When a vehicle tends to make a turn along a curved road, there is a probability of it to skid. For making a safe turn, the vehicle requires a centripetal force. 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.

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

A rotating body feels an attraction towards the center of rotation along the radius of the circular path. This force is called the centripetal force. Is a body of mass m is moving with speed v along a circle of radius r, the magnitude of centripetal force is,

\[F_{c}\] = \[\frac{mv^{2}}{r}\]

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 in two perpendicular components.

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The vertical component \[N_{y}\] 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),

\[\nu\] = \[\sqrt{grtan\theta}\]

This is the expression of the maximum velocity of an object to remain in the curved path.

The friction force f 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),

\[\frac{sin\theta + \mu cos \theta}{cos\theta + \mu sin\theta}\] = \[\frac{\nu^{2}}{gr}\]

\[\nu\] = \[\sqrt{gr(\frac{tan\theta + \mu}{1 - \mu tan \theta}})\]

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,

\[\nu\] = \[\sqrt{gr(\frac{tan\theta - \mu}{1 + \mu tan \theta}})\]

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 = \[\frac{mv^{2}}{r}\]

\[\nu\] = \[\sqrt{\mu gr}\]

This is the expression of maximum possible velocity on a curved flat road.

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.

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,

\[\nu_{m}\] = 72 km/hr = 20 m/s

The radius of banking r=100 m

Gravitational acceleration g=10m/s2

From the expression of maximum velocity,

\[\nu_{m}\] = \[\sqrt{grtan\theta}\]

tan \[\theta\] = \[\frac{\nu_{m}^{2}}{gr}\]

= \[\frac{20^{2}}{10 \times 1000}\]

= 0.04

\[\theta\] = 2.290

The banking angle is 2.290 .

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,

\[\nu\] = \[\sqrt{grtan\theta}\]

Here, g is the gravitational acceleration, ris the radius of curvature, and is the banking angle. Since the two turns have the same banking angle, the ratio of maximum velocities v_{1} and v_{2} is,

\[\frac{\nu_{1}}{\nu_{2}}\] = \[\sqrt{\frac{r_{1}}{r_{2}}}\]

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,

\[\frac{\nu_{1}}{\nu_{2}}\] = \[\frac{1}{\sqrt{5}}\]

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.

FAQ (Frequently Asked Questions)

1. What is Banking is Physics?

The outer edge of a curved road is kept higher than the inner edge (towards the center of curvature) such that the surface of the road makes an angle with the horizontal. This is called banking. The normal reaction of the inclined surface on a vehicle now has a horizontal component, which combined with friction force, serves as the centripetal force for the vehicle’s rotation. Such banked turns of vehicles can prevent skidding and overturning, ensuring a safe turn along the curved road.

2. What is the Angle of Banking? What Does the Motion of a Car on a Banked Road Depend on?

The angle, made by the surface of the road with the horizontal, is called the angle of banking. The maximum velocity, which an object can take during a banked turn, depends on the banking angle.

The maximum and minimum velocity of the car depends on the banking angle, the radius of curvature, and the coefficient of friction and not on its mass.