Chapter B, Problem B.72P

To determine

(a)

The principal mass moment of inertia at the origin $O$.

Answer to Problem B.72P

The principal mass moment of inertia at the origin $O$ with respect to principal axis 1 is $0.1639\left(\frac{W{a}^2}{g}\right)$.

The principal mass moment of inertia at the origin $O$ with respect to principal axis 2 is $1.054\left(\frac{W{a}^2}{g}\right)$.

The principal mass moment of inertia at the origin $O$ with respect to principal axis 3 is $1.115\left(\frac{W{a}^2}{g}\right)$.

Explanation of Solution

Given information:

The thin bent plate with uniform density, the weight of both the plates is $W$, the origin is $O$, the plate is square with dimension $a$ unit, the centroidal axis for plate 1 is $x^{\prime }$, $y^{\prime }$, $z^{\prime }$, the centroidal axis for plate 2 is $x^{″}$, $y^{″}$, $z^{″}$, the perpendicular distance between the centroidal point M and $\text{y}$ -axis is $0$, the perpendicular distance between centroidal point N and $x$ -axis is $0$.

The figure below illustrate different centroidal axis.

Figure-(1)

Write the expression for the mass of plate 1.

$m_1=\frac{W}{2g}$   ....(I)

Here, the weight of both the plates is $W$, the acceleration due to gravity is $g$, the mass of the plate 1 is $m_1$.

Write the expression for the mass of plate 2.

$m_2=\frac{W}{2g}$   ....(II)

Here, the weight of both the plates is $W$, the acceleration due to gravity is $g$, the mass of the plate 1 is $m_2$.

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

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

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 the magnitude of position vector $\overrightarrow{OA}$.

$\left|\overrightarrow{OA}\right|=\sqrt{x_1^2+y_1^2+z_1^2}$   ....(IV)

Here, the magnitude of the position vector $\overrightarrow{OA}$ is $\overrightarrow{\left|OA\right|}$.

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

${\hat{\eta }}_{\text{OA}}=\frac{\overrightarrow{OA}}{\left|\overrightarrow{OA}\right|}$   ....(V)

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.

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

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.

${\left(I_{\text{x}}\right)}_2=\frac{m_2}{12}\left(L_2^2+L_2^2\right)+m_2\left({\left(\frac{L_2}{2}\right)}^2+{\left(\frac{L_2}{2}\right)}^2\right)$   ....(VII)

Here, the mass moment of inertia of section 2 about $x$ -axis is ${\left(I_{\text{x}}\right)}_2$, the dimension of square plate section 2 is $L_2$.

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

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

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.

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

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.

${\left(I_{\text{y}}\right)}_2=\frac{m_2}{12}\left(L_2^2\right)+m_2\left(L_2^2+{\left(\frac{L_2}{2}\right)}^2\right)$   ....(X)

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.

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

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

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

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

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.

${\left(I_{\text{z}}\right)}_2=\frac{m_1{L}_2^2}{12}+m_1\left(L_2+{\left(\frac{L_2}{2}\right)}^2\right)$   ....(XIII)

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.

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

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 $z^{\prime }$.

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

Here, the product mass moment of inertia in $x^{\prime }$ - $y^{\prime }$ plane is ${\left(I_{\text{<msup><mo> x</mo><mo>′</mo></msup><msup><mo> y</mo><mo>′</mo></msup>}}\right)}_1$, the product mass moment of inertia in $y^{\prime }$ - $z^{\prime }$ plane is ${\left(I_{\text{<msup><mo> y</mo><mo>′</mo></msup><msup><mo> z</mo><mo>′</mo></msup>}}\right)}_1$, the product mass moment of inertia in $z^{\prime }$ - $x^{\prime }$ plane is ${\left(I_{\text{<msup><mo> z</mo><mo>′</mo></msup><msup><mo> x</mo><mo>′</mo></msup>}}\right)}_1$.

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

$\begin{array}{c}{\left(I_{\text{<msup><mo> x</mo><mo>″</mo></msup><msup><mo> y</mo><mo>″</mo></msup>}}\right)}_2={\left(I_{\text{<msup><mo> y</mo><mo>″</mo></msup><msup><mo> z</mo><mo>″</mo></msup>}}\right)}_2\\ ={\left(I_{\text{<msup><mo> z</mo><mo>″</mo></msup><msup><mo> x</mo><mo>″</mo></msup>}}\right)}_2\\ =0\end{array}$   ....(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 product mass moment of inertia is $\left(I_{\text{xy}}\right)$, the perpendicular distance between $x^{\prime }$ -axis and $x$ axis $\overline{x_1}$, the perpendicular distance between $y^{\prime }$ axis and $y$ -axis is $\overline{y_1}$, the perpendicular distance between $x^{″}$ -axis and $x$ axis $\overline{x_2}$, the perpendicular distance between $y^{″}$ axis and $y$ -axis 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 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 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 principal mass moment of inertia $\left(K\right)$ of the thin plate.

$K^3-\left[\begin{array}{l}\left(I_{\text{x}}+I_{\text{y}}+I_{\text{z}}\right)K^2+\left(I_{\text{x}}{I}_{\text{y}}+I_{\text{y}}{I}_{\text{z}}+I_{\text{z}}{I}_{\text{x}}-I_{\text{xy}}^2-I_{\text{yz}}^2-I_{\text{zx}}^2\right)K\\ -\left(I_{\text{x}}{I}_{\text{y}}{I}_{\text{z}}-I_{\text{x}}{I}_{\text{yz}}^2-I_{\text{y}}{I}_{\text{zx}}^2-I_{\text{z}}{I}_{\text{xy}}^2-2I_{\text{xy}}{I}_{\text{yz}}{I}_{\text{zx}}\right)\end{array}\right]=0$   ....(XX)

Here, the principal moment of inertia is $K$.

Calculation:

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

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

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

$\begin{array}{c}\left|\overrightarrow{OA}\right|=\sqrt{a^2+a^2+a^2}\\ =\sqrt{3a^2}\\ =\sqrt{3}a\end{array}$

Substitute $a\hat{i}+a\hat{j}+a\hat{k}$ for $\overrightarrow{OA}$, $\sqrt{3}a$ for $\left|\overrightarrow{OA}\right|$ in Equation (V).

$\begin{array}{c}{\hat{\eta }}_{\text{OA}}=\frac{a\hat{i}+a\hat{j}+a\hat{k}}{\sqrt{3}a}\\ =\frac{a\left(\hat{i}+\hat{j}+\hat{k}\right)}{\sqrt{3}a}\\ =\frac{1}{\sqrt{3}}\hat{i}+\frac{1}{\sqrt{3}}\hat{j}+\frac{1}{\sqrt{3}}\hat{k}\end{array}$

Substitute $a$ for $L_1$ in Equation (VI).

