Chapter 9.6, Problem 9.183P

(a)

To determine

Find the principal mass moment of inertia of the cylinder at the origin O.

Answer to Problem 9.183P

The principal moment of inertia are $\underset{\_}{K_1=2.26\frac{\gamma t}{g}{a}^4}$, $\underset{\_}{K_2=17.27\frac{\gamma t}{g}{a}^4}$, and $\underset{\_}{K_3=19.08\frac{\gamma t}{g}{a}^4}$.

Explanation of Solution

Given information:

Refer Problem 9.168.

Show the moment of inertia as follows:

$\begin{array}{l}{I}_x=18.91335\frac{\gamma t}{8}{a}^4\\ I_y=7.68953\frac{\gamma t}{8}{a}^4\\ I_z=12.00922\frac{\gamma t}{8}{a}^4\end{array}$

$\begin{array}{l}{I}_{xy}=1.33333\frac{\gamma t}{8}{a}^4\\ I_{yz}=7.14159\frac{\gamma t}{8}{a}^4\\ I_{zx}=0.66667\frac{\gamma t}{8}{a}^4\end{array}$

Calculation:

Show the Equation 9.56 as follows:

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

Substitute $18.91335\frac{\gamma t}{8}{a}^4$ for $I_x$, $7.68953\frac{\gamma t}{8}{a}^4$ for $I_y$, $12.00922\frac{\gamma t}{8}{a}^4$ for $I_z$, $1.33333\frac{\gamma t}{8}{a}^4$ for $I_{xy}$, $7.14159\frac{\gamma t}{8}{a}^4$ for $I_{yz}$, and $0.66667\frac{\gamma t}{8}{a}^4$ for $I_{zx}$.

$\left[\begin{array}{l}{K}^3-\left(18.91335\frac{\gamma t}{8}{a}^4+7.68953\frac{\gamma t}{8}{a}^4+12.00922\frac{\gamma t}{8}{a}^4\right)K^2+\\ \left(\begin{array}{l}18.91335\frac{\gamma t}{8}{a}^4\times 7.68953\frac{\gamma t}{8}{a}^4+7.68953\frac{\gamma t}{8}{a}^4\times 12.00922\frac{\gamma t}{8}{a}^4\\ +12.00922\frac{\gamma t}{8}{a}^4\times 18.91335\frac{\gamma t}{8}{a}^4-{\left(1.33333\frac{\gamma t}{8}{a}^4\right)}^2-{\left(7.14159\frac{\gamma t}{8}{a}^4\right)}^2\\ -{\left(0.66667\frac{\gamma t}{8}{a}^4\right)}^2\end{array}\right)K\\ -\left(\begin{array}{l}18.91335\frac{\gamma t}{8}{a}^4\times 7.68953\frac{\gamma t}{8}{a}^4\times 12.00922\frac{\gamma t}{8}{a}^4\\ -18.91335\frac{\gamma t}{8}{a}^4\times {\left(7.14159\frac{\gamma t}{8}{a}^4\right)}^2-7.68953\frac{\gamma t}{8}{a}^4\times {\left(0.66667\frac{\gamma t}{8}{a}^4\right)}^2\\ -12.00922\frac{\gamma t}{8}{a}^4\times {\left(1.33333\frac{\gamma t}{8}{a}^4\right)}^2-2\left(1.33333\frac{\gamma t}{8}{a}^4\right)\left(7.14159\frac{\gamma t}{8}{a}^4\right)\times \left(0.66667\frac{\gamma t}{8}{a}^4\right)\end{array}\right)\end{array}\right]=0$

Consider the value of $K=K\frac{\gamma t}{8}{a}^4$.

$K^3-38.1210K^2+411.69009K-744.47027=0$

Solve the above Equation and get the value of $K_1=2.25890$, $K_2=17.27274$, and $K_3=19.08046$.

The principal moment of inertia are $K_1=2.26\frac{\gamma t}{g}{a}^4$, $K_2=17.27\frac{\gamma t}{g}{a}^4$, and $K_3=19.08\frac{\gamma t}{g}{a}^4$.

Thus, The principal moment of inertia are $\underset{\_}{K_1=2.26\frac{\gamma t}{g}{a}^4}$, $\underset{\_}{K_2=17.27\frac{\gamma t}{g}{a}^4}$, and $\underset{\_}{K_3=19.08\frac{\gamma t}{g}{a}^4}$.

