What is the Coefficient of Rolling Friction for a Tire Under Low Pressure?

[tmilla]

Homework Statement

Two bicycle tires are set rolling with the same initial speed of 3.60 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 18.7m ; the other is at 105 psi and goes 93.3m . Assume that the net horizontal force is due to rolling friction only.

What is the coefficient of rolling friction for the tire under low pressure?

Homework Equations

Vf^2=Vi^2 +2ax
Ur=a/g ?? (Ur= coefficient of rolling friction

The Attempt at a Solution

1st Tire:
-(3.60)^2/(2*18.7) = a = -.3465

.3465/9.81=Ur

This is wrong, but I honestly have no idea what to do. Help is GREATLY appreciated! :)

 

[LowlyPion]

Welcome to PF.

I would start off by recalculating your expression for tire 1.

Reread the problem statement. The distance of 18.7 is when |v| = 1/2*|Vo|, not when it goes to 0.

 

[tmilla]

ahh, missed that little fact. Got the right answer now :)
Thank you!