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.​

Answer 1

Answer:

Banking of road means raising the outer edge of the road with respect to inner edge so that the road makes an angle with the horizontal.

When a vehicle moves along a curve, the force of friction provides the necessary centripetal force. But this force has a limit = μmg.When the speed of the vehicle increases beyond this maximum value, the banking of roads is necessary so that the component of normal reaction of the road towards the centre provides the necessary centripetal force.

From the figure, for vertical equilibrium :

Total upwards force = total downwards force

Ncosθ=mg+fsinθ

mg=Ncosθ−fsinθ...(1)

As the forces Nsinθ,fcosθ provide the necessary centripetal force.

r/mv ²

=Nsinθ+fcosθ...(2)

Thus dividing (2) by (1),

mg

r/mv ² =

Ncosθ−fsinθ

Nsinθ+fcosθ

As f=μ

s

N

Thus,

rg

v

2

=

cosθ−μ

s

sinθ

sinθ+μ

s

cosθ

=

1−μ

s

tanθ

tanθ+μ

s

=

1−tanλtanθ

tanθ+tanλ

=tan(θ+λ)

v=

rgtan(θ+λ)

where tanλ=μ

s

Answer 2

CORRECT 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.

Refer to the attachment ❤️