(b)

To determine

Find the angles made by the principal axis of inertia at O with the coordinate axis.

Sketch the body and show the orientation of the principal axis of inertia relative to x, y, and z axis.

Answer to Problem 9.183P

The angles made by the principal axis of inertia at O with the coordinate axis is,

${\left(\theta x\right)}_1=85°$, ${\left(\theta y\right)}_1=36.8°$, ${\left(\theta z\right)}_1=53.7°$

${\left(\theta x\right)}_2=81.7°$, ${\left(\theta y\right)}_2=54.7°$, ${\left(\theta z\right)}_1=143.4°$

${\left(\theta x\right)}_3=9.70°$, ${\left(\theta y\right)}_3=99°$, ${\left(\theta z\right)}_3=86.3°$

Explanation of Solution

Given information:

Consider the direction cosines of each principal axis are denoted by ${\lambda }_x,{\lambda }_y,{\lambda }_z$.

Calculation:

Refer Part (a).

Consider K1.

Show the Equation 9.54 as follows:

$\begin{array}{l}\left(I_x-K_1\right){\left({\lambda }_x\right)}_1-I_{xy}{\left({\lambda }_y\right)}_1-I_{zx}{\left({\lambda }_z\right)}_1=0\\ -I_{xy}{\left({\lambda }_x\right)}_1-I_{yz}{\left({\lambda }_z\right)}_1+\left(I_y-K_1\right){\left({\lambda }_y\right)}_1=0\end{array}\}$ (1)

Substitute $18.91335\frac{\gamma t}{8}{a}^4$ for $I_x$, $7.68953\frac{\gamma t}{8}{a}^4$ for $I_y$, $1.33333\frac{\gamma t}{8}{a}^4$ for $I_{xy}$, $7.14159\frac{\gamma t}{8}{a}^4$ for $I_{yz}$, and $0.66667\frac{\gamma t}{8}{a}^4$ for $I_{zx}$ and $2.26\frac{\gamma t}{g}{a}^4$ for $K_1$.

$\begin{array}{l}\left(18.91335\frac{\gamma t}{8}{a}^4-2.26\frac{\gamma t}{g}{a}^4\right){\left({\lambda }_x\right)}_1-1.33333\frac{\gamma t}{8}{a}^4{\left({\lambda }_y\right)}_1-0.66667\frac{\gamma t}{8}{a}^4{\left({\lambda }_z\right)}_1=0\\ -1.33333\frac{\gamma t}{8}{a}^4{\left({\lambda }_x\right)}_1-7.14159\frac{\gamma t}{8}{a}^4{\left({\lambda }_z\right)}_1+\left(7.68953\frac{\gamma t}{8}{a}^4-2.26\frac{\gamma t}{g}{a}^4\right){\left({\lambda }_y\right)}_1=0\end{array}\}$ (2)

Solve Equation (2).

Get the value of ${\left({\lambda }_z\right)}_1=6.74653{\left({\lambda }_x\right)}_1,{\left({\lambda }_y\right)}_1=9.11761{\left({\lambda }_x\right)}_1$.

Show the Equation 9.57 as follows:

$\begin{array}{l}{\left({\lambda }_x\right)}_1^2+{\left({\lambda }_y\right)}_1^2+{\left({\lambda }_z\right)}_1^2=1\\ {\left({\lambda }_x\right)}_1^2+{\left[9.11761{\left({\lambda }_x\right)}_1\right]}^2+{\left[6.74653{\left({\lambda }_x\right)}_1\right]}^2=1\end{array}$

Solve above Equation and get the value of ${\left({\lambda }_x\right)}_1=0.087825,{\left({\lambda }_y\right)}_1=0.80075,{\left({\lambda }_z\right)}_1=0.59251$.

