Two bicycle tires are set rolling with the same initial speed of 3.40 m / s alon
ID: 2138433 • Letter: T
Question
Two bicycle tires are set rolling with the same initial speed of 3.40m/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 17.1m ; the other is at 105 psiand goes a distance of 92.1m . Assume that the net horizontal force is due to rolling friction only and take the free-fall acceleration to be g = 9.80m/s2
What is the coefficient of rolling friction ?r for the tire under low pressure?
Please expain :)
Explanation / Answer
coefficient of roling friction u = Frotational/Fnormal = ma/mg = a/g
s = (v^2-u^2)/2a so a = (v^2-u^2)/2s
a) at 40 psi, u = (3.4^2-1.7^2)/(2*17.1*9.8) = 0.02587
b) at 105 psi, u = (3.4^2-1.7^2)/(2*92.1*9.8) = 0.004803