Two bicycle tires are set rolling with the same initial speed of 3.50 m/s along

Question

Two bicycle tires are set rolling with the same initial speed of 3.50 m/s along a long, straight road, and the distance each travels before its speed is reduced by half is measured. One tire is inflated to a pressure of 40 psi and goes a distance of 17.1 m ; the other is at105 psi and goes a distance of 94.0 m . Assume that the net horizontal force is due to rolling friction only and take the free-fall acceleration to be g = 9.80 m/s2 .

A/ What is the coefficient of rolling friction r for the tire under low pressure?

b/What is the coefficient of rolling friction r for the second tire (the one inflated to 105 psi)?

Explanation / Answer

Here ,

initial speed , u = 3.50 m/s

A) as the acceleration is given as

a = u * g

Now , for final speed , v = 3.5/2

v = 1.75 m/s

Using third equation of motion

v^2 - u^2 = 2 *a * d

3.5^2 - 1.75^2 = 2 * u * 9.8 * 17.1

solving for u

u = 0.027

rolling friction at low pressure is 0.027

b)

at high pressure

Using third equation of motion

v^2 - u^2 = 2 *a * d

3.5^2 - 1.75^2 = 2 * u * 9.8 *94

solving for u

u = 0.00499

rolling friction at low pressure is 0.00499

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