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Normal force is a self adjusting force. It is responsible for circular motion on a frictionless banked road. But for a given radius, there is a specific value for velocity at which circular motion can occur. If we increase the speed than that value, the car slides upwards and similarly it slides downwards if we decrease the speed. Why normal force is not adjusted in the way that even if we increase the speed, then the car continues to move in the circular path of same radius?

omcric
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2 Answers2

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On a banked road, the normal force from the road has two components. On a friction-less road the vertical component supports the weight of the car. There is only one speed at which the horizontal component will supply the required centripetal force. You can't change one component without changing the other.

R.W. Bird
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In the absence of friction there are only two forces acting on the car - its weight $mg$ and the normal force $N$. If the road is at an angle $\theta$ to the horizontal then the horizontal force on the car is $N \sin \theta$. If the car is travelling at speed $v$ around a circle of horizontal radius $r$ then we must have

$$\displaystyle N \sin \theta = \frac {mv^2}{r} \\ \displaystyle \Rightarrow N = \frac {mv^2}{r\sin \theta}$$

If $m$, $r$ and $\theta$ are fixed then $N$ is proportional to $v^2$. If $v$ is too large then the vertical component of $N$, which is $N \sin \theta$, is greater than the car’s weight and the car will move up the slope. If $v$ is too small then $N \sin \theta$ is less than the car’s weight and the car will move down the slope. Only if $v$ is such that $N \sin \theta =mg$ will the car neither move up nor down.

gandalf61
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  • N sin theta is the horizontal component of normal force – omcric Jan 10 '22 at 05:28
  • NASCAR refers to steeply banked tracks as "self-cleaning" of slow or stopped cars... – DJohnM Jan 10 '22 at 07:32
  • @gandalf61 I am unable to understand – omcric Jan 10 '22 at 14:09
  • @DJohnM What do you mean? – omcric Jan 10 '22 at 14:10
  • @omcric Can you be more specific ? Which part exactly are you unable to understand ? – gandalf61 Jan 10 '22 at 15:54
  • It is the same thing which I have written in my question – omcric Jan 10 '22 at 15:56
  • @omcric It might help if you think in the other direction. The car is travelling in a circle. Therefore the horizontal component of the normal force must take a particular value, which depends on the car's speed. If the vertical component of the normal force is less than the car's weight the car will slide down the track, and vice versa. – gandalf61 Jan 10 '22 at 17:28