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 `405"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 no slipping is
`v_("max")=sqrt(mu_(s)rg)=sqrt(0.1xx3xx9.8)=1.72ms^(-1)`
(i) If `v=4.5"km h"^(-1)=4.5xx(5)/(18)=(5)/(4)ms^(-1)`
`=1.25ms^(-1)`, he will not slip.
(ii) If `v=9"km h"^(-1)=9xx(5)/(18)=(5)/(2)ms^(-1)=2.5ms^(-1)`, he will slip.