Question
Obtain an expression for maximum safety speed with which a vehicle can be safely driven along a curved banked road.
✓
Answer
i. The vertical section of a vehicle on a curved road (considering friction) of radius ' $r$ ' banked at an angle ' $\theta$ ' with the horizontal is shown in the figure. If the vehicle is running exactly at the optimum speed, then the forces acting on the vehicle are a. weight $mg$ acting vertically downwards b. normal reaction $N$ acting perpendicular to the road. ii. But in practice, vehicles never travel exactly at this speed. iii. Hence, for speeds other than this, the component of the force of static friction between the road and the tires helps us, up to a certain limit. iv. For maximum possible speed, The component $N \sin \theta$ is less than the centrifugal force $\frac{ mv ^2}{r}$. $\therefore \frac{ mv ^2}{r}> N \sin \theta$
Banked road: upper-speed limit
$v$. In this case, the direction of the force of static friction $\left(f_s\right)$ between the road and the tires is directed along with the inclination of the road, downwards.
vi. The horizontal component $\left(f_s \cos \theta\right)$ is parallel to $N \sin \theta$. These two forces take care of the necessary centripetal force (or balance the centrifugal force).
$\therefore \frac{m v ^2}{r}=N \sin \theta+f_s \cos \theta \text {. }$
vii. The vertical component, $N \cos \theta$ balances the component $f_s \sin \theta$ and weight ' $mg$ '.
$ \therefore N \cos \theta= f _{ s } \sin \theta+ mg$
$\therefore mg = N \cos \theta- f _{ s } \sin \theta $
viii. For maximum possible speed, $fs _{ s }$ is maximum and equal to $\mu_{ s } N$. From equations ( 1 ) and (2),
$v _{\max }=\sqrt{r g\left(\frac{\tan \theta+\mu_s}{1-\mu_s \tan \theta}\right)}$
This is an expression for maximum safety speed with which a vehicle can be safely driven along a curved banked road (considering friction).
x. From equation (3) and equation (4) we can write,
$ \left(\frac{\tan \theta+\mu_s}{1-\mu_s \tan \theta}\right)=\frac{V_{\max }^2}{r g} \ldots . . . .(5)$
$\text { and } \tan \theta=\frac{V_{\max }^2}{r g}$
$\therefore \theta=\tan ^{-1}\left(\frac{V_{\max }^2}{r g}\right) . . . \ldots \ldots \ldots \ldots . . .(6) $
From equation (5) and equation (6), angle of banking is independent of mass of vehicle.
Banked road: upper-speed limit$v$. In this case, the direction of the force of static friction $\left(f_s\right)$ between the road and the tires is directed along with the inclination of the road, downwards.
vi. The horizontal component $\left(f_s \cos \theta\right)$ is parallel to $N \sin \theta$. These two forces take care of the necessary centripetal force (or balance the centrifugal force).
$\therefore \frac{m v ^2}{r}=N \sin \theta+f_s \cos \theta \text {. }$
vii. The vertical component, $N \cos \theta$ balances the component $f_s \sin \theta$ and weight ' $mg$ '.
$ \therefore N \cos \theta= f _{ s } \sin \theta+ mg$
$\therefore mg = N \cos \theta- f _{ s } \sin \theta $
viii. For maximum possible speed, $fs _{ s }$ is maximum and equal to $\mu_{ s } N$. From equations ( 1 ) and (2),
$v _{\max }=\sqrt{r g\left(\frac{\tan \theta+\mu_s}{1-\mu_s \tan \theta}\right)}$
This is an expression for maximum safety speed with which a vehicle can be safely driven along a curved banked road (considering friction).
x. From equation (3) and equation (4) we can write,
$ \left(\frac{\tan \theta+\mu_s}{1-\mu_s \tan \theta}\right)=\frac{V_{\max }^2}{r g} \ldots . . . .(5)$
$\text { and } \tan \theta=\frac{V_{\max }^2}{r g}$
$\therefore \theta=\tan ^{-1}\left(\frac{V_{\max }^2}{r g}\right) . . . \ldots \ldots \ldots \ldots . . .(6) $
From equation (5) and equation (6), angle of banking is independent of mass of vehicle.
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