Question

A tire-pressure monitoring system warns you with a dashboard alert when one of your car tires is significantly under-inflated.
How does this pressure monitoring system work? Many cars have sensors that measure the period of rotation of each wheel. Based on the period of rotation, the car’s computer determines the relative size of the tire when the car is moving at a certain speed.
(a) Derive a mathematical equation that relates the period T of the wheel rotation of the tire radius r and the speed v of the car.
(b) The figure below shows a properly inflated tire (left) and an under-inflated tire (right) of the same car. Estimate the percent change of the period of the underinflated tire compared to the properly inflated tire.
The inflated tire has a radius of 303 mm and the underinflated tire has a radius of 293mm. (Radius is from the ground to center of the wheel.)
(c) Estimate the distance this car would have to travel in order for one wheel to make one more complete rotation than the other. Which will undergo more turns, the under-inflated or the properly inflated tire?

Answer 1

Answer:

The answers to the question are

(a) T = 2·π·r/v

(b) 3.3 % change in period of the under-inflated tire compared to the properly inflated tire

(c) Therefore the distance this car would have to travel in order for under-inflated tire to make one more complete rotation than the inflated tire is 57.685 meters.

Explanation:

(a) The period T = 2π/ω

The velocity v = ωr or ω = v/r

Therefore T = 2π/(v/r) = 2πr/v

T = 2·π·r/v

(b) Period of properly inflated tire with radius = 303 mm is 2π303/v

Period of under-inflated tire with radius = 293 mm is 2π293/v

Therefore we have percentage change in period of  of the under-inflated tire compared to the properly inflated tire is given by

(2π303/v -2π293/v)/(2π303/v) = 2π10/v/(2π303/v) = 10/303 × 100 = 3.3 %

(c) The period of the under-inflated tire is 10/303 less than that of the inflated tire. Therefore for the under-inflated tire to make one complete turn more than the inflated tire, we have 1/(10/303)  = 303/10 or 30.3 revolutions of either tire which is 30.3×2×π×303 = 57685.296 mm = 57.685 meters

Therefore the distance this car would have to travel in order for under-inflated tire to make one more complete rotation than the inflated tire is 57.685 meters

At 30.3 revolutions the distance covered by the under-inflated

= 55781.49 mm

Subtracting the two distances gives

1903.805 mm

The circumference of the inflated tire = 2×π×303 = 1903.805 mm

Answer 2

A) The mathematical equation that relates the period T of the wheel rotation of the tire radius r and the speed v of the car is; T = 2πr/v

B) The percent change of the period of the underinflated tire compared to the properly inflated tire is; 3.3 %

C) The distance this car would have to travel in order for under-inflated tire to make one more complete rotation than the inflated tire is; 57685.296 mm

Relationship between motion of Cars and it's tires

(A) The formula for period is;

T = 2π/ω

angular velocity is;

ω = v/r

Thus; T = 2πr/v

where T is period, v is velocity and r is radius

(b) We are told that radius is 303 mm. Thus Period of inflated tire with radius is;

T =  2π ×  303/v

Period of under-inflated tire with radius of 293 mm is;

T = 2π× 293/v

Percentage Change is;

((2π ×  303/v) - (2π× 293/v))/(2π ×  303/v) * 100% = 10/303 * 100% = 3.3 %

(c) From B above we see that the period of the under-inflated tire is 10/303 less than that of the inflated tire.

Thus,  for the under-inflated tire to make one complete turn more than the inflated tire, we have 1/(10/303)  = 30.3 revolutions of either tire.

Thus, the distance is;

d = 30.3 × 2π × 303

d = 57685.296 mm

Using the same concept, at 30.3 revolutions, the distance covered by the under-inflated tire is calculated to be; 55781.49 mm

Difference in distance = 57685.296 mm -  55781.49 mm

Difference in distance = 1903.805 mm

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