# Rolling Friction-Bicycle Tires

by toxsic
Tags: frictionbicycle, rolling, tires
 P: 3 1. The problem statement, all variables and given/known data 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 18.6 m; the other is at 105 psi and goes 93.5 m. Assume that the net horizontal force is due to rolling friction only. What is the coefficient of rolling friction mu_r for the tire under low pressure? 2. Relevant equations 3. The attempt at a solution Honestly, I have no idea where to begin. The professor has never explained anything like it and this is a first. It's on mastering physics and it's well above my head
 PF Gold P: 362 Do you have a mass for each of the tires?
PF Gold
P: 362
 Quote by toxsic 1. The problem statement, all variables and given/known data 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 18.6 m; the other is at 105 psi and goes 93.5 m. Assume that the net horizontal force is due to rolling friction only. What is the coefficient of rolling friction mu_r for the tire under low pressure? 2. Relevant equations 3. The attempt at a solution Honestly, I have no idea where to begin. The professor has never explained anything like it and this is a first. It's on mastering physics and it's well above my head
In my previous post, I asked if you had the mass of the tires (plus wheel, of course) but you can get pretty far without that information. Here's the basic idea: The initial kinetic energy of each tire is going to go into work to overcome friction. That work is equal to the frictional force times the distance. So let's set this up:

Let the coefficient of kinetic friction for the tire inflated to 40 psi be denoted by $$\mu_1$$.

And let the coefficient of kinetic friction for the other tire be denoted by $$\mu_2$$

Then the basic relationship is

$$K.E. = F_f d$$

The frictional force for the first tire is, of course, given by $$F_f = \mu_1 N$$
with a similar equation for the other tire.

That implies that, if we label the distance the tire travels by $$d_1$$

$$K.E. = \mu_1 N d_1$$

for the first tire and a similar relationship for the second tire. We'll assume the two tires have the same mass and we know that they start with the same kinetic energy, so you can either set the two expressions equal to each other (or divide one of the equations by the other). At any rate the K.E and the normal forces cancel out leaving you with the two coefficients of friction and the two distances. You can then express one coefficient of friction in terms of the other.

Got that?

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