Two bicycle tires are set rolling with the same initial speed of 3.40 m/s along
ID: 1699116 • Letter: T
Question
Two bicycle tires are set rolling with the same initial speed of 3.40 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 { m psi} and goes a distance of 19.0 m; the other is at 105 { m psi} and goes a distance of 92.2 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/s^2. What is the coefficient of rolling friction for the tire under low pressure? for the tire of higher pressure?Explanation / Answer
given
intial speed =v1 = 3.40m/s
final speed = vo = 3.40m/s/2 = 1.7m/s
Friction coefficient = F(rf)/F(normal) = ma/mg = a/g
from the kinematic equation
s = (v0^2-v1^2)/(2a)
a = (v0^2-v1^2)/(2s)
A. At 40 psi, = (3.4^2-1.7^2)/(2*19*9.8) = 0.02328
B. At 105 psi, = (3.4^2-1.7^2)/(2*92.2*9.8) = 0.004797