msup><mo> y</mo><mo>′</mo></msup><msup><mo> z</mo><mo>′</mo></msup>}}\right)}_1$, $0$ for ${\left(I_{\text{<msup><mo> y</mo><mo>′</mo></msup><msup><mo> z</mo><mo>′</mo></msup>}}\right)}_2$, $\frac{4\gamma a^{2}t}{g}$ for $m_1$, $\frac{\gamma \pi r^{2}t}{2g}$ for $m_2$, $a$ for $\overline{y_1}$, $a$ for $\overline{z_1}$, $2a$ for $\overline{y_2}$, $a$ for $\overline{z_2}$, in Equation (XVIII).

$\begin{array}{c}\left(I_{\text{yz}}\right)=\left(0+\left(\frac{4\gamma a^{2}t}{g}\right)a\left(a\right)\right)+\left(0+\frac{\gamma \pi r^{2}t}{2g}\left(2a\right)\left(a\right)\right)\\ =\left(\frac{\gamma a^{4}t}{g}\right)\left(4+\pi \right)\\ =7.14159\left(\frac{\gamma a^{4}t}{g}\right)\end{array}$

Substitute $0$ for ${\left(I_{\text{<msup><mo> z</mo><mo>′</mo></msup><msup><mo> x</mo><mo>′</mo></msup>}}\right)}_1$, $0$ for ${\left(I_{\text{<msup><mo> z</mo><mo>′</mo></msup><msup><mo> x</mo><mo>′</mo></msup>}}\right)}_2$, $\frac{4\gamma a^{2}t}{g}$ for $m_1$, $\frac{\gamma \pi r^{2}t}{2g}$ for $m_2$, $\frac{4a}{3\pi }$ for $\overline{x_2}$, $0$ for $\overline{x_1}$, $2a$ for $\overline{z_1}$, $a$ for $\overline{z_2}$ in Equation (XIX).

$\begin{array}{c}\left(I_{\text{zx}}\right)=\left(0+\frac{4\gamma a^{2}t}{g}\left(0\right)\left(2a\right)\right)+\left(0+\frac{\gamma \pi r^{2}t}{2g}\left(\frac{4a}{3\pi }\right)\left(a\right)\right)\\ =\frac{\gamma \pi a^{2}t}{2g}\left(\frac{4a^2}{3\pi }\right)\\ =\left(\frac{\gamma a^{4}t}{g}\right)0.666667\end{array}$

Substitute $18.91335\left(\frac{\gamma t{a}^4}{g}\right)$ for $I_{\text{x}}$, $7.68937\left(\frac{\gamma a^{4}t}{g}\right)$ for $I_{\text{y}}$, $12.00922\left(\frac{\gamma a^{4}t}{g}\right)$ for $I_{\text{z}}$, $\left(\frac{\gamma a^{4}t}{g}\right)2$ for $\left(I_{\text{zx}}\right)$, $7.14159\left(\frac{\gamma a^{4}t}{g}\right)$ for $\left(I_{\text{yz}}\right)$ and $1.3333\left(\frac{\gamma a^{2}t}{g}\right)$ for $\left(I_{\text{xy}}\right)$ in Equation (XX).

$\begin{array}{l}{K}_2=17.27\left(\frac{\gamma a^{4}t}{g}\right)\\ K_3=19.08\left(\frac{\gamma a^{4}t}{g}\right)\end{array}$

Conclusion:

The principal mass moments of inertia at origin are $2.26\left(\frac{\gamma a^{4}t}{g}\right)$ , $17.27\left(\frac{\gamma a^{4}t}{g}\right)$ and $19.08\left(\frac{\gamma a^{4}t}{g}\right)$.

To determine

(b)

The principal axes of inertia at the origin.

Answer to Problem B.73P

The principal axes of inertia at the origin at $K_1$ are ${\left({\theta }_x\right)}_1=85°$, ${\left({\theta }_y\right)}_1=36.8°$ and ${\left({\theta }_z\right)}_1=53.7°$.

