Two bicycle tires are set rolling with the same initial speed of 3.30 m/s along
ID: 1417328 • Letter: T
Question
Two bicycle tires are set rolling with the same initial speed of 3.30 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 18.4 m ; the other is at 105 psi and goes a distance of 93.7 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 second tire (the one inflated to 105 psi)?
Explanation / Answer
simplest way to do this is to use this formula to get the acceleration.
vf^2 = vi^2 + 2*a*d
vi = 3.30 m/s
vf = 1/2 3.30 m/s = 1.65 m/s
a = ???
s = 18.4 m
1.65^2 = 1.65^2 + 2*a*18.4
4 = 16 + 2a*17.8
-12 = 35.6 * a
a = - 0.222 m/s^2
m*a = mu * Normal
Normal = m*9.81
m*-0.222 = - mu* m*9.81 The m/s cancel and for force of friction always goes in the opposite direction of motion.
mu = 0.222/9.81 = 0.022
calculation may be wrong but concept is right