Chapter 6.2, Problem 39E

a.

To determine

To find: the number of times the wheel turns.

Answer to Problem 39E

2017

Explanation of Solution

Given information:

  $\begin{array}{c}\text{Diameter }\left(2\pi r\right)=30\text{ inches}\\ \text{Distance covered by tire}=3\text{ miles}\end{array}$

Calculation:

The revolution of one wheel $=30\pi \text{ inches}$

Total revolutions in 3 miles will be evaluated as follows:

  $\begin{array}{c}\text{Total revolutions}=\frac{3\text{ miles}}{30\pi \text{ inches}}\\ =\frac{3\text{ miles}}{30\pi \text{ inches}}\times \frac{5280\text{ feet}}{\text{1 mile}}\times \frac{\text{12 inches}}{\text{1 foot}}\left[\text{Changing miles into inches}\right]\\ =2016.81\\ \approx 2017\text{ revolutions}\end{array}$

Thus the total number of revolutions is 2017.

b.

To determine

To find: the linear speed in miles per hour of a point on a tire.

Answer to Problem 39E

  $14.7\text{ miles per hour}$

Explanation of Solution

Given information:

  $\begin{array}{c}\text{Diameter }\left(2\pi r\right)=30\text{ inches}\\ \omega =2.75\text{ revolutions per second}\end{array}$

Calculation:

The radius of the wheel is evaluated as follows:

  $\begin{array}{c}2\pi r=30\text{ inches}\\ r=\frac{30\text{ inches}}{2\pi }\\ =\frac{15\text{ inches}}{\pi }\end{array}$

Thus the radius is $\frac{15}{\pi }\text{inches}$ .

The linear velocity $\left(\nu \right)$ is evaluated as follows:

  $\begin{array}{c}v=r\omega \\ =\left(\frac{15}{\pi }\text{inches}\right)\left(\frac{\text{2}\text{.75}rev}{1\text{ second}}\right)\left[\text{Put the value of }r\text{ and }\omega \right]\\ =\left(\frac{15\text{inches}}{\pi }\right)\left(\frac{\text{2}\text{.75}rev}{1\text{ second}}\right)\left(\frac{2\pi \text{ radians}}{\text{1 revolution}}\right)\left[\text{Convert revolutions into radians}\right]\\ =\left(15\text{inches}\right)\times \frac{\left(\text{2}\text{.75}\right)\left(2\right)\left(3600\right)}{1\text{ second}}\\ =\left(\frac{\left(15\right)\left(\text{2}\text{.75}\right)\left(2\right)\left(3600\right)\text{inches}}{1\text{ second}}\right)\left(\frac{3600\text{ seconds}}{\text{1 hour}}\right)\left[\text{Convert second into hour}\right]\\ =\frac{933053.02\text{ inches}}{1\text{ hour}}\\ =\frac{933053.02\text{ inches}}{1\text{ hour}}\times \frac{1\text{ foot}}{12\text{ inches}}\times \frac{1\text{ mile}}{\text{5280 feet}}\left[\text{Convert inches into miles}\right]\\ =\left(\frac{933053.02}{12\times \text{5280}}\right)\left(\frac{1\text{ mile}}{1\text{ hour}}\right)\\ =14.7\text{ miles per hour}\end{array}$

Thus the linear velocity is $14.7\text{ miles per hour}$ .