A high-speed flywheel in a motor is spinning at 500 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and diameter 75.0 cm. The power is off for 30.0 s, and during this time the flywheel slows due to friction in its axle bearings. During the time the power is off, the flywheel makes 200 complete revolutions. (a) At what rate is the flywheel spinning when the power comes back on? (b) How long after the beginning of the power failure would it have taken the flywheel to stop if the power had not come back on, and how many revolutions would the wheel have made during this time?

The initial angular speed is ${\omega }_{0_z}=500\hspace{0.33em}rpm$.

The angular displacement is $\theta -{\theta }_0=200\hspace{0.33em}rev$.

The time is t=30 s or t =0.5 m in..

Derive the required expression to find the rate of spinning of flywheel as follows.

$$\begin{gathered} \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\theta }\mathbf{-}{\mathbf{\theta }}_{\mathbf{0}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\left({\omega }_z+{\omega }_{0_z}\right)\mathbf{t} \\ \mathbf{2}\left(\theta -{\theta }_0\right)\mathbf{=}\left({\omega }_z+{\omega }_{0_z}\right)\mathbf{t} \\ \frac{\mathbf{2}\left(\theta -{\theta }_0\right)}{\mathbf{t}}\mathbf{=}{\mathbf{\omega }}_{\mathbf{z}}\mathbf{+}{\mathbf{\omega }}_{{\mathbf{0}}_{\mathbf{z}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathbf{\omega }}_{\mathbf{z}}\mathbf{=}\frac{\mathbf{2}\left(\theta -{\theta }_0\right)}{\mathbf{t}}\mathbf{-}{\mathbf{\omega }}_{{\mathbf{0}}_{\mathbf{z}}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(1\right) \end{gathered}$$

Substitute ${\omega }_{0_z}=500\hspace{0.33em}rpm$, $\theta -{\theta }_0=200\hspace{0.33em}rev$, and t =0.5 min in equation (1).

$$\begin{gathered} {\mathrm{\omega }}_{\mathrm{z}}=\frac{2\left(200\mathrm{rav}\right)}{0.5\min }-500\mathrm{rpm} \\ =800\mathrm{rpm}-500\mathrm{rpm} \\ =300\mathrm{rpm} \end{gathered}$$

Therefore, the rate of spinning when the power comes back is 300 rpm.

From the above results, it can be observed that the speed is decreasing which means the flywheel is decelerated.

Use the following expression to find the angular acceleration.

$$\begin{gathered} {\omega }_z={\omega }_{0_z}-{\alpha }_{z}t \\ {\alpha }_z=\frac{{\omega }_{0_z}-{\omega }_z}{t}......\left(2\right) \end{gathered}$$

Substitute ${\omega }_{0_z}=500\hspace{0.33em}rpm$, t=30 s , and${\omega }_z=300\hspace{0.33em}rpm$in equation ${\omega }_z^2={{\omega }_{o_z}}^2+2{\alpha }_z\theta$.

$$\begin{gathered} {\alpha }_z=\frac{500\mathrm{rpm}-300\mathrm{rpm}}{30\mathrm{s}} \\ =\frac{200\mathrm{rpm}}{30\mathrm{s}} \\ =\frac{1}{30\mathrm{s}}\left(200\frac{\mathrm{rev}}{\min }\times \frac{1\min }{60\sec }\right) \\ =0.11\mathrm{rev}/{\mathrm{s}}^2 \end{gathered}$$

At the end the flywheel is stopped, so take ${\omega }_z=0$.

Substitute ${\omega }_{0_z}=500\hspace{0.33em}rpm$,${\alpha }_z=0.11\hspace{0.33em}rev/s^2$ , and${\omega }_z=0\hspace{0.33em}rpm$in equation (2) to find time.

$$\begin{gathered} t=\frac{\left(500\frac{\mathrm{rev}}{\min }\times \frac{1\min }{60\sec }\right)-0}{0.11\mathrm{rev}/{\mathrm{s}}^2} \\ =\frac{8.33\mathrm{rev}/\mathrm{s}}{0.11\mathrm{rev}/{\mathrm{s}}^2} \\ =75.7\sec \end{gathered}$$

Therefore, the required time is 75.7 sec.

Use the following expression to find the angular displacement.

$\Delta \theta ={\omega }_{0_z}t-1/2{\alpha }_z{t}^2\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}......\left(3\right)$

Substitute ${\omega }_{0_z}=8.33\hspace{0.33em}rev/s$,${\alpha }_z=0.11\hspace{0.33em}rev/s^2$ , and t=75.7 sec in equation (3).

$$\begin{gathered} \mathrm{\Delta }\theta =(8.33\mathrm{rev}/\mathrm{s})\left(75.7\sec \right)-\frac{1}{2}\left(0.11\mathrm{rev}/{\mathrm{s}}^2\right)(75.7\sec {)}^2 \\ =630\mathrm{rev}-315\mathrm{rev} \\ =315\mathrm{rev} \end{gathered}$$

Therefore, the angular displacement is 315 rev.