Physics, asked by mohammedalishareef3, 9 months ago

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

Answers

Answered by krishnamoorthy2006
0

Refraction occurs in between two transparent mediums

Answered by shadowsabers03
0

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.

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