The principal axes of inertia at the origin at $K_2$ are ${\left({\theta }_x\right)}_2=81.7°$, ${\left({\theta }_y\right)}_2=54.7°$ and ${\left({\theta }_z\right)}_2=143.4°$.

The principal axes of inertia at the origin at $K_3$ are ${\left({\theta }_x\right)}_3=9.7°$, ${\left({\theta }_y\right)}_3=99°$ and ${\left({\theta }_z\right)}_3=86.3°$.

Explanation of Solution

Write the expression for direction cosine in terms of $K_1$.

$\left(I_x-K_1\right){\left({\lambda }_x\right)}_1-I_{xy}{\left({\lambda }_y\right)}_1-I_{zx}{\left({\lambda }_z\right)}_1=0$   ....(XXI)

Write the expression for direction cosine in terms of $K_1$.

$-I_{xy}{\left({\lambda }_x\right)}_1+\left(I_y-K_1\right){\left({\lambda }_y\right)}_1-I_{yz}{\left({\lambda }_z\right)}_1=0$   ....(XXII)

Write the expression for the direction cosines at x, y and z axis.

${\left({\lambda }_x\right)}_1^2+{\left({\lambda }_y\right)}_1^2+{\left({\lambda }_z\right)}_1^2=1$   ....(XXIII)

Write the expression for the principal axes of the inertia at x-axis.

${\left({\theta }_x\right)}_1={\cos }^{-1}{\left({\lambda }_x\right)}_1$   ....(XXIV)

Write the expression for the principal axes of the inertia at y-axis.

${\left({\theta }_y\right)}_1={\cos }^{-1}{\left({\lambda }_y\right)}_1$   ....(XXV)

Write the expression for the principal axes of the inertia at z-axis.

${\left({\theta }_z\right)}_1={\cos }^{-1}{\left({\lambda }_z\right)}_1$   ....(XXVI)

Write the expression for direction cosine in terms of $K_2$.

$\left(I_x-K_2\right){\left({\lambda }_x\right)}_2-I_{xy}{\left({\lambda }_y\right)}_2-I_{zx}{\left({\lambda }_z\right)}_2=0$   ....(XXVII)

Write the expression for direction cosine in terms of $K_2$.

$-I_{xy}{\left({\lambda }_x\right)}_2+\left(I_y-K_2\right){\left({\lambda }_y\right)}_2-I_{yz}{\left({\lambda }_z\right)}_2=0$   ....(XXVIII)

Write the expression for the direction cosines at x, y and z axis.

${\left({\lambda }_x\right)}_2^2+{\left({\lambda }_y\right)}_2^2+{\left({\lambda }_z\right)}_2^2=1$   ....(XXIX)

Write the expression for the principal axes of the inertia at x-axis.

${\left({\theta }_x\right)}_2={\cos }^{-1}{\left({\lambda }_x\right)}_2$   ....(XXX)

Write the expression for the principal axes of the inertia at y-axis.

${\left({\theta }_y\right)}_2={\cos }^{-1}{\left({\lambda }_y\right)}_2$   ....(XXXI)

Write the expression for the principal axes of the inertia at z-axis.

${\left({\theta }_z\right)}_2={\cos }^{-1}{\left({\lambda }_z\right)}_2$   ....(XXXII)

Write the expression for direction cosine in terms of $K_3$.

$\left(I_x-K_3\right){\left({\lambda }_x\right)}_3-I_{xy}{\left({\lambda }_y\right)}_3-I_{zx}{\left({\lambda }_z\right)}_3=0$   ....(XXXIII)

Write the expression for direction cosine in terms of $K_3$.

$-I_{xy}{\left({\lambda }_x\right)}_3+\left(I_y-K_3\right){\left({\lambda }_y\right)}_3-I_{yz}{\left({\lambda }_z\right)}_3=0$   ....(XXXIV)

Write the expression for the direction cosines at x, y and z axis.

${\left({\lambda }_x\right)}_3^2+{\left({\lambda }_y\right)}_3^2+{\left({\lambda }_z\right)}_3^2=1$   ....(XXXV)

Write the expression for the principal axes of the inertia at x-axis.

