Derive expression for the maximum speed of a vechile on the banked road . State th factor on which the optimum of speed depend

• Motion of a car on a level road:

If R₁ and R₂ are the normal reactions of the ground on the two tyres of a car of weight Mg, going around on a circular turn of radius r, with velocity v, on a level road, then

F₁ = µR₁ and F₂ = µR₂

Where, µ is the coefficient of friction between the tyres and the road

The total force of friction provides the necessary centripetal force, i.e.,

The total normal reaction balances the weight of the car, i.e.,

From equations (i) and (ii), we have

The above equation gives the maximum velocity with which the car can take a turn of radius r_s, when the coefficient of friction between the tyres and the road is equal to r.

• Motion of a car on a banked road:

For the vehicle to go round the curved track at a reasonable speed without skidding, the greater centripetal force is managed for it by raising the outer edge of the track a little above the inner edge. It is called banking of circular tracks.

Consider a vehicle of weight Mg, moving round a curved path of radius r, with a speed v, on a road banked through angleθ.

The vehicle is under the action of the following forces:

• The weight Mg acting vertically downwards

• The reaction R of the ground to the vehicle, acting along the normal to the banked road OA in the upward direction

The vertical component R cos θ of the normal reaction R will balance the weight of the vehicle and the horizontal component R sin θ will provide the necessary centripetal force to the vehicle. Thus,

R cosθ = Mg …(i)

R sinθ = …(ii)

On dividing equation (ii) by equation (i), we get

As the vehicle moves along the circular banked road OA, the force of friction between the road and the tyres of the vehicle, F = μR, acts in the direction AO.

The frictional force can be resolved into two components:

• μ R sinθ in the downward direction

• μ R cosθ in the inward direction

Since there is no motion along the vertical,

R cos θ = Mg + μ R sinθ ……. (iii)

Let v_max be the maximum permissible speed of the vehicle. The centripetal force is now provided by the components R sinθ and μ Mg cosθ, i.e.,

R sin θ + μ R cosθ = …….. (iv)

From equation(iii),we have

Mg = R cosθ (1−μ tanθ)…(v)

Again from equation (iv), we have

 = R cosθ (μ + tanθ) …(vi)

On dividing equation (iv) by (v), we have

Maximum optimum speed depends on:
1) Radius of the curved path,
2) Coefficient of friction
3) angle on inclination