Question

. What do you mean by banking of roads? What is its need? Make a diagram and derive the

expression of angle of banking when friction is negligible.
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Answer 1

To make the turning of vehicles on the curved road safer, the outer edge of the road is raised above the inner edge making some inclination with the horizontal.This is known as banking of roads

Answer 2

Answer:  Banking of roads- The phenomenon of raising the outer edge of road with respect to the inner edge is called banking of roads.

Need of banking of roads- Friction isn't a reliable source of centripetal acceleration. Thus, roads are banked so as to remove the dependency of centripetal acceleration upon the frictional force.

Consider a vehicle/car of mass 'm' moving on a horizontal circular path of radius 'R' over a banked road of inclination θ with horiontal with uniform speed 'V'.

Forces acting on vehicle are:

1. Weight 'Mg' vertically downward

2. Normal reaction 'N'

3. Frictional force 'f' down the inclined plane

N has 2 components: N cos θ acting as vertical component and N sin θ acting as horizontal component along the centre.

f has 2 components: f sin θ acting vertically downwards and f cos θ acting along the centre.

N cos θ = f sin θ + Mg --- (1)

f = μN

N cos θ = μN sin θ + Mg

N (cos θ - μ sin θ) = Mg

$N = \dfrac{Mg}{cos \theta - \mu sin \theta}$

$Also, \: N sin \theta + f cos \theta = \dfrac{MV^2}{R}$  --- (2)

$N sin \theta + \mu N cos \theta = \dfrac{MV^2}{R}$

$N (sin \theta + \mu cos \theta) = \dfrac{MV^2}{R}\\\\Mg \dfrac{sin \theta + \mu cos \theta}{cos \theta - \mu sin \theta} = \dfrac{MV^2}{R}$

Solving the equation above, we get,

$V = \sqrt {\dfrac{gR(sin \theta + \mu cos \theta}{cos \theta - \mu sin \theta}}$

When friction is negligible, μ = 0.

$V = \sqrt{\dfrac{gR\: sin \theta}{cos \theta}}$

$V = \sqrt{gR \: tan \theta}$

Squaring both sides, we get,

V² = gR tan θ

$tan \: \theta = \dfrac{V^2}{gR}$

∴ $\boxed{\theta = tan^{-1} \dfrac{V^2}{gR}}$ is the angle of banking when friction is

negligible.