Banking of Roads - Meaning, Formula, FAQs

When we are in a car or even on a bicycle, our speed tends to decrease just as we near a corner, this happens because as we turn, the likelihood of our tyres slipping increases with vehicle speed hence the need to slowly down; hence the adaptation of brake pads that adjusts speed. It is noticeable that, anytime it rains, you are forced to decelerate while going around a corner than under normal circumstances; this is due to a decrease in frictional force assisting in making turns.

In this article, we will cover the concept of banking on the road. This concept falls under the chapter Law of Motion, which is a crucial chapter in Class 11 physics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), National Eligibility Entrance Test (NEET), and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE and more. Over the last ten years of the JEE Main exam (from 2013 to 2023), a total of two questions have been asked on this concept. And for NEET one questions were asked from this concept.

What is Meant by Banking of Roads Meaning?

Banking of roads is the phenomenon in which the outer borders of why are curved roads banked are raised above the inner edge to provide the necessary centripetal force to the cars in order for them to make a safe turn.

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Centripetal Force

It is the force that causes angular or circular motion by pulling or pushing an object toward the centre of a circle as it travels.

Another phrase used is banked turn, which is described as a turn or direction change in which the vehicle inclines towards the inside. The bank angle is the angle at which the vehicle is inclining. At the longitudinal and horizontal axes, there is an inclination.

Illustration of Banked Road

The outer border of a road is pushed up such that it is higher than the inner edge during a turn, and the road's surface seems to be a slightly inclined plane. This is referred to as road banking. The banking angle is the angle formed by the surface with the horizontal, i.e. the angle of inclination. When driving down such a curvy road, it's important to keep your eyes on the road.

There is a horizontal component to the normal force pressing on the vehicle. This component produces the centripetal force required to keep the vehicle from skidding.

Frictionless Banking of Road Formula

The vertical component of the road's normal force on a vehicle balances its weight in the absence of friction, while the horizontal component provides the centripetal force towards the road's centre of curvature. The applied load N on a body of mass m travelling with velocity v along a curved road with a banking angle formula can be split into two perpendicular components.

The weight mg is balanced by the vertical component Nvertical of N. (g is the gravitational acceleration).

Nvertical = mg

Ncosθ = mg (1)

The centripetal force is provided by the horizontal component Nhorizontal of N. If r is the radius of curvature,

Horizontal =Fc

Nsinθ=mv²/r.........(2)

Equation (2) is divided by (1),

$v=\sqrt{g * r *(\tan \theta)}$

This is the greatest velocity at which an object can remain in a curved path.

Also read :

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Without Friction

From figure,

$\begin{aligned}
&R \cos \theta=m g .... (i)\\
&\begin{aligned}
& R \sin \theta=\frac{m v^2}{r} ... (ii)\\
& \tan \theta=\frac{v^2}{r g} \\
& \tan \theta=\frac{\omega^2 r}{g}=\frac{V \omega}{g}=\frac{h}{l}
\end{aligned}
\end{aligned}$

$\mathrm{h}=$ height of outer edge from the ground level
$l=$ width of the road
$\mathrm{r}=$ radius

If Friction is Also Present

$
\frac{V^2}{r g}=\frac{\mu+\tan \theta}{1-\mu \tan \theta}
$

Where $\theta=$ angle of banking
$\mu=$ coefficient of friction
$V=$ velocity

• Maximum speed on a banked frictional road

$V=\sqrt{\frac{r g(\mu+\tan \theta)}{1-\mu \tan \theta}}$

Zero Banking Angle

A vehicle must turn on a flat surface if the banking angle is zero. Because the normal force is vertical and balances the vehicle's weight, it can no longer contribute to the centripetal force. There is no way to turn if the surface is absolutely smooth. The friction force can only provide the centripetal force on a rough surface. The forces' vertical components counterbalance each other.

Necessity of Banking

1. Banking is a technique for giving a vehicle the necessary centripetal force to make a safe turn on a curved route.

2. Skidding can be avoided by banking.

3. Overturning or toppling is less likely when roads are banked.

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Solved Examples Based on Banking of Roads

Example 1: A circular road of radius 30m has banking at an angle of 45⁰. The maximum safe speed of the car (in m/s ) having a mass of 1000kg will be if the coefficient of friction between road and tyre is 0.5 [g = 10m/s²]

1) 30

2) 20

3) 35

4) 45

Solution :

Maximum speed on a banked frictional road

$
V=\sqrt{\frac{r g(\mu+\tan \theta)}{1-\mu \tan \theta}}
$
maximum speed with the banked road with the friction is
$
\begin{aligned}
& v^2=r g\left[\frac{\mu+\tan \theta}{1-\mu \tan \theta}\right]=900 \mathrm{~m} / \mathrm{s} \\
\Rightarrow & v=30 \mathrm{~m} / \mathrm{s}
\end{aligned}
$

Hence, the answer is option (1).

Example 2: Statement I: A cyclist is moving on an unbanked road with a speed of 7 kmh^-1 and takes a sharp circular turn along the path of a radius of 2m without reducing the speed. The static friction coefficient is 0.2. The cyclist will not slip and pass the curve ( g= 9.8 m/s²)

Statement II: If the road is banked at an angle of $45^{\circ}$, the cyclist will not slip and pass the curve of a $2 \mathrm{~m}$ radius with a speed of $18.5 \mathrm{~km}-1$ without slipping.

1) Statement I is correct and Statement II is incorrect.

2) Both statement I and statement II are false.

3) Both statement I and statement II are true.

4) Statement I is incorrect and statement II is correct.