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

Substitute $a$ for $L_2$ in Equation (VII).

$\begin{array}{c}{\left(I_{\text{x}}\right)}_2=\frac{m_2}{12}\left(a^2+a^2\right)+m_2\left({\left(\frac{a}{2}\right)}^2+{\left(\frac{a^2}{2}\right)}^2\right)\\ =\frac{m_2{a}^2}{6}+\frac{m_2{a}^2}{2}\\ =\frac{3m_2{a}^2+m_2{a}^2}{6}\\ =\frac{2m_2{a}^2}{3}\end{array}$

Substitute $\frac{m_1{a}^2}{3}$ for ${\left(I_{\text{x}}\right)}_1$, $\frac{2m_2{a}^2}{3}$ for ${\left(I_{\text{x}}\right)}_2$, $\frac{W}{2g}$ for $m_1$, $\frac{W}{2g}$ for $m_2$ in Equation (VIII).

$\begin{array}{c}{I}_{\text{x}}=\frac{W{a}^2}{\left(2g\right)3}+\frac{2W{a}^2}{\left(2g\right)3}\\ =\frac{W}{2g}{a}^2\end{array}$

Substitute $a$ for $L_1$ in Equation (IX).

$\begin{array}{c}{\left(I_{\text{y}}\right)}_1=\frac{m_1}{12}\left(a^2+a^2\right)+m_1\left({\left(\frac{a}{2}\right)}^2+{\left(\frac{a}{2}\right)}^2\right)\\ =\frac{m_1{a}^2}{6}+\frac{m_1{a}^2}{2}\\ =\frac{2m_1{a}^2}{3}\end{array}$

Substitute $a$ for $L_2$ in Equation (X).

$\begin{array}{c}{\left(I_{\text{y}}\right)}_2=\frac{m_2}{12}\left(a^2\right)+m_2\left(a^2+{\left(\frac{a}{2}\right)}^2\right)\\ =\frac{m_2}{12}\left(a^2\right)+\frac{5m_1{a}^2}{4}\\ =\frac{m_2{a}^2+15m_2{a}^2}{12}\\ =\frac{4m_2{a}^2}{3}\end{array}$

Substitute $\frac{2m_1{a}^2}{3}$ for ${\left(I_{\text{y}}\right)}_1$, $\frac{4m_2{a}^2}{3}$ for ${\left(I_{\text{y}}\right)}_2$, $\frac{W}{2g}$ for $m_1$, $\frac{W}{2g}$ for $m_2$ in Equation (XI).

$\begin{array}{c}{I}_{\text{y}}=\frac{2a^{2}W}{\left(2g\right)3}+\frac{4W{a}^2}{\left(2g\right)3}\\ =\frac{6W{a}^2}{\left(2g\right)3}\\ =\frac{W{a}^2}{g}\end{array}$

Substitute $a$ for $L_1$ in Equation (XII).

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

Substitute $a$ for $L_2$ in Equation (XIII).

$\begin{array}{c}{\left(I_{\text{z}}\right)}_2=\frac{m_2}{12}\left(a^2\right)+m_2\left(a^2+{\left(\frac{a}{2}\right)}^2\right)\\ =\frac{m_2}{12}\left(a^2\right)+\frac{5m_1{a}^2}{4}\\ =\frac{m_2{a}^2+15m_2{a}^2}{12}\\ =\frac{4m_2{a}^2}{3}\end{array}$

Substitute $\frac{m_1{a}^2}{3}$ for ${\left(I_{\text{y}}\right)}_1$, $\frac{4m_2{a}^2}{3}$ for ${\left(I_{\text{y}}\right)}_2$, $\frac{W}{2g}$ for $m_1$, $\frac{W}{2g}$ for $m_2$ in Equation (XIV).

$\begin{array}{c}{I}_{\text{z}}=\frac{a^{2}W}{\left(2g\right)3}+\frac{4W{a}^2}{\left(2g\right)3}\\ =\frac{5W{a}^2}{\left(2g\right)3}\\ =\frac{5W{a}^2}{6g}\end{array}$

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

$\begin{array}{c}\left(I_{\text{xy}}\right)=\left(0+0\left(\frac{W}{2g}\left(\frac{a}{2}\right)\right)\right)+\left(0+\frac{W}{2g}\left(a\right)\left(\frac{a}{2}\right)\right)\\ =\left(0+\frac{W}{2g}\left(a\right)\left(\frac{a}{2}\right)\right)\\ =\frac{W{a}^2}{4g}\end{array}$

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

$\begin{array}{c}\left(I_{\text{yz}}\right)=\left(0+\frac{W}{2g}0\left(\frac{a}{2}\right)\right)+\left(0+\frac{W}{2g}\left(\frac{a}{2}\right)\left(\frac{a}{2}\right)\right)\\ =\frac{W}{2g}\left(\frac{a^2}{4}\right)\\ =\frac{W{a}^2}{8g}\end{array}$

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

$\begin{array}{c}\left(I_{\text{zx}}\right)=\left(0+\frac{W}{2g}\left(\frac{a}{2}\right)\left(\frac{a}{2}\right)\right)+\left(0+\frac{W}{2g}\left(a\right)\left(\frac{a}{2}\right)\right)\\ =\left(\frac{W{a}^2}{8g}\right)+\left(\frac{W{a}^2}{4g}\right)\\ =\left(\frac{W{a}^2+2W{a}^2}{8g}\right)\\ =\frac{3W{a}^2}{8g}\end{array}$

Substitute $\frac{W}{2g}{a}^2$ for $I_{\text{x}}$, $\frac{W{a}^2}{g}$ for $I_{\text{y}}$, $\frac{5W{a}^2}{6g}$ for $I_{\text{z}}$, $\frac{3W{a}^2}{8g}$ for $\left(I_{\text{zx}}\right)$, $\frac{W{a}^2}{8g}$ for $\left(I_{\text{yz}}\right)$, $\frac{W{a}^2}{4g}$ for $\left(I_{\text{xy}}\right)$ in Equation (XX).

