Question

A circular road of radius $50m$ has the angle of banking equal to $30^\circ$. At what speed should a vehicle go on this road so that friction is not used.
(A) $7m{s^{ - 1}}$
(B) $17m{s^{ - 1}}$
(C) $22m{s^{ - 1}}$
(D) $27m{s^{ - 1}}$

Answer 1

This problem is solved by the concept of banking of road, where to make the turning of a vehicle safer in a circular surface the outer edge of the road is raised above the inner edge to provide necessary centripetal force by making some inclination with the horizontal which is called angle of banking.

Formula used: $v = \sqrt {rg\tan \theta }$

Complete step by step answer:
$\;N$ is the normal force, ‘$f$’ being the frictional force between the road and tires of the vehicle. ‘$mg$’ is the force due to gravity acting downwards. $\dfrac{{m{v^2}}}{r}$ is the centripetal force acting towards the center of the road? $\Theta$ is the angle of banking.
In this problem friction is not used so, $f = 0$.
In equilibrium net force acting on the vehicle should be zero. Equating components along with vertical and horizontal directions respectively.
$mg = N\cos \theta$....................................$\left( 1 \right)$
$\dfrac{{m{v^2}}}{r} = N\sin \theta$...........................................$\left( 2 \right)$
Dividing equation $\left( 2 \right){\text{ }}by{\text{ }}\left( 1 \right)$
$= >$ $\dfrac{{{v^2}}}{{rg}} = \dfrac{{\sin \theta }}{{\cos \theta }}$
= > $$$$v = \sqrt {rg\tan \theta }. Expression for safe velocity on banked roads.
Given,
$r = 50m,g = 9.8m/{s^{ - 2}}$,$\theta = 30^\circ$ , $\tan 30 = 1/\sqrt 3$

$$v = \sqrt {\dfrac{{50 \times 9.8}}{{\sqrt 3 }}} \\\ = 16.99m/s \approx 17\dfrac{m}{s} \\\$$

So the answer is $\left( B \right)$ $17m/s$

Note: $1.$ The vehicle can be driven with maximum safe speed by making an angle of banking $= 45^\circ {\text{ }}i.e{\text{ }}tan\theta = 1.$
$2.$ When the force $f \ne 0.$Then velocity becomes $v = \sqrt {\dfrac{{rg\tan \theta + {\mu _s}}}{{1 - {\mu _s}\tan \theta }}}$.