Question

What 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.

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Answer 1

Answer- The above question is from the chapter 'Dynamics'.

Given question: What 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:  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.

(See the diagram attached.)

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.

Answer 2

Answer:

$\huge \fbox\red{Answer}$

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.

Consider a vehicle of mass ‘m’ with moving speed ‘v’ on the banked road with radius ‘r’. Let ϴ be the angle of banking, with frictional force f acting between the road and the tyres of the vehicle.

Total upwards force = Total downward force

NcosΘ=mg+fsinΘ

Where,

NcosΘ : one of the components of normal reaction along the verticle axis

mg: weight of the vehicle acting vertically downward

fsinΘ : one of the components of frictional force along the verticle axis

mg: weight of the vehicle acting vertically downward

fsinΘ : one of the components of frictional force along the verticle axis

therefore, mg=NcosΘ−fsinΘ (eq.1)

mv2r=NsinΘ+fcosΘ (eq.2)

Where,

NsinΘ : one of the components of normal reaction along the horizontal axis

fcosΘ : one of the components of frictional force along the horizontal axis

therefore, v2rg=NsinΘ+fcosΘNcosΘ−fsinΘ

Frictional force f=μsN v2rg=NsinΘ+μsNcosΘNcosΘ−μsNsinΘ v2rg=N(sinΘ+μscosΘ)N(cosΘ−μssinΘ) v2rg=(sinΘ+μscosΘ)(cosΘ−μssinΘ) v2rg=(tanΘ+μs)(1−μstanΘ)

therefore, v=rg(tanΘ+μs)(1−μstanΘ)−−−−−−−−√ vmax=rgtanΘ−−−−−−√ tanΘ=v2rg Θ=tan−1v2rg

Above is the expression for the angle of banking.