Two bicycle tires are set rolling with the same initial speed of 3.80 m/s along

Question

Two bicycle tires are set rolling with the same initial speed of 3.80 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 at 105 psi and goes a distance of 92.7 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 . What is the coefficient of rolling friction r for the tire under low pressure? What is the coefficient of rolling friction r for the tire under high pressure?

Explanation / Answer

Here ,

for the low pressure tyre

initial speed , u = 3.8 m/s

final speed , v = 3.8/2 = 1.9 m/s

as the acceleration is - ur * g

using third equation of motion

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

1.9^2 - 3.8^2 = 2 * (-ur * 9.8 ) * 17.1

sovling for ur

ur = 0.0323

the coefficient of rolling friction r for the tire under low pressure is 0.0323

Now , for the high pressure

initial speed , u = 3.8 m/s

final speed , v = 3.8/2 = 1.9 m/s

as the acceleration is - ur * g

using third equation of motion

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

1.9^2 - 3.8^2 = 2 * (-ur * 9.8 ) * 92.7

sovling for ur

ur = 0.00591

the coefficient of rolling friction r for the tire under high pressure is 0.00591

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