Here , we have a car on a banking road. Considering two cases :

enter image description here Frictionless and friction surface of the road.

I give an initial velocity to the car by a push for a second. Then , the cases discuss about the direction of friction and motion of the car.

Case 1: Frictionless road surface

We will have an $N*sin$$\theta$ force acting towards the circle. Initial direction of motion was along the road but isn’t it that due to the banking of road and frictionless road. The car would slowly start to slip down $? $Is this case also uniform circular motion $? $

non uniform circular motion is not possible here since friction is not present. Therefore , no acceleration.

Also , can we say no centripetal and tangential acceleration is present$ ?$

_____ 1)N cos $\theta$ = mg

_____ 2)N sin $\theta$ = $\frac{Mv^2}{r}$ . From this , we can say the car moves down $?$.

Case 2: Friction is present between road surface and car.

All directions are same as the above except for friction. So , can I say that friction also acts in the same direction as $Nsin$$\theta$ ? in case of uniform circular motion.

In non uniform , friction force will be at an angle alpha to the centripetal acceleration $?$

By the $?$, I wish to understand what would happen.

  • $\begingroup$ If there is no friction will you need to push it or will gravity and the slope take over? If you left your car without handbrake on or in gear and on a slope would it move? $\endgroup$
    – Solar Mike
    May 15, 2021 at 15:26
  • $\begingroup$ Yes. One components of Mg and other of N would make the car slip down. $\endgroup$
    – S.M.T
    May 15, 2021 at 15:27
  • $\begingroup$ @SolarMike Yes. Since no friction and only if there was some initial force. It would move in like a diagonal direction and finally slip down. $\endgroup$
    – S.M.T
    May 15, 2021 at 15:35

3 Answers 3


Frictionless surface.

Assuming the initial velocity is perfectly tangential to the circular motion to begin with :

  • greater than $u_0 >\sqrt{g R \sin\theta}$, then the car will spiral outwards
  • less than $u_0 < \sqrt{g R \sin\theta}$, then the car will spiral inwards
  • exactly equal to $u_0 =\sqrt{g R \sin\theta}$, then the car will continue to move on the circle. (Note that there is no way to accelerate or decelerate if there is no friction).

If the velocity has any component that is not tangential then the car will either move outwards on inwards.

surface with friction

The direction of the friction would be depended no the magnitude of $\frac{v^2}{R}$. If:

  • $\frac{v^2}{R}>N\sin\theta$, the Friction will have the same direction as $N\sin\theta$. (i.e. it will be preventing -as much as possible- the car from spiraling outwards)
  • $\frac{v^2}{R}<N\sin\theta$, the Friction will have the opposite direction as $N\sin\theta$ (i.e. it will be preventing -as much as possible- the car from spiraling inwards)
  • $\frac{v^2}{R}=N\sin\theta$, there will be no lateral Friction (there will be friction only along the motion of travel.).
  • $\begingroup$ Hmm Ok. I got it now. Thanks a lot $\endgroup$
    – S.M.T
    May 16, 2021 at 6:29
  • $\begingroup$ You're assuming that this car has wheels that always point in the azimuthal direction, and that rolling resistance is negligible compared with sliding friction, right? That's OK, except in the diagram, there are no wheels! $\endgroup$ May 16, 2021 at 7:44
  • $\begingroup$ When there is no friction it doesn't matter that there are no wheels. Altgough you have a valid point that I shouldn't refer to the mass as a car. $\endgroup$
    – NMech
    May 16, 2021 at 10:45
  • $\begingroup$ @NMech Indeed, it was the with-friction case I had in mind. If there are no wheels, it seems to me that, when the car (or rather, brick) is moving, there is no way to transmit information about the slope of the road (i.e. about the in-plane component of gravity) to the mating surfaces; therefore, the only special in-plane direction about which the mating surfaces have information is the direction of motion of the car (brick), and the frictional force can only be directed oppositely to the instantaneous velocity of the car (brick). $\endgroup$ May 16, 2021 at 11:54
  1. No friction

Two outcomes after the initial push: a) run out on the curve, and b) slide down the slope.

  1. With friction

The outward centrifugal force is overcome by the inward friction force $\mu N or Wsin\theta$, and the car stays on its course.

  • $\begingroup$ Can it stay on the bank road without slip and without friction. $\endgroup$
    – S.M.T
    May 16, 2021 at 3:50
  • $\begingroup$ No. Without friction is similar to standing on a sloping ice rink with regular walking shoes. You can move your legs but won't go forward, and you can't stop the motion to slide down the slope. Similarly, the car wheels will spin with no forward motion but sliding even with the brake fully engaged. $\endgroup$
    – r13
    May 16, 2021 at 3:58
  • $\begingroup$ Ok. Thanks a lot. $\endgroup$
    – S.M.T
    May 16, 2021 at 6:29

If the road width is wide enough:

  • No friction and no air friction: the car will seek a wider curve up the bank and circle around or go around an ellipse if not initially pointed straight, if the initial velocity, Vi is greater than,$$V_i> \sqrt {rgtan\theta} $$ Or else a lower orbit.

  • If there is friction:

  • If $V_i> \sqrt{\frac{rg (sin\theta+\mu cos\theta)}{cos\theta-\mu sin\theta}} $ Then the car will skid up and rotate in a higher orbit likely an ellipse similar to a rocket changing orbit.

  • If $Vi< \sqrt{\frac{rg (sin\theta-\mu cos\theta)}{cos\theta+\mu sin\theta}}$

the car will skid down and rotate in a lower orbit.

Otherwise, it will stay the course.


Just for the record the accepted answer is wrong.

For example in frictionless situation as $\theta$ approaches 90 the speed, V, must approach infinity, otherwise the car will slide down due to lack of friction. but $sin90=1 $ which implies $ V = \sqrt{Rg}.$


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