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Friction in non-uniform circular motion of a car

  1. May 10, 2015 #1
    Hi guys...

    So i searched the net for 2 whole days and found a couple of topics on how static friction is responsible for creating the force required to keep the body moving in a circle.. Friction points towards the center of the circle ONLY when the car is free wheeling i.e moving at a constant speed. But didn't found much on what is its direction when car is actually accelerating during the circular motion.

    First of all i want to see if my concepts area clear.. hope you guys can help..

    1) When a car is moving linearly with an acceleration... the torque produced by the engine is delivered to the wheels, they push the ground backwards as a result the friction pushes forward. This means that friction is actually opposing the rotation of the wheels but the torque produced is enough to overcome it, hence a net force in the forward direction.. Now if the person removes his foot from the gas pedal.. the car stops after travelling a certain distance. This is due to the rolling resistance caused by the deformation produced when 2 bodies are in contact. Note ( static friction is still pointing forward but together with rolling resistance helps to stops the rotation of the wheel). Is this right?

    2) Now when the car makes a turn and is in Uniform Circular motion and not sliding/slipping ( constant speed ) the car tends to move tangentially ( hence radially outward ) due to inertia, hence friction acts towards the center producing the centripetal force. Now i found a diagram ( given below ) and i wanna know if this is correct or not.

    According to it This net friction inwards can be resolved into the friction force that opposes sliding outwards from the curve ( blue vector) and and the friction as a result of the tires pushing backwards ( red vector)..

    Is the blue vector opposite to the tangential direction the car tends to move in? ( i think yes )

    3) Now if we suppose that the car is in non uniform circular motion, hence accelerating then where does the friction point? since the net acceleration point in between the centripetal and tangential, i guess that's where the friction points, in the direction of the net acceleration.
  2. jcsd
  3. May 11, 2015 #2
    If you know the acceleration parallel to the instantaneous velocity of the car, and the mass of the car then you must know the friction force required to produce such acceleration? If you know additionally the radius of the path the car takes and the instantaneous velocity of the car than you must know the friction force required to keep the car moving in a circular path at that instantaneous velocity? Now add the two perpendicular force vectors, won't the instantaneous acceleration of the car point in the same direction?
  4. May 11, 2015 #3


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    The diagram seems to be incorrect. Assuming a steady speed and ignoring rolling resistance, when a car is turning, the friction force is a Newton third law pair of forces, the outwards force exerted by the tires onto the pavement (which is attached to the earth), and the inwards force that the pavement exerts onto the tires. The outwards force acts upon the earth, producing a very tiny reaction, while the inwards force acts upon the car, resulting in the car's centripetal acceleration.
  5. May 12, 2015 #4
    I know that the net acceleration points in between the tangential and centripetal acceleration...
    What i wanted to confirm was that the friction force also acts in the same direction? and not towards the center right

    Yes that happens while its not accelerating but when the car is accelerating then the friction would point in the direction of net acceleration right ? ( between centripetal and tangential)

    and what seems wrong with the diagram?
  6. May 13, 2015 #5
    Hi, I think you are getting a little confused by trying to use friction as a directional force as you would with thrust, drag, momentum and centrifugal force.
    What are you trying to achieve?
    Friction is static so can't be represented as a force on a vector diagram like this, friction opposes a force created by relative movement of objects.
    momentum is what keeps the car moving, the direction of which is being provided by the direction the tyres are facing if there is enough friction to overcome the resultant force.

    If you want to find the resultant forces friction has to compete against in a particular set of circumstances to find the friction required then you plot the forces acting on the car combine them to find a resulting vector and compare that to the static friction available between the tyres and the road. This resultant vector will be equal the friction required to stop the car from sliding in that case but not necessarily the maximum friction available. The reverse of the force vector would be the resultant drag, part of which would be due to friction.
  7. May 13, 2015 #6
    What do you mean by the direction the tires are facing? You said momentum is what keeps the car moving but how do you get that velocity for the momentum? It comes from the static friction between the tires and the road. That's what i think.. so i don't think there's any wrong representing friction as a force on vector diagram..
  8. May 13, 2015 #7
    Sorry, badly worded, doing too many things at once and trying to get out the door for an appointment.
    Momentum and as you mentioned torque if you're adding acceleration to the equation keeps the car moving. The friction of the front tyres makes the car keep turning in a circle.
    Obviously you can represent frictional force on a vector diagram as opposing the force creating the movement.
    I meant you will get confused, or i would, if you try to combine frictional force directions without looking at the other forces involved, there are too many other forces acting on the car.
    For example the angle of the friction opposing the centrifugal force (blue line) would be more at 90 degrees to the direction the front wheels are facing and not straight backwards in relation to the car, so your resultant will be incorrect. If you then add an acceleration to this it would depend on if the car is front or rear wheel drive, FWD would tend to act more towards the direction the front wheels are pointing, if its a RWD they would be pushing the car more on a tangent in the direction the car is facing not the direction of the front wheels so the resultant force direction would change dramatically.
    Working out all the known forces first will help. A change in weight distribution and body roll will mean more friction is supplied by the outside wheel, etc.
    It would depend on what purpose you were putting this data to as to how many forces and variables you put into the equation.
  9. May 13, 2015 #8


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    OK, assuming this is a front wheel drive car, and that the car is accelerating, the diagram is still wrong. The diagram has the base of the vectors together, which is used for vector subtraction. In this case, the base of one of the vectors should be at the (arrow) end of the other vector, then the sum of the two vectors goes from the base of the first vector to the end of the second vector. That vector will point forward and inwards, not backwards as shown in the diagram. Also those force vectors are the forces that the pavement exerts onto the tires. The friction force is a Newton third law pair of forces, the force the tires exert onto the pavement (which is connected to the earth), and the force the pavement exerts onto the tires (which are connected to the car).
  10. May 13, 2015 #9
    Keep it simple.. this means that if the car is accelerated during the turn the net frictional force points in the direction of the net acceleration vector?
  11. May 13, 2015 #10


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    Yes, if you ignore aerodynamic drag, then the wheels are the only source of external horizontal forces.
  12. May 13, 2015 #11
    Thanks man,
  13. May 13, 2015 #12


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    Technically, it's the pavement attached to the earth that is the source of the external (to the car) force, the tire and wheels just transmit this external force back onto the rest of the car.
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