Turnings on roads and banked roads

  • Context: Undergrad 
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SUMMARY

The discussion centers on the physics of turning on roads and banked roads, specifically the role of static friction in providing the necessary centripetal force for a car's circular motion. The maximum velocity a car can maintain without leaving the circular path is defined by the formula √μsrg, where μ represents the coefficient of static friction. On a banked road, the horizontal component of friction and the normal force contribute to centripetal force, with the optimal velocity to minimize tire wear being v0 = √rg tanθ. If the car's velocity is below v0, friction acts up the banked plane, preventing the car from slipping down.

PREREQUISITES
  • Understanding of centripetal force and its application in circular motion
  • Knowledge of static friction and its role in vehicle dynamics
  • Familiarity with banked road design and its physics
  • Basic grasp of the relationship between velocity, radius, and angle in motion
NEXT STEPS
  • Study the derivation of the centripetal force formula in circular motion
  • Learn about the effects of different coefficients of friction on vehicle stability
  • Research the principles of banked curves in road engineering
  • Explore tire dynamics and the impact of friction on vehicle performance
USEFUL FOR

Physics students, automotive engineers, and anyone interested in understanding vehicle dynamics and road design principles.

takando12
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The centripetal force required for the circular motion of the car during a turning is provided by the static friction between the road and the tire and the maximum velocity it can go at without leaving the circular path is √μsrg. But why is it static friction? Isn't the car moving? How then do we say static friction?

For a banked road we have the horizontal component of the friction and the normal reaction providing the necessary centripetal force.. The velocity at which the car must be driven to minimise wear and tear of the tires( ie μ=0, is v0 =√rg tanθ.
My textbook says that if the velocity is lesser than v0, the force of friction will act up the banked plane.
How is this possible? And what will happen? will the car slip down the banked plane?
 
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takando12 said:
But why is it static friction?

Because the tires aren't sliding (well, maybe a little). The static friction is the torque that causes the wheel to rotate.
 

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