A cyclist speeding at `4.5"km h"^(-1)` on a level road takes a sharp circular turn of radius 3 m without reducing the speed The coefficient of static friction between the road and the tyres is `0.1` will the cyclist slip while taking the turn (i) with a speed of `4.5"km h"^(-1)` and (ii) with a speed of `9 "km k^(-1)`?

Explanation

Frictional force provides the necessary centripetal force . He will slip if the turn is too sharp .(i.e., too small a radius ) or if his speed is too large .
Maximum speed for not slipping is
` V_(max) = sqrt(mu_(s) Rg )`
` = sqrt ( 0.1 xx 3 xx 9.8)`
` = 1. 71 ` m/s
(i) If v = 4.5 Km/h = `5/4 m//s = 1.25 m//s`
(ii) If v = 9 km/h = `5/2 m//s = 2.5 m//s`, he will slip.