Coefficient of Rolling Friction

In summary: Just remember that without the normal force, there is no friction.In summary, the problem involves two bicycle tires with the same initial speed rolling on a straight road. The distance traveled before the speed is halved is measured for each tire, with one having a pressure of 40 and going a distance of 17.2, and the other having a pressure of 105 and going a distance of 92.3. The net horizontal force is due to rolling friction only and the free-fall acceleration is 9.80. The task is to find the coefficient of rolling friction for the tire under low pressure. Using Newton's second law and the equation for friction force, which is the coefficient of friction multiplied by the normal force, the
  • #1
stokes
27
0

Homework Statement


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?



Homework Equations





The Attempt at a Solution



I found the acceleration to be -0.19 I am stuck at that part I don't know what to do from there.
 
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  • #2
Yes, good so far, don't forget your units. Now that you have solved for a, why not try Newton 2?
 
  • #3
Ive tried that but Fnet=ma... they don't give mass in the problem. I can't find the coefficient without mass. I am really stuck...I know I have to use Newtons second law but I don't know how to start...
 
  • #4
stokes said:
Ive tried that but Fnet=ma... they don't give mass in the problem. I can't find the coefficient without mass. I am really stuck...I know I have to use Newtons second law but I don't 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?
 
  • #5
Friction force= coefficient of friction * Normal force...?
 
  • #6
stokes said:
Friction force= coefficient of friction * Normal force...?

Yes, and what is the normal force in this problem?
 
  • #7
Hmmm that is what I am stuck on calculating normal force. Is there a way to calculate normal force with the information provided?
 
Last edited:
  • #8
stokes said:
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..
 
  • #9
Fn= 0?
 
Last edited:
  • #10
stokes said:
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?
 
  • #11
Force of gravity. 9.8m/s^2
 
  • #12
stokes said:
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?
 
  • #13
Sorry I don't really know...
 
  • #14
stokes said:
Sorry I don't 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 ?
 
  • #15
n=mg...
 
  • #16
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...
 
  • #17
I believe that's where I am stuck... I can't seem to continue from there. I would be missing the friction force to calculate the coefficient of friction.
 
  • #18
stokes said:
I believe that's where I am stuck... I can't 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?
 
  • #19
Thank you for your help. I don't know why I couldn't put that all together. Thanks again.
 
  • #20
stokes said:
Thank you for your help. I don't know why I couldn't put that all together. Thanks again.
It's OK, you toughed it out.
 

1. What is the coefficient of rolling friction?

The coefficient of rolling friction is a dimensionless quantity that represents the amount of resistance or force that is generated when one object rolls over another. It is denoted by the symbol μr and is typically measured in units of N/N or m/m.

2. How is the coefficient of rolling friction determined?

The coefficient of rolling friction can be determined through experimentation, by measuring the force required to keep an object rolling at a constant velocity over a flat surface. The ratio of this force to the weight of the object gives the value of the coefficient of rolling friction.

3. What factors affect the coefficient of rolling friction?

The coefficient of rolling friction is affected by various factors such as the nature of the surfaces in contact, the smoothness of the surfaces, the weight and shape of the object, and the speed at which the object is rolling. It is also influenced by external factors such as temperature and humidity.

4. How does the coefficient of rolling friction differ from the coefficient of static friction?

The coefficient of rolling friction is typically lower than the coefficient of static friction. This is because rolling motion involves less resistance than sliding motion, as only a small portion of the object is in contact with the surface at any given time, while in sliding motion, the entire object is in contact with the surface.

5. Why is the coefficient of rolling friction important?

The coefficient of rolling friction is an important concept in science and engineering as it helps in understanding the forces involved in the movement of objects over surfaces. It is also crucial in the design of various systems and machines, such as vehicles, bearings, and conveyor belts, where minimizing friction is essential for efficient operation.

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