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 suffecient to provide the necessary centripetal force, and the cyclist slips . The condition for the cyclist not to slip is given by Eq. (5.18) :
`nu^(2)` `le` `mu_s`Rg
Now, R = 3 m , g = 9.8 `ms^(-2)` , `mu_s` = 0.01. That is `mu_s`Rg = 2.94 `m^(2)s^(-2)` , `nu` = 18 km/h = 5 `ms^(-1)`, i.e., `nu^(2)` = 25 `m^(2)s^(-2)` . The condition is not obeyed . The cycllist will slip while taking the circular turn.