$\left[\begin{array}{l}{K}^3-\left(\frac{1}{2}+1+\frac{5}{6}\right)\left(\frac{W{a}^2}{g}\right)K^2\\ +\left(\left(\begin{array}{l}\frac{1}{2}\left(1\right)+\left(1\right)\frac{5}{6}+\left(\frac{5}{6}\right)+\\ \left(\frac{1}{2}\right)-{\left(\frac{1}{4}\right)}^2-{\left(\frac{1}{8}\right)}^2-\\ {\left(\frac{3}{8}\right)}^2\end{array}\right)\right){\left(\frac{W{a}^2}{g}\right)}^2\left(K\right)\\ -\left(\begin{array}{l}\left(\frac{1}{2}\right)\left(1\right)\left(\frac{5}{6}\right)-\left(\frac{1}{2}\right){\left(\frac{1}{8}\right)}^2-\left(1\right){\left(\frac{3}{8}\right)}^2-\left(\frac{5}{6}\right){\left(\frac{1}{4}\right)}^2\\ -2\left(\frac{1}{4}\right)\left(\frac{1}{8}\right)\left(\frac{3}{8}\right)\end{array}\right){\left(\frac{W{a}^2}{g}\right)}^3\end{array}\right]=0$

$K^3-\left[\begin{array}{l}\left(2.3333\right)\left(\frac{W{a}^2}{g}\right)K^2\\ +\left(1.5312\right){\left(\frac{W{a}^2}{g}\right)}^2\left(K\right)-\left(0.1927\right){\left(\frac{W{a}^2}{g}\right)}^3\end{array}\right]=0$

Solving above cubic Equation to get the values of $K_1$, $K_2$, $K_3$.

$\begin{array}{l}{K}_1=0.1639\left(\frac{W{a}^2}{g}\right)\\ K_2=1.054\left(\frac{W{a}^2}{g}\right)\\ K_3=1.115\left(\frac{W{a}^2}{g}\right)\end{array}$

Conclusion:

The principal mass moment of inertia at the origin $O$ with respect to principal axis 1 is $0.1639\left(\frac{W{a}^2}{g}\right)$.

The principal mass moment of inertia at the origin $O$ with respect to principal axis 2 is $1.054\left(\frac{W{a}^2}{g}\right)$.

The principal mass moment of inertia at the origin $O$ with respect to principal axis 3 is $1.115\left(\frac{W{a}^2}{g}\right)$.

To determine

(b)

The determine the orientation of principal axis at origin $O$ with respect to $x$ -axis, $y$ -axis and $z$ -axis.

Answer to Problem B.72P

The orientation of first principal axis at origin $O$ with respect to $x$ -axis is $36.727°$.

The orientation of first principal axis at origin $O$ with respect to $y$ -axis is $71.604°$.

The orientation of first principal axis at origin $O$ with respect to $z$ -axis is $59.475°$.

The orientation of second principal axis at origin $O$ with respect to $x$ -axis is $74.896°$.

The orientation of second principal axis at origin $O$ with respect to $y$ -axis is $54.5424°$.

The orientation of second principal axis at origin $O$ with respect to $z$ -axis is $140.509°$.

The orientation of third principal axis at origin $O$ with respect to $x$ -axis is $36.85°$.

The orientation of third principal axis at origin $O$ with respect to $y$ -axis is $102.830°$.

The orientation of third principal axis at origin $O$ with respect to $z$ -axis is $123.85°$.

Explanation of Solution

The figure below represents the orientation of principal axis with respect to the Cartesian coordinates.

Figure-(1)

Write the equation of direction cosines formed by principal axes 1.

$\left(I_{\text{x}}-K_1\right){\left({\lambda }_{\text{x}}\right)}_1-I_{\text{xy}}{\left({\lambda }_{\text{y}}\right)}_1-I_{\text{zx}}{\left({\lambda }_{\text{z}}\right)}_1=0$   ....(XXI)

$-\left(I_{\text{xy}}\right){\left({\lambda }_{\text{x}}\right)}_1+\left(I_{\text{y}}-K_1\right){\left({\lambda }_{\text{y}}\right)}_1-I_{\text{yz}}{\left({\lambda }_{\text{z}}\right)}_1=0$   ....(XXII)

${\left({\lambda }_{\text{x}}\right)}_1^2+{\left({\lambda }_{\text{y}}\right)}_1^2+{\left({\lambda }_{\text{z}}\right)}_1^2=1$   ....(XXIII)

Here, the principal mass moment of inertia about principal axis 1 is $K_1$.

Write the expression of orientation of direction cosine about principal axis 1 in $x$ -direction.

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

Here, the orientation of direction cosine about principal axis 1 in $x$ -direction is ${\left({\theta }_{\text{x}}\right)}_1$.

Write the expression of orientation of direction cosine about principal axis 2 in $y$ -direction.

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

Here, the orientation of direction cosine about principal axis 1 in $y$ -direction is ${\left({\theta }_{\text{y}}\right)}_1$.

Write the expression of orientation of direction cosine about principal axis 1 in $z$ -direction.

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

Here, the orientation of direction cosine about principal axis 1 in $z$ -direction is ${\left({\theta }_{\text{z}}\right)}_1$.

Write the equation of direction cosines formed by principal axes 2.

$\left(I_{\text{x}}-K_2\right){\left({\lambda }_{\text{x}}\right)}_2-I_{\text{xy}}{\left({\lambda }_{\text{y}}\right)}_2-I_{\text{zx}}{\left({\lambda }_{\text{z}}\right)}_2=0$   ....(XXVII)

$-\left(I_{\text{xy}}\right){\left({\lambda }_{\text{x}}\right)}_2+\left(I_{\text{y}}-K_2\right){\left({\lambda }_{\text{y}}\right)}_2-I_{\text{yz}}{\left({\lambda }_{\text{z}}\right)}_2=0$   ....(XXVIII)

${\left({\lambda }_{\text{x}}\right)}_2^2+{\left({\lambda }_{\text{y}}\right)}_2^2+{\left({\lambda }_{\text{z}}\right)}_2^2=1$   ....(XXIX)

Here, the principal mass moment of inertia about principal axis in $x$ -direction 2 is $K_2$.

Write the expression of orientation of direction cosine about principal axis 2 in $x$ -direction.

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

Here, the orientation of direction cosine about principal axis 2 is ${\left({\theta }_{\text{x}}\right)}_2$.

Write the expression of orientation of direction cosine about principal axis 2 in $y$ -direction.

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

Here, the orientation of direction cosine about principal axis 2 in $y$ -direction is ${\left({\theta }_{\text{y}}\right)}_2$.

Write the expression of orientation of direction cosine about principal axis 2 in $z$ -direction.

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

Here, the orientation of direction cosine about principal axis 2 in $z$ -direction is ${\left({\theta }_{\text{z}}\right)}_2$.

Write the equation of direction cosines formed by principal axes 3.

$\left(I_{\text{x}}-K_3\right){\left({\lambda }_{\text{x}}\right)}_3-I_{\text{xy}}{\left({\lambda }_{\text{y}}\right)}_3-I_{\text{zx}}{\left({\lambda }_{\text{z}}\right)}_3=0$   ....(XXXIII)

$-\left(I_{\text{xy}}\right){\left({\lambda }_{\text{x}}\right)}_3+\left(I_{\text{y}}-K_2\right){\left({\lambda }_{\text{y}}\right)}_3-I_{\text{yz}}{\left({\lambda }_{\text{z}}\right)}_3=0$   ....(XXXIV)

${\left({\lambda }_{\text{x}}\right)}_3^2+{\left({\lambda }_{\text{y}}\right)}_3^2+{\left({\lambda }_{\text{z}}\right)}_3^2=1$   ....(XXXV)

Here, the principal mass moment of inertia about principal axis 3 is $K_3$.