Show the direction cosines ${\left({\theta }_x\right)}_1,{\left({\theta }_y\right)}_1,{\left({\theta }_z\right)}_1$ using the relation:

$\begin{array}{c}\cos {\left({\theta }_x\right)}_1={\left({\lambda }_x\right)}_1=0.087825\\ {\left({\theta }_x\right)}_1={\cos }^{-1}\left(0.087825\right)\\ =85°\end{array}$

$\begin{array}{c}\cos {\left({\theta }_z\right)}_1={\left({\lambda }_z\right)}_1=0.59251\\ {\left({\theta }_z\right)}_1={\cos }^{-1}\left(0.59251\right)\\ =53.7°\end{array}$

$\begin{array}{c}\cos {\left({\theta }_y\right)}_1={\left({\lambda }_y\right)}_1=0.80075\\ {\left({\theta }_y\right)}_1={\cos }^{-1}\left(0.80075\right)\\ =36.8°\end{array}$

Consider K2.

Show the Equation 9.54 as follows:

$\begin{array}{l}\left(I_x-K_2\right){\left({\lambda }_x\right)}_2-I_{xy}{\left({\lambda }_y\right)}_2-I_{zx}{\left({\lambda }_z\right)}_2=0\\ -I_{xy}{\left({\lambda }_x\right)}_2-I_{yz}{\left({\lambda }_z\right)}_2+\left(I_y-K_2\right){\left({\lambda }_y\right)}_2=0\end{array}\}$ (3)

Substitute $18.91335\frac{\gamma t}{8}{a}^4$ for $I_x$, $7.68953\frac{\gamma t}{8}{a}^4$ for $I_y$, $1.33333\frac{\gamma t}{8}{a}^4$ for $I_{xy}$, $7.14159\frac{\gamma t}{8}{a}^4$ for $I_{yz}$, and $0.66667\frac{\gamma t}{8}{a}^4$ for $I_{zx}$ and $17.27\frac{\gamma t}{g}{a}^4$ for $K_2$.

$\begin{array}{l}\left(18.91335\frac{\gamma t}{8}{a}^4-17.27\frac{\gamma t}{g}{a}^4\right){\left({\lambda }_x\right)}_2-1.33333\frac{\gamma t}{8}{a}^4{\left({\lambda }_y\right)}_2-0.66667\frac{\gamma t}{8}{a}^4{\left({\lambda }_z\right)}_2=0\\ -1.33333\frac{\gamma t}{8}{a}^4{\left({\lambda }_x\right)}_2-7.14159\frac{\gamma t}{8}{a}^4{\left({\lambda }_z\right)}_2+\left(7.68953\frac{\gamma t}{8}{a}^4-17.27\frac{\gamma t}{g}{a}^4\right){\left({\lambda }_y\right)}_2=0\end{array}\}$ (4)

Solve Equation (4).

Get the value of ${\left({\lambda }_z\right)}_2=-5.58515{\left({\lambda }_x\right)}_2,{\left({\lambda }_y\right)}_2=4.02304{\left({\lambda }_x\right)}_2$.

Show the Equation 9.57 as follows:

$\begin{array}{l}{\left({\lambda }_x\right)}_2^2+{\left({\lambda }_y\right)}_2^2+{\left({\lambda }_z\right)}_2^2=1\\ {\left({\lambda }_x\right)}_2^2+{\left[4.02304{\left({\lambda }_x\right)}_2\right]}^2+{\left[-5.58515{\left({\lambda }_x\right)}_2\right]}^2=1\end{array}$

Solve above Equation and get the value of ${\left({\lambda }_x\right)}_2=0.14377,{\left({\lambda }_y\right)}_2=\mathrm{0.0.57839},{\left({\lambda }_z\right)}_2=-0.80298$.

Show the direction cosines ${\left({\theta }_x\right)}_2,{\left({\theta }_y\right)}_2,{\left({\theta }_z\right)}_2$ using the relation:

$\begin{array}{c}\cos {\left({\theta }_x\right)}_2={\left({\lambda }_x\right)}_2=0.14377\\ {\left({\theta }_x\right)}_2={\cos }^{-1}\left(0.14377\right)\\ =81.7°\end{array}$

$\begin{array}{c}\cos {\left({\theta }_z\right)}_2={\left({\lambda }_z\right)}_2=-0.80298\\ {\left({\theta }_z\right)}_2={\cos }^{-1}\left(-0.80298\right)\\ =143.4°\end{array}$

$\begin{array}{c}\cos {\left({\theta }_y\right)}_2={\left({\lambda }_y\right)}_2=\mathrm{0.0.57839}\\ {\left({\theta }_y\right)}_2={\cos }^{-1}\left(\mathrm{0.0.57839}\right)\\ =54.7°\end{array}$

Consider K3.

