I have had this doubt for a while now.

Let there be a circular banked road of some inclination with the ground.

For the sake of this question let's assume it is frictionless.

If an object with a constant velocity is travelling on it, it is in turn having a centripetal force acting on it towards the centre of its path.

This centripetal force is caused by a component of normal force which I have understood well.

But on increasing the velocity of the object, the centripetal force isn't sufficient for that radius and hence the object moves up the banked road to its corresponding radius.

However I couldn't account for any force 'up the banked road' that causes it to move that way.

Normal forces are perpendicular to that path so they don't seem to have an effect. Gravity pulls it further down so it doesn't seem like that plays a role too.

What am I missing here?


1 Answer 1


If the speed is increased, the horizontal component of the normal force must increase to provide the extra centripetal acceleration. If the normal force increases, then its vertical component will exceed the weight of the car.

  • $\begingroup$ Oh so it is the split second in the air that makes it go up? $\endgroup$ Nov 16, 2020 at 18:31
  • $\begingroup$ No, but I do need to re-think my answer. $\endgroup$
    – R.W. Bird
    Nov 17, 2020 at 14:16
  • $\begingroup$ A moving car has momentum in the forward direction. If it comes into a friction-less, banked curve with too much speed, it will go up and over the edge of the embankment. It is the vertical component of the normal force which does the lifting, but there is no loss of contact with the road. $\endgroup$
    – R.W. Bird
    Nov 19, 2020 at 19:31
  • $\begingroup$ Okay but what if it was just a sloped but straight frictionless surface? The vertical component does the lifting, but what causes the horizontal movement so that it stays on the inclined plane? $\endgroup$ Nov 20, 2020 at 6:20
  • $\begingroup$ Some of the initial kinetic energy is transferred to the increasing gravitational potential. $\endgroup$
    – R.W. Bird
    Nov 20, 2020 at 14:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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