What Factors Cause a Car to Skid on a Circular Track with Kinetic Friction?

AI Thread Summary
A car skids on a circular track when the required centripetal force exceeds the available frictional force. This occurs when the car's velocity surpasses the threshold defined by the equation v > √(µgr), where µ is the coefficient of static friction, g is the acceleration due to gravity, and r is the radius of the track. On a level, unbanked road, the frictional force provides the necessary centripetal force to keep the car in circular motion. If the car's speed increases beyond this limit, it will lose traction and skid. Understanding these dynamics is crucial for safe driving on curves.
Nishikino Maki
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Homework Statement


This isn't a problem from a textbook or homework but just a general question.

Say there's a car traveling in a circle and that the track has some coefficient of kinetic friction µ. What would make the car skid?

Homework Equations

The Attempt at a Solution


I'm thinking that it skids when the centripetal is greater than the friction force? Not sure why but that's just my intuition.
 
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If it is a level unbanked road, the centripetal force is the static friction force. What does that tell you about when the car will skid?
 
That actually makes sense because on a level curve the friction and centripetal forces are the only ones pointing to the center.

Then I suppose it would be when the velocity of the car is greater than \sqrt{µgr}?
 
Nishikino Maki said:
That actually makes sense because on a level curve the friction and centripetal forces are the only ones pointing to the center.
the friction force is the centripetal force.
Then I suppose it would be when the velocity of the car is greater than \sqrt{µgr}?
yes, where u is the coefficient of static friction, not kinetic friction.
 

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