Write the expression of orientation of direction cosine about principal axis 3 in $x$ -direction.

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

Here, the orientation of direction cosine about principal axis 3 in $x$ -direction is ${\left({\theta }_{\text{x}}\right)}_3$.

Write the expression of orientation of direction cosine about principal axis 2 in $y$ -direction.

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

Here, the orientation of direction cosine about principal axis 3 in $y$ -direction is ${\left({\theta }_{\text{y}}\right)}_3$.

Write the expression of orientation of direction cosine about principal axis 2 in $z$ -direction.

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

Here, the orientation of direction cosine about principal axis 3 in $z$ -direction is ${\left({\theta }_{\text{z}}\right)}_3$.

Calculation:

Substitute $\frac{W}{2g}{a}^2$ for $I_{\text{x}}$, $\frac{3W{a}^2}{8g}$ for $\left(I_{\text{zx}}\right)$, $\frac{W{a}^2}{4g}$ for $\left(I_{\text{xy}}\right)$, $0.1639\left(\frac{W{a}^2}{g}\right)$ for $K_1$ in Equation (XXI).

$\begin{array}{l}\left(\frac{W}{2g}{a}^2-0.1639\left(\frac{W{a}^2}{g}\right)\right){\left({\lambda }_{\text{x}}\right)}_1-\left(\frac{W{a}^2}{4g}\right){\left({\lambda }_{\text{y}}\right)}_1-\left(\frac{3W{a}^2}{8g}\right){\left({\lambda }_{\text{z}}\right)}_1=0\\ \left(\frac{W{a}^2}{g}\right)\left(\left(0.5-0.1639\right){\left({\lambda }_{\text{x}}\right)}_1-\left(0.25\right){\left({\lambda }_{\text{y}}\right)}_1-\left(0.375\right){\left({\lambda }_{\text{z}}\right)}_1\right)=0\\ \left(0.5-0.1639\right){\left({\lambda }_{\text{x}}\right)}_1-\left(0.25\right){\left({\lambda }_{\text{y}}\right)}_1-\left(0.375\right){\left({\lambda }_{\text{z}}\right)}_1=0\\ 0.3361{\left({\lambda }_{\text{x}}\right)}_1-\left(0.25\right){\left({\lambda }_{\text{y}}\right)}_1-\left(0.375\right){\left({\lambda }_{\text{z}}\right)}_1=0\end{array}$

$\begin{array}{l}\frac{0.3361}{\left(0.25\right)}{\left({\lambda }_{\text{x}}\right)}_1-{\left({\lambda }_{\text{y}}\right)}_1-\frac{\left(0.375\right)}{\left(0.25\right)}{\left({\lambda }_{\text{z}}\right)}_1=0\\ 1.34444{\left({\lambda }_{\text{x}}\right)}_1-{\left({\lambda }_{\text{y}}\right)}_1-1.5{\left({\lambda }_{\text{z}}\right)}_1=0\end{array}$   ....(XXXIX)

Substitute $\frac{W}{g}{a}^2$ for $I_{\text{y}}$, $\frac{W{a}^2}{8g}$ for $\left(I_{\text{yz}}\right)$, $\frac{W{a}^2}{4g}$ for $\left(I_{\text{xy}}\right)$, $0.1639\left(\frac{W{a}^2}{g}\right)$ for $K_1$ in Equation (XXII).

$\begin{array}{l}-\left(\frac{W{a}^2}{4g}\right){\left({\lambda }_{\text{x}}\right)}_1+\left(\frac{W}{g}{a}^2-\left(0.1639\left(\frac{W{a}^2}{g}\right)\right)\right){\left({\lambda }_{\text{y}}\right)}_1-\frac{W{a}^2}{8g}{\left({\lambda }_{\text{z}}\right)}_1=0\\ \frac{W}{g}{a}^2\left(-\left(\frac{1}{4}\right){\left({\lambda }_{\text{x}}\right)}_1+{\left({\lambda }_{\text{y}}\right)}_1-0.1639{\left({\lambda }_{\text{y}}\right)}_1-\left(\frac{1}{8}\right){\left({\lambda }_{\text{z}}\right)}_1\right)=0\\ \frac{W}{g}{a}^2\left(\left(-0.25\right){\left({\lambda }_{\text{x}}\right)}_1+0.8361{\left({\lambda }_{\text{y}}\right)}_1-0.125{\left({\lambda }_{\text{z}}\right)}_1\right)=0\\ \left(-0.25\right){\left({\lambda }_{\text{x}}\right)}_1+0.8361{\left({\lambda }_{\text{y}}\right)}_1-0.125{\left({\lambda }_{\text{z}}\right)}_1=0\end{array}$

$\begin{array}{l}\frac{\left(-0.25\right)}{0.8361}{\left({\lambda }_{\text{x}}\right)}_1+{\left({\lambda }_{\text{y}}\right)}_1-\frac{0.125}{0.8361}{\left({\lambda }_{\text{z}}\right)}_1=0\\ -0.2990{\left({\lambda }_{\text{x}}\right)}_1+{\left({\lambda }_{\text{y}}\right)}_1-0.1495{\left({\lambda }_{\text{z}}\right)}_1=0\end{array}$   ....(XL)

Add Equation (XXXIX) and (XL).

$\begin{array}{l}\left[\begin{array}{l}\left(1.34444{\left({\lambda }_{\text{x}}\right)}_1-{\left({\lambda }_{\text{y}}\right)}_1-1.5{\left({\lambda }_{\text{z}}\right)}_1\right)+\\ \left(-0.2990{\left({\lambda }_{\text{x}}\right)}_1+{\left({\lambda }_{\text{y}}\right)}_1-0.1495{\left({\lambda }_{\text{z}}\right)}_1\right)\end{array}\right]=0\\ \left(1.0454{\left({\lambda }_{\text{x}}\right)}_1-1.6495{\left({\lambda }_{\text{z}}\right)}_1\right)=0\\ 1.6495{\left({\lambda }_{\text{z}}\right)}_1=1.0454{\left({\lambda }_{\text{x}}\right)}_1\\ {\left({\lambda }_{\text{z}}\right)}_1=\frac{1.0454}{1.6495}{\left({\lambda }_{\text{x}}\right)}_1\end{array}$

${\left({\lambda }_{\text{z}}\right)}_1=0.63376{\left({\lambda }_{\text{x}}\right)}_1$   ....(XLI)

Subtract Equation (XXXIX) and (XL).

