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Turnings on roads and banked roads

  1. Sep 3, 2015 #1
    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?
     
  2. jcsd
  3. Sep 3, 2015 #2
    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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