Question

what do u mean by banking of roads? what is its need? make diagram and derive the expression of angle of banking when friction of negligible​

Answer 1

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.