Two bicycle tires are set rolling with the same initial speed of 3.30 m/s along
ID: 1543893 • 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 17.3 m ; the other is at 105 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 .
Part A: What is the coefficient of rolling friction r for the tire under low pressure?
Explanation / Answer
I guess the 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 = 17.3 m
1.65^2 = 3.3^2 + 2*a*17.3
2.7225 = 10.89 + 2a*17.3
-8.1675 = 34.6 * a
a = - 0.236 m/s^2
m*a = mu * Normal
Normal = m*9.81
m*-0.236 = - mu* m*9.81 The m/s cancel and for force of friction always goes in the opposite direction of motion.
mu = 0.236/9.81 = 0.024