Solving For Rolling Friction in Bike Tires

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Homework Help Overview

The discussion revolves around calculating the coefficient of rolling friction for bicycle tires under different inflation pressures. The original poster presents data from an experiment where two tires, one at 40 psi and the other at 105 psi, are rolled and their distances before speed reduction are measured.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to apply kinematic equations to determine acceleration and subsequently the coefficient of rolling friction. Some participants question the directionality of forces and acceleration in the context of rolling motion.

Discussion Status

Participants are actively engaging with the original poster's calculations, offering insights into the nature of rolling friction and the importance of force direction. There is an exploration of the underlying physics concepts, but no consensus has been reached regarding the calculations or assumptions made.

Contextual Notes

Participants note the complexity of rolling friction, which may involve multiple factors, and question the availability of experimental data on rolling friction coefficients. The discussion reflects a mix of theoretical understanding and practical experimentation.

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Homework Statement



Two bicycle tires are set rolling with the same initial speed of 3.60m/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 18.0m ; the other is at 105 psi and goes a distance of 92.6m . Assume that the net horizontal force is due to rolling friction only and take the free-fall acceleration to be g = 9.80m/s^2.

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

Homework Equations



The equations I used are:

1) F=m*a (This would apply to the force in the x direction)
2) (FV)^2= (IV)^2 + 2*(a)*(d)
3) Ff = μ*Fn
4) Fn= m*g

The Attempt at a Solution



I solved for the acceleration in the x direction using equation 2.
(FV)^2= (IV)^2 + 2*(a)*(d)
(1.8)^2 = (3.6)^2 + 2*(a)*(18)
a= -0.27

So I then plugged this into equations 1 and 3.
F=m*a= μ*Fn
m*a=μ*m*g
m cancels
μ=a/g, so μ=-0.27/-9.8=0.276

I know μ is supposed to be positive, so that is why I made acceleration in the equation negative. I don't quite know where I am going wrong with this work. Any help would be greatly appreciated.
 
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You want to draw a diagram of the forces. Crux here is that ##\vec a## (hence ##\vec F##) and ##\vec d## point in opposite directions.
In other words: once you established a positive x-direction (18 m, 3.6 m/s) it follows that ## |\vec F| = - \mu |\vec F_N|## -- friction force is always opposite to moving force/moving direction
 
Rolling friction is the resistive force that slows down the motion of a rolling ball or wheel. This type of friction is typically a combination of several friction forces at the point of contact between the wheel and the ground or other surface.

When the materials are both hard, static friction and molecular friction slow down the rolling. When the wheel or tire is soft, its distortion slows down the motion. When the other surface is soft, the plowing effect is a major force in slowing the motion.

Because of the various factors, the coefficient of rolling friction is usually determined experimentally.
 
Took you a while to post :smile: ...
Do these experimenters make their results public somewhere accessible ?
I see a trend towards fatter tyres for trekking bicycles and always thought narrow and hard was better than wide and soft.
 

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