${\left({\theta }_x\right)}_3={\cos }^{-1}{\left({\lambda }_x\right)}_3$   ....(XXXVI)

Write the expression for the principal axes of the inertia at y-axis.

${\left({\theta }_y\right)}_3={\cos }^{-1}{\left({\lambda }_y\right)}_3$   ....(XXXVII)

Write the expression for the principal axes of the inertia at z-axis.

${\left({\theta }_z\right)}_3={\cos }^{-1}{\left({\lambda }_z\right)}_3$   ....(XXXVIII)

Calculation:

Substitute $2.26\left(\frac{\gamma a^{4}t}{g}\right)$ for $K_1$, $18.91335\left(\frac{\gamma t{a}^4}{g}\right)$ for $I_{\text{x}}$, $\left(\frac{\gamma a^{4}t}{g}\right)2$ for $\left(I_{\text{zx}}\right)$ and $1.3333\left(\frac{\gamma a^{2}t}{g}\right)$ for $\left(I_{\text{xy}}\right)$ in Equation (XXI).

$\begin{array}{l}\left[\begin{array}{l}\left(18.91335\left(\frac{\gamma t{a}^4}{g}\right)-2.26\left(\frac{\gamma a^{4}t}{g}\right)\right){\left({\lambda }_x\right)}_1-1.3333\left(\frac{\gamma a^{2}t}{g}\right){\left({\lambda }_y\right)}_1-\\ \left(\frac{\gamma a^{4}t}{g}\right)2{\left({\lambda }_z\right)}_1=0\end{array}\right]\\ \left[\begin{array}{l}\left(18.9133-2.26\right)\left(\frac{\gamma t{a}^4}{g}\right){\left({\lambda }_x\right)}_1-1.3333\left(\frac{\gamma t{a}^4}{g}\right){\left({\lambda }_y\right)}_1-\\ 0.6666\left(\frac{\gamma t{a}^4}{g}\right){\left({\lambda }_z\right)}_1=0\end{array}\right]\\ \left[12.49087{\left({\lambda }_x\right)}_1-{\left({\lambda }_y\right)}_1-0.5{\left({\lambda }_z\right)}_1\right]\left(\frac{\gamma t{a}^4}{g}\right)\left(1.3333\right)=0\\ 12.49087{\left({\lambda }_x\right)}_1-{\left({\lambda }_y\right)}_1-0.5{\left({\lambda }_z\right)}_1\end{array}$   ....(XXXIX)

Substitute $2.26\left(\frac{\gamma a^{4}t}{g}\right)$ for $K_1$, $7.68937\left(\frac{\gamma a^{4}t}{g}\right)$ for $I_{\text{y}}$, $7.14159\left(\frac{\gamma a^{4}t}{g}\right)$ for $\left(I_{\text{yz}}\right)$ and $1.3333\left(\frac{\gamma a^{2}t}{g}\right)$ for $\left(I_{\text{xy}}\right)$ in Equation (XXII).

$\begin{array}{l}\left[\begin{array}{l}-\left(1.3333\left(\frac{\gamma a^{2}t}{g}\right)\right){\left({\lambda }_x\right)}_1+\left(7.68937\left(\frac{\gamma a^{4}t}{g}\right)-2.26\left(\frac{\gamma a^{4}t}{g}\right)\right){\left({\lambda }_y\right)}_1\\ -7.14159\left(\frac{\gamma a^{4}t}{g}\right){\left({\lambda }_z\right)}_1=0\end{array}\right]\\ \left[\begin{array}{l}-1.3333\left(\frac{\gamma a^{2}t}{g}\right){\left({\lambda }_x\right)}_1+\left(7.6895-2.2589\right)\left(\frac{\gamma a^{4}t}{g}\right){\left({\lambda }_y\right)}_1-\\ 7.1415\left(\frac{\gamma a^{4}t}{g}\right){\left({\lambda }_z\right)}_1=0\end{array}\right]\\ \left(-0.2452{\left({\lambda }_x\right)}_1+{\left({\lambda }_y\right)}_1-1.3150{\left({\lambda }_z\right)}_1\right)\left(\frac{\gamma a^{4}t}{g}\right)5.4306=0\\ -0.2452{\left({\lambda }_x\right)}_1+{\left({\lambda }_y\right)}_1-1.3150{\left({\lambda }_z\right)}_1=0\end{array}$   ....(XL)

Add Equation (XXXIX) and Equation (XL).

