When a car is turning along a circular path with a constant speed, it requires a centripetal force to keep it moving along that path. This force is a frictional force which points toward the center of the circle which the car traces. But why exactly does friction have to point radially towards the center? From my understanding, friction opposes the car's tangential motion, which means that friction should also be tangential but opposite to the velocity. I already understand that static friction will prevent the car from sliding along the path which its tangential velocity indicates, but I don't understand why the friction has to be perpendicular to the car. Please, if anyone knows, explain this using a vector diagram showing where exactly this perpendicular friction vector comes from. I would highly appreciate it! Thanks.
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Well any object wants to move in a straight line (or be stationary) in the absence of an external force. If the force of friction pushed the car in any other direction but radially inwards it wouldn't remain in circular motion. – Charlie Dec 23 '19 at 19:11
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2Does this answer your question? Why does friction act radially inwards when a car turns? – Charlie Dec 23 '19 at 19:15
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Well, yes I understand that the only place where the centripetal force can come from is through friction. But I would like to see, through a diagram, how exactly there is a component of the friction which is perpendicular to the car. – Stewie Dec 23 '19 at 19:22
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As the car turns inward there is a centrifugal force that pushes it outward, this force is opposed by the inward frictional force. The car is being pushed outwards because of it's motion and friction is what's holding it in so the direction must be inwards. Are you also wondering how friction works? – Charlie Dec 23 '19 at 19:29
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I thought that the centrifugal force was something that only passengers inside the car would feel, since the car can only feel a centripetal force. I don't understand how static friction would counter that centrifugal force. If the centrifugal force were high enough, the car would slide radially outwards? – Stewie Dec 23 '19 at 19:39
2 Answers
This force is a frictional force which points toward the center of the circle which the car traces. But why exactly does friction have to point radially towards the center?
It points to the center because the centripetal force is needed to keep the vehicle on a circular path.
From my understanding, friction opposes the car's tangential motion, which means that friction should also be tangential but opposite to the velocity.
The centripetal friction does not oppose the car's tangential motion and is not, therefore, opposite to the velocity. That would be the case if, for example, when applying the brakes to a car moving in a straight line. The friction force of the brakes (and tires, if skidding occurs) opposes the direction of the vehicle causing it to decelerate, but the direction of motion is unchanged.
When the car is on a circular path, the tangential velocity is constantly changing direction. To keep the car on a circular path the centripetal force continually acts perpendicular to the tangential velocity, pointing towards the center.
I already understand that static friction will prevent the car from sliding along the path which its tangential velocity indicates, but I don't understand why the friction has to be perpendicular to the car.
The force has to be perpendicular to the car in order to continually change the direction of the car to keep it on a circular path. Without it the car would travel a straight line due to Newton's first law which which tells us that and object in uniform motion in a straight line will continue to move in a straight line unless acting upon by an external force.
Please, if anyone knows, explain this using a vector diagram showing where exactly this perpendicular friction vector comes from.
See the diagram below.
The green arrow shows the direction that the car would travel without the centripetal force due to its inertia. The red arrow shows the direction of the centripetal force acting on the car. The blue arrow shows the change in direction of the instantaneous velocity due to the centripetal force.
Hope this helps.
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First of all, thanks for answering :) This looks great, but intuitively I wouldn't be able to understand this. I have always thought that friction opposes motion, in this case the tangential motion of the car, but somehow the friction is actually pointing perpendicular to the velocity. I understand that, in this case, only the frictional force can take the place of the centripetal force, but still, I feel like that is a rather limited explanation. – Stewie Dec 23 '19 at 20:05
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@It does oppose motion. Without it the car would slide radially outward opposite the direction of the centripetal force, when viewed in the non-inertial rotating frame of the car. That is the centrifugal force, not shown because it is a pseudo force. Although it would slide outward in the rotating frame, in the inertial frame it goes in a straight line. – Bob D Dec 23 '19 at 20:13
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Alright that makes sense. Is this how you always conceptualized it? Is there no vector diagram which can show how the perpendicular friction arises? I just want to be sure. Thanks for the answer! – Stewie Dec 23 '19 at 20:27
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@Stewie won’t be able to get to my computer for a while. But all you need to do is draw a vector opposite the direction of the red centripetal force and that would be the centrifugal force. It is not a real force but appears to be there because if there were no centripetal force the car would appear to accelerate away from the center – Bob D Dec 23 '19 at 21:05
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@Stewie it is easier to see this if you imagine yourself at the center holding a string with a ball on the end twirling it around. It feels like a force is pulling on you. But obviously there is nothing physically pulling outward on the ball. That’s the fictitious centrifugal force. You are really feeling the inertia of the ball – Bob D Dec 23 '19 at 21:10
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Right, so the friction would act as the force opposing the centrifugal force, which in turn depends on the centripetal force required by the car's velocity. So, the faster the car goes around the circle, the more centripetal force required and so the higher the centrifugal force. If the apparent centrifugal force were to be higher than the friction, the car would be released radially outward (according to the rotating frame of reference). In other words, the static friction on the tires is trying to counter the skidding of the car radially out. Correct me if I'm wrong. Thanks! – Stewie Dec 23 '19 at 21:34
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Also, is it valid if we view this case from just one reference frame like the one which is most convenient for this explanation (the non-inertial one)? Cause from the outside reference frame, the friction wouldn't be opposing any centrifugal force. – Stewie Dec 23 '19 at 21:36
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@Stewie yes and yes. The centrifugal force only “exists” in the non inertial (rotating) frame. The diagram I gave is the view from the inertial frame. From that frame it is clear that the only force acting on the car is the centripetal force and that is the net force that causes the car to change direction. In the rotating frame the centrifugal force equals the centripetal force to prevent acceleration in radial direction – Bob D Dec 23 '19 at 22:05
But why exactly does friction have to point radially towards the center?
Actually that's only the case if the car is traveling at a constant speed in a circular arc -- basically the object goes in a constant speed circle if and only if the acceleration is pointed straight at the center. If the driver has their foot on the gas and maintains circular motion, then the net force will point ahead of the center; if the car is being allowed to slow down, then the net force will point behind center.
but I don't understand why the friction has to be perpendicular to the car
It doesn't have to be perpendicular to the car -- just to the line of motion.
Get on YouTube and watch videos of race cars going around corners for examples of cars that are turning around a center that's not on a line perpendicular to the car.
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