Question

A vehicle is moving with a velocity 'V' on a curved road of width 'b'and radius of curvature 'R'.For counter acting the centrifugal force on the vehicle the difference in elevation reguired in between the outer and inner edge of the road is​

Answer 1

Answer:

Explanation:

Given

Speed of vehicle is V

Width of road is b

radius of curvature is R

From Free body diagram we can write

$N\sin \theta =mg---1$

Normal reaction sin component will provide the centripetal force to move in circular path

$N\sin \theta =\dfrac{mv^2}{R}----2$

Divide 1 and 2 we get

$\dfrac{\sin \theta }{\cos \theta }=\dfrac{v^2}{Rg}$

$\tan \theta =\dfrac{v^2}{Rg}$

and we can write $\tan \theta as \dfrac{h}{b}$

$\dfrac{h}{b}=\dfrac{v^2}{Rg}$

$h=\dfrac{bv^2}{Rg}}$