Question

What is the angle of banking necessary for a curved frictionless road of radius $30\text{m}$ for safe driving at a speed of $20\text{m/s}$? (Consider value of $g=10{\text{m/s}}^2$)

• A. ${\tan }^{-1}\left(\frac{4}{3}\right)$
• B. ${\tan }^{-1}\left(\frac{3}{4}\right)$
• C. ${\tan }^{-1}\left(1\right)$
• D. ${60}^{\circ }$

Answer 1

The correct option is A ${\tan }^{-1}\left(\frac{4}{3}\right)$

In case of banking of road, the levelled road is elevated at some angle $\theta$, so that component of normal reaction ($N\sin \theta$) becomes available to meet the requirement of centripetal force at the turn/curve. Thus, eliminating the dependency on frictional force.

FBD of vehicle moving on banked road:

Applying equation of dynamics in horizontal ($X$) and vertical direction ($Y$):


$N\sin \theta =\frac{M{v}^2}{r}$..........$\left(i\right)$
$N\cos \theta =Mg$.......$\left(ii\right)$

Dividing Eq. $\left(i\right)$ and Eq.$\left(ii\right)$ gives:

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

Putting the values $r=30\text{m},v=20\text{m/s}$:
$\tan \theta =\frac{{20}^2}{30\times 10}=\frac{400}{300}$

$\therefore \theta ={\tan }^{-1}\left(\frac{4}{3}\right)$