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Centripetal acceleration question: car moving around banked curve

  1. Oct 10, 2012 #1
    If the wheels and tires of a car are rolling without slipping or sliding when turning, the bottom of the tire is rest against the road at each instant, so the force of friction is the static friction. Essentially if you are moving around a banked curve and the car is not skidding, then friction will be calculated by using coefficient of static friction. Why though? Isn't the car moving?
  2. jcsd
  3. Oct 10, 2012 #2
    The point of the tire in contact isn't moving- it is constantly changing as the tire rotates, but it is not skidding so it uses static rather than kinetic friction.
  4. Oct 10, 2012 #3
    I still don't understand. Could you elaborate a bit more? Alright the car is not skidding, but I still don't understand.
  5. Oct 10, 2012 #4
    Like schaefera said, you will use the static friction coefficient as the point of the tire in contact with road is not moving against the road. The wheels TURN and so the point in contact changes. Of course the previous point in contact moves but not against the road. So since the point is not moving but only changing its position that too not against the road, each point will have μ as the static friction co-efficient.
    Last edited: Oct 10, 2012
  6. Oct 11, 2012 #5


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    Not quite, The coefficient of static friction just tells you what the maximal static friction force is, that the contact could transmit. The actual transmitted force of static friction can be less than that.

    No skidding = no relative horizontal movement between the contact patches. The tire contact point moves on a cycloid. Around contact time it moves only vertically:

  7. Oct 11, 2012 #6
    Static friction occurs when there is no slipping between the surfaces. If the tyres don't slip, then this is what you use.
  8. Oct 11, 2012 #7


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    Compare it with walking: When your feet touch the surface, they do not move (relative to the surface), but you can move forward as you constantly switch between both feet.
  9. Oct 11, 2012 #8


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