2
$\begingroup$

My information is very limited. All I know is that there is a normal force and gravity acting on the car. I know what a banked road is, I know a centripetal force is a force that tries to pull the car towards the center. However, I have tried researching and I can't seem to understand so please make me comprehend on this topic. I need it to be simple otherwise it'll be futile. English isn't my native language. Thanks in advance.

Can someone just tell me why friction isn't needed?

$\endgroup$
2
  • $\begingroup$ Duplicate? physics.stackexchange.com/q/298151/104696 $\endgroup$
    – Farcher
    Jan 1, 2017 at 13:27
  • $\begingroup$ As I said I have tried researching and I can't seem to understand. That seems complicated, and I a m trying to get as simple answers as I can. $\endgroup$ Jan 1, 2017 at 13:29

3 Answers 3

2
$\begingroup$

The frictionless banked curve exerts a normal force $F_{n}$ perpendicular to its surface. The downward force of the gravity $F_{g}$ is present. The two forces add as vectors and the resultant or net force $F_{net}$ points toward the center of the circle. This is the centripetal force.

enter image description here

When the forces are resolved into their components, you will find that $F_{net,y}=F_{n}\cos\theta-F_{g}=0$. Hence, $F_{n}=\frac{F_{g}}{\cos\theta}$. You will also find that $F_{net,x}=F_{n}\sin\theta$. You know that this is equal to the radial force, $F_{r}=m\frac{v^2}{r}$.

$$F_{r}=F_{net,x}\Longrightarrow\frac{mv^2}{r}=F_{n}\sin\theta\Longrightarrow\frac{mv^2}{r}=\frac{F_{g}}{\cos\theta}\sin\theta\Longrightarrow\frac{mv^2}{r}=mg\tan\theta\Longrightarrow\frac{v^2}{r}=g\tan\theta$$

After simplifying, you will find that $v=\sqrt{rg\tan\theta}$, where $\theta$ is the angle that will allow a car to travel on a frictionless curve of radius $r$ with constant speed $v$. A banked curve is designed for one specific speed. Traveling at a speed higher than $v$ means the car will slide out, up, and over the edge. Traveling at a speed lower than $v$ means the car will slide in, down, and off the bank.

$\endgroup$
3
  • $\begingroup$ What do you mean by a frictionless curve? And how can the car move without any fricton? $\endgroup$ Jan 2, 2017 at 13:22
  • $\begingroup$ A frictionless curve is the same as a circular, banked road without friction. The car can stay on the road at a specific speed, as derived above. $\endgroup$
    – V.Z.
    Jan 3, 2017 at 22:21
  • $\begingroup$ @Okama Ksakas , For any body to undergo circular motion, there should exist some force to provide enough centripetal force. At a " specific speed " , the component of normal reaction itself is self sufficient to provide the required centripetal force in-order to make the car move along the banked road ( of given angle of banking and radius). The car has to travel at that specific speed along the given banked road to avoid friction. A small increase or decrease in speed makes friction come into play. This article is sure at help to you : en.wikipedia.org/wiki/Banked_turn $\endgroup$ May 31, 2017 at 5:20
1
$\begingroup$

Cars are complicated; don't let the complications distract you. Go in your kitchen and get a big mixing bowl and an ice cube.

bowls

Hold the ice cube in your hand a moment until it goes from sticky-cold to damp-cold. Now it's melting, and it'll separate from whatever surface it touches by a thin layer of water. This is basically an ideal low-friction interface.

Give the bowl a shake, and you can get the ice cube to slide around the bowl in a circle. Once it's going, it'll go for a pretty long time in a level circle. The faster the ice cube is going once you get it level, the higher up in the bowl it'll ride --- because the bowl is more steeply sloped up near the rim than it is near the base.

This should make it clear that an object can use a banked curve to follow a circular path even with negligible friction. You have other answers that get deeper into the the vector arithmetic of why.

$\endgroup$
0
$\begingroup$

For movement of a car on banked road, all you need is a component of any force towards center of circle.

If you find that friction is not present that means that a component of normal reaction acting on the wheels of the vehicle is present in the radial direction and is enough to sustain the circular motion of the vehicle without the need of friction.

$\endgroup$
4
  • $\begingroup$ What do you mean by the radial direction? $\endgroup$ Jan 1, 2017 at 13:25
  • $\begingroup$ And isn't friction the stopping force ? I am confused how friction is related to this at all, I get everything but the friction force and why is it mentioned :S please be as simple as you can be $\endgroup$ Jan 1, 2017 at 13:27
  • $\begingroup$ First of all radial direction means pointing towards the centre of circle, that is along the radius. Now what i am saying is friction may or may not be present, not that it is definately present. The reason why friction may be present is if the car is slipping or skidding on the road, that again depends on the speed at which the car is moving, if the speed is too high, friction comes into play to keep it in a circle. $\endgroup$ Jan 1, 2017 at 14:02
  • $\begingroup$ I don't get how friction is not needed when there's always friction between the car wheels and the road? $\endgroup$ Jan 2, 2017 at 13:57

Not the answer you're looking for? Browse other questions tagged or ask your own question.