Question

what do you mean by banking of roads? what is it's need ?Make a diagram and derive the expression of angle of banking when friction is negligible​

Answer 1

Answer:

. Banking of roads is defined as the phenomenon in which the edges are raised for the curved roads above the inner edge to provide the necessary centripetal force to the vehicles so that they take a safe turn. ... The angle at which the vehicle is inclined is defined as the bank angle.When the road is banked, the horizontal component of the normal reaction provides the necessary centripetal force required for circular motion of vehicle. vi. Thus, to provide the necessary centripetal force at the curved road, banking of road is necessary.Angle of Banking Formula

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

The expression for the safe velocity on the banked road is given by v_{max}=\sqrt{rg\,\tan\Theta}

Centripetal Force F = mv2/r = mω2r

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.