Rolling Friction of Bicycle Tires

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SUMMARY

The discussion focuses on calculating the coefficient of rolling friction (μr) for bicycle tires inflated to different pressures. A tire at 40 psi travels 17.3 meters before its speed reduces to half, while a tire at 105 psi travels 93.0 meters. The user attempted to find the acceleration using the equation v² = v₀² + 2a(x - x₀) and derived an incorrect value for μr. The correct approach involves ensuring proper calculations for acceleration and understanding the implications of a negative coefficient of friction.

PREREQUISITES
  • Understanding of Newton's second law (Fx = ma)
  • Familiarity with the equations of motion (v² = v₀² + 2a(x - x₀))
  • Knowledge of the relationship between normal force and weight (n = mg)
  • Basic grasp of friction concepts, specifically rolling friction
NEXT STEPS
  • Review the principles of rolling friction and its calculation methods
  • Study the effects of tire pressure on rolling resistance
  • Learn how to accurately apply kinematic equations in physics problems
  • Explore the concept of negative coefficients in friction and their physical implications
USEFUL FOR

This discussion is beneficial for physics students, mechanical engineers, and anyone interested in the dynamics of bicycle performance and tire mechanics.

neongoats
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Homework Statement


Two bicycle tires are set rolling with the same initial speed of 3.30m/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 17.3m ; the other is at 105 psi and goes a distance of 93.0m . Assume that the net horizontal force is due to rolling friction only and take the free-fall acceleration to be g = 9.80m/s2
What is the coefficient of rolling friction μr for the tire under low pressure?

Homework Equations


Fx=ma
Fx=μn
n=mg
v2=v02+2a(x−x0)

The Attempt at a Solution



First I solved for the acceleration:
1.65^2 = 3.3^2 + 2a(17.3)
a= -.24

Then I set Fx=ma and Fx=μn to be equal, and substituted n=mg in for n
ma=μmg
The masses cancel out so I get
a=μg
I tried solving for this
-.24=u(9.8)
and got -.024, but this answer was incorrect.

I feel good about this process but don't know I'm doing wrong? I found another question exactly like this on here but when I tried to do the calculations for acceleration I kept getting a different number, so I don't know if that's where my problem lies but if so I don't understand how my math is wrong.
 
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Have a think about the concept of a negative coefficient of friction.
 

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