A computerized spin balance machine rotates a 35-inch-diameter tire at 460 revolutions per minute. (a) Find the road speed (in miles per hour) at which the tire is being balanced. (Round your answer to two decimal places.) 442.53 Incorrect: Your answer is incorrect. mph (b) At what rate should the spin balance machine be set so that the tire is being tested for 50 miles per hour

a. Road speed = 47.90 miles per hour

b. Rate = 480.20 revolution per minute

Step-by-step explanation:

Diameter D = 35 inch

Radius r= D/2 = 35/2 = 17.5 inch

"1 ft = 12 inch and 1 mile = 5280 feet". R = 17.5*(1/12)*(1/5280) mile

Revolution n =460 rev./minute , n=460*60 rev./hour

Angular speed=2*pi*n  =2*pi*n radian per hour  =2*pi*460*60

a)  Road speed = r*angular speed  

Road speed = 17.5*(1/12)*(1/5280)*(2*pi*460*60)

Road speed = 47.8973879951

Road speed = 47.90 miles per hour

b. Speed v = 50 miles per hour

V = R * Angular speed

50 = 17.5*(1/12)*(1/5280)*angular speed

Angular speed = (50*12*5280) / 17.5

Angular speed = 181029 radian per hour

Angular speed = 181029 / 60 radian per minute

Angular speed = 3017.15 per minute

On revolution

w = 2*pi*n

3017.15 = 2*pi*n

3017.15 / 2 = pi * n

1508.58 = pi * n

n = 1508.58 / pi

n = 480.1959

n = 480.20 revolution per minute

y > 1/3x

y < = -x + 4

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