Part A) What is the coefficient of rolling friction ? r for the tire under low p

Question

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

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

Two bicycle tires are set rolling with the same initial speed of 3.30m/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 18.9m ; the other is at 105 psi and goes a distance of 93.0m . Assume that the net horizontal force is due to rolling friction only and take the free-fall acceleration to be g = 9.80m/s2 .

Explanation / Answer

v2-u2 = -2aS (bcoz retardation)

v=u/2 (given)

a=3u2/(4*2*S)

a1=3(3.3*3.3)/(8*18.9) = 0.216 m/s2

a2=3(3.3*3.3)/(8*93) = 0.0439 m/s2

force of friction = Coefficient of friction * Normal reaction

[ normal reaction = mg]

applying second law of motion F=ma

f1= r1*mg = m*a1

=> r1 = a1/g = 0.022

similarly r2 = a2/g = 0.00448

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