A flywheel turns through $\mathbf{40}\mathbf{}\mathbf{rev}\mathbf{}$ as it slows from an angular speed of $\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{rad}\mathbf{/}\mathbf{s}\mathbf{}$to a stop.

(a) Assuming a constant angular acceleration, find the time for it to come to rest.

(b) What is its angular acceleration?

(c) How much time is required for it to complete the first $\mathbf{20}\mathbf{}$ of the $\mathbf{40}$ revolutions?

The initial angular speed of theflywheel,${\mathrm{\omega }}_{\mathrm{i}}=1.5\frac{\text{rad}}{\text{s}}$

The final angular speed of the engine,$\mathrm{\omega }=0\frac{\text{rad}}{\text{s}}$

The angular displacement, $\mathrm{\theta }=40\text{\hspace{0.17em}rev}=251.3\text{rad}$

Use the kinematic equation for constant angular acceleration to calculate the time and angular acceleration. The time for 20 revolutions can be calculated by using the value of angular acceleration.

$\mathrm{\omega }={\mathrm{\omega }}_0+\mathrm{\alpha t}$

${\mathrm{\omega }}^2={\mathrm{\omega }}_0^2+2\mathrm{\alpha \theta }$

$\mathrm{\theta }=\frac{1}{2}(\mathrm{\omega }+{\mathrm{\omega }}_0)\mathrm{t}$

The kinematic equation for angular motion as

$\mathrm{\theta }=\frac{1}{2}(\mathrm{\omega }+{\mathrm{\omega }}_0)\mathrm{t}$

By rearranging equation for time solve further as:

$\begin{array}{c}\mathrm{t}=\frac{2\mathrm{\theta }}{\mathrm{\omega }+{\mathrm{\omega }}_0}\\ =\frac{2\times 251.3}{0+1.5}\\ =335\text{s}\\ \approx 340\text{s}\end{array}$

The time for the flywheel to come to rest, $\mathrm{t}=340\text{s}$.

Consider the formula for the angular acceleration as:

$\begin{array}{c}\mathrm{\omega }={\mathrm{\omega }}_0+\mathrm{\alpha t}\\ \mathrm{\alpha }=\frac{\mathrm{\omega }-{\mathrm{\omega }}_0}{\mathrm{t}}\end{array}$

Substitute the values and solve as:

$\begin{array}{c}\mathrm{\alpha }=\frac{0-1.5}{335}\\ \mathrm{\alpha }=-4.5\times {10}^{-3}\frac{\text{rad}}{{\text{s}}^2}\end{array}$

Angular acceleration of the flywheel, $\mathrm{\alpha }=-4.5\times {10}^{-3}\frac{\text{rad}}{{\text{s}}^2}$

Now, for $\mathrm{\theta }=20\text{rev}=125.6637\text{rad}$, consider the formula:

$\begin{array}{c}\mathrm{\theta }={\mathrm{\omega }}_0\mathrm{t}+\frac{1}{2}{\mathrm{\alpha t}}^2\\ 125.6637=1.5\mathrm{t}-\left(\frac{1}{2}\times 4.5\times {10}^{-3}\times {\mathrm{t}}^2\right)\\ 2.25\times {10}^{-3}{\mathrm{t}}^2-1.5\mathrm{t}+125.6637=0\end{array}$

Solution to the equation is:

$\mathrm{t}=98.2576\text{s}\text{\hspace{0.17em}or}568.409\text{sec}$

The practical value of t should be smaller, therefore,

$\begin{array}{c}\text{t}=98.2576\text{s}\\ \approx 98\text{s}\end{array}$

The time required for the flywheel to complete first 20 revolutions is $98\mathrm{s}$.