$\begin{array}{l}\left(-0.2452{\left({\lambda }_x\right)}_1+{\left({\lambda }_y\right)}_1-1.3150{\left({\lambda }_z\right)}_1\right)+\left(12.49087{\left({\lambda }_x\right)}_1-{\left({\lambda }_y\right)}_1-0.5{\left({\lambda }_z\right)}_1\right)\\ \left(-0.2452{\left({\lambda }_x\right)}_1-1.3150{\left({\lambda }_z\right)}_1\right)+\left(12.49087{\left({\lambda }_x\right)}_1-0.5{\left({\lambda }_z\right)}_1\right)\\ {\left({\lambda }_z\right)}_1=6.7465{\left({\lambda }_x\right)}_1\end{array}$ (XLI)

Substitute $6.7465{\left({\lambda }_x\right)}_1$ for ${\left({\lambda }_z\right)}_1$ in Equation (XL).

$\begin{array}{l}-0.2452{\left({\lambda }_x\right)}_1+{\left({\lambda }_y\right)}_1-1.3150\left(6.7465{\left({\lambda }_x\right)}_1\right)=0\\ -0.2452{\left({\lambda }_x\right)}_1+{\left({\lambda }_y\right)}_1-8.8716{\left({\lambda }_x\right)}_1=0\\ {\left({\lambda }_y\right)}_1=9.1176{\left({\lambda }_x\right)}_1\end{array}$   ....(XLII)

Substitute $6.7465{\left({\lambda }_x\right)}_1$ for ${\left({\lambda }_z\right)}_1$ and $9.1176{\left({\lambda }_x\right)}_1$ for ${\left({\lambda }_y\right)}_1$ in Equation (XXIII).

$\begin{array}{l}{\left({\lambda }_x\right)}_1^2+{\left(9.1176{\left({\lambda }_x\right)}_1\right)}^2+{\left(6.7465{\left({\lambda }_x\right)}_1\right)}^2=1\\ {\left({\lambda }_x\right)}_1^2+\left(83.1306{\left({\lambda }_x\right)}_1^2\right)+\left(45.5152{\left({\lambda }_x\right)}_1^2\right)=1\\ {\left({\lambda }_x\right)}_1^2\left(1+\left(83.1306\right)+\left(45.5152\right)\right)=1\\ {\left({\lambda }_x\right)}_1=0.08782\end{array}$

Substitute $0.08782$ for ${\left({\lambda }_x\right)}_1$ in Equation (XLI).

$\begin{array}{c}{\left({\lambda }_z\right)}_1=6.7465\left(0.08782\right)\\ =0.5925\end{array}$

Substitute $0.08782$ for ${\left({\lambda }_x\right)}_1$ in Equation (XLII).

$\begin{array}{c}{\left({\lambda }_y\right)}_1=9.1176\left(0.08782\right)\\ =0.8007\end{array}$

Substitute $0.08782$ for ${\left({\lambda }_x\right)}_1$ in Equation (XXIV).

$\begin{array}{c}{\left({\theta }_x\right)}_1={\cos }^{-1}\left(0.08782\right)\\ =84.9617°\\ \approx 85°\end{array}$

Substitute $0.8007$ for ${\left({\lambda }_y\right)}_1$ in Equation (XXV).

$\begin{array}{c}{\left({\theta }_y\right)}_1={\cos }^{-1}\left(0.8007\right)\\ =36.8°\end{array}$

Substitute $0.5925$ for ${\left({\lambda }_z\right)}_1$ in Equation (XXVI).

