Can Static Friction Exist on a Moving Bicycle on a Flat Track?

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SUMMARY

The discussion centers on calculating the smallest radius of an unbanked track for a bicyclist traveling at 29 km/h with a coefficient of static friction of 0.29. The key equations used include the relationship between static friction and normal force, as well as the net force equation, Fnet = m * (v^2 / r). It is established that static friction can exist on a moving bicycle due to the instantaneous rest condition at the contact point of the tires with the track, allowing for the use of static friction in calculations.

PREREQUISITES
  • Understanding of static and kinetic friction concepts
  • Familiarity with Newton's laws of motion
  • Knowledge of basic physics equations related to circular motion
  • Ability to convert units (e.g., km/h to m/s)
NEXT STEPS
  • Study the principles of circular motion in physics
  • Learn about the role of friction in different types of motion
  • Explore the effects of varying coefficients of friction on motion
  • Investigate the dynamics of bicycle handling and stability
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Physics students, educators, and anyone interested in understanding the mechanics of motion and friction in cycling scenarios.

A_lilah
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Homework Statement



What is the smallest radius of an unbanked (flat) track around which a bicyclist can travel if her speed is 29 km/h and the coefficient of static friction between tires and track is 0.29?



Homework Equations



coefficient of static friction = (f of maximum static friction / Normal force)

fnet = ma = m * v^2 / r

The Attempt at a Solution



If an object is moving, isn't their just kinetic friction? I get that the normal force is the acceleration, but how can there possibly be a maximum static friction at this point?

Here's what I've got:

29km/h = 8.0556 m/s

.29 = fstatmax / a, therefore a = fstatmax / .29

Fnet = m* a = m * (v^2 / r)
Fnet = a = v^2 / r
fstatmax / .29 = (8.0556)^2 / r

Thanks for any help!
 
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If an object is moving, isn't their just kinetic friction?

In case of tyres, the bottom-most point is instantaneously at rest. That's why static friction.

f=0.29Xmg.
[tex]f=\frac{m{v^2}}{r}[/tex]

solve to get r.
 
Ah...
thank you for the help!
 

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