# Maximum velocity on Unbanked roads and why should friction be more than centripetal force? [closed]

Why is maximum velocity on unbanked circular road given by $\sqrt{urg}$ where $u$ is coefficient of friction, $r$ is radius of circular track and $g$ is acceleration due to gravity? Why should frictional force be more than centripetal force? What will happen if centripetal force is more than friction? I was absent when this topic was taught in the class so there might be some mistakes in the question.

## closed as off-topic by Rob Jeffries, ACuriousMind♦, JamalS, Prahar, DanuNov 19 '14 at 13:04

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• This question appears to be off-topic because it shows insufficient prior research – Rob Jeffries Nov 19 '14 at 12:22

If there is not enough friction to keep the vehicle in its circular path, it will skid. The force needed for the circular path is the centripetal force: friction (the force keeping the car on the road) must be greater that that. Now the no-slip condition (centripetal force < friction) implies = $\frac{mv^2}{r} < mg\mu$ . Your equation follows by simple manipulation.