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?
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?
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 Equation
$v^{2} \leq \mu_{s} R g$
Now, $R=3 \,m g=9.8 \,m s ^{-2}, \mu_{s}=0.1 .$ That $1\, s$
$\mu_{s} R g=2.94\, m ^{2}\, s ^{-2}, v=18 \,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.
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