0
$\begingroup$

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.

$\endgroup$

closed as off-topic by Rob Jeffries, ACuriousMind, JamalS, Prahar, Danu Nov 19 '14 at 13:04

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – ACuriousMind, Prahar, Danu
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ This question appears to be off-topic because it shows insufficient prior research $\endgroup$ – Rob Jeffries Nov 19 '14 at 12:22
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

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