Question

Explain circular motion of car on a level road for 5 marks

Answer 1

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
f₁ =μ_sR₁ and f₂ = μ_sR₂
Where μs is the coefficient of static friction between the tyre and the road
Hence, total frictional force
f = μ_sR₁ + μ_sR₂ = μ_s(R₁ +R₂)= μ_sR
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 ≤ μ_sR
This frictional force will provide the necessary centripetal force f i.e., f = (mv²/r)
Hence, mv²/r ≤ μ_sR ……………….(1)
As total normal reaction balances the weight of the car, hence R =mg …………..(2)
From equation (1) and (2), we get
mv²/r ≤ μ_smg or
v² ≤ μ_srg or
v ≤ √μ_srg

Therefore, the maximum velocity (Vmax) with which a car can travel on level curved road is given by
V _max = √(μ_srg)
If the car is driven at a speed greater than V_max, 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.