$\begin{array}{c}{\left({\theta }_z\right)}_1={\cos }^{-1}\left(0.5925\right)\\ =53.665°\\ \approx 53.7°\end{array}$

Substitute $17.27\left(\frac{\gamma a^{4}t}{g}\right)$ for $K_2$, $18.91335\left(\frac{\gamma t{a}^4}{g}\right)$ for $I_{\text{x}}$, $\left(\frac{\gamma a^{4}t}{g}\right)2$ for $\left(I_{\text{zx}}\right)$ and $1.3333\left(\frac{\gamma a^{2}t}{g}\right)$ for $\left(I_{\text{xy}}\right)$ in Equation (XXVII).

$\begin{array}{l}\left[\begin{array}{l}\left(18.91335\left(\frac{\gamma t{a}^4}{g}\right)-17.27\left(\frac{\gamma a^{4}t}{g}\right)\right){\left({\lambda }_x\right)}_2-1.3333\left(\frac{\gamma a^{2}t}{g}\right){\left({\lambda }_y\right)}_2-\\ \left(\frac{\gamma a^{4}t}{g}\right)2{\left({\lambda }_z\right)}_2=0\end{array}\right]\\ \left[\begin{array}{l}\left(18.9133-17.27\right)\left(\frac{\gamma t{a}^4}{g}\right){\left({\lambda }_x\right)}_2-1.3333\left(\frac{\gamma t{a}^4}{g}\right){\left({\lambda }_y\right)}_2-\\ 0.6666\left(\frac{\gamma t{a}^4}{g}\right){\left({\lambda }_z\right)}_2=0\end{array}\right]\\ \left[1.2304{\left({\lambda }_x\right)}_2-{\left({\lambda }_y\right)}_2-0.5{\left({\lambda }_z\right)}_2\right]\left(\frac{\gamma t{a}^4}{g}\right)\left(1.3333\right)=0\\ 1.2304{\left({\lambda }_x\right)}_2-{\left({\lambda }_y\right)}_2-0.5{\left({\lambda }_z\right)}_2=0\end{array}$   ....(XLIII)

Substitute $17.27\left(\frac{\gamma a^{4}t}{g}\right)$ for $K_2$, $7.68937\left(\frac{\gamma a^{4}t}{g}\right)$ for $I_{\text{y}}$, $7.14159\left(\frac{\gamma a^{4}t}{g}\right)$ for $\left(I_{\text{yz}}\right)$ and $1.3333\left(\frac{\gamma a^{2}t}{g}\right)$ for $\left(I_{\text{xy}}\right)$ in Equation (XXVIII).

$\begin{array}{l}\left[\begin{array}{l}-\left(1.3333\left(\frac{\gamma a^{2}t}{g}\right)\right){\left({\lambda }_x\right)}_2+\left(7.68937\left(\frac{\gamma a^{4}t}{g}\right)-17.27\left(\frac{\gamma a^{4}t}{g}\right)\right){\left({\lambda }_y\right)}_2\\ -7.14159\left(\frac{\gamma a^{4}t}{g}\right){\left({\lambda }_z\right)}_2=0\end{array}\right]\\ \left[\begin{array}{l}-1.3333\left(\frac{\gamma a^{2}t}{g}\right){\left({\lambda }_x\right)}_2+\left(7.6895-17.27\right)\left(\frac{\gamma a^{4}t}{g}\right){\left({\lambda }_y\right)}_2-\\ 7.1415\left(\frac{\gamma a^{4}t}{g}\right){\left({\lambda }_z\right)}_2=0\end{array}\right]\\ \left(0.1391{\left({\lambda }_x\right)}_2+{\left({\lambda }_y\right)}_2+0.7452{\left({\lambda }_z\right)}_2\right)\left(\frac{\gamma a^{4}t}{g}\right)\left(-9.5805\right)=0\\ 0.1391{\left({\lambda }_x\right)}_2+{\left({\lambda }_y\right)}_2+0.7452{\left({\lambda }_z\right)}_2=0\end{array}$   ....(XLIV)

Add Equation (XLIII) and Equation (XLIV).

