Rolling Friction and Bicycle Tires

AI Thread Summary
Two bicycle tires, one at 40 psi and the other at 105 psi, were tested for rolling friction, with the lower pressure tire traveling 18.4 m and the higher pressure tire 94.0 m before their speeds halved. The initial attempt to calculate the coefficient of rolling friction using the formula ur = v/[(d/v) x g] yielded an incorrect result. Participants clarified that the distance measured was not the stopping distance but the distance before speed reduction, and emphasized the need for the correct kinematic equations. The normal reaction force (R) is derived from mass and gravitational force, which cancels out in the calculations. Ultimately, the discussion highlighted the importance of using the correct equations, specifically suggesting v² = u² + 2as for accurate results.
pureouchies4717
Messages
98
Reaction score
0
my teacher Never went over this in class. ever. please help, thanks

Two bicycle tires are set rolling with the same initial speed of 3.70 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 18.4 m; the other is at 105 psi and goes a distance of 94.0 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/s^2. What is the coefficient of rolling friction ur for the tire under low pressure?
 
Physics news on Phys.org
i tried to use this formula:

ur= v/[(d/v) x g]

and got

ur=.075, but it said i was wrong

You are off by a constant. Be sure you are using the right equation of motion. Also, note that the distance given is not the stopping distance, but rather the distance that the object travels before its speed is reduced by half.

^ what exactly does it mean to be off by a constant?
 
Last edited:
anyone?

:frown:
 
Use the kinematic formula v = u + at to obtain an acceleration. Then equate this with the force in F_r = \mu R
 
Hootenanny said:
Use the kinematic formula v = u + at to obtain an acceleration. Then equate this with the force in F_r = \mu R

hmm but the problem is that i don't have u; that is actually what I am looking for
 
Yes you do, u refers to the intial velocity. \mu is the coeffiecent of friction. Sorry for the different symbols, its just the notation I'm used to.
 
Hootenanny said:
Yes you do, u refers to the intial velocity. \mu is the coeffiecent of friction. Sorry for the different symbols, its just the notation I'm used to.

o i see, but i still don't have time. and i don't really get what that "R" refers to. if its radius, i didnt get it in this problem
 
R is the normal reaction force, which is the product of the mass and the gravitational field strength.
 
i still don't understand since I am not given mass
 
  • #10
Masses will cancel, this is the same process you used for the boy/swing/inclined plane question.
ma = \mu mg
 
  • #11
so which equations did you use?
 
  • #12
How do you know what time is? when using v = v + at..
 
  • #13
cpark43 said:
How do you know what time is? when using v = v + at..
Looking back on this thread from two years ago, it would have been more appropriate to use the equation v2 = u2+2as. I don't know why I suggested the original one.
 
Back
Top