$\begin{array}{l}\left[\begin{array}{l}\left(1.34444{\left({\lambda }_{\text{x}}\right)}_1-{\left({\lambda }_{\text{y}}\right)}_1-1.5{\left({\lambda }_{\text{z}}\right)}_1\right)-\\ \left(-0.2990{\left({\lambda }_{\text{x}}\right)}_1+{\left({\lambda }_{\text{y}}\right)}_1-0.1495{\left({\lambda }_{\text{z}}\right)}_1\right)\end{array}\right]=0\\ \left(1.6434{\left({\lambda }_{\text{x}}\right)}_1-2{\left({\lambda }_{\text{y}}\right)}_1-1.3505{\left({\lambda }_{\text{z}}\right)}_1\right)=0\end{array}$ …... (XLII)

Substitute $0.63376{\left({\lambda }_{\text{x}}\right)}_1$ for ${\left({\lambda }_{\text{z}}\right)}_1$ in Equation (XLII).

$\begin{array}{l}\left(1.6434{\left({\lambda }_{\text{x}}\right)}_1-2{\left({\lambda }_{\text{y}}\right)}_1-\left(1.3505\right)\left(0.63376\right){\left({\lambda }_{\text{x}}\right)}_1\right)=0\\ \left(1.6434{\left({\lambda }_{\text{x}}\right)}_1-2{\left({\lambda }_{\text{y}}\right)}_1-\left(0.85589\right){\left({\lambda }_{\text{x}}\right)}_1\right)=0\\ 0.78750{\left({\lambda }_{\text{x}}\right)}_1-2{\left({\lambda }_{\text{y}}\right)}_1=0\\ {\left({\lambda }_{\text{y}}\right)}_1=\frac{0.78750}{2}{\left({\lambda }_{\text{x}}\right)}_1\end{array}$

${\left({\lambda }_{\text{y}}\right)}_1=0.39375{\left({\lambda }_{\text{x}}\right)}_1$   ....(XLIII)

Substitute $0.39375{\left({\lambda }_{\text{x}}\right)}_1$ for ${\left({\lambda }_{\text{y}}\right)}_1$, $0.63376{\left({\lambda }_{\text{x}}\right)}_1$ for ${\left({\lambda }_{\text{z}}\right)}_1$ in Equation (XXIII).

$\begin{array}{l}{\left({\lambda }_{\text{x}}\right)}_1^2+{\left(0.39375{\left({\lambda }_{\text{x}}\right)}_1\right)}^2+{\left(0.63376{\left({\lambda }_{\text{x}}\right)}_1\right)}^2=1\\ {\left({\lambda }_{\text{x}}\right)}_1^2\left(1+{\left(0.39375\right)}^2+{\left(0.63376\right)}^2\right)=1\\ {\left({\lambda }_{\text{x}}\right)}_1^2\left(1+0.155039+0.40165\right)=1\\ {\left({\lambda }_{\text{x}}\right)}_1^2\left(1.55668\right)=1\end{array}$

$\begin{array}{l}{\left({\lambda }_{\text{x}}\right)}_1^2=\frac{1}{\left(1.55668\right)}\\ {\left({\lambda }_{\text{x}}\right)}_1=\sqrt{\frac{1}{\left(1.55668\right)}}\\ {\left({\lambda }_{\text{x}}\right)}_1=\sqrt{0.64238}\\ {\left({\lambda }_{\text{x}}\right)}_1=0.80149\end{array}$

Substitute $0.80149$ for ${\left({\lambda }_{\text{x}}\right)}_1$ in Equation (XLI).

$\begin{array}{l}{\left({\lambda }_{\text{z}}\right)}_1=\left(0.63376\right)\left(0.80149\right)\\ {\left({\lambda }_{\text{z}}\right)}_1=0.5079\end{array}$

Substitute $0.80149$ for ${\left({\lambda }_{\text{x}}\right)}_1$ in Equation (XLIII).

$\begin{array}{l}{\left({\lambda }_{\text{y}}\right)}_1=\left(0.39375\right)\left(0.80149\right)\\ {\left({\lambda }_{\text{y}}\right)}_1=0.315586\end{array}$

Substitute $0.80149$ for ${\left({\lambda }_{\text{x}}\right)}_1$ in Equation (XXIV).

$\begin{array}{l}{\left({\theta }_{\text{x}}\right)}_1={\cos }^{-1}\left(0.80149\right)\\ {\left({\theta }_{\text{x}}\right)}_1=36.727°\end{array}$

Substitute $0.315586$ for ${\left({\lambda }_{\text{y}}\right)}_1$ in Equation (XXV).

$\begin{array}{l}{\left({\theta }_{\text{y}}\right)}_1={\cos }^{-1}\left(0.315586\right)\\ {\left({\theta }_{\text{y}}\right)}_1=71.604°\end{array}$

Substitute $0.5079$ for ${\left({\lambda }_{\text{z}}\right)}_1$ in Equation (XXVI).

$\begin{array}{l}{\left({\theta }_{\text{z}}\right)}_1={\cos }^{-1}\left(0.5079\right)\\ {\left({\theta }_{\text{z}}\right)}_1=59.475°\end{array}$

Substitute $\frac{W}{2g}{a}^2$ for $I_{\text{x}}$, $\frac{3W{a}^2}{8g}$ for $\left(I_{\text{zx}}\right)$, $\frac{W{a}^2}{4g}$ for $\left(I_{\text{xy}}\right)$, $1.054\left(\frac{W{a}^2}{g}\right)$ for $K_2$ in Equation (XXVII).

$\begin{array}{l}\left[\begin{array}{l}\left(\left(\frac{W}{2g}{a}^2\right)-\left(1.054\left(\frac{W{a}^2}{g}\right)\right)\right){\left({\lambda }_{\text{x}}\right)}_2\\ -\left(\frac{W{a}^2}{4g}\right){\left({\lambda }_{\text{y}}\right)}_2-\left(\frac{3W{a}^2}{8g}\right){\left({\lambda }_{\text{z}}\right)}_2\end{array}\right]=0\\ \left(\frac{W{a}^2}{g}\right)\left(\left(0.5-1.054\right){\left({\lambda }_{\text{x}}\right)}_2-\left(0.25\right){\left({\lambda }_{\text{y}}\right)}_2-\left(0.375\right){\left({\lambda }_{\text{z}}\right)}_2\right)=0\\ -0.554{\left({\lambda }_{\text{x}}\right)}_2-0.25{\left({\lambda }_{\text{y}}\right)}_2-0.375{\left({\lambda }_{\text{z}}\right)}_2=0\\ 0.554{\left({\lambda }_{\text{x}}\right)}_2+0.25{\left({\lambda }_{\text{y}}\right)}_2+0.375{\left({\lambda }_{\text{z}}\right)}_2=0\end{array}$

$\begin{array}{l}\frac{0.554}{0.25}{\left({\lambda }_{\text{x}}\right)}_2+{\left({\lambda }_{\text{y}}\right)}_2+\frac{0.375}{0.25}{\left({\lambda }_{\text{z}}\right)}_2=0\\ 2.216{\left({\lambda }_{\text{x}}\right)}_2+{\left({\lambda }_{\text{y}}\right)}_2+1.5{\left({\lambda }_{\text{z}}\right)}_2=0\end{array}$   ....(XLIV)

Substitute $\frac{W}{g}{a}^2$ for $I_{\text{y}}$, $\frac{W{a}^2}{8g}$ for $\left(I_{\text{yz}}\right)$, $\frac{W{a}^2}{4g}$ for $\left(I_{\text{xy}}\right)$, $1.054\left(\frac{W{a}^2}{g}\right)$ for $K_2$ in Equation (XXVIII).

