What is meant by Banking of Roads?
When a vehicle tends to make a turn along a curved road, there is a probability of it skidding. For making a safe turn, the vehicle requires a centripetal force. The banking of a road is done to provide that centripetal force. During a “banked” or inclined turn, the chances of skidding reduce.
A turn is made inclined with the horizontal such that the outer edge is lifted up. For a particular angle of inclination, the maximum allowed speed of a vehicle is restricted. This maximum speed is independent of the mass of the vehicle. It depends on the banking angle, the coefficient of friction, and the radius of curvature.
Illustration of a Banked Turn
Along a turn, the outer edge of a road is lifted up such that it is higher than the inner edge and the surface of the road looks like a slightly inclined plane. This is called banking of a road. The angle made by the surface with the horizontal, i.e. the angle of inclination, is referred to as the banking angle. While moving through such a curved road, the normal force acting on the vehicle has a horizontal component. This component provides the centripetal force to avoid skidding.
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Banking circular motion
Centripetal Force
A rotating body feels an attraction towards the centre of rotation along the radius of the circular path. This force is called the centripetal force. If a body of mass m is moving with speed v along a circle of radius r, the magnitude of centripetal force is,
\[Fc = \frac{mv^{2}}{r}\]
Frictionless Banking of Road Derivation
In the absence of friction, the vertical component of the road’s normal force on a vehicle balances its weight and the horizontal component gives the centripetal force towards the center of curvature of the road. If a body of mass m is moving with velocity v along a curved road with a banking angle, the normal force N on it can be decomposed into two perpendicular components.
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Frictionless Banking
The vertical component Ny of N balances the weight mg (g is the gravitational acceleration).
\[N_{y} = mg\]
\[N cos \theta = mg\] (1)
The horizontal component \[N_{x}\] of N provides the centripetal force. If the radius of curvature is r,
\[N_{x} = F_{c}\]
\[N sin \theta = \frac{mv^{2}}{r}\] (2)
Dividing equation (2) by (1),
\[v = \sqrt{gr tan\theta}\]
Banking of Road Formula with Friction
The friction force acts along the inclined plane towards its inner edge. In the scenario mentioned above, the horizontal component of the friction force acts along the center while its vertical component acts downwards. If the coefficient of friction is, the maximum friction force is related to the normal force as,
\[f = \mu N\]
Since there is no vertical acceleration, the vertical components of the forces cancel each other i.e.
\[N_{y} = mg + f sin \theta\]
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Banking of road with friction
\[N cos \theta = mg + \mu N sin \theta\]
\[N(cos \theta - \mu sin \theta) = mg\] (3)
The total horizontal force provides the centripetal force,
\[N_{x} + f cos \theta = F_{c}\]
\[N sin \theta + \mu N cos \theta = \frac{mv^{2}}{r}\]
\[N (sin \theta + \mu cos \theta) = \frac{mv^{2}}{r}\] (4)
Dividing equation (4) by (3),
\[\frac{sin \theta + \mu cos \theta}{cos \theta + \mu sin \theta} = \frac{v^{2}}{gr}\]
\[v = \sqrt{gr(\frac{tan \theta + \mu}{1 - \mu tan \theta})}\]
This is the expression of maximum velocity to remain in the curved road. If the direction of the friction force is towards the outer edge of the road, a similar analysis gives the minimum required velocity,
\[v = \sqrt{gr(\frac{tan \theta - \mu}{1 + \mu tan \theta})}\]
Zero Banking Angle
If the banking angle is zero, a vehicle has to make a turn on a flat surface. The normal force can no longer contribute to the centripetal force since it is vertical and balances the weight of the vehicle. If the surface is perfectly smooth, there is no way to make a turn. Only on a rough surface, the friction force can provide the centripetal force. The vertical components of the forces balance each other,
N = mg
Friction force \[f = \mu N = \mu mg\] contributes to the centripetal force i.e,
\[f = F_{c}\]
\[\mu mg = \frac{mv^{2}}{r}\]
\[v = \sqrt{\mu gr}\]
This is the expression of the maximum possible velocity on a curved flat road.
Importance of Banking
Banking is a way of providing the required centripetal force to a vehicle to make a safe turn along a curved road.
Banking helps to avoid skidding.
Banking of roads helps to prevent overturning or toppling.
Solved Examples
1. A vehicle can have a maximum velocity of 72 km/hr on a smooth turn of radius 100 m. What is the banking angle?
Solution: Maximum velocity,
\[V_{m}\] = 72 km/hr = 20 m/s
The radius of banking r = 100 m
Gravitational acceleration g = 10 m/s^{2}
From the expression of maximum velocity.
\[V_{m} = \sqrt{gr tan \theta}\]
\[tan \theta = \frac{v^{2}m}{gr}\]
\[ = \frac{20^{2}}{10 \times 1000}\]
= 0.04
\[\theta = 2.29^{0}\]
The banking angle is \[2.29^{0}\]
2. Two turns of a smooth road are banked with the same angle. What is the ratio of maximum velocities for the two turns if the ratio of radii of curvature is 1:5?
Solution: The expression of maximum velocity v,
\[v = \sqrt{gr tan \theta}\]
Here, g is the gravitational acceleration, ris the radius of curvature, and the banking angle. Since the two turns have the same banking angle, the ratio of maximum velocities \[v_{1}\] and \[v_{2}\] is,
\[\frac{v_{1}}{v_{2}} = \sqrt{\frac{r_{1}}{r_{2}}}\]
\[r_{1}\] and \[r_{2}\] are the radii of curvature with their ratio being
\[\frac{r_{1}}{r_{2}} = \frac{1}{5}\]
Hence the ratio of velocities is,
\[\frac{v_{1}}{v_{2}} = \frac{1}{\sqrt{5}}\]
Did You Know?
