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I have been reading though a textbook on banked corners (which I get to shortly) and centripetal force, and I have a few question as some of the things mentioned don't make sense to me. So to start, I was told that the horizontal force component of the tension on the conical pendulum provides the centripetal force in the system (and please correct me if I am wrong). I understand that there must be a force acting, even if the ball is at a constant speed, because the direction is changing, thus velocity. A change in velocity is acceleration (centripetal acceleration), thus we get a force. If we draw a draw two velocity vectors at $90$ degrees to each other, and subtract them to find the change in velocity the resulting vector acts inwards, thus acceleration and therefore the centripetal force acts in the same direction as the horizontal component of the tension force.

So far this makes sense to me. However:

I was reading about banked corners. They increase the centripetal force to allow cars to travel around a corner at a high speed. The textbook says: "If a banked corner was frictionless, to prevent the car sliding down the bank from gravity, or excessive speed resulting in the car going up the bank, the horizontal component of the reaction force (to me what looks like the equivalent of the horizontal tension force) must be equal to the $F_c = mv^2/r$" This is called the critical speed. In addition to this, I thought that a car turning around a corner had its centripetal force provided by friction. This is frictionless, so is the car not turning?

It is hard to explain, but I am confused with this concept, that it is possible that the horizontal component of the reaction force doesn't have to equal the centripetal force. I thought that the horizontal component of the friction force was the centripetal force.

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  • $\begingroup$ By "tension force" do you mean "normal force"? If you do mean tension, where are the ropes in your system? $\endgroup$ Apr 21 '20 at 0:41
  • $\begingroup$ Haha, sorry, yes I do mean normal force. $\endgroup$
    – Angus
    Apr 21 '20 at 0:55
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Banked Curve Diagram

I've drawn a diagram here that should illustrate what we're talking about. Hopefully you can see from the picture that the reaction force, normal to the road surface, has a horizontal component that point towards the "center" of the banked curve (imagine it as being part of a larger conic shape). I drew the normal force components in red to make it clear. The horizontal component of $F_N$ is the only force acting in this direction, thus it acts as the centripetal force.

In the case of a car driving on an unbanked road, this reaction force is still normal to the road surface, but now the road is flat. Therefore, the normal force only has a nonzero vertical component, and friction would indeed be the cause for the centripetal force.

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  • $\begingroup$ Thanks for your response. I understand the first part of this. The horizontal component provides the centripetal force. If the velocity of the car increases, what happened to the size of the horizontal component? Would it increase or decrease? I think this would help me build up a better picture of what is happening. I understand that increasing the velocity results in an increase in the turning radius of the car, so it would go up to the bank more. Also is this force only provided by the bank, and not the car turning? (when it is on the bank?) $\endgroup$
    – Angus
    Apr 20 '20 at 23:40
  • $\begingroup$ @Angus A more general analysis of the frictionless banked curve problem is pretty involved, which is why introductory physics texts just focus on fixed velocity solutions. $\endgroup$ Apr 21 '20 at 0:45
  • $\begingroup$ Awesome, thanks. So, in a banked system, friction acts in the opposite direction of the motion, in this case it acts up the bank to stop the car from sliding down (right?). So how does friction provide a centripetal force when a car turns? physics.stackexchange.com/questions/306151/… This post looks at the answer but I can't get my heard around the explanation that JMac gave as his answer. $\endgroup$
    – Angus
    Apr 21 '20 at 0:57

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