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 coefficient 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.