I know that friction helps car to move forward. The engine of the car turns the wheels, and the tyres have to grip the road in order for the car to start moving forward. If there is no friction, tyres cannot grip the road! It is similar to a geared wheel moving on the ground enter image description here

But I can't imagine how does friction act as a centripetal force when the car is turning on a roundabout? What is happening at the point of contact between the wheel and the road ?

  • $\begingroup$ Your geared wheel diagram is / was used on steep inlined railways... $\endgroup$ – Solar Mike Jun 13 '17 at 6:33
  • $\begingroup$ @SolarMike ... properly called "funiculars" $\endgroup$ – Carl Witthoft Jun 13 '17 at 15:31
  • $\begingroup$ @CarlWitthoft yes, I am in Switzerland - we've got a few here! Also used in some mines as well. $\endgroup$ – Solar Mike Jun 13 '17 at 16:00

The key phrase here is slip angle

This is not a very easy concept to visualise but in essence comes down to the fact that when you turn the wheels the sidewalls of the tyre flex a bit relative to the contact patch between the tyre and the road so the whole tyre twists a bit. Of to put it another way the wheel it self wants to turn but the tread of the tyre wants to keep going straight which creates an elastic force in the sidewall. It is this force which pulls the car towards the centre of the turn.

An easy way to see this in practice is with a bicycle (it is more obvious if you lower the tyre pressure a bit). With the bike stationary lean your weight over the handlebars and turn gently and you should be able to both see the tyre flex and feel the resistance through the handlebars.

  • $\begingroup$ Do you mean that , if a car is moving straight forward and then it turns right ,, the part of the tyre not touching the road will turn right , but the part of the tyre touching the road will remain "approximately "straight due to friction ? $\endgroup$ – Eman.suradi Jun 13 '17 at 12:16
  • $\begingroup$ Not necessarily straight ahead but yes closer to straight ahead than the angle of the rest of the wheel. $\endgroup$ – Chris Johns Jun 13 '17 at 12:32
  • $\begingroup$ So the friction force acts outward ? Since it is preventing the part of the tyre touching the car from turning ? $\endgroup$ – Eman.suradi Jun 13 '17 at 12:58
  • $\begingroup$ @Eman.suradi: I think you misunderstood: when the wheel turns, it always rolls in direction of its tread. Which after turning is not the same as the car. Car inertia pushes the tire diagonally off its rolling direction, but its flexibility opposes this force, pushing the car sideways. Friction just prevents the tire from slipping, and in this context acts on the tire in the direction of the center of turn (centripetal), and on the road surface in direction opposite. $\endgroup$ – SF. Jun 13 '17 at 14:22
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    $\begingroup$ @Eman.suradi: That is only true to full extent when the car is immobile. Normally, in motion, the amount the segment in contact with surface is trailing behind the rim is negligibly small (as any part that was held back by friction is lifted up by the wheel rolling, replaced by a segment that was up in the air and directly flush with the rim, the torsional distortion removed nearly instantaneously. What causes the car to turn is a shearing stress; the part of the tire in contact with ground trying to move in a circle tighter than the rest of the car. $\endgroup$ – SF. Jun 17 '17 at 13:54

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