The maximum velocity of a vehicle (marginal value before skidding) is proportional to the banking angle. So, to turn at a high speed, the banking angle should also be large. This is why race bikers get inclined with much larger angles than regular bikers. The race tracks are banked with larger angles to allow greater speeds.
The maximum velocity for a particular banking angle does not depend on the mass of an object moving on the curved path.
It is not possible to turn on a perfectly smooth flat road.
Banking is also done in railway tracks. For aircraft to make turns, the wings get inclined about their horizontal position.
FAQs on Banking of Roads
1. What is the banking of roads in Physics?
In Physics, the banking of roads is a technique where the outer edge of a curved road is raised higher than its inner edge. This creates an inclined surface, forming an angle with the horizontal. This design is crucial for vehicles navigating turns, as it helps provide the necessary force to move in a circular path safely.
2. Why are curved roads banked?
Curved roads are banked for several important reasons related to vehicle safety and dynamics:
- To provide centripetal force: The primary purpose is to use a component of the vehicle's normal reaction force to provide the required centripetal force for the turn.
- To reduce reliance on friction: On a flat curve, a vehicle relies entirely on the friction between its tyres and the road to turn. Banking reduces this dependency, which is especially important in wet or icy conditions where friction is low.
- To prevent skidding: By providing the necessary inward force, banking significantly lowers the risk of a vehicle skidding outwards off the road.
- To increase safe speed: It allows vehicles to navigate the curve at a higher speed safely than they could on a flat road of the same radius.
3. What is the formula for the ideal angle of banking on a road?
The formula to calculate the ideal angle of banking (θ) for a specific speed (v) and curve radius (r) is given by: tan θ = v² / rg. In this formula:
- θ is the angle of banking.
- v is the ideal speed of the vehicle.
- r is the radius of the curved road.
- g is the acceleration due to gravity (approximately 9.8 m/s²).
This formula represents the ideal condition where no friction is needed to make the turn.
4. How does banking a road help in providing the necessary centripetal force?
When a road is banked, the normal force (N) exerted by the road on the vehicle is no longer vertical; it is perpendicular to the inclined surface. This normal force can be resolved into two components: a vertical component (N cos θ) that balances the vehicle's weight, and a horizontal component (N sin θ) that points towards the centre of the curve. This horizontal component provides the necessary centripetal force (mv²/r) required for the vehicle to turn, making the turn safer and more stable.
5. What is the difference between a car turning on a level road versus a banked road?
The key difference lies in the source of the centripetal force. On a level road, the centripetal force required to make a turn is provided entirely by the force of friction between the tyres and the road surface. If this friction is insufficient, the car will skid. On a banked road, the centripetal force is primarily provided by the horizontal component of the normal reaction force. Friction still plays a role, but the vehicle is no longer solely dependent on it, allowing for a safer turn at higher speeds.
6. What factors determine the ideal speed for a vehicle on a banked curve?
The ideal speed for a vehicle on a banked curve, often called the 'design speed', is the speed at which the vehicle can navigate the turn without any assistance from friction. This speed is determined by two main factors:
- The angle of banking (θ): A steeper angle allows for a higher ideal speed.
- The radius of the curve (r): A larger radius (a gentler curve) also allows for a higher ideal speed for the same angle of banking.
7. What happens if a car travels much faster or slower than the ideal speed on a banked road?
If a car's speed deviates from the ideal speed, the force of friction is required to prevent it from sliding.
- Faster than ideal speed: The car has a tendency to skid upwards along the bank. The force of static friction acts down the slope, towards the inner edge of the road, to provide the additional centripetal force needed and prevent skidding.
- Slower than ideal speed: The car has a tendency to slide downwards along the bank. The force of static friction acts up the slope, towards the outer edge, to counteract the excess inward pull and prevent the car from slipping down.
8. How is the formula for the maximum safe speed on a banked road derived, considering friction?
To derive the maximum safe speed (v_max), we consider the point where the car is about to skid upwards. At this point, both the horizontal component of the normal force (N sin θ) and the horizontal component of the maximum static friction (f_s cos θ) act towards the centre. These forces together provide the centripetal force. By setting up and solving the force balance equations for the vertical and horizontal directions, we arrive at the formula: v_max = √[rg (tanθ + μs) / (1 - μs tanθ)], where μs is the coefficient of static friction.
9. Can you give a real-world example where the banking of roads is critical?
A classic real-world example is a highway exit ramp or a high-speed interchange. These ramps are often sharply curved and must be designed to handle vehicles moving at relatively high speeds. Banking the ramp is critical to ensure that cars and trucks can safely navigate the turn without skidding or overturning. Another prominent example is in motor racing on tracks called velodromes or speedways, where extremely steep banking allows vehicles to maintain very high speeds through the corners.
10. What is meant by a 'zero banking angle'?
A 'zero banking angle' refers to a completely flat or level road (θ = 0°). On such a surface, the normal force on a vehicle is purely vertical and has no horizontal component. Therefore, it cannot contribute to the centripetal force required for turning. The vehicle must rely entirely on the force of friction between its tyres and the road to navigate any curve. This is the standard condition for most straight roads and slow-speed city turns.
