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Why is a road banked on a circular turn? Obtain the maximum velocity of a vehicle on such a turn. If we assume that the friction on road is negligible, then obtain the banking angle.

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Banking of roads : The value of maximum velocity for a vehicle to take a circular turn (without skidding) on a level road is quite low. This limiting value of the velocity decreases further due to decrease in the value of the coefficient of friction p on a slippery road and for a vehicle, whose tyres have worn out. Therefore especially in hilly areas, where the vehicle has to move constantly along the curved tracks, the maximum speed at which it can be run, will be very low. If any attempt is made to run it at greater speed, the vehicle is likely to skid and go off the track. In order that the vehicle can go round the curved tracks at reasonable speed without skidding, the sufficient centripetal force is managed for it by raising the outer edge of the track a little above the inner one. It is called the banking of the circular tracks.

Consider a vehicle of weight ‘mg’ moving round a curved path of radius r with speed v on a road banked through angle θ. If OA is banked road and OX is a horizontal line, then ∠AOX = θ is called angle of banking,

The vehicle is under the action of following forces :

(i) Weight ‘mg’ of the vehicle acting vertically downwards.

(ii) Normal reaction W of the ground to the vehicle acting along normal to the banked road OA in the upward direction.

(iii) Force of friction F’ between the banked road and the tyres, acting along OA 

‘N’ can be resolved into two rectangular components :

(a) N cosθ, along vertically upward direction.

(b) N sinθ, along the horizontal, towards the centra of the curved round.

F’ can also be resolved into two rectangular components :

(a) Fsinθ, along vertically downward direction

(b) Fcosθ, along the horizontal, towards the centre of curved road.
As there is no acceleration along the vertical direction, the net force along this direction must be zero.
Therefore, N cosθ = mg + Fsinθ ………… (1)
The horizontal component N sinθ and Fcosθ will provide the necessary centripetal force to the vehicle,
Thus N sinθ + F cosθ = \(\frac{m v^{2}}{r}\) …………….. (2)
But F ≤ μsN, where μs is coefficient of static friction between the banked road and tyres. To obtain vmax, we Put F = μsN in equation (1) and (2)
∴ Ncosθ = mg + μsNsinθ …………. (3)
N sinθ + μsN cosθ = \(\frac{m v^{2}}{r}\) ……………….. (4)
From equation (3)
N (cosθ – μs sinθ) = mg
⇒ N = \(\frac{m g}{\left(\cos \theta-\mu_{s} \sin \theta\right)}\) …………… (5)
From equation (4) N (sinθ – μs cosθ) = \(\frac{m v^{2}}{r}\)

Equation (6) represents the maximum velocity of vehicle on a banked road.

Discussion :
(i) If μs = 0 i. e., if banked road is perfectly smooth, then from equation (6)
vmax = (rgtanθ)1/2 ………… (7)
This is the speed at which a banked road can be rounded even when there is no friction. Driving at this speed on banked road will cause no wear and tear of the tyres.
From equation (7) \(v_{\max }^{2}\) = rgtanθ
or tan = \(\frac{v_{\max }^{2}}{r g}\) …………….. (8)
(ii) If h is the height AB of outer edge of the road above the inner edge and l = OA is breadth of the road then from the figure.

Roads are usually banked for the average speed of vehicles passing over them. However, if the speed of a vehicle is some what less or more than this, the self adjusting static friction will operate between the tyres and the road, and the vehicle will not skid.

On the same basis, curved railway tracks are also banked. The level of outer rail is raised a little above the level of inner rail, while lying a curved railway track.

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