$\begin{array}{l}\left(1.2304{\left({\lambda }_x\right)}_2-{\left({\lambda }_y\right)}_2-0.5{\left({\lambda }_z\right)}_2\right)+\left(0.1391{\left({\lambda }_x\right)}_2+{\left({\lambda }_y\right)}_2+0.7452{\left({\lambda }_z\right)}_2\right)\\ \left(1.2304{\left({\lambda }_x\right)}_2-0.5{\left({\lambda }_z\right)}_2\right)+\left(0.1391{\left({\lambda }_x\right)}_2+0.7452{\left({\lambda }_z\right)}_2\right)\\ {\left({\lambda }_z\right)}_2=-5.5851{\left({\lambda }_x\right)}_2\end{array}$ (XLV)

Substitute $-5.5851{\left({\lambda }_x\right)}_2$ for ${\left({\lambda }_z\right)}_2$ in Equation (XLIV).

$\begin{array}{l}0.1391{\left({\lambda }_x\right)}_2+{\left({\lambda }_y\right)}_2+0.7452\left(-5.5851{\left({\lambda }_x\right)}_2\right)=0\\ 0.1391{\left({\lambda }_x\right)}_2+{\left({\lambda }_y\right)}_2+0.7452\left(-4.1620{\left({\lambda }_x\right)}_2\right)=0\\ {\left({\lambda }_y\right)}_2=4.0230{\left({\lambda }_x\right)}_2\end{array}$   ....(XLVI)

Substitute $-5.5851{\left({\lambda }_x\right)}_2$ for ${\left({\lambda }_z\right)}_2$ and $4.0230{\left({\lambda }_x\right)}_2$ for ${\left({\lambda }_y\right)}_2$ in Equation (XXIX).

$\begin{array}{l}{\left({\lambda }_x\right)}_1^2+{\left(4.0230{\left({\lambda }_x\right)}_2\right)}^2+{\left(-5.5851{\left({\lambda }_x\right)}_2\right)}^2=1\\ {\left({\lambda }_x\right)}_2^2+\left(16.1845{\left({\lambda }_x\right)}_2^2\right)+\left(31.1933{\left({\lambda }_x\right)}_2^2\right)=1\\ {\left({\lambda }_x\right)}_2^2\left(1+\left(16.1845\right)+\left(31.1933\right)\right)=1\\ {\left({\lambda }_x\right)}_2=0.1437\end{array}$

Substitute $0.1437$ for ${\left({\lambda }_x\right)}_2$ in Equation (XLV).

$\begin{array}{c}{\left({\lambda }_z\right)}_2=-5.5851\left(0.1437\right)\\ =-0.8029\end{array}$

Substitute $0.1437$ for ${\left({\lambda }_x\right)}_2$ in Equation (XLVI).

$\begin{array}{c}{\left({\lambda }_y\right)}_2=4.0230\left(0.1437\right)\\ =0.5783\end{array}$

Substitute $0.1437$ for ${\left({\lambda }_x\right)}_2$ in Equation (XXX).

$\begin{array}{c}{\left({\theta }_x\right)}_2={\cos }^{-1}\left(0.1437\right)\\ =81.7°\end{array}$

Substitute $0.5783$ for ${\left({\lambda }_y\right)}_2$ in Equation (XXXI).

$\begin{array}{c}{\left({\theta }_y\right)}_2={\cos }^{-1}\left(0.5783\right)\\ =54.7°\end{array}$

Substitute $-0.8029$ for ${\left({\lambda }_z\right)}_2$ in Equation (XXXII).

$\begin{array}{c}{\left({\theta }_z\right)}_2={\cos }^{-1}\left(-0.8029\right)\\ =143.4°\end{array}$

Substitute $19.08\left(\frac{\gamma a^{4}t}{g}\right)$ for $K_3$, $18.91335\left(\frac{\gamma t{a}^4}{g}\right)$ for $I_{\text{x}}$, $\left(\frac{\gamma a^{4}t}{g}\right)2$ for $\left(I_{\text{zx}}\right)$ and $1.3333\left(\frac{\gamma a^{2}t}{g}\right)$ for $\left(I_{\text{xy}}\right)$ in Equation (XXXIII).

