Chapter 20, Problem 1P

The propeller of an airplane is rotating at a constant speed ω_xi, while the plane is undergoing a turn at a constant rate ω_t. Determine the angular acceleration of the propeller. If (a) the turn is horizontal, i.e., ω_t k, and (b) the turn is vertical, downward, i.e., ω_t j.

Prob. 20-1

To determine

1. (a) The angular acceleration of the turn is horizontal ${\omega }_t\overrightarrow{\text{k}}$.
2. (b) The angular acceleration of the turn is vertical, downward ${\omega }_t\overrightarrow{\text{j}}$.

Answer to Problem 1P

1. (a) The angular acceleration of the turn is horizontal ${\omega }_t\overrightarrow{\text{k}}$ is $\underset{\_}{\alpha ={\omega }_x{\omega }_t\text{<mover accent="true">j<mo>→</mo></mover>}}$.
2. (b) The angular acceleration of the turn is vertical, downward ${\omega }_t\text{<mover accent="true">j<mo>→</mo></mover>}$ is $\underset{\_}{\alpha =-{\omega }_x{\omega }_t\text{<mover accent="true">k<mo>→</mo></mover>}}$.

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:

1. (a) 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 $\underset{\_}{\alpha ={\omega }_x{\omega }_t\text{<mover accent="true">j<mo>→</mo></mover>}}$.

1. (b) 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 turn is vertical, downward ${\omega }_t\text{<mover accent="true">j<mo>→</mo></mover>}$ is $\underset{\_}{\alpha =-{\omega }_x{\omega }_t\text{<mover accent="true">k<mo>→</mo></mover>}}$.