An electric fan is turned off, and its angular velocity decreases uniformly from 500 rev/min to 200 rev/min in 4.00 s. (a) Find the angular acceleration in and the number of revolutions made by the motor in the 4.00-s interval. (b) How many more seconds are required for the fan to come to rest if the angular acceleration remains constant at the value calculated in part (a)?

The initial angular speed is ${\mathrm{\omega }}_{0\mathrm{z}}=500\mathrm{rev}/\min$.

The final angular speed is ${\mathrm{\omega }}_{\mathrm{z}}=200\mathrm{rev}/\min$.

The time is t = 4 s.

Angular velocity at time t of a rigid body with constant angular acceleration is given by,

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

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

Angular position at time t of a rigid body with constant angular acceleration is given by,

role="math" localid="1667980841517" $\mathit{\theta }\mathbf{=}{\mathit{\theta }}_{\mathbf{0}}\mathbf{+}{\mathit{\omega }}_{\mathbf{0}\mathbf{z}}\mathit{t}\mathbf{\pm }\frac{\mathbf{1}}{\mathbf{2}}{\mathit{\alpha }}_{\mathbf{z}}{\mathit{t}}^{\mathbf{2}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{2}\mathbf{\right)}$

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

Convert the initial angular speed in .

$$\begin{gathered} {\mathrm{\omega }}_{0\mathrm{z}}=\left(500\frac{\mathrm{rev}}{\min }\right)\left(\frac{1\min }{60\sec }\right) \\ =8.33\mathrm{rev}/\mathrm{s} \end{gathered}$$

Convert the final angular speed in .

$$\begin{gathered} {\mathrm{\omega }}_{\mathrm{z}}=\left(200\frac{\mathrm{rev}}{\min }\right)\left(\frac{1\min }{60\sec }\right) \\ =3.33\mathrm{rev}/\mathrm{s} \end{gathered}$$

From the initial and final angular speed, it can be concluded that the motion of the body is deceleration.

Substitute ${\mathrm{\omega }}_{0\mathrm{z}}=8.33\mathrm{rev}/\mathrm{s},\mathrm{t}=4\mathrm{s}$and${\mathrm{\omega }}_{\mathrm{z}}=3.33\mathrm{rev}/\mathrm{s}$ in equation${\omega }_z={\omega }_z-{\alpha }_{z}t$ .

$$\begin{gathered} 3.33\mathrm{rev}/\mathrm{s}=8.33\mathrm{rev}/\mathrm{s}-{\mathrm{\alpha }}_{\mathrm{z}}\left(4\mathrm{s}\right) \\ {\mathrm{\alpha }}_{\mathrm{z}}=\frac{8.33\mathrm{rev}/\mathrm{s}-3.33\mathrm{rev}/\mathrm{s}}{4\mathrm{s}} \\ =\frac{5\mathrm{rev}/\mathrm{s}}{4\mathrm{s}} \\ =1.25\mathrm{rev}/{\mathrm{s}}^2 \end{gathered}$$

Therefore, the angular acceleration is$1.25\mathrm{rev}/{\mathrm{s}}^2$ .

Substitute ${\mathrm{\theta }}_0=0,{\mathrm{\omega }}_{0\mathrm{z}}=8.33\mathrm{rev}/\mathrm{s},\mathrm{t}=4\mathrm{s},$and${\mathrm{\alpha }}_{\mathrm{z}}=1.25\mathrm{rev}/{\mathrm{s}}^2$ in equation .

$$\begin{gathered} \mathrm{\theta }={\mathrm{\theta }}_0+{\mathrm{\omega }}_{0\mathrm{z}}\mathrm{t}-\frac{1}{2}{\mathrm{\alpha t}}^2 \\ \mathrm{\theta }=0+\left(8.33\mathrm{rev}/\mathrm{s}\right)\left(4\mathrm{s}\right)-\frac{1}{2}\left(1.25\mathrm{rev}/{\mathrm{s}}^2\right){\left(4\mathrm{s}\right)}^2 \\ =23.32\mathrm{rev} \end{gathered}$$

Therefore, the number of revolutions are 23.32 rev.

When the body stops then ${\omega }_z=0$.

Substitute role="math" localid="1667981776572" ${\mathrm{\omega }}_{0_{\mathrm{z}}}=8.33\mathrm{rev}/\mathrm{s},{\mathrm{\alpha }}_{\mathrm{z}}=1.25\mathrm{rev}/{\mathrm{s}}^2$,and${\mathrm{\omega }}_{\mathrm{z}}=0\mathrm{rev}/\mathrm{s}$ in equation .

${\mathrm{\omega }}_{\mathrm{z}}={\mathrm{\omega }}_{0_{\mathrm{z}}}-{\mathrm{\alpha }}_{\mathrm{z}}\mathrm{t}$

$$\begin{gathered} 0\mathrm{rev}/\mathrm{s}=8.33\mathrm{rev}/\mathrm{s}-\left(1.25\mathrm{rev}/{\mathrm{s}}^2\right)\mathrm{t} \\ \mathrm{t}=\frac{8.33\mathrm{rev}/\mathrm{s}}{1.25\mathrm{rev}/{\mathrm{s}}^2} \\ =6.66\mathrm{rev}/{\mathrm{s}}^2 \end{gathered}$$

Find the time it takes to stop after t = 4 s .

$$\begin{gathered} {\mathrm{t}}_{\mathrm{rest}}=6.66\mathrm{s}-4\mathrm{s} \\ =2.66\mathrm{s} \end{gathered}$$

Therefore, the required time is 2.66 s.