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It is given here that the maximum safe speed of a vehicle on a banked road having coefficient of friction $0$, is $V =\sqrt{R g \tan \theta}$, where $\theta$ is the angle of banking. I can't understand how circular motion is possible without friction.

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    $\begingroup$ A banked road without friction has a component of gravity towards the center of the circle. Do a free body diagram and it will become obvious. $\endgroup$ – ja72 Apr 2 '14 at 18:42
  • $\begingroup$ Gravity has a horizontal component? $\endgroup$ – DJohnM Apr 2 '14 at 21:57
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    $\begingroup$ No, the normal force exerted by the road has a horizontal component. $\endgroup$ – Shivam Sarodia Apr 2 '14 at 22:55
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What you need for circular motion is Centripetal Force. Definition:

Centripetal force is a force that makes a body follow a curved path: its direction is always orthogonal to the velocity of the body, toward the fixed point of the instantaneous center of curvature of the path. Centripetal force is generally the cause of circular motion.

If the road is flat, the centripetal force is provided by Friction between the tyres of the car and the road. This image show how: enter image description here

If the there is no friction and the road is flat, the car would not be able to turn, it would keep sliding in the same direction.

So, if there is no friction, there has to be some force that can provide that necessary centripetal force. If the road is banked, the horizontal component of the Normal vector of the car that is going towards the centre of turn can act as the centripetal force. Hence, the car can turn on a banked road even without friction. Here is how: enter image description here

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  • $\begingroup$ Are you IIT ian? $\endgroup$ – user43704 Apr 2 '14 at 19:00
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    $\begingroup$ Haha no. All Indians who know a bit of Physics are not in the IITs. I am doing a Bachelors in Science, not Engineering. $\endgroup$ – user42733 Apr 2 '14 at 19:02
  • $\begingroup$ Good night parth $\endgroup$ – user43704 Apr 2 '14 at 19:05
  • $\begingroup$ I have still doubt about, why there is not gravitational acceleration component along surface of road ? $\endgroup$ – user43704 Apr 3 '14 at 1:58
  • $\begingroup$ You mean down the slope? $mgSin\theta$ is there of course. Normal force exists because of gravity only, it's a reaction to gravity (Newton's Third Law). $\endgroup$ – user42733 Apr 3 '14 at 5:15

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