$\begin{array}{l}-\left(\frac{W{a}^2}{4g}\right){\left({\lambda }_{\text{x}}\right)}_2+\left(\left(\frac{W}{g}{a}^2\right)-\left(1.054\left(\frac{W{a}^2}{g}\right)\right)\right){\left({\lambda }_{\text{y}}\right)}_2-\left(\frac{W{a}^2}{8g}\right){\left({\lambda }_{\text{z}}\right)}_2=0\\ \left(\frac{W}{g}{a}^2\right)\left(-\left(0.25{\left({\lambda }_{\text{x}}\right)}_2\right)+\left(1-1.054\right){\left({\lambda }_{\text{y}}\right)}_2-\left(0.125\right){\left({\lambda }_{\text{z}}\right)}_2\right)=0\\ -0.25{\left({\lambda }_{\text{x}}\right)}_2-0.054{\left({\lambda }_{\text{y}}\right)}_2-0.125{\left({\lambda }_{\text{z}}\right)}_2=0\\ 0.25{\left({\lambda }_{\text{x}}\right)}_2+0.054{\left({\lambda }_{\text{y}}\right)}_2+0.125{\left({\lambda }_{\text{z}}\right)}_2=0\end{array}$

$\begin{array}{l}\frac{0.25}{0.054}{\left({\lambda }_{\text{x}}\right)}_2+{\left({\lambda }_{\text{y}}\right)}_2+\frac{0.125}{0.054}{\left({\lambda }_{\text{z}}\right)}_2=0\\ 4.6296{\left({\lambda }_{\text{x}}\right)}_2+{\left({\lambda }_{\text{y}}\right)}_2+2.3148{\left({\lambda }_{\text{z}}\right)}_2=0\end{array}$   ....(XLV)

Add Equation (XLIV) and(XLV).

$\begin{array}{l}\left[\begin{array}{l}\left(2.216{\left({\lambda }_{\text{x}}\right)}_2+{\left({\lambda }_{\text{y}}\right)}_2+1.5{\left({\lambda }_{\text{z}}\right)}_2\right)+\\ \left(4.6296{\left({\lambda }_{\text{x}}\right)}_2+{\left({\lambda }_{\text{y}}\right)}_2+2.3148{\left({\lambda }_{\text{z}}\right)}_2\right)\end{array}\right]=0\\ 6.8456{\left({\lambda }_{\text{x}}\right)}_2+2{\left({\lambda }_{\text{y}}\right)}_2+3.8148{\left({\lambda }_{\text{z}}\right)}_2=0\end{array}$   ....(XLVI)

Subtract Equation (XLIV) and(XLV).

$\begin{array}{l}\left[\begin{array}{l}\left(2.216{\left({\lambda }_{\text{x}}\right)}_2+{\left({\lambda }_{\text{y}}\right)}_2+1.5{\left({\lambda }_{\text{z}}\right)}_2\right)-\\ \left(4.6296{\left({\lambda }_{\text{x}}\right)}_2+{\left({\lambda }_{\text{y}}\right)}_2+2.3148{\left({\lambda }_{\text{z}}\right)}_2\right)\end{array}\right]=0\\ -2.4132{\left({\lambda }_{\text{x}}\right)}_2+0-0.8142{\left({\lambda }_{\text{z}}\right)}_2=0\\ 0.8142{\left({\lambda }_{\text{z}}\right)}_2=-2.4132{\left({\lambda }_{\text{x}}\right)}_2\\ {\left({\lambda }_{\text{z}}\right)}_2=\frac{-2.4132}{0.8142}{\left({\lambda }_{\text{x}}\right)}_2\end{array}$

$\begin{array}{l}{\left({\lambda }_{\text{z}}\right)}_2=\frac{-2.4132}{0.8142}{\left({\lambda }_{\text{x}}\right)}_2\\ {\left({\lambda }_{\text{z}}\right)}_2=-2.9617{\left({\lambda }_{\text{x}}\right)}_2\end{array}$   ....(XLVII)

Substitute $-2.9617{\left({\lambda }_{\text{x}}\right)}_2$ for ${\left({\lambda }_{\text{z}}\right)}_2$ in Equation (XLVI).

$\begin{array}{l}6.8456{\left({\lambda }_{\text{x}}\right)}_2+2{\left({\lambda }_{\text{y}}\right)}_2+\left(3.8148\right)\left(-2.9617\right){\left({\lambda }_{\text{x}}\right)}_2=0\\ 6.8456{\left({\lambda }_{\text{x}}\right)}_2+2{\left({\lambda }_{\text{y}}\right)}_2-11.2983{\left({\lambda }_{\text{x}}\right)}_2=0\\ -4.452725{\left({\lambda }_{\text{x}}\right)}_2+2{\left({\lambda }_{\text{y}}\right)}_2=0\\ 2{\left({\lambda }_{\text{y}}\right)}_2=4.452725{\left({\lambda }_{\text{x}}\right)}_2\end{array}$

$\begin{array}{l}{\left({\lambda }_{\text{y}}\right)}_2=\frac{4.452725}{2}{\left({\lambda }_{\text{x}}\right)}_2\\ {\left({\lambda }_{\text{y}}\right)}_2=2.22636{\left({\lambda }_{\text{x}}\right)}_2\end{array}$   ....(XLVIII)

Substitute $2.22636{\left({\lambda }_{\text{x}}\right)}_2$ for ${\left({\lambda }_{\text{y}}\right)}_2$, $-2.9617{\left({\lambda }_{\text{x}}\right)}_2$ for ${\left({\lambda }_{\text{z}}\right)}_2$ in Equation (XXIX).

