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Question

What is banking of roads? Obtain an expression for the maximum safety speed of a vehicle moving along a curved horizontal road.


Solution

Banking of roads : To avoid risk of skidding as well as to reduce the wear and tear of the car tyres, the road surface at a bend is tilted inwards, i.e., the outer side of road is raised above its inner side. This is called 'banking of roads'.
Consider a car taking a left turn along a road of radius $$r$$ banked at an angle $$\theta$$ for a designed optimum speed $$V$$. Let $$m$$ be the mass of the car. In general, the forces acting on the car are:
(a) Its weight $$\vec {mg}$$, acting vertically down
(b) The normal reaction of the road $$\vec {N}$$, perpendicular to the road surface
(c) The frictional force $$\vec {f_{s}}$$ along the inclined surface of the road.
Resolve $$\vec {N}$$ and $$\vec {f_{s}}$$ into two perpendicular components $$N\cos \theta$$ vertically up and $$\vec {f_{s}} \sin \theta$$ vertically down, $$N\sin \theta$$ and $$\vec {f_{s}} \cos \theta$$ horizontally towards the centre of the circular path.
If $$v_{max}$$ is the maximum safe speed without skidding.

$$\dfrac {mv_{max}^{2}}{r} = N\sin \theta + f_{s}\cos \theta$$

               $$= N\sin \theta + \mu_{s} N\cos \theta$$

$$\dfrac {mv_{max}^{2}}{r}= N(\sin \theta + \mu_{s} \cos \theta) .... (1)$$

and 
$$N\cos \theta = mg + f_{s} \sin \theta$$
               $$= mg + \mu_{s} N\sin \theta$$
   $$\therefore mg = N(\cos \theta - \mu_{s}\sin \theta) ... (2)$$

Dividing eq. $$(1)$$ by eq. $$(2)$$,
$$\dfrac {mv_{max}^{2}}{r.mg} = \dfrac {N(\sin \theta + \mu_{s}\cos \theta)}{N(\cos \theta - \mu_{s}\sin \theta)}$$

$$\therefore \dfrac {v_{max}^{2}}{rg} = \dfrac {\sin \theta + \mu_{s}\cos \theta}{\cos \theta - \mu_{s}\sin \theta} = \dfrac {\tan \theta + \mu_{s}}{1 - \mu_{s}\tan \theta}$$

$$\therefore v_{max} = \sqrt {\dfrac {rg(\tan \theta + \mu_{s})}{1 - \mu_{s}\tan \theta}} ...,. (3)$$

This is the expression for the maximum safe speed on a banked road.
At the optimum speed, the friction between the car tyres and the road surface is not called into play. Hence, by setting $$\mu_{s} = 0$$ in eq. $$(3)$$, the optimum speed on a banked circular road is
$$v = \sqrt {rg \tan \theta} ... (4)$$
$$\therefore \tan \theta = \dfrac {v^{2}}{rg}$$ or $$\theta = \tan^{-1}\left (\dfrac {v^{2}}{rg}\right )$$
From this eq. we see that $$\theta$$ depends upon $$v, r$$ and $$g$$. The angle of banking is independent of the mass of a vehicle negotiating the curve.

629002_601456_ans_e364791c789d46428ec8f90b1b967836.png

Physics

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