Question

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?

Short answer

1. The time fortheflywheel to come to rest, $\mathrm{t}=340\text{s}$
2. Angular acceleration oftheflywheel, $\mathrm{\alpha }=-4.5\times {10}^{-3}\frac{\text{rad}}{{\text{s}}^2}$
3. The time required for the flywheel to complete first 20 revolutions, $\mathrm{t}=98\mathrm{s}$.

Step-by-step solution

Listing the given quantities 

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

Understanding the concept of angular speed and displacement  

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.

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

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

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

(a) Determine the time for the flywheel to come at rest

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, $\mathrm{t}=340\text{s}$.

(b) Determine the angular acceleration of the flywheel

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, $\mathrm{\alpha }=-4.5\times {10}^{-3}\frac{\text{rad}}{{\text{s}}^2}$

(c) Consider the time required to complete first 20 revolutions is determined as:

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 $98\mathrm{s}$.