Two bicycle tires are set rolling with the same initial speed of 3.20 m/s along
ID: 584560 • Letter: T
Question
Two bicycle tires are set rolling with the same initial speed of 3.20 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 18.7 m ; the other is at 105 psi and goes a distance of 93.8 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 tire under low pressure?
r =Explanation / Answer
Vi = 3.2 m/s
Vf = 1.6 m/s
s = 18.7 m
For 40 psi pressure -
Vf^2 = Vi^2 - 2*a*s
1.6^2 = 3.2^2 - 2*a*18.7
a1 = 0.205 m/s^2
For 105 psi pressure -
Vf^3 = Vi^2 - 2*a*s
1.6^2 = 3.2^2- 2*a*93.8
a2 = 0.0409 m/s^2
Friction Force, Fs = r*m*g
m*a = r*m*g
a = r*g
r = a/g
Coefficient of friction for 40 psi pressure tyre -
r = 0.205 / 9.8
r = 0.0209
Similarly Coefficient of friction for 105 psi pressure tyre -
r = 0.0409 / 9.8
r = 0.00417