Why are circular roads banked? Draw a diagram showing forces acting on a car moving
on a circular banked road. Obtain an expression for the maximum speed vmax with which
a vehicle can safely negotiate a curved road banked at an angle ?, taking into account
the friction between the tyres of the vehicle and the road. The coefficient of static friction
between the tyres and the road is ?s. And hence find the maximum driving speed vo of the
vehicle on the banked road which will cause little wear and tear of the tyres. Also show
that vehicle can be parked on the banked road only if tan ? is less or equal to ?

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 = …(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

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