$\begin{array}{l}{\left({\lambda }_{\text{x}}\right)}_2^2+{\left(2.22636{\left({\lambda }_{\text{x}}\right)}_2\right)}^2+{\left(-2.9617{\left({\lambda }_{\text{x}}\right)}_2\right)}^2=1\\ {\left({\lambda }_{\text{x}}\right)}_2^2\left(1+{\left(2.22636\right)}^2+{\left(-2.9617\right)}^2\right)=1\\ {\left({\lambda }_{\text{x}}\right)}_2^2\left(1+4.9566+8.77166\right)=1\\ {\left({\lambda }_{\text{x}}\right)}_2^2\left(14.7282\right)=1\end{array}$

$\begin{array}{l}{\left({\lambda }_{\text{x}}\right)}_2^2=\frac{1}{\left(14.7282\right)}\\ {\left({\lambda }_{\text{x}}\right)}_2=\sqrt{\frac{1}{\left(14.7282\right)}}\\ {\left({\lambda }_{\text{x}}\right)}_2=0.26056\end{array}$

Substitute $0.26056$ for ${\left({\lambda }_{\text{x}}\right)}_1$ in Equation (XLVII).

$\begin{array}{l}{\left({\lambda }_{\text{z}}\right)}_2=\left(-2.9617\right)\left(0.26056\right)\\ {\left({\lambda }_{\text{z}}\right)}_2=-0.771729\end{array}$

Substitute $0.26056$ for ${\left({\lambda }_{\text{x}}\right)}_1$ in Equation (XLVIII).

$\begin{array}{l}{\left({\lambda }_{\text{y}}\right)}_2=\left(2.22636\right)\left(0.26056\right)\\ {\left({\lambda }_{\text{y}}\right)}_2=0.58010\end{array}$

Substitute $0.26056$ for ${\left({\lambda }_{\text{x}}\right)}_2$ in Equation (XXX).

$\begin{array}{l}{\left({\theta }_{\text{x}}\right)}_2={\cos }^{-1}\left(0.26056\right)\\ {\left({\theta }_{\text{x}}\right)}_2=74.896°\end{array}$

Substitute $0.58010$ for ${\left({\lambda }_{\text{y}}\right)}_2$ in Equation (XXXI).

$\begin{array}{l}{\left({\theta }_{\text{y}}\right)}_2={\cos }^{-1}\left(0.58010\right)\\ {\left({\theta }_{\text{y}}\right)}_2=54.5424°\end{array}$

Substitute $-0.771729$ for ${\left({\lambda }_{\text{z}}\right)}_2$ in Equation (XXXII).

$\begin{array}{l}{\left({\theta }_{\text{z}}\right)}_2={\cos }^{-1}\left(-0.771729\right)\\ {\left({\theta }_{\text{z}}\right)}_2=140.509°\end{array}$

Substitute $\frac{W}{2g}{a}^2$ for $I_{\text{x}}$, $\frac{3W{a}^2}{8g}$ for $\left(I_{\text{zx}}\right)$, $\frac{W{a}^2}{4g}$ for $\left(I_{\text{xy}}\right)$, $1.115\left(\frac{W{a}^2}{g}\right)$ for $K_3$ in Equation (XXXIII).

$\begin{array}{l}\left(\left(\frac{W}{2g}{a}^2\right)-\left(1.115\left(\frac{W{a}^2}{g}\right)\right)\right){\left({\lambda }_{\text{x}}\right)}_3-\left(\frac{W{a}^2}{4g}\right){\left({\lambda }_{\text{y}}\right)}_3-\left(\frac{3W{a}^2}{8g}\right){\left({\lambda }_{\text{z}}\right)}_3=0\\ \left(\frac{W{a}^2}{g}\right)\left(\left(0.5-1.115\right){\left({\lambda }_{\text{x}}\right)}_3-0.25{\left({\lambda }_{\text{y}}\right)}_3-0.375{\left({\lambda }_{\text{z}}\right)}_3\right)=0\\ -0.615{\left({\lambda }_{\text{x}}\right)}_3-0.25{\left({\lambda }_{\text{y}}\right)}_3-0.375{\left({\lambda }_{\text{z}}\right)}_3=0\\ 0.615{\left({\lambda }_{\text{x}}\right)}_3+0.25{\left({\lambda }_{\text{y}}\right)}_3+0.375{\left({\lambda }_{\text{z}}\right)}_3=0\end{array}$

$\begin{array}{l}\frac{0.615}{0.25}{\left({\lambda }_{\text{x}}\right)}_3+{\left({\lambda }_{\text{y}}\right)}_3+\frac{0.375}{0.25}{\left({\lambda }_{\text{z}}\right)}_3=0\\ 2.4615{\left({\lambda }_{\text{x}}\right)}_3+{\left({\lambda }_{\text{y}}\right)}_3+1.5{\left({\lambda }_{\text{z}}\right)}_3=0\end{array}$   ....(XLIX)

Substitute $\frac{W}{g}{a}^2$ for $I_{\text{y}}$, $\frac{W{a}^2}{8g}$ for $\left(I_{\text{yz}}\right)$, $\frac{W{a}^2}{4g}$ for $\left(I_{\text{xy}}\right)$, $1.115\left(\frac{W{a}^2}{g}\right)$ for $K_3$ in Equation (XXXIV).

$\begin{array}{l}-\left(\frac{W{a}^2}{4g}\right){\left({\lambda }_{\text{x}}\right)}_3+\left(\left(\frac{W}{g}{a}^2\right)-\left(1.115\left(\frac{W{a}^2}{g}\right)\right)\right){\left({\lambda }_{\text{y}}\right)}_3-\left(\frac{W{a}^2}{8g}\right){\left({\lambda }_{\text{z}}\right)}_3=0\\ \left(\frac{W{a}^2}{g}\right)\left(\left(-0.25\right){\left({\lambda }_{\text{x}}\right)}_3+\left(1-1.115\right){\left({\lambda }_{\text{y}}\right)}_3-0.125{\left({\lambda }_{\text{z}}\right)}_3\right)=0\\ \left(-0.25\right){\left({\lambda }_{\text{x}}\right)}_3-\left(0.115\right){\left({\lambda }_{\text{y}}\right)}_3-0.125{\left({\lambda }_{\text{z}}\right)}_3=0\\ 0.25{\left({\lambda }_{\text{x}}\right)}_3+\left(0.115\right){\left({\lambda }_{\text{y}}\right)}_3+0.125{\left({\lambda }_{\text{z}}\right)}_3=0\end{array}$

$\begin{array}{l}\frac{0.25}{\left(0.115\right)}{\left({\lambda }_{\text{x}}\right)}_3+\frac{\left(0.115\right)}{\left(0.115\right)}{\left({\lambda }_{\text{y}}\right)}_3+\frac{0.125}{\left(0.115\right)}{\left({\lambda }_{\text{z}}\right)}_3=0\\ 2.1739{\left({\lambda }_{\text{x}}\right)}_3+{\left({\lambda }_{\text{y}}\right)}_3+1.08695{\left({\lambda }_{\text{z}}\right)}_3=0\end{array}$   ....(L)

Add Equation (XLIX) and (L).