Show the Equation 9.54 as follows:

$\begin{array}{l}\left(I_x-K_3\right){\left({\lambda }_x\right)}_3-I_{xy}{\left({\lambda }_y\right)}_3-I_{zx}{\left({\lambda }_z\right)}_3=0\\ -I_{xy}{\left({\lambda }_x\right)}_3-I_{yz}{\left({\lambda }_z\right)}_3+\left(I_y-K_3\right){\left({\lambda }_y\right)}_3=0\end{array}\}$ (5)

Substitute $18.91335\frac{\gamma t}{8}{a}^4$ for $I_x$, $7.68953\frac{\gamma t}{8}{a}^4$ for $I_y$, $1.33333\frac{\gamma t}{8}{a}^4$ for $I_{xy}$, $7.14159\frac{\gamma t}{8}{a}^4$ for $I_{yz}$, and $0.66667\frac{\gamma t}{8}{a}^4$ for $I_{zx}$ and $19.08\frac{\gamma t}{g}{a}^4$ for $K_3$.

$\begin{array}{l}\left(18.91335\frac{\gamma t}{8}{a}^4-19.08\frac{\gamma t}{g}{a}^4\right){\left({\lambda }_x\right)}_2-1.33333\frac{\gamma t}{8}{a}^4{\left({\lambda }_y\right)}_2-0.66667\frac{\gamma t}{8}{a}^4{\left({\lambda }_z\right)}_2=0\\ -1.33333\frac{\gamma t}{8}{a}^4{\left({\lambda }_x\right)}_2-7.14159\frac{\gamma t}{8}{a}^4{\left({\lambda }_z\right)}_2+\left(7.68953\frac{\gamma t}{8}{a}^4-19.08\frac{\gamma t}{g}{a}^4\right){\left({\lambda }_y\right)}_2=0\end{array}\}$ (6)

Solve Equation (6).

Get the value of ${\left({\lambda }_z\right)}_3=-0.06522{\left({\lambda }_x\right)}_3,{\left({\lambda }_y\right)}_3=-0.15794{\left({\lambda }_x\right)}_3$.

Show the Equation 9.57 as follows:

$\begin{array}{l}{\left({\lambda }_x\right)}_3^2+{\left({\lambda }_y\right)}_3^2+{\left({\lambda }_z\right)}_3^2=1\\ {\left({\lambda }_x\right)}_3^2+{\left[-0.15794{\left({\lambda }_x\right)}_3\right]}^2+{\left[-0.06522{\left({\lambda }_x\right)}_3\right]}^2=1\end{array}$

Solve above Equation and get the value of ${\left({\lambda }_x\right)}_3=0.98571,{\left({\lambda }_y\right)}_3=-0.15568,{\left({\lambda }_z\right)}_3=0.06429$.

Show the direction cosines ${\left({\theta }_x\right)}_3,{\left({\theta }_y\right)}_3,{\left({\theta }_z\right)}_3$ using the relation:

$\begin{array}{c}\cos {\left({\theta }_x\right)}_3={\left({\lambda }_x\right)}_3=0.98571\\ {\left({\theta }_x\right)}_3={\cos }^{-1}\left(0.98571\right)\\ =9.7°\end{array}$

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

$\begin{array}{c}\cos {\left({\theta }_y\right)}_3={\left({\lambda }_y\right)}_3=-0.15568\\ {\left({\theta }_y\right)}_3={\cos }^{-1}\left(-0.15568\right)\\ =86.3°\end{array}$

The angles made by the principal axis of inertia at O with the coordinate axis is,

${\left(\theta x\right)}_1=85°$, ${\left(\theta y\right)}_1=36.8°$, ${\left(\theta z\right)}_1=53.7°$

${\left(\theta x\right)}_2=81.7°$, ${\left(\theta y\right)}_2=54.7°$, ${\left(\theta z\right)}_1=143.4°$

${\left(\theta x\right)}_3=9.70°$, ${\left(\theta y\right)}_3=99°$, ${\left(\theta z\right)}_3=86.3°$

Sketch the body and show the orientation of the principal axis of inertia relative to x, y, and z axis as shown in Figure 1.