Chapter 4.1, Problem 67E

a.

To determine

To find: the road speed (in miles per hour) at which the tire is being balanced.

Answer to Problem 67E

  $\text{35}\text{.70 mi/h}$

Explanation of Solution

Given:

Length of tire $d=25\text{ in}\text{.}$

No. of revolution per minute = 480

Concept Used:

If $\theta$ is the angle (in radian) corresponding to the arc length s , then the linear speed v is:

  $v=\frac{s}{t}$

The circumference of the tire,

  $\begin{array}{c}C=\pi d\\ =25\pi \text{ in}\text{.}\\ \text{=}\frac{25\pi }{63360}\text{ miles}\end{array}$

The above value is for 1 revolution but total number of revolution per minute is 480.

So, in 480 revolutions, so arc length s here is,

  $s=480\times \frac{25\pi }{63360}\text{ miles}$ .

So, the road speed which is the linear speed is,

  $\begin{array}{c}v=\frac{s}{t}\\ =\frac{480\times \frac{25\pi }{63360}\text{ miles}}{1\text{ minute}}\\ =\frac{480\times \frac{25\pi }{63360}\text{ miles}}{\frac{1}{60}\text{ hour}}\\ \approx \text{35}\text{.70 miles per hour}\end{array}$

b.

To determine

To find: the rate at which the spin balance machine should be set so that the tire is being tested for 55 miles per hour.

Answer to Problem 67E

Approximately $\text{739}\text{.50 revolutions/min}$

Explanation of Solution

Given that the tire is being tested for 55 miles per hour.

Convert it to miles per minute.

  $\begin{array}{c}v=\frac{55\text{ mi}}{h}\\ =\frac{55\text{ mi}}{h}\times \frac{h}{60\text{ min}}\\ =\frac{\text{55 mi}}{60\min }\end{array}$

From part (a), circumference of tire is,

  $\begin{array}{c}C=\pi d\\ =25\pi \text{ in}\text{.}\\ \text{=}\frac{25\pi }{63360}\text{ mi}\end{array}$

The required rate is,

  $\begin{array}{c}\frac{v}{C}=\frac{\frac{\text{55 mi}}{60\min }}{\frac{25\pi }{63360}\text{ mi}}\\ \approx 739.50\text{ revolutions per minute}\end{array}$