Question

The optimum speed for turning on a banked road is

• A. $\sqrt{rg\tan \theta }$
• B. $\sqrt{rgcot\theta }$
• C. $\sqrt{rg\sin \theta }$
• D. $\sqrt{rgsec\theta }$

Answer 1

The correct option is A

$\sqrt{rg\tan \theta }$

Banking of road

1. The outside end of the road is raised higher than the inner end due to road banking. Simply put, banking on the road means slanting the road's surface.
2. A vehicle must make a circular motion when turning on a flat curved road. As a result, the vehicle must accelerate in a centripetal direction. The frictional force between the car and the road provides this acceleration.
3. The frictional force, on the other hand, is restricted. When the road is banked, the ground's typical reaction contributes to the centripetal force as well. As a result, the maximum speed for a safe turn rises.

Optimum speed

1. Optimum speed is the speed with which one can take turns on banked roads without wear and tear.
2. Blanked roads are those that have a tilt to them. To increase safety when making a turn, roads must be banked. The friction between the car tires and the road surface is negligible at the optimal speed.

Consider the figure. The car of mass m moves on a smooth circular road of radius r

Here, $F_{net}$ is the net force applied, $m$ is the mass, $v$ is the velocity, $\theta$ is the angle of inclination, $g$ is the acceleration due to gravity, ${\mu }_S$ coefficient of static friction and $N$ is the normal reaction.

From the free body diagram: $N\sin \theta =\frac{m{V}^2}{r}andN\cos \theta =mg$

$$\begin{gathered} \mathrm{ta}n\theta =\frac{\sin \theta }{\cos \theta } \\ \tan \theta =\frac{{\displaystyle \frac{m{V}^2}{rN}}}{{\displaystyle \frac{mg}{N}}} \\ \tan \theta =\frac{m{V}^2}{rN}\times \frac{N}{mg} \\ \tan \theta =\frac{V^2}{rg} \end{gathered}$$

$V=\sqrt{rg\tan \theta }[whereV=optimumspeed,r=radius,g=gravitationalcons\tan t,\theta =angleofbanking]$

Hence, option A is the correct answer.