## Banking of Road Formula :

The velocity of a vehicle on a curved banked road: v= √(rg(tanΦ+μ_{s}))/1-μ_{s} tanΦ

For a given pair of roads, and Tyre μ_{s }= tanλ, then the velocity of a vehicle on a curved banked road is: v = √rg tan (Φ+λ)

The safe velocity on an unbanked road is: v_{max} = √ μ * r*g

The expression for the angle of banking of road is: θ = tan^{-1 }[v_{0}^{2} / Rg]

Expression for the safe velocity on the banked road is: v_{max} = √rg tan Φ

Centripetal Force F= mv^{2 }/ r = mw^{2}r

## Sample Problems :

## Problem1 :

Calculate the maximum speed of a car with which it can be safely driven along a curve of radius 100m and the coefficient of friction between tires and road is 0.2. (take g = 9.8 m/s^{2})

### Solution :

Given ,

Radius (r) = 100m

Coefficient of friction (μ) = 0.2

Gravity (g) = 9.8 m/s^{2}

Formula: v= √(μrg)

v = √(μrg)

= √(0.2 * 100 * 9.8)

= √196

v = 14 m/s

Thus maximum speed of car is 14 m/s.

## Problem2 :

Calculate the angle which the bicycle and its rider make with the vertical when going at 18 km/hr around a curved road of radius 10m on level ground. (take g = 9.8 m/s^{2})

### Solution :

Given ,

Radius (r) = 10 m

Max speed of rider (v) = 18 km/hr

Gravity (g) = 9.8 m/s^{2}

Formula: tanθ = v^{2} / rg

v = 18 km/hr

= 18 * 1000 / 3600

= 5 m/s

Tangent angle of banking is tanθ = v^{2} / rg

= 5 * 5 / 10 * 9.8

= 5 / 2 * 9.8

tanθ = 0.251

θ = tan^{-1} (0.2551)

θ = 14° 19^{‘}

Thus the angle which the bicycle and rider make is 14° 19^{‘} .