(a) Derive Eq. (9.12) by combining Eqs. (9.7) and (9.11) to eliminate t. (b) The angular velocity of an airplane propeller increases from 12.0 rad/s to 16.0 rad/s while turning through 7.00 rad. What is the angular acceleration in ?

The initial angular speed is${\omega }_{0_z}=12rad/s$ .

The final angular speed is${\omega }_z=16rad/s$ .

The angular displacement is$\theta =7rad$ .

The equation (9.7) is given by,

${\mathit{\omega }}_{\mathbf{z}}\mathbf{=}{\mathit{\omega }}_{{\mathbf{0}}_{\mathbf{z}}}\mathbf{+}{\mathit{\alpha }}_{\mathbf{z}}\mathit{t}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(1\right)$

Here,${\mathit{\alpha }}_{\mathbf{z}}$ is the angular acceleration,${\mathit{\omega }}_{{\mathbf{0}}_{\mathbf{z}}}$ is the angular velocity of the body at time 0, t is the time.

The equation (9.11) is given by,

$\mathit{\theta }\mathbf{=}{\mathit{\theta }}_{\mathbf{0}}\mathbf{+}{\mathit{\omega }}_{{\mathbf{0}}_{\mathbf{z}}}\mathit{t}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}{\mathit{\alpha }}_{\mathbf{z}}{\mathit{t}}^{\mathbf{2}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(2\right)$

Here,${\mathit{\theta }}_{\mathbf{0}}$ is the Angular position of the body at time 0.

Rewrite the equation (1) as follows.

$$\begin{gathered} {\omega }_z={\omega }_{0_z}+{\alpha }_{z}t \\ {\alpha }_{z}t={\omega }_z-{\omega }_{0_z} \\ t=\frac{{\omega }_z-{\omega }_{0_z}}{{\alpha }_z} \end{gathered}$$

Substitute$t=\frac{{\omega }_z-{\omega }_{0_z}}{{\alpha }_z}$ in equation (2).

$$\begin{gathered} \theta ={\theta }_0+{\omega }_{0_z}\left(\frac{{\omega }_z-{\omega }_{0_z}}{{\alpha }_z}\right)+\frac{1}{2}{\alpha }_z{\left(\frac{{\omega }_z-{\omega }_{0_z}}{{\alpha }_z}\right)}^2 \\ =\frac{1}{{\alpha }_z}\left[{\omega }_{0_z}\left({\omega }_z-{\omega }_{0_z}\right)+\frac{1}{2}{\left({\omega }_z-{\omega }_{0_z}\right)}^2\right] \\ =\frac{\left({\omega }_z-{\omega }_{0_z}\right)}{{\alpha }_z}\left[{\omega }_{0_z}+\frac{1}{2}\left({\omega }_z-{\omega }_{0_z}\right)\right] \\ =\frac{\left({\omega }_z-{\omega }_{0_z}\right)\left({\omega }_z+{\omega }_{0_z}\right)}{2{\alpha }_z} \end{gathered}$$

Simplify further.

$$\begin{gathered} \theta =\frac{\left({\omega }_z^2-{\omega }_{0_2}^2\right)}{2{\alpha }_z} \\ {\omega }_z^2-{\omega }_{0_2}^2=2{\alpha }_z\theta \\ {\omega }_z^2={\omega }_{0_2}^2+2{\alpha }_z\theta \end{gathered}$$

Therefore, the required expression is${\omega }_z^2={\omega }_{0_2}^2+2{\alpha }_z\theta$ .

Substitute ${\omega }_{0_z}^2=12rad/s,\theta =7rad,and{\omega }_z=16rad/s$and in equation${\omega }_z^2={\omega }_{0_z}^2+2{\alpha }_z\theta$ .

$$\begin{gathered} {\left(16rad/s\right)}^2={\left(12rad/s\right)}^2+2{\alpha }_z\left(7rad\right) \\ {\alpha }_z=\frac{{\left(16rad/s\right)}^2-{\left(12rad/s\right)}^2}{2\left(7rad\right)} \\ =8rad/s^2 \end{gathered}$$

Therefore, the angular acceleration is$8rad/s^2$ .