$\begin{array}{l}\left[\begin{array}{l}2.4615{\left({\lambda }_{\text{x}}\right)}_3+{\left({\lambda }_{\text{y}}\right)}_3+1.5{\left({\lambda }_{\text{z}}\right)}_3+\\ \left(2.1739{\left({\lambda }_{\text{x}}\right)}_3+{\left({\lambda }_{\text{y}}\right)}_3+1.08695{\left({\lambda }_{\text{z}}\right)}_3\right)\end{array}\right]=0\\ 4.6354{\left({\lambda }_{\text{x}}\right)}_3+2{\left({\lambda }_{\text{y}}\right)}_3+2.58695{\left({\lambda }_{\text{z}}\right)}_3=0\\ \frac{4.6354}{2}{\left({\lambda }_{\text{x}}\right)}_3+{\left({\lambda }_{\text{y}}\right)}_3+\frac{2.58695}{2}{\left({\lambda }_{\text{z}}\right)}_3=0\\ 1.1781{\left({\lambda }_{\text{x}}\right)}_3+{\left({\lambda }_{\text{y}}\right)}_3+1.2934{\left({\lambda }_{\text{z}}\right)}_3=0\end{array}$   ....(LI)

Substract Equation (XLIX) and (L).

$\begin{array}{l}\left[\begin{array}{l}2.4615{\left({\lambda }_{\text{x}}\right)}_3+{\left({\lambda }_{\text{y}}\right)}_3+1.5{\left({\lambda }_{\text{z}}\right)}_3-\\ \left(2.1739{\left({\lambda }_{\text{x}}\right)}_3+{\left({\lambda }_{\text{y}}\right)}_3+1.08695{\left({\lambda }_{\text{z}}\right)}_3\right)\end{array}\right]=0\\ 0.2876{\left({\lambda }_{\text{x}}\right)}_3+0+0.41305{\left({\lambda }_{\text{z}}\right)}_3=0\\ 0.41305{\left({\lambda }_{\text{z}}\right)}_3=-0.2876{\left({\lambda }_{\text{x}}\right)}_3\\ {\left({\lambda }_{\text{z}}\right)}_3=\frac{-0.2876{\left({\lambda }_{\text{x}}\right)}_3}{0.41305}\end{array}$

${\left({\lambda }_{\text{z}}\right)}_3=-0.69628{\left({\lambda }_{\text{x}}\right)}_3$   ....(LII)

Substitute $-0.69628{\left({\lambda }_{\text{x}}\right)}_3$ for ${\left({\lambda }_{\text{z}}\right)}_3$ in Equation (LI).

$\begin{array}{l}1.1781{\left({\lambda }_{\text{x}}\right)}_3+{\left({\lambda }_{\text{y}}\right)}_3+\left(1.2934\right)\left(-0.69628{\left({\lambda }_{\text{x}}\right)}_3\right)=0\\ 1.1781{\left({\lambda }_{\text{x}}\right)}_3+\left({\lambda }_{\text{y}}\right)-{0.90056}_3{\left({\lambda }_{\text{x}}\right)}_3=0\\ {\left({\lambda }_{\text{y}}\right)}_3=-0.27753{\left({\lambda }_{\text{x}}\right)}_3\end{array}$   ....(LIII)

Substitute $-0.27753{\left({\lambda }_{\text{x}}\right)}_3$ for ${\left({\lambda }_{\text{y}}\right)}_3$, $-0.69628{\left({\lambda }_{\text{x}}\right)}_3$ for ${\left({\lambda }_{\text{z}}\right)}_3$ in Equation (XXXV).

$\begin{array}{l}{\left({\lambda }_{\text{x}}\right)}_3^2+{\left(-0.27753{\left({\lambda }_{\text{x}}\right)}_3\right)}^2+{\left(-0.69628{\left({\lambda }_{\text{x}}\right)}_3\right)}^2=1\\ {\left({\lambda }_{\text{x}}\right)}_3^2\left(1+0.07702+0.48480\right)=1\\ {\left({\lambda }_{\text{x}}\right)}_3^2=\frac{1}{\left(1+0.07702+0.48480\right)}\\ {\left({\lambda }_{\text{x}}\right)}_3=\sqrt{\frac{1}{1.5618}}\end{array}$

${\left({\lambda }_{\text{x}}\right)}_3=0.80017$

Substitute $0.80017$ for ${\left({\lambda }_{\text{x}}\right)}_3$ in Equation (LII).

$\begin{array}{l}{\left({\lambda }_{\text{z}}\right)}_3=\left(-0.69628\right)\left(0.80017\right)\\ {\left({\lambda }_{\text{z}}\right)}_3=-0.55714\end{array}$

Substitute $0.80017$ for ${\left({\lambda }_{\text{x}}\right)}_3$ in Equation (LIII).

$\begin{array}{l}{\left({\lambda }_{\text{y}}\right)}_3=\left(-0.27753\right)\left(0.80017\right)\\ {\left({\lambda }_{\text{y}}\right)}_3=-0.22207\end{array}$

Substitute $0.80017$ for ${\left({\lambda }_{\text{x}}\right)}_3$ in Equation (XXXVI).

$\begin{array}{l}{\left({\theta }_{\text{x}}\right)}_3={\cos }^{-1}\left(0.80017\right)\\ {\left({\theta }_{\text{x}}\right)}_3=36.85°\end{array}$

Substitute $-0.22207$ for ${\left({\lambda }_{\text{y}}\right)}_3$ in Equation (XXXVII).

$\begin{array}{l}{\left({\theta }_{\text{y}}\right)}_3={\cos }^{-1}\left(-0.22207\right)\\ {\left({\theta }_{\text{y}}\right)}_3=102.830°\end{array}$

Substitute $-0.55714$ for ${\left({\lambda }_{\text{z}}\right)}_3$ in Equation (XXXVIII).

$\begin{array}{l}{\left({\theta }_{\text{z}}\right)}_3={\cos }^{-1}\left(-0.55714\right)\\ {\left({\theta }_{\text{z}}\right)}_3=123.85°\end{array}$

Conclusion:

The orientation of first principal axis at origin $O$ with respect to $x$ -axis is $36.727°$.

The orientation of first principal axis at origin $O$ with respect to $y$ -axis is $71.604°$.

The orientation of first principal axis at origin $O$ with respect to $z$ -axis is $59.475°$.

The orientation of second principal axis at origin $O$ with respect to $x$ -axis is $74.896°$.

The orientation of second principal axis at origin $O$ with respect to $y$ -axis is $54.5424°$.

The orientation of second principal axis at origin $O$ with respect to $z$ -axis is $140.509°$.

The orientation of third principal axis at origin $O$ with respect to $x$ -axis is $36.85°$.

The orientation of third principal axis at origin $O$ with respect to $y$ -axis is $102.830°$.

The orientation of third principal axis at origin $O$ with respect to $z$ -axis is $123.85°$.