Question

Why are circular roads banked?
Deduce an expression for the angle of banking.

Answer 1

Explanation:

Anthe fill kia hai kya............

Answer 2

Roads are banked to avoid sliding of the vehicle and thus avoid accidents. The roads are slightly raised and tilted towards the inner side. Hence, the horizontal and vertical components of forces get balanced thus balancing the vehicle.

Here is how we can deduce an expression for the angle of banking;

N: Normal reaction acting on the vehicle

f: frictional force acting on the vehicle

mg: weight of the vehicle

$\frac{mv^{2} }{r}$: centripetal force acting on the vehicle

$\theta$: angle of banking

Balancing the horizontal components,

we get,

NSinФ = $\frac{mv^{2} }{r}$- fCosФ

Balancing the vertical components,

we get,

NCosФ = fSinФ + mg

Further,

$\frac{mv^{2} }{r}$= NSinФ + fCosФ                                     .......... (1)

mg = NCosФ - mg                                          .......... (2)

Dividing equations 1 and 2,

we get,

$\frac{\frac{mv^{2} }{r} }{mg} = \frac{NSin\theta+fCos\theta}{NCos\theta-fSin\theta}$

so,

$\frac{v^{2} }{rg} =\frac{NSin\theta+fCos\theta}{NCos\theta-fSin\theta}$

we know,

Frictional force, $f=\mu N$

$\frac{v^{2} }{rg} =\frac{NSin\theta+\mu N Cos\theta}{NCos\theta-\mu N Sin\theta}$

$\frac{v^{2} }{rg} =\frac{N(Sin\theta+\mu Cos\theta)}{N(Cos\theta-\mu Sin\theta)}$

$\frac{v^{2} }{rg} =\frac{Sin\theta+\mu Cos\theta}{Cos\theta-\mu Sin\theta}$

Dividing the RHS with $Cos\theta$,

$\frac{v^{2} }{rg} =\frac{Tan\theta+\mu }{1-\mu Tan\theta}$

$v=\sqrt{\frac{rg(Tan\theta+ \mu)}{1- \mu Tan\theta} }$

$v_{max} =\sqrt{rgTan\theta}$

$Tan\theta = \frac{v^{2} }{rg}$

$\theta= Tan^{-1} \frac{v^{2} }{rg}$

This is the expression for angle of banking i.e $\theta= Tan^{-1} \frac{v^{2} }{rg}$

#SPJ3