Question

Explain what is meant by 'banking of curves'? Hence obtain an expression
for the speed with which a car negotiates a curve

Answer 1

Refraction occurs in between two transparent mediums

Answer 2

To avoid the wear and tear of the tyres of vehicles, a small inclination is given to the outer edge of the curved roads by a certain small angle over the inner edge. This is called banking of curves.

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Consider a car or any other vehicle of mass m moving along a banked road. Then the car experiences,

• weight mg acting vertically downwards.

• normal reaction N, acting upwards perpendicular to the banked curve.

• frictional force f acting along the banked road towards the center of the circular curve.

Thus, we know that,

$f=\mu N\\\\\\\mu=\dfrac {f}{N}\quad\longrightarrow\quad (1)$

Here the normal reaction and the frictional force can be resolved into their components according to the angle of inclination of the banked road, θ.

Since the car moves on a circular road, it experiences centripetal force which is completely provided by the resultant of the horizontal components of the frictional force and the normal reaction, i.e.,

$N\sin\theta+f\cos\theta=\dfrac {mv^2}{R}\quad\longrightarrow\quad (2)$

Here, from the fig., the resultant of the vertical components of N and f balances the weight of the car, i.e.,

$N\cos\theta=f\sin\theta+mg\\\\N\cos\theta-f\sin\theta=mg\quad\longrightarrow\quad (3)$

Then, dividing (2) by (3),

$\dfrac {N\sin\theta+f\cos\theta}{N\cos\theta-f\sin\theta}=\dfrac {\left (\dfrac {mv^2}{R}\right)}{mg}\\\\\\\dfrac {\left (\dfrac {N\sin\theta+f\cos\theta}{N\cos\theta}\right)}{\left (\dfrac {N\cos\theta-f\sin\theta}{N\cos\theta}\right)}=\dfrac {v^2}{Rg}\\\\\\\dfrac {\tan\theta+\frac {f}{N}}{1-\frac {f}{N}\tan\theta}=\dfrac {v^2}{Rg}$

But from (1),

$\dfrac {v^2}{Rg}=\dfrac {\tan\theta+\mu}{1-\mu\tan\theta}\\\\\\\boxed {v=\sqrt {Rg\left (\dfrac {\tan\theta+\mu}{1-\mu\tan\theta}\right)}}$

Let $\mu=\tan\phi.$ Then,

$v=\sqrt {Rg\left (\dfrac {\tan\theta+\tan\phi}{1-\tan\theta\tan\phi}\right)}\\\\\\v=\sqrt {Rg\tan(\theta+\phi)}$

This is just to simplify the expression.