$\begin{array}{l}\left[\begin{array}{l}\left(18.91335\left(\frac{\gamma t{a}^4}{g}\right)-19.08\left(\frac{\gamma a^{4}t}{g}\right)\right){\left({\lambda }_x\right)}_3-1.3333\left(\frac{\gamma a^{2}t}{g}\right){\left({\lambda }_y\right)}_3-\\ \left(\frac{\gamma a^{4}t}{g}\right)2{\left({\lambda }_z\right)}_3=0\end{array}\right]\\ \left[\begin{array}{l}\left(18.9133-19.08\right)\left(\frac{\gamma t{a}^4}{g}\right){\left({\lambda }_x\right)}_3-1.3333\left(\frac{\gamma t{a}^4}{g}\right){\left({\lambda }_y\right)}_3-\\ 0.6666\left(\frac{\gamma t{a}^4}{g}\right){\left({\lambda }_z\right)}_3=0\end{array}\right]\\ \left[0.1253{\left({\lambda }_x\right)}_3-{\left({\lambda }_y\right)}_3-0.5{\left({\lambda }_z\right)}_3\right]\left(\frac{\gamma t{a}^4}{g}\right)\left(1.3333\right)=0\\ -0.1253{\left({\lambda }_x\right)}_3-{\left({\lambda }_y\right)}_3-0.5{\left({\lambda }_z\right)}_3=0\end{array}$   ....(XLVII)

Substitute $19.08\left(\frac{\gamma a^{4}t}{g}\right)$ for $K_3$, $7.68937\left(\frac{\gamma a^{4}t}{g}\right)$ for $I_{\text{y}}$, $7.14159\left(\frac{\gamma a^{4}t}{g}\right)$ for $\left(I_{\text{yz}}\right)$ and $1.3333\left(\frac{\gamma a^{2}t}{g}\right)$ for $\left(I_{\text{xy}}\right)$ in Equation (XXXIV).

$\begin{array}{l}\left[\begin{array}{l}-\left(1.3333\left(\frac{\gamma a^{2}t}{g}\right)\right){\left({\lambda }_x\right)}_3+\left(7.68937\left(\frac{\gamma a^{4}t}{g}\right)-19.08\left(\frac{\gamma a^{4}t}{g}\right)\right){\left({\lambda }_y\right)}_3\\ -7.14159\left(\frac{\gamma a^{4}t}{g}\right){\left({\lambda }_z\right)}_3=0\end{array}\right]\\ \left[\begin{array}{l}-1.3333\left(\frac{\gamma a^{2}t}{g}\right){\left({\lambda }_x\right)}_3+\left(7.6895-19.08\right)\left(\frac{\gamma a^{4}t}{g}\right){\left({\lambda }_y\right)}_3-\\ 7.1415\left(\frac{\gamma a^{4}t}{g}\right){\left({\lambda }_z\right)}_3=0\end{array}\right]\\ \left(0.1170{\left({\lambda }_x\right)}_3+{\left({\lambda }_y\right)}_3+0.6269{\left({\lambda }_z\right)}_3\right)\left(\frac{\gamma a^{4}t}{g}\right)\left(-11.3905\right)=0\\ 0.1170{\left({\lambda }_x\right)}_3+{\left({\lambda }_y\right)}_3+0.6269{\left({\lambda }_z\right)}_3=0\end{array}$ … (XLVIII)

Add Equation (XLVII) and Equation (XLVIII).

