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

AI Thread Summary
The discussion revolves around calculating the coefficient of rolling friction for bicycle tires at different pressures. One tire, inflated to 40 psi, travels 18.7 meters before its speed is halved, while the 105 psi tire covers 93.3 meters. The initial attempt at solving the problem incorrectly applied the formula, but the user later realized that the distance should be measured when the speed is reduced to half, not to zero. After correcting this misunderstanding, the user successfully found the right answer. The importance of careful interpretation of problem statements in physics is emphasized.
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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! :)
 
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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.
 
ahh, missed that little fact. Got the right answer now :)
Thank you!
 

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