Will a Car Skid? | Friction and Centripetal Acceleration

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SUMMARY

A car traveling around a circular path with a diameter of 500m at a constant speed of 25 m/s will not skid, as the static friction force exceeds the required centripetal force. The static friction coefficient is 0.3, providing a maximum static friction force of 2.94m, while the required centripetal force is 2.5m. The kinetic friction coefficient of 0.2 is irrelevant until skidding occurs, as it only applies when the tires lose grip. Understanding the distinction between static and kinetic friction is crucial for analyzing the car's motion.

PREREQUISITES
  • Understanding of centripetal acceleration and its formula: a = v²/r
  • Knowledge of static and kinetic friction coefficients
  • Basic principles of Newton's second law (F = ma)
  • Familiarity with the concept of circular motion
NEXT STEPS
  • Research the effects of different friction coefficients on vehicle dynamics
  • Learn about the role of centripetal force in circular motion
  • Explore real-world applications of friction in automotive safety
  • Study the physics of skidding and tire performance under various conditions
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Physics students, automotive engineers, and anyone interested in vehicle dynamics and safety mechanisms.

Joe Cruz
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1. A car travels around a circle with a diameter of 500m at a constant speed of 25 m/s. The static friction coefficient is 0.3 and the kinetic friction coefficient is 0.2. Will the car skid?

2. F = ma. Ffr = μmg. Centripetal acceleration = v2/radius3. So the math for this particular problem was really simple, my problem seems to be more of a conceptual one.

The force acting on the car = F = m(252/250) = m(2.5)
The static friction force = Fsfr = m(.3)(9.8) = m(2.94)
The kinetic friction force = Fkfr = m(.2)(9.8) = m(1.96)

My first assumption is that the car is going to skid since the car is moving and the force acting on the car exceeds the kinetic friction force, but since Fkfr < F < Fsfr, I'm not entirely sure if this is the correct answer. Anyone want to help clarify this problem for me?
 
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What do you think it means to skid? What is the difference in the frictional forces acting on a car that is skidding and those on a car that is not skidding? At what point will a car become uncontrollable and begin to skid?
 
Ask yourself: If the car was not to skid, what kind of friction must act?
 
First, just want to say Mastodon is the dang. Second, I'm assuming skidding means the force acting on the object exceeds the force of friction.

To answer you Doc, I would say the kinetic friction. My problem is that the force of friction is acting perpendicular to the motion of the car itself, and if only the kinetic friction was determining whether the car would skid, why would he include the static friction coefficient as well? Unless it was just there to distract me...
 
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Joe Cruz said:
I'm assuming skidding means the force acting on the object exceeds the force of friction.
What kind of friction?

Joe Cruz said:
To answer you Doc, I would say the kinetic friction.
That would be incorrect. Realize that when the tire skids the surfaces are slipping against each other--and that means kinetic friction. So only if skidding occurs will kinetic friction be involved.

Joe Cruz said:
My problem is that the force of friction is acting perpendicular to the motion of the car itself, and if only the kinetic friction was determining whether the car would skid, why would he include the static friction coefficient as well? Unless it was just there to distract me...
The force that prevents the car from skidding outward--the force that keeps it moving in a circle--must be static friction. The question is: Is there enough static friction to provide the needed centripetal force?
 
Doc Al said:
That would be incorrect. Realize that when the tire skids the surfaces are slipping against each other--and that means kinetic friction. So only if skidding occurs will kinetic friction be involved.

The car being in motion really confused me, but that cleared everything up for me. Thank you.
 

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