Banking of Roads And Circular Motion.

PPT on circular motion and banking of roads. Detailed Information on banking of roads with friction and derivation using diagram.

1. BANKING OF ROADS. Made by – Hrishank Chhatbar (103038) Ayush Bulchandani (103018) Sahil Chawla (103036) Dhruv Chechani (103037) Ishan Daga (103047)

2. Circular motion When a car is moving along a curved road, it is performing circular motion. For circular motion it is not necessary that the car should complete a full circle; an arc of a circle is also treated as a circular path . We know that centripetal force (CPF) is necessary for circular motion. If CPF is not present, the car cannot travel along a circular path and will instead travel along a tangential path . The centripetal force for circular motion of the car can be provided in two ways: Frictional force between the tyres of the car and the road. Banking of Roads

3. What is banking of road? The process of raising the outer edge of a road over the inner edge through a certain angle is known as banking of road . Purpose of Banking of Roads : To provide the necessary centripetal force for circular motion. To reduce wear and tear of tyres due to friction. To avoid skidding. To avoid overturning of vehicles. When a vehicle goes round a curved road, it requires some centripetal force. While rounding the curve, the wheels of the vehicle have a tendency to leave the curved path and regain the straight line path. Force of friction between wheels and the roads opposes this tendency of the wheels. This force of friction therefore, acts towards the center of circular track and provides the necessary centripetal force .

4. DEFINITION AND DERIVATION The angle made by the surface of the road with the horizontal surface is called as angle of banking . Consider a car moving along a banked road. Let: m = mass of the car V = speed of the car Θ is angle of banking. The forces acting on the car are: Its weight mg acting vertically downwards. The normal reaction R acting perpendicular to the surface of the road. The normal reaction can be resolved (broken up) into two components: R cosθ is the vertical component. R sinθ is the horizontal component.

5. Since the vehicle has no vertical motion, the weight is balanced by the vertical component. R cosθ = mg (1 ) ( weight is balanced by vertical component means weight is equal to vertical component) The horizontal component is the unbalanced component . This horizontal component acts towards the center of the circular path . This component provides the centripetal force for circular motion. R sin θ = (2) By dividing both (1) and (2). = tan θ = θ = tan -1  

6. Therefore, the angle of banking is independent of the mass of the vehicle. The maximum speed with which the vehicle can safely travel along the curved road is: V = Question: A curve has a radius of 50 meters and a banking angle of 15 o . What is the ideal , or critical, speed (the speed for which no friction is required between the car's tires and the surface) for a car on this curve ? Solution: Here , radius of curve, r = 50 m, banking angle, θ = 15 o , a=g= 9.8 m/s 2 To find: The ideal speed v (the speed for which no friction is required between the car's tires and the surface ).   F Net  = F Centripetal mg tanθ =   Therefore, V = Substituting the values we get   V = ( 50 m) (9.8 ) (tan 15 o ) V = 11 m/s If the car has a speed of about 11 m/s, it can negotiate the curve without any friction.  

7. Max speed of a vehicle on a banked road with friction Consider a vehicle moving along a curved banked road. Let m = mass of vehicle r = radius of curvature of road θ = angle of banking F = frictional force between tyres and road. The forces acting on the vehicle are shown in the diagram . The forces acting on the vehicle are: Weight of the vehicle mg , acting vertically downwards. Normal reaction N acting on vehicle, perpendicular to the surface of the road. Friction force between tyres and road. Forces N and frictional force f are now resolved into two components.

8. The normal reaction N is resolved into 2 components: N cosθ is vertical component of N. N sinθ is horizontal component of N. The frictional force f is resolved into 2 components: f cosθ is horizontal component of f. f sinθ is vertical component of f. The vertical component N cos θ is balanced by the weight of the vehicle and the component f sin θ. ∴ N cos θ = mg + f sin θ ∴ mg = N cos θ - f sin θ (1) Resolving vectors.

9. The horizontal component N sinθ and f cosθ provide the centripetal force for circular motion. ∴ N sin θ + f cos θ = ∴ = N sin θ + f cos θ (2) Dividing (2) by (1), we get = = Let V max be the maximum speed of the vehicle. Frictional force at this speed will be : f m = μ s N  

10. = But f m = μ s N ∴ = Dividing numerator and denominator of RHS by cos θ, we get: = ∴ V max = This is the maximum velocity with which a vehicle can travel on a banked road with friction. For a frictionless road, μ s = ∴ V max =  

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