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Our physics teacher told us that a car moving in a circular track experiences static friction whose perpendicular component to the tangential velocity acts as centripetal force. But how can static friction act on a moving body? Shouldn't it be kinetic?

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    $\begingroup$ Arent the tires stationary with respect to the road? $\endgroup$
    – R. Emery
    Dec 26, 2020 at 10:26
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    $\begingroup$ Does this answer your question? Directions of static & kinetic friction? See answer by Jmac here $\endgroup$ Dec 26, 2020 at 12:54
  • $\begingroup$ There should be some force to keep the car move in circle. Otherwise it would go in straight motion.This force is the static friction motion. The tires are static relative to the street because there is no motion directed away from the center. $\endgroup$
    – Aven Desta
    Dec 26, 2020 at 19:13
  • $\begingroup$ Because by definition, there is no movement perpendicular to the tangential velocity. $\endgroup$
    – Florian F
    Dec 26, 2020 at 22:18

5 Answers 5

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That is because the bottom of the tyre is moving at exactly an equal speed but in the opposite direction to that of the car, so the point at which the tyre is in contact with the ground has a net velocity of 0 and is therefore stationary relative to the ground. This can happen because the wheels can rotate freely in the axis that is parallel to the direction of the wheels with a much lower internal friction i.e. the friction of the parts moving inside the car. Note that the internal friction within the car is kinetic friction, but because of lubricants and the materials used, the internal friction is quite low

Since the car is moving in a circular track, it is constantly changing direction. The change in direction happens when the front wheels are turned inward with respect to the circular track, so that they are at an angle to the direction of the velocity of the car, which would be in a straight line along the tangent of the circular path without any external forces. This creates a perpendicular outward component to the momentum of the car relative to the direction of the front wheels and also the circular track. But since there is friction and the wheels can’t rotate freely in the perpendicular axis, the friction acts in the opposite direction, that is inward, and this ultimately causes the car to turn in the direction of the front wheels. In both the parallel friction, which causes the car to move forward, and perpendicular friction, which causes the car to turn, the point at which the wheel is in contact with the ground is stationery relative to the ground, so the friction is static.

However if there were enough acceleration, or if the wheels were turned enough at a high enough velocity, the static friction would no longer be enough to counteract the force of the car, and the car would skid instead i.e. the wheels would slide against the road. In that case static friction would no longer apply, and instead there would be kinetic friction

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    $\begingroup$ In normal driving but it could be drifting. $\endgroup$
    – badjohn
    Dec 26, 2020 at 11:17
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    $\begingroup$ In that case static friction would not apply. That is also why cars have a higher tendency to skid while drifting $\endgroup$
    – Neelim
    Dec 26, 2020 at 11:23
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    $\begingroup$ @Neelim whose outward component are you talking about ? $\endgroup$
    – Ankit
    Dec 26, 2020 at 18:02
  • $\begingroup$ The outward component of the momentum of the car relative to the turned front wheels and also the circular path $\endgroup$
    – Neelim
    Dec 27, 2020 at 5:55
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It depends on what you mean by “moving”. The static friction force prevents relative motion between the tires and the road in the form of skidding or sliding. On the other hand it is the static friction that enables the car to move in a circular path without skidding. It also enables a car to accelerate in a straight line without skidding.

In the accelerating (non inertial) reference frame of the car It prevents the car from skidding radially outward due to the centrifugal force. It acts opposite to the centrifugal force.

In the inertial frame of the road the static friction force (centripetal force) acts radially inward resulting in circular motion without skidding (losing traction), which it would otherwise do because of its inertia (Newton’s first law).

But wouldn't there be a (kinetic)frictional force opposite to the tangential motion?

First of all, kinetic friction only occurs when the maximum possible static friction force is exceeded and the tires begin to skid. Thereafter the kinetic friction force is generally considered constant acting in a direction that opposes the skidding motion.

If the tangential velocity is constant (constant circular speed) then there is neither a static nor kinetic friction force acting opposite to the tangential motion. As along as the speed is not too great, the static friction force acting radially inward will keep the car in circular motion.

