Motion of a car on a circular level road

Motion of a car on a circular level road

1. Consider a car of weight mg going around on a circular turn of radius r with constant velocity v on a level road as shown in following figure. While taking the round, the tyres of the car tend to leave the road and move away from the centre of curve. Hence the forces of friction f1 and f2 will act inward at the inner and the outer tyres respectively. If R1 and R2 are the normal reactions of ground on the tyres, then
f1sR1 and f2 = μsR2
Where μs is the coefficient of static friction between the tyre and the road
Hence, total frictional force
f = μsR1 + μsR2 = μs(R1 +R2)= μsR
Where R is the reaction of the ground on the car.

2. Because the total frictional force f cannot exceed μsR, therefore we may write
f ≤ μsR
This frictional force will provide the necessary centripetal force f i.e., f = (mv2/r)
Hence, mv2/r ≤ μsR ……………….(1)
As total normal reaction balances the weight of the car, hence R =mg …………..(2)
From equation (1) and (2), we get
mv2/r ≤ μsmg or
v2 ≤ μsrg or
v ≤ √μsrg

Therefore, the maximum velocity (Vmax) with which a car can travel on level curved road is given by
V max = √(μsrg)
If the car is driven at a speed greater than Vmax, then the car will skip and go off the road in a circle of radius greater than r. This is because even the maximum available friction will be in adequate to provide the necessary centripetal force.