Solution :

$\begin{aligned}
V_{\max } & =\sqrt{\mu \mathrm{rg}}=\sqrt{(0.2) \times 2 \times 9.8} \\
\mathrm{~V}_{\max } & =1.97 \mathrm{~m} / \mathrm{s}=7.12 \mathrm{~km} / \mathrm{h}
\end{aligned}$

A cyclist is moving on an unbanked road with a speed of 7 kmh^-1

$\text { Speed is lower than } V_{\max } \text {, hence it can take a safe turn. }$

Since the cyclist is riding at a safer speed of 7km/h, Statement I is correct.

For statement II

$\begin{aligned}
& \mathrm{V}_{\max }=\sqrt{\mathrm{rg}\left[\frac{\tan \theta+\mu}{1-\mu \tan \theta}\right]}=\sqrt{2 \times 9.8\left[\frac{1+0.2}{1-0.2}\right]} \\
& \mathrm{V}_{\max }=5.42 \mathrm{~m} / \mathrm{s}=19.5 \mathrm{~km} / \mathrm{h} \\
& V_{\min }=\sqrt{\frac{r g(\mu-\tan \theta)}{1+\mu \tan \theta}} \\
& \mathrm{V}_{\min }=3.615 \mathrm{~m} / \mathrm{s}=13.01 \mathrm{~km} / \mathrm{h}
\end{aligned}$

The speed of the cyclist is $18.5 \mathrm{~km} / \mathrm{h}=5.14 \mathrm{~m} / \mathrm{s}$

The speed of cyclist is between the safe limit of $V_{\min }$ and $V_{\max }$, Statement II is also correct.

So, both statement I and statement II are true.

Hence, the answer is the option(3).

Example 3: A turn of radius 20 m is banked for the vehicle going to a speed of 5 m/s If the width of a road is 8 m then what should be the height (in m) of the outer edge w.r.t inner edge of the road

1) 1

2) 0.5

3) 0.75

4) 0.25

Solution :

From figure,

$
\begin{aligned}
& N \sin \theta=\frac{m v^2}{R} \\
& N \cos \theta=m g \\
& \tan \theta=\frac{v^2}{R g}
\end{aligned}
$

For small angle $\theta \quad \tan \theta \approx \sin \theta \approx \frac{h}{l}$
$
\begin{aligned}
& \frac{h}{l}=\frac{v^2}{R g} \Rightarrow=\frac{5 \times 5}{20 \times 10} \\
& \frac{h}{l}=\frac{1}{8} \\
& \Rightarrow h=l / 8 \Rightarrow 8 / 8=1 \mathrm{~m}
\end{aligned}
$

Hence, the answer is option (1).

Example 4: The normal reaction ' $\mathrm{N}$ ' for a vehicle of $800 \mathrm{~kg}$ mass, negotiating a turn on a $30^{\circ}$ banked road at maximum possible speed without skidding is ____ $\qquad$ $\times 10^3 \mathrm{~kg} \mathrm{~m} / \mathrm{s}^2$

1) 10.2

2) 7.2

3) 12.4

4) 6.96

Solution:

For maximum possible speed, $\mathrm{f}_{\mathrm{s}}$ will be equal to $\mathrm{f}_{\mathrm{L} \text { i.e. }} \mathrm{f}_{\mathrm{L}}=\mu \mathrm{N}$
Along X-axis: $N \cos \theta=m g+f_L \sin \theta \rightarrow$ (1)
Along Y-axis: $f_L \cos \theta+N \sin \theta=\frac{m v_{\max }^2}{R} \rightarrow 2$
From Eqn (1)
$
N \cos \theta=m g+\mu N \sin \theta
$

$\begin{aligned}
& \mu=0.2 \\
& N \cos \theta=\mathrm{mg}+(0.2) N \sin \theta \\
& \theta=30^{\circ} \\
& N\left[\cos 30^{\circ}-0.2 \sin 30^{\circ}\right]=\mathrm{mg} \\
& N=\frac{800 \times 10}{\left[\frac{\sqrt{3}}{2}-0.2 \times \frac{1}{2}\right]}=\frac{800 \times 10}{(0.866-0.1)} \\
& \quad=\frac{8000}{0.766} \simeq 10.2 \times 10^3
\end{aligned}$

Hence, the answer is option (1).

Example 5: A car of mass 2000kg going round a banked curved a radius 40m on a frictionless road. If the banking angle is 45⁰, the speed of the car is :

1) 10m/s

2) 15m/s

3) 20m/s

4) 25m/s

Solution:

Banking of Road -

From figure

$\begin{aligned}
& R \cos \theta=m g ....(i)\\
& R \sin \theta=\frac{m v^2}{r}.....(ii) \\
& \tan \theta=\frac{v^2}{r g} \\
& \tan \theta=\frac{\omega^2 r}{g}=\frac{V \omega}{g}=\frac{h}{l}
\end{aligned}$

$\begin{aligned}
& \mathrm{h}=\text { highest of the outer edge } \\
& l \text { =length of rod } \\
& \mathrm{r}=\text { radius } \\
& \tan \theta=\frac{v^2}{r g} \\
& v=\sqrt{r g \tan \theta}=\sqrt{40 \times 10 \times \tan 45^{\circ}}=20 \mathrm{~m} / \mathrm{s}
\end{aligned}$

Hence, the answer is option (3).

Summary

Banking of roads refers to the sideways tilting of the road surface while negotiating curves. This facilitates safer vehicle turning. It reduces the possibility of skidding or sliding off the road. Banking counteracts centrifugal force, thus keeping them on course. The necessity of banking is more vital on highways and especially at racing tracks. Banking allows for greater grip as well as more stable turns.

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