However, if the speed becomes too great, or if the car is accelerating while cornering and that acceleration becomes too great, the maximum possible static friction force will be exceeded and the tires will skid both laterally and tangentially. Once skidding begins, the friction changes from static to a generally constant kinetic force. It is at that point where it is the kinetic friction force that opposes skidding motion both laterally and tangentially.

Hope this helps.

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  • $\begingroup$ But wouldn't there be a (kinetic)frictional force opposite to the tangential motion? $\endgroup$
    – Wormhole
    Dec 26, 2020 at 18:51
  • $\begingroup$ @RavinshuGupta I've updated my answer to respond to your follow up comment. $\endgroup$
    – Bob D
    Dec 26, 2020 at 20:51
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Okay first of all, we know that circular motion is impossible without a centre - seeking force (i.e centripetal force).

A body will always tend to go straight. So when it tries to make a turn, it slips(skids) away from the road and this slipping is prevented by the Static friction force.

And slipping away from the centre can be prevented only if the force acts towards the centre and hence it is clear that the static friction force acts as the centripetal force.

enter image description here

In the above image, it is clear that the car alongwith turning is also going away from the track and this is the point where the static friction comes into existence. This force prevents the outward skid.

However there is a certain limit in the magnitude of this force and after you exceed a certain velocity, you skid away from the track and the static friction is replaced by the kinetic friction.

Hope it helps 🙂.

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  • $\begingroup$ @Ravinshu Gupta was that helpful ? $\endgroup$
    – Ankit
    Dec 26, 2020 at 18:59
  • $\begingroup$ But wouldn't there be a (kinetic)frictional force opposite to the tangential motion? $\endgroup$
    – Wormhole
    Dec 26, 2020 at 19:38
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Kinetic friction is only experienced when objects are sliding against each other, i.e. rubbing. For a car, that would mean the car was skidding which can occur, but is typically not something you'd like to have occur.

Wheels on a car are a bit hard to conceptualize at first, but you should think of them(*) as being like an infinite number of infinitely tiny "feet" all arranged in a circle, and the rolling of the wheel is these feet "walking". And in walking, only static friction forces act, unless you lose your footing, i.e. skid, such as on ice and with bad shoes. Your forward motion is instead due to you putting your center of mass off balance until your next foot catches you while your previous foot is withdrawn, and the wheel does the same thing.

That said, rolling wheels do also dissipate energy, but the reason they do so involves a third type of friction called rolling friction, which results from the wheel and/or road's propensity to squish in some way. The squishier the wheel and road, the more rolling friction there is.

(Note that because it's not "rubbing", some might say it is better to call it "rolling resistance" instead of "rolling friction", but given that it slows things down, I like to call it "friction". Meh.)


(*) Insofar as classical mechanics is concerned.

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    $\begingroup$ I like the analogy of an infinite number of infinitesimally tiny feet. Imagine each tiny foot with a tiny rubber sole that compresses under the weight of the car at it takes its turn to meet the road. The place it meets the road is ahead of the axle and it will meet maximum compression when directly under the axle. This tiny time, over this tiny distance, being compressed this tiny height, creates a tiny backward force. Multiply this tiny force over all those tiny feet and it's not so tiny any more. Some of this is returned when the sole decompresses behind the axle, the losses are heat. $\endgroup$
    – MacGuffin
    Dec 27, 2020 at 0:55
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Unless the car is drifting, then the friction will be static.

Consider the point of the wheel that touches the ground. That is the point of contact. The force of friction acts through this point of contact.

What's important to grasp is that the velocity of the point of contact with respect to the ground is zero.

The reason is the following:

Consider a wheel that is only translating (i.e. no rotation about the center of mass). I have illustrated this below -- the red dot is the center of mass, and notice that every point in this wheel is moving with the same velocity $\vec v$.

Now consider a wheel that is only rotating. Every point on the outer edge of the wheel will have speed $v$ (assuming no slipping).

See the following sketch. The red dot is the center of mass.

enter image description here

Now, to view the 'full motion,' we can 'add' the rolling motion and the translating motion and get this:

enter image description here

Notice how the velocity of the bottom-most point is zero? That is, the contact point (through which friction would act) is instantaneously at rest!

Since it's not moving wrt the ground, then the friction must be static.

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