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A cyclist speeding at 18 km/h on a level road takes a sharp circular turn of radius 3 m without reducing the speed. The co-efficient of static friction between the tyres and the road is 0.1. Will the cyclist slip while taking the turn?

Answer» <html><body><p></p>Solution :On an unbanked road, frictional force alone can provide the centripetal force needed to keep the cyclist moving on a circular <a href="https://interviewquestions.tuteehub.com/tag/turn-1429043" style="font-weight:bold;" target="_blank" title="Click to know more about TURN">TURN</a> withoutslipping. If the <a href="https://interviewquestions.tuteehub.com/tag/speed-1221896" style="font-weight:bold;" target="_blank" title="Click to know more about SPEED">SPEED</a> is too large, or if the turn is too <a href="https://interviewquestions.tuteehub.com/tag/sharp-641613" style="font-weight:bold;" target="_blank" title="Click to know more about SHARP">SHARP</a> (i.e. of too small a radius) or both, the frictional force is not sufficient to provide the necessary centripetal force, and the cyclist slips. The condition for the cyclist not to slip is given by <a href="https://interviewquestions.tuteehub.com/tag/eq-446394" style="font-weight:bold;" target="_blank" title="Click to know more about EQ">EQ</a>. <br/> `v^2 lt= mu_s Rg` <br/>Now `R = <a href="https://interviewquestions.tuteehub.com/tag/3m-310999" style="font-weight:bold;" target="_blank" title="Click to know more about 3M">3M</a> , g = 9.8 ms^(-2) , mu_s = 0.1`. That is`mu_s Rg = 2.94 m^2 s^(-2) .v = 19 km//h = 5 ms^(-1) , i.e., v^2 = 25 m^2 s^(-2)` .The condition is not obeyed. The cyclist will slip while taking the circular turn.</body></html>


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