Question

A circular road of radius 1000 m has a banking angle $45^\circ$. If the coefficient of friction between tyre and road is 0.5, then the maximum safe speed of a car having mass 2000 kg will be
(A) 172 m/s
(B) 124 m/s
(C) 99 m/s
(D) 86 m/s

Answer 1

Friction is desirable and necessary for providing traction and facilitating movement on land. For acceleration, deceleration, and changing direction, most ground vehicles rely on friction. Sudden traction loss can result in a loss of control and an accident. Friction is not a basic force in and of itself. This is a notion that we use in this problem.
Formula used:
$v = \sqrt {\dfrac{{rg\left( {\tan \theta - {\mu _s}} \right)}}{{1 + {\mu _s}\tan \theta }}}$
V = velocity
R = radius
G = acceleration due to gravity
${\mu _s}$= coefficient of friction

Complete answer:
A banked turn (also known as a banking turn) is a turn or direction change in which the vehicle banks or inclines towards the inside of the turn. This is generally owing to the roadbed having a transverse down-slope towards the inside of the bend on a road or railroad. The bank angle is the angle at which the vehicle's longitudinal axis is inclined in relation to the horizontal. When examining the system's impacts of friction, it's important to remember which way the friction force is pointing. When determining our car's maximum speed, friction will point down the gradient and towards the circle's centre.
As a result, the horizontal component of friction must be added to the normal force. Our new net force in the direction of the turn's centre (centripetal force) is the sum of these two forces:
$v = \sqrt {\dfrac{{rg\left( {\tan \theta - {\mu _s}} \right)}}{{1 + {\mu _s}\tan \theta }}}$
Upon squaring on both the sides we get
${v^2} = \dfrac{{rg\left( {\tan \theta - {\mu _s}} \right)}}{{1 + {\mu _s}\tan \theta }}$
Substituting all the values obtained from the question
R = 1000 m
G = $9.8{\text{ }}m{s^{ - 2}}$
$\theta = {45^o}$
Hence $\tan {45^o} = 1$
${\mu _s} = 0.5$
$\therefore {v^2} = 9.8 \times 1000 \times \left( {\dfrac{{0.5 \times 1}}{{1 - 0.5 \times 1}}} \right)$
$\Rightarrow v = 172m{s^{ - 1}}$

Hence option A is correct.

Note:
When considering the direction of friction for the automobile's lowest velocity, the discrepancy in the latter analysis arises (towards the outside of the circle). As a result, when inserting friction into equations for forces in the centripetal and vertical directions, the opposite procedures are done.