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, Danu Nov 19 '14 at 13:04

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    $\begingroup$ This question appears to be off-topic because it shows insufficient prior research $\endgroup$ – 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.

  • $\begingroup$ If centripetal force is more than frictional force it will skid off while if frictional force is greater than centripetal force the car will stay on road. Correct me if I am wrong. $\endgroup$ – pcforgeek Nov 19 '14 at 3:28
  • $\begingroup$ You are correct and this is what I said? I admit I wrote it in a confusing manner. In one line I said "skid when x", in the next I wrote "not x implies y"... $\endgroup$ – Floris Nov 19 '14 at 11:56

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