$\begin{array}{l}\left(-0.1253{\left({\lambda }_x\right)}_3-{\left({\lambda }_y\right)}_3-0.5{\left({\lambda }_z\right)}_3\right)+\left(0.1170{\left({\lambda }_x\right)}_3+{\left({\lambda }_y\right)}_3+0.6269{\left({\lambda }_z\right)}_3\right)\\ \left(-0.1253{\left({\lambda }_x\right)}_3-0.5{\left({\lambda }_z\right)}_3\right)+\left(0.1170{\left({\lambda }_x\right)}_3+0.6269{\left({\lambda }_z\right)}_3\right)\\ {\left({\lambda }_z\right)}_3=0.06522{\left({\lambda }_x\right)}_3\end{array}$ (XLIX)

Substitute $0.06522{\left({\lambda }_x\right)}_3$ for ${\left({\lambda }_z\right)}_3$ in Equation (XLVIII).

$\begin{array}{l}0.1170{\left({\lambda }_x\right)}_3+{\left({\lambda }_y\right)}_3+0.6269\left(0.06522{\left({\lambda }_x\right)}_3\right)=0\\ 0.1170{\left({\lambda }_x\right)}_3+{\left({\lambda }_y\right)}_3+\left(0.04088{\left({\lambda }_x\right)}_3\right)=0\\ {\left({\lambda }_y\right)}_3=-0.1579{\left({\lambda }_x\right)}_3\end{array}$   ....(L)

Substitute $0.06522{\left({\lambda }_x\right)}_3$ for ${\left({\lambda }_z\right)}_3$ and $-0.1579{\left({\lambda }_x\right)}_3$ for ${\left({\lambda }_y\right)}_3$ in Equation (XXXV).

$\begin{array}{l}{\left({\lambda }_x\right)}_1^2+{\left(-0.1579{\left({\lambda }_x\right)}_3\right)}^2+{\left(0.06522{\left({\lambda }_x\right)}_3\right)}^2=1\\ {\left({\lambda }_x\right)}_3^2+\left(0.02493{\left({\lambda }_x\right)}_3^2\right)+\left(0.06522{\left({\lambda }_x\right)}_3^2\right)=1\\ {\left({\lambda }_x\right)}_3^2\left(1+\left(0.02493\right)+\left(0.06522\right)\right)=1\\ {\left({\lambda }_x\right)}_3=0.9857\end{array}$

Substitute $0.9857$ for ${\left({\lambda }_x\right)}_3$ in Equation (XLIX).

$\begin{array}{c}{\left({\lambda }_z\right)}_3=0.06522\left(0.9857\right)\\ =0.06429\end{array}$

Substitute $0.9857$ for ${\left({\lambda }_x\right)}_3$ in Equation (L).

$\begin{array}{c}{\left({\lambda }_y\right)}_3=-0.1579\left(0.9857\right)\\ =-0.1556\end{array}$

Substitute $0.9857$ for ${\left({\lambda }_x\right)}_3$ in Equation (XXXVI).

$\begin{array}{c}{\left({\theta }_x\right)}_3={\cos }^{-1}\left(0.9857\right)\\ =9.7°\end{array}$

Substitute $-0.1556$ for ${\left({\lambda }_y\right)}_3$ in Equation (XXXVII).

$\begin{array}{c}{\left({\theta }_y\right)}_3={\cos }^{-1}\left(-0.1556\right)\\ =99°\end{array}$

Substitute $0.06429$ for ${\left({\lambda }_z\right)}_3$ in Equation (XXXVIII).

$\begin{array}{c}{\left({\theta }_z\right)}_3={\cos }^{-1}\left(0.06429\right)\\ =86.3°\end{array}$

Conclusion:

The principal axes of inertia at the origin at $K_1$ are ${\left({\theta }_x\right)}_1=85°$, ${\left({\theta }_y\right)}_1=36.8°$ and ${\left({\theta }_z\right)}_1=53.7°$.

The principal axes of inertia at the origin at $K_2$ are ${\left({\theta }_x\right)}_2=81.7°$, ${\left({\theta }_y\right)}_2=54.7°$ and ${\left({\theta }_z\right)}_2=143.4°$.

The principal axes of inertia at the origin at $K_3$ are ${\left({\theta }_x\right)}_3=9.7°$, ${\left({\theta }_y\right)}_3=99°$ and ${\left({\theta }_z\right)}_3=86.3°$.