91Ó°ÊÓ

The initial angular speed is ${\mathrm{Ó\neg }}_{0_z}=500 rpm$.

The angular displacement is $\mathrm{θ}-{\mathrm{θ}}_0=200 rev$.

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.

$$\begin{gathered} \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{θ}\mathbf{−}{\mathbf{θ}}_{\mathbf{0}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\left({\mathrm{Ó\neg }}_z+{\mathrm{Ó\neg }}_{0_z}\right)\mathbf{t} \\ \mathbf{2}\left(\mathrm{θ}−{\mathrm{θ}}_0\right)\mathbf{=}\left({\mathrm{Ó\neg }}_z+{\mathrm{Ó\neg }}_{0_z}\right)\mathbf{t} \\ \frac{\mathbf{2}\left(\mathrm{θ}−{\mathrm{θ}}_0\right)}{\mathbf{t}}\mathbf{=}{\mathbf{Ó\neg }}_{\mathbf{z}}\mathbf{+}{\mathbf{Ó\neg }}_{{\mathbf{0}}_{\mathbf{z}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathbf{Ó\neg }}_{\mathbf{z}}\mathbf{=}\frac{\mathbf{2}\left(\mathrm{θ}−{\mathrm{θ}}_0\right)}{\mathbf{t}}\mathbf{−}{\mathbf{Ó\neg }}_{{\mathbf{0}}_{\mathbf{z}}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(1\right) \end{gathered}$$

Substitute ${\mathrm{Ó\neg }}_{0_z}=500 rpm$, $\mathrm{θ}-{\mathrm{θ}}_0=200 rev$, and t =0.5 min in equation (1).

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

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.

$$\begin{gathered} {\mathrm{Ó\neg }}_z={\mathrm{Ó\neg }}_{0_z}-{\mathrm{Î\pm }}_{z}t \\ {\mathrm{Î\pm }}_z=\frac{{\mathrm{Ó\neg }}_{0_z}-{\mathrm{Ó\neg }}_z}{t}......\left(2\right) \end{gathered}$$

Substitute ${\mathrm{Ó\neg }}_{0_z}=500 rpm$, t=30 s , and${\mathrm{Ó\neg }}_z=300 rpm$in equation ${\mathrm{Ó\neg }}_z^2={{\mathrm{Ó\neg }}_{o_z}}^2+2{\mathrm{Î\pm }}_z\mathrm{θ}$.

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

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

Substitute ${\mathrm{Ó\neg }}_{0_z}=500 rpm$,${\mathrm{Î\pm }}_z=0.11 rev/s^2$ , and${\mathrm{Ó\neg }}_z=0 rpm$in equation (2) to find time.

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

Therefore, the required time is 75.7 sec.

Use the following expression to find the angular displacement.

$\mathrm{Δ}\mathrm{θ}={\mathrm{Ó\neg }}_{0_z}t-1/2{\mathrm{Î\pm }}_z{t}^2         ......\left(3\right)$

Substitute ${\mathrm{Ó\neg }}_{0_z}=8.33 rev/s$,${\mathrm{Î\pm }}_z=0.11 rev/s^2$ , and t=75.7 sec in equation (3).

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

Therefore, the angular displacement is 315 rev.