Thus to obtain v max we put f N . Then Eqs. ( .19a) and ( .19b) become N cos θ = mg + s N sin θ ( .20a) N sin θ + s N cos θ = mv / R ( .20b) From Eq. ( .20a), we obtain – mg N cos sin Substituting value of N in Eq.
( .20b), we get max – mg sin cos mv cos sin R + or max – tan v Rg tan + = ( . ) Comparing this with Eq. ( . ) we see that maximum possible speed of a car on a banked road is greater than that on a flat road.
For µ s = in Eq. ( . ), v o = ( R g tan θ ) ½ ( . ) At this speed, frictional force is not needed at all to provide the necessary centripetal force.
Driving at this speed on a banked road will cause little wear and tear of the tyres. The same equation also tells you that for v < v o , frictional force will be up the slope and that a car can be parked only if tan θ ≤ µ s . Example . A cyclist speeding at km/h on a level road takes a sharp circular turn of radius m without reducing the speed.
The co-efficient of static friction between the tyres and the road is . . Will the cyclist slip while taking the turn? Answer 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. ( .