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?

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 necessary centripetal force, and the cyclist slips. The condition for the cyclist not to slip is given by Eq.
`v^2 lt= mu_s Rg`
Now `R = 3m , 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.