Chapter 4.1, Problem 108E

(a)

To determine

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

Answer to Problem 108E

  $941.7\text{rev}/\min$

Explanation of Solution

Given:

Diameter of the tire is 25 inch and number of revolution per min is 480.

Concept Used:

1 revolution = circumference of the tire.

Circumference = $\text{π}·\text{d}$

Calculation:

Given: Diameter of the tire is 25 inch&number of revolution per min is 480.

Circumference = $\text{π}·\text{d}$

  $=\text{25}\text{π}\text{inch}$

Now multiply with revaluation per minute = Circumference $·$ rev/min

= $\text{25}\text{π}·480=12000\text{π}\text{inch}/\min =37,680\text{inch}/\min [\pi =3.14]$

Now convert inch/min to mile / min = $\frac{37,680\text{inch}}{1\min }·\frac{1\text{ft}}{12\text{in}}·\frac{1\text{mile}}{5280\text{ft}}=0.594696\text{mile/min}$

Now convert mile/min to mile/hr

  $=\frac{0.594696\text{mile}}{1\min }·\frac{60\text{min}}{1\text{hr}}=35.68\text{miles/hr}\approx 35.7\text{miles}\text{/}\text{hr}$

Thus the road speed is $35.7\text{miles}\text{/}\text{hr}$

Thus,

(b)

To determine

Find the rate of the spin balance set so that the tire for 70 miles / hr

Answer to Problem 108E

Explanation of Solution

Given:

The diameter of the tire is 25 inch and the tested rate is 70 miles / hr

Concept Used:

1 revolution = circumference of the tire.

Circumference = $\text{π}·\text{d}$

Calculation:

Given: Diameter of the tire is 25 inch& tested speed is 70 miles/ hr

Circumference = $\text{π}·\text{d}$

  $=\text{25}\text{π}\text{inch}$

Convert the rate from miles / hr to miles/minute = $\frac{70\text{mile}}{1\text{hr}}·\frac{1\text{hr}}{60\text{min}}=1.667\text{miles}\text{/}\text{hr}$

Convert the rate from miles / min to inches/min

  $=\frac{1.667\text{miles}}{1\min }·\frac{5280\text{ft}}{1\text{mile}}·\frac{12\text{inch}}{1\text{ft}}=73920\text{inch/min}$

Revolution per minute = $\frac{\text{Rate}\text{in}\text{inch}\text{per}\text{min}}{\text{Circumference}}$ = $=\frac{\text{73920}\text{inch/min}}{\text{25}\text{π}\text{inch}}=941.7\text{rev}/\min$

Thus, the rate of spin balance is $941.7\text{rev}/\min$