Coefficient of Rolling Friction

[stokes]

1. The problem statement, all variables and given/known data
Two bicycle tires are set rolling with the same initial speed of 3.00 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 and goes a distance of 17.2 ; the other is at 105 and goes a distance of 92.3 . Assume that the net horizontal force is due to rolling friction only and take the free-fall acceleration to be = 9.80 .

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



2. Relevant equations



3. The attempt at a solution

I found the acceleration to be -0.19 I am stuck at that part I dont know what to do from there.

[PhanthomJay]

Yes, good so far, don't forget your units. Now that you have solved for a, why not try Newton 2?

[stokes]

Ive tried that but Fnet=ma... they dont give mass in the problem. I cant find the coefficient without mass. Im really stuck...I know I have to use newtons second law but I dont know how to start...

[PhanthomJay]

Ive tried that but Fnet=ma... they dont give mass in the problem. I cant find the coefficient without mass. Im really stuck...I know I have to use newtons second law but I dont know how to start...
You may not need to know the mass. What's the formula for the friction force, which is given as Fnet in this problem?

[stokes]

Friction force= coefficient of friction * Normal force...?

[PhanthomJay]

Friction force= coefficient of friction * Normal force...?

Yes, and what is the normal force in this problem?

[stokes]

Hmmm that is what I am stuck on calculating normal force. Is there a way to calculate normal force with the information provided?

[PhanthomJay]

Hmmm that is what I am stuck on calculating normal force.To calculate the normal force, look in the y direction for all forces, and apply Newton 1, since there is no acceleration in the y direction..

[stokes]

Fn= 0?

[PhanthomJay]

Fnet= 0?Yes, Fnet = 0 in the y direction. The normal force acts up on the tire in the y direction. What other force acts in the y direction?

[stokes]

Force of gravity. 9.8m/s^2

[PhanthomJay]

Force of gravity. 9.8m/s^2
The acceleration of gravity is 9.8m/s^2;the force of gravity is the tires weight, which is what?

[stokes]

Sorry I dont really know...

[PhanthomJay]

Sorry I dont really know...Oh,you should be familiar with the equation for weight : W = mg. So if the tire weighs 'mg', then the normal force must be ????

[stokes]

n=mg...

[PhanthomJay]

Yes! Now you've got everything you need. Go back into the horizontal direction and solve for the friction coefficient. The mass term should cancel out......

[stokes]

I believe thats where Im stuck... I cant seem to continue from there. I would be missing the friction force to calculate the coefficient of friction.

[PhanthomJay]

I believe thats where Im stuck... I cant seem to continue from there. I would be missing the friction force to calculate the coefficient of friction.No, you already stated what the friction force, F_f is: it's the coef of friction, u, times the normal force, N, that is F_f = Fnet = u*N, and since N=mg, then the friction force is u*mg. And since F_net = ma, can you now solve for u?

[stokes]

Thank you for your help. I dont know why I couldnt put that all together. Thanks again.

[PhanthomJay]

Thank you for your help. I dont know why I couldnt put that all together. Thanks again.It's OK, you toughed it out.