Chapter 20.3, Problem 1P

(a)

To determine

The angular acceleration of the propeller when the turn is horizontal.

Answer to Problem 1P

The angular acceleration of the propeller when the turn is horizontal is $\underset{\_}{\alpha ={\omega }_x{\omega }_t\overrightarrow{\text{j}}}$.

Explanation of Solution

Write the expression of angular acceleration at constant speed.

    ${\left({\dot{\omega }}_x\right)}_{XYZ}={\left({\dot{\omega }}_x\right)}_{xyz}+\Omega \times {\omega }_x$        (I)

Write the expression of angular acceleration turning at constant rate.

    ${\left({\dot{\omega }}_t\right)}_{XYZ}={\left({\dot{\omega }}_t\right)}_{xyz}+\Omega \times {\omega }_t$        (II)

Here, $\dot{\omega }$ for the angular acceleration, $x,y,z$ for the translating-rotating frame of reference, $X,Y,Z$ for the fixed frame of reference, and $\Omega$ for angular velocity.

Write the expression of angular acceleration.

    $\alpha ={\left({\dot{\omega }}_x\right)}_{XYZ}+{\left({\dot{\omega }}_t\right)}_{XYZ}$        (III)

Conclusion:

Substitute ${\omega }_t\overrightarrow{\text{k}}$ for $\Omega$, $0$ for ${\left({\dot{\omega }}_x\right)}_{xyz}$, and ${\dot{\omega }}_s\overrightarrow{\text{i}}$ for ${\dot{\omega }}_s$ in Equation (I).

    $\begin{array}{c}{\left({\dot{\omega }}_x\right)}_{XYZ}=0+\left({\dot{\omega }}_t\overrightarrow{\text{k}}\right)\times \left({\dot{\omega }}_x\overrightarrow{\text{i}}\right)\\ ={\omega }_x{\omega }_t\overrightarrow{\text{j}}\end{array}$

Substitute $0$ for $\Omega$, and $0$ for ${\left({\dot{\omega }}_x\right)}_{xyz}$ in Equation (II).

    $\begin{array}{c}{\left({\dot{\omega }}_t\right)}_{XYZ}=0+0\\ =0\end{array}$

Substitute ${\omega }_x{\omega }_t\overrightarrow{\text{j}}$ for ${\left({\dot{\omega }}_x\right)}_{XYZ}$ and $0$ for ${\left({\dot{\omega }}_t\right)}_{XYZ}$ in Equation (III).

    $\begin{array}{c}\alpha ={\omega }_x{\omega }_t\overrightarrow{\text{j}}+0\\ ={\omega }_x{\omega }_t\overrightarrow{\text{j}}\end{array}$

Thus, the angular acceleration of the turn is horizontal ${\omega }_t\overrightarrow{\text{k}}$ is $\underset{\_}{\alpha ={\omega }_x{\omega }_t\overrightarrow{\text{j}}}$.

(b)

To determine

The angular acceleration of the propeller when the turn is vertical, downward.

Answer to Problem 1P

The angular acceleration of the propeller when the turn is vertical, downward is $\underset{\_}{\alpha =-{\omega }_x{\omega }_t\overrightarrow{\text{k}}}$.

Explanation of Solution

Substitute $-{\omega }_t\overrightarrow{\text{j}}$ for $\Omega$, $0$ for ${\left({\dot{\omega }}_x\right)}_{xyz}$, and ${\dot{\omega }}_s\overrightarrow{\text{i}}$ for ${\dot{\omega }}_s$ in Equation (I).

    $\begin{array}{c}{\left({\dot{\omega }}_x\right)}_{XYZ}=0+\left(-{\dot{\omega }}_t\overrightarrow{\text{j}}\right)\times \left({\dot{\omega }}_x\overrightarrow{\text{i}}\right)\\ =-{\omega }_x{\omega }_t\overrightarrow{\text{k}}\end{array}$

Substitute $0$ for $\Omega$, and $0$ for ${\left({\dot{\omega }}_x\right)}_{xyz}$ in Equation (II).

    $\begin{array}{c}{\left({\dot{\omega }}_t\right)}_{XYZ}=0+0\\ =0\end{array}$

Substitute $-{\omega }_x{\omega }_t\overrightarrow{\text{k}}$ for ${\left({\dot{\omega }}_x\right)}_{XYZ}$ and $0$ for ${\left({\dot{\omega }}_t\right)}_{XYZ}$ in Equation (III).

    $\begin{array}{c}\alpha =-{\omega }_x{\omega }_t\overrightarrow{\text{k}}+0\\ =-{\omega }_x{\omega }_t\overrightarrow{\text{k}}\end{array}$

Thus, the angular acceleration of the propeller when the turn is vertical, downward is $\underset{\_}{\alpha =-{\omega }_x{\omega }_t\overrightarrow{\text{k}}}$.