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Friction Force And Circular Motion

  1. Jun 5, 2012 #1
    I was reading that the friction force involved in circular motion was in the same direction as the centripetal force; but I thought that the frictional was suppose to act in the opposite direction of the intended motion, is this not true? Also, I am having a hard time figuring out how a car could move in the first place, let alone in a circular fashion. What is the action and reaction pairs? I thought the tires pushing against the ground and the ground (friction) pushing against the tires were the force pairs? What is the external force that admits the car to move?
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  3. Jun 5, 2012 #2
    I've also been reading that static friction is the centripetal force, how is that so?
  4. Jun 5, 2012 #3


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    Either you or your book is confused here. Friction doesn't "know" anything about what sort of motion (circular or whatever) is involved. Dynamic friction always acts in the opposite direction to the relative motion of the sliding objects. Static friction can act in any direction (with magnitude <= a limiting value) to oppose other forces and maintain static equilibrium.

    The obvious answer is "the engine applying torque to the wheels", but there are other possibilities (e.g. gravity if the car is rolling down hill with the engine off).
  5. Jun 5, 2012 #4
    Yes. More specifically,
    cylinder <--force pair--> burning gas
    burning gas <--force pair--> piston
    piston <--force & torque pair--> rod
    piston <--force & torque pair--> crank
    crank <--torque pair--> axle
    axle <--torque pair--> wheel & tire
    tire <--force pair--> ground

    The car & the engine & the Earth oscillate up and down w.r.t. each other due to gas expansion and contraction (also gravity, normal forces, etc.); the car & the wheel rotate w.r.t. each another; the wheel and the Earth rotate w.r.t. each another.
    Last edited: Jun 5, 2012
  6. Jun 5, 2012 #5
    Well, I have been reading several different sources on why static friction is the reason why there is centripetal force, why is acts in the same direction as the centripetal acceleration, and why it is even static friction, but I simply don't understand. Isn't a car moving? So shouldn't it be kinetic friction?
  7. Jun 5, 2012 #6
    I was reading some where that there is static friction in circular motion because the tire isn't actually moving relative...to...something--I am not entirely sure what it said.
  8. Jun 5, 2012 #7
    You can have static friction force while nothing is moving relative to one another. The tire is not moving relatively to the ground at the contact point, but the rest of it (& the car) is.
  9. Jun 5, 2012 #8
    What exactly do you mean by the portion of the sentence in red?

    So, do you mean by this that the part of the tire that is in contact with the road is not moving relative to the ground, and this is true because at every instant the tire rotates allowing a new part of the tire to be in contact with the ground? And so the tire at every instant has to overcome this static friction, which means a constant force is being applied at every instant. But the tire as a whole is moving relative to the ground, it is just when we look at one portion of it, it is not moving? If this is true, I can understand why there is static friction, but I just have a hard time understanding why it is directed towards the center of a circle.
  10. Jun 5, 2012 #9
    Okay, I believe I am starting to understand this static friction force more, now. Although, It would be appreciated in someone could still confirm my thoughts in post #8. Okay, so I've been thinking about what would qualify the static friction force to be centripetal force, and one way to think of it was its reaction force. I believe its reaction force is the inertia of the car, that it wants to move in a straight line path relative to the circle; and so this inertia want the pull against the tires that are turned inward. Is this right?
  11. Jun 5, 2012 #10
    Since I am discussing things that pertain to cars, it seems, from what I am reading, friction isn't the reaction pair to a tire rotating, but it is the axle. So, static friction is just an external force.
  12. Jun 5, 2012 #11


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    Yes. In rolling contact, the instantaneous centre of rotation of the tire is the point of contact.
    Not sure what you mean by "has to overcome". It would have to overcome it to skid.
    Static friction opposes the relative motion that would occur otherwise. Circling at a steady rate, if the static friction were to suddenly disappear the vehicle would continue in a straight line tangentially to the circle. The difference between that path and the circular path, in the first fraction of a second, is radial. So the relative motion to be opposed is in a radial direction.
  13. Jun 5, 2012 #12


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    It is the external force (ignoring drag). The acceleration of the car is the result. That is true whether the acceleration is forwards (the everyday meaning of acceleration), centripetal, or retardant (braking).
  14. Jun 5, 2012 #13
    Well, how does a tire rotate, wouldn't it have to overcome some resistive force? And before it can experience any kinetic friction due to it overcoming the static friction, the tire would have rotated and that part of the tire that overcome static friction would not be in contact with the road anymore.
  15. Jun 5, 2012 #14
    Okay, that sounds good. So does the friction have an equal and opposite force? Or do not all forces have to have equal and opposite forces?
  16. Jun 5, 2012 #15
    I am not really sure what you mean about, "how does a tire rotate," but obviously it must have a torque applied to it to get some speed. Once the car is going around in a circle no work is needed to keep it going in a circle. The frictional force is parallel to the tire axles and thus creates no torque on them (car doesn't slow down.)

    The frictional force is what makes the car go in a circle (at which the acceleration vector is constantly changing.) The reaction force could then be the surface of the earth wanting to drift outward. It would actually be noticeable if the earth did not have much mass.
    Last edited: Jun 5, 2012
  17. Jun 5, 2012 #16


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    The tire rotates by moving the car forwards. On the flat, not accelerating, no losses in the bearings etc., there is no resistive force to be overcome. In practice, there are always losses. E.g. rolling resistance: as a tire rotates it changes shape; that takes energy - in effect, the tire is going uphill even when the road is level.
    The conceptual difficulty arises because we naturally think of a wheel as rotating about its centre. Try thinking of the tire instead as a set of spokes, one in contact with the ground at a time. At any moment, the spoke touching the ground is rotating about its point of contact with the ground. A moment later, the next spoke makes contact, lifting the first one straight up from the ground. It never has to slide.
    There are two ways to think about action and reaction. If the car is accelerating, in whichever sense, there may well be an opposing force but it won't be equal. Or, you can think of the car's resistance to the acceleration as an inertial force equal and opposite. Just don't mix the two ways of thinking!
    At a steady speed on a straight level road, the forces will be in balance. The opposing forces will be drag, bearing friction and rolling resistance.
  18. Jun 5, 2012 #17
    Not that I am judging this as incorrect thinking, but I like to think that because of this existing resistance to a change in motion, force can then be defined. I figure that is what you are saying, but it gets a bit confusing!
  19. Jun 5, 2012 #18
    It's not that hard to explain. When the rear tires (friction forces) push on the ground, they also push on the front tires (rigid body constraint through the chassis, axles, etc.). The front tires are turned at an angle, however; so unless you're doing something unsafe, the friction forces on the front tires constrain them and the motion of the entire car. This constraint is only perpendicular to the front tires, however, and they, as well as the rest of the car, are free to move along the angle they are turned to. So the car turns. If you're doing something risky given the driving conditions, however, the car might skid, tilt, or even flip over.

    The tire rotate because of the axle, the crank, and ultimately the combustion of gasoline inside the cylinder.

    In Newtonian mechanics/model, all forces are paired. The friction pushes on both the tire and the Earth, but in opposite directions.
    Last edited: Jun 5, 2012
  20. Jun 6, 2012 #19
    In circular motion we need force to accelerate the car to the center.
    On banking curve, the normal force supply the force.
    On level curve, the only force is from static force(since radius remains constant).
    It has limited supply of force = μsMg
    If more force is needed due to increase in speed or reduction of radius and static friction has reached its maximum value then the force to keep the car in circular motion not met thus 'overshoot' occurs.
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