Banking of Road Formula

Kinematics is a branch of physics that lets us understand our actions. Every action can be mathematically proved with the help of kinematics and laws of motion. The banking of road formula is also one of the interesting phenomena that we experience in our day to day lives. Whenever we are travelling through a curved road in a vehicle, we observe that on the side of a curved path there will be a kind of slope that has been constructed, in other words, at the curved or circular path a part will be constructed above the ground and another part will be slightly flattened.

This kind of slope is constructed at the curved path in order to maintain the speed of the vehicle, and this construction is known as the banking of the road and the formula used to calculate the accurate banking angle is known as the banking of road formula.

Banking Angle Formula

The main aim of banking is to provide a necessary centripetal force to the vehicle travelling on a curved surface. The banking of the road will also reduce maximum frictional wear and frictional tyres, which further will avoid skidding and overturning of the vehicles. 

The roads are usually constructed with banking for the average speed of vehicles passing over them, the average speed of the vehicles on a particular road will be predetermined. For example, we see the speed boards that say the speed limit is 20km/h, which implies that the road and the banking are constructed for the average speed of 20km/h. Anyway, if the average speed of a vehicle is lesser or more than this, the self-adjusting state friction will operate between tyre and road and the vehicle will not skid, which will prevent the vehicle from accidents.

Banking Formula

Let us have a look at the banking formula followed by a small derivation. We know that whenever a vehicle takes a turn on a straight road, then its safe driving mainly depends on the coefficient of friction. And we know that the velocity of such motion is given by the formula:

⇒\[ v = \sqrt{\mu r g}\]........(1)

Where,

μ-  The coefficient of friction

r - The radius of the turn taken by the vehicle

ｇ-  The acceleration due to gravity

Now let us derive the banking angle formula, consider a vehicle (say a car) of mass m moving with velocity v on the banked road of radius r. Let be the angle of inclination with the banking with a frictional force between the surface of the road and the tyres. 

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From the free body diagram, we can determine the forces acting on the vehicle and hence the equation of motion is given by:

\[\Rightarrow N cos \theta = mg + f_{r}sin \theta\] ……..(2)

Where,

N - The normal force acting on the vehicle in an upward direction 

θ - The angle of banking

m - The mass of the vehicle

\[f_{r}\] - The frictional force

N cosθ -  The horizontal component of the normal force

\[f_{r}\cosθ\] - The horizontal component of the frictional force

Similarly, in order to balance the rest of the forces, a pseudo force will be applied in the direction that will help the vehicle to have turned on the road, and hence we get:

⇒ Fp = N sin θ + fr cos θ

⇒\[ \frac{mv^{2}}{r}\] = N sin θ + fr cos θ………(3)

Where,

N -The normal force acting on the vehicle in an upward direction 

θ -The angle of banking

m -The mass of the vehicle

\[f_{r}\] - The frictional force

N cosθ - The vertical component of the normal force

\[f_{r}\cosθ\] - The vertical component of the frictional force

Now our aim is to arrive at the banking of road formula, divide equation (3) by (2) we get:

⇒ \[\frac{mv^{2}}{r(mg)} = \frac{Nsin\theta+f_{r}cos\theta }{Ncos\theta -f_{r}sin\theta } = \frac{Nsin\theta+\mu Ncos\theta }{Ncos\theta -\mu Nsin\theta }\]

(we know that the frictional force is given by \[f_{r}=\mu N\]

\[\Rightarrow \frac{v^{2}}{rg}= \frac{N(sin\theta+\mu cos\theta )}{N(cos\theta -\mu sin\theta } = \frac{(sin\theta+\mu cos\theta )}{(cos\theta -\mu sin\theta }\]

\[\Rightarrow \frac{v^{2}}{rg}= \frac{(tan\theta + \mu )cos\theta }{cos\theta (1-\mu tan\theta )}\]

\[\Rightarrow v^{2}=\frac{rg(tan\theta+\mu  )}{(1-\mu tan\theta )}\]

\[\Rightarrow v = \sqrt{\frac{rg(tan\theta+\mu  )}{1-\mu tan\theta }}\]........ (4)

By using equation (4), we can determine the maximum speed of the vehicle. If we can not adjust the desired coefficient of friction value we go for the banking of roads. For banking of roads, we assume that the coefficient of friction is zero i.e., μ=0Then equation from equation (4) we get:

\[\Rightarrow v = \sqrt{rgtan\theta }\]

\[\Rightarrow v^{2} = rg tan\theta\]

\[\Rightarrow tanθ = \frac{v^{2}}{rg}\]

⇒ θ = \[\tan^{-1} \frac{v^{2}}{rg}\].......(5)

Equation (5) is known as the banking angle formula and it can be used to calculate the banking angle if the required values are known.

Example

1. Consider a curved road such that its curve has a radius of 30 meters and a banking angle of  15°. What is the ideal speed or critical speed or the speed for which no friction is required between the car's tires and the surface for a car on this curved path?

Sol:

Given,

The radius of the curved road = r = 30 m

The angle of banking of the road =\[ \theta\]  = 15°

The acceleration due to gravity = ց = 10 m/s2( for simplicity)

The value of the coefficient of friction is assumed to be zero for an ideal condition. Thus we write μ=0. We are asked to calculate the ideal speed or critical speed required for the car moving on the banked road. We know that the velocity or the speed of the vehicle for the frictionless road is given by:

⇒\[v_{max}=\sqrt{rg\, tan\theta }\].......(1)

Where,

r - The radius of the turn taken by the vehicle

θ - The angle of banking

ｇ - The acceleration due to gravity

Substituting the given values in equation (1) we write:

⇒\[v_{max} = \sqrt{(30)(10)tan15}\]

⇒\[v_{max} = \sqrt{(30)(10)tan15}\]= 8.9 m/s

Therefore, the ideal speed of the car moving on the curved road will be 9m/s.