Perhaps most of us have noticed that when a road is straight, it is horizontal as well. But, when a sharp turn comes, the surface of the road will not remain horizontal. This process is known as banking on the roads. We may have seen that overturned trucks lying on the road. The heavily loaded trucks, who could have overturned them. Overturning occurs on the roads when the trucks try to change their directions while taking sharp turns. This article will help a student to get the concept of banking of roads as well as banking road formula with examples. Let us learn it!

**Concept of Banking Roads**

This is a phenomenon in which the edges of the curved roads are raised above its inner edge for providing the necessary centripetal force to the vehicles for taking a safe turn. For banking of roads, many terms and computations are important. These are as follows:

- Banked Turn – It is the turn or change of direction with which the vehicle inclines towards inside.
- Bank Angle – It is the angle of inclination of the vehicle. Therefore, at this angle, the vehicle is inclined about its longitudinal axis with respect to the plane of its curved path.

Source:en.wikipedia.org

If the force of friction is not strong enough, then the vehicle may skid. Even if there is very little force of friction the vehicle can still go around the curve without the tendency of skid. This is the reason that roads are banked, also it is due to the inertia of vehicles driving on the road. Roads are most often banked for the average speed of vehicles passing over them. Nevertheless, if the 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.

**Some Formulae for Banking Roads**

- The velocity of a vehicle on a curved banked road: \(v=\sqrt{\frac{(rg(tan\Theta+\mu_s))}{1-\mu_S\,tan\theta}}\)

- For a given pair of roads, and tyre \(\mu_s =tan \lambda\), then the velocity of a vehicle on a curved banked road is: \(v=\sqrt{rg\,tan(\Theta+\lambda)}\)

- The safe velocity on an unbanked road is: \(v_{max}=\sqrt{\mu\times r\times g}\)

- The expression for the angle of banking of road is: \(\Theta=\tan^{-1}\frac{v^2}{rg}\)

- Expression for the safe velocity on the banked road is: \(v_{max}=\sqrt{rg\,\tan\Theta}\)

- The expression for the angle of banking: \(\Theta=\tan^{-1}\frac{v^2}{rg}\)

v | Velocity on a banked curve |

\(\theta\) | Turn angle |

\(v_{max}\) | Safe velocity |

g | Acceleration constant |

\(\lambda\) | Angle of friction |

r | Radius of curve |

\(\mu\) | Coefficient of friction |

**Solved Examples for Banking Road Formula**

Q.1: A curve has a radius of 50 meters with a banking angle of 15 ^{\circ}. What will be the ideal or critical velocity for a car on this curve using Banking Road Formula?

Solution: Here, a radius of the curve, r = 50 m

Banking angle, \(\theta = 15 ^{\circ}\)

Free-fall acceleration, \(g = 9.8 m s^{-2}\)

We have to compute the ideal speed v.

Thus, \(F_{net} = F_{centripetal}\)

\(mg tan \theta = \frac {mv ^ 2} {2}\)

\(v = \sqrt {(50 \times 9.8 ) \times (tan 15 ^{\circ})}\\\)

\(= \sqrt {(50 \times 9.8 ) \times 0.27 )}\\\) \(= \sqrt {132.5}\\\)

\(= 11.5 m s^{-1} .\)

Therefore, if the car has a speed of approximately 11 meters per sec, it can manage the curve without any friction.

Typo Error>

Speed of Light, C = 299,792,458 m/s in vacuum

So U s/b C = 3 x 10^8 m/s

Not that C = 3 x 108 m/s

to imply C = 324 m/s

A bullet is faster than 324m/s

I have realy intrested to to this topic

m=f/a correct this

Interesting studies

It is already correct f= ma by second newton formula…