Draw a diagram showing all components of force acting on a vehicle moving on a curved banked road. Write the necessary equation for maximum safety, speed and state the significance of each term involved in it.

Dear student,

• 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 = $\frac{M{v}^2}{r}$ …(ii)

On dividing equation (ii) by equation (i), we get
$$\begin{gathered} \frac{R \sin \theta }{R \cos \theta } =\frac{M{v}^2/r}{Mg} \\ \tan \theta =\frac{v^2}{rg} \end{gathered}$$

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 = $\frac{M{v^2}_{max}}{r}$ …….. (iv)

From equation(iii),we have

Mg = R cosθ (1−μ tanθ)…(v)

Again from equation (iv), we have

$\frac{M{v^2}_{max}}{r}$ = R cosθ (μ + tanθ) …(vi)

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

$$\begin{gathered} \frac{{v^2}_{max}}{gr} =\frac{\mu +\tan \theta }{1-\tan \theta } \\ {v}_{max} ={\left[\frac{gr\left(\mu +\tan \theta \right)}{1-\mu \tan \theta }\right]}^{1/2} \end{gathered}$$

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

Significance of the terms involved:
a. The maximum safety speed of a vehicle on a curved road depends upon friction between tyres and roads.
b​​​​​​. It also depends on the angle through which road is banked.
Also absence of term ‘m’ indicates, it is independent of mass of the vehicle.

Regards