Question

What is meant by banking of roads? Obtain an expression for the maximum speed with which a vehicle can safely negotiate a curved road banked at an angle $\theta$, considering the friction is negligible.

Answer 1

Hint: Banking of roads refers to when there is an angle between the surface of the road with the horizontal, that is, the road is inclined. This incline in the road helps in turning since the inclined direction of the normal force has a component which provides the centripetal force necessary for completing a curve.

Formula used:
For a body of mass $ M $ going in a circle of radius $ R $ at a constant speed $ v $ , the centripetal force $ {{F}_{C}} $ required is
 $ {{F}_{C}}=\dfrac{M{{v}^{2}}}{R} $
For a body in static equilibrium, the summation of all forces $ \overrightarrow{F} $ on the body is zero. That is,
 $ \sum{\overrightarrow{F}}=0 $

Complete step-by-step answer:
When a car goes on a curved road, it requires a certain centripetal force to stay on the curved path and complete the turn. This centripetal force acts towards the center of the circle of the curve and constantly changes the direction of the car to help it to stay on the circular path.
Banking of roads refers to the concept that the surface of the road is inclined with the horizontal. This angle is known as the angle of inclination or the angle of banking.
When the road is banked the normal force provided by the road on the car provides a component which is in the direction of the centripetal force. Hence, when the road is banked, it is easier for a car to take the turn at a higher speed.
To obtain the expression for the maximum speed with which a vehicle can negotiate a curve on a banked road with angle $ \theta $, let us draw a free body diagram.

Let the mass of the car be $ M $ , and the normal force exerted by the car on the road be $ N $. The car maneuvers a turn of radius $ R $ for which the centripetal force acts in the positive – X direction. The angle of inclination of the road is $ \theta $ . The car goes about the circle at a constant speed $ v $.
For a body of mass $ M $ going in a circle of radius $ R $ at a constant speed $ v $, the centripetal force $ {{F}_{C}} $ required is
 $ {{F}_{C}}=\dfrac{M{{v}^{2}}}{R} $ --(1)
For a body in static equilibrium, the summation of all forces $ \overrightarrow{F} $ on the body is zero. That is, $ \sum{\overrightarrow{F}}=0 $ --(2)
First, let us analyze the forces in the Y-direction with the forces in the positive direction of the axis being considered positive. Since, the car does not move in the y – direction, the summation of the forces must be zero.
Therefore, according to the free body diagram, using (2), we get,
$ N\cos \theta -Mg=0 $
$ \Rightarrow N\cos \theta =Mg $
$ \Rightarrow N=\dfrac{Mg}{\cos \theta } $ --(3)

Now, in the positive X – direction, the summation of forces must be equal to the centripetal force for the turn. Forces in the positive X – direction are considered positive.
Therefore, according to the free body diagram and using (1), we get,
$ N\sin \theta ={{F}_{C}} $ (Since, the weight of the body is in the vertical direction, it has no component in the horizontal direction, the only force in that direction is the component of the normal force)
$ \Rightarrow N\sin \theta =M\dfrac{{{v}^{2}}}{R} $ [Using (1)]
$ \Rightarrow \dfrac{Mg}{\cos \theta }\sin \theta =M\dfrac{{{v}^{2}}}{R} $ [Using (3)]
$ \Rightarrow g\tan \theta =\dfrac{{{v}^{2}}}{R} $
$ \Rightarrow {{v}^{2}}=Rg\tan \theta $
Square rooting both sides, we get,
 $ \Rightarrow \sqrt{{{v}^{2}}}=\sqrt{Rg\tan \theta } $
 $ \Rightarrow v=\sqrt{Rg\tan \theta } $
Therefore, the maximum speed at which the car can negotiate the turn will be $v=\sqrt{Rg\tan \theta }$

Note: It is always good to approach problems of mechanics by drawing a free body diagram and apply some rules of geometry. This helps in understanding the problem a lot and devising relations between different variables, sometimes even by geometry. Drawing a proper free body diagram is an essential skill that cannot be understated.
Students might be thinking that if $ \theta =0 $, $ \tan \theta =0,v=0 $ , hence the car will not be able to turn on a flat road. However, this is not true as the friction between the tyres and the road provides the necessary centripetal force. Even when the roads are banked, a component of the friction helps in providing the centripetal force. Thus, it is easier to negotiate a turn at a higher speed on a banked road with good friction than on a banked road with less friction. The question asked to neglect friction, therefore, it does not take part in the calculations above.