A cyclist speeding at 18 km/h or 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 ?

Explanation

On an unbanked road , frictional force alone can provide the centripetal force needed to keep the cyclist moving on a circular turn without slipping . If the speed is too large , or if the turn is too sharp (i.e of too small a radius ) or both , the frictional force is not sufficient to provide the necessaryy centripeetal force , and the cyclist slips . The condition for the cyclist not to slip is given by
`v_(max) = sqrt(mu_(s)R_(g)) v_(2) le mu_(s) Rg`
Now , R = 3m , `g = 9.8 " m"s^(-2) , mu_(s) = 0.1 `
That is `mu_(s) " R g " = 2 .94 m^(2)s^(-2) v = 18 km//h = 5 m s^(-1)` i.e ., ` = 25 v^(2)` . The condition is not obeyed . The cyclist will slip while taking the circular turn .