I have immense interest in bikes. From a very young age, I have taken interest in different kinds of bikes available in the market. I was so fascinated about them that whenever I came across a bike picture, I would cut it and paste it in my bike collection scrapbook.

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My dad owns a motorcycle and it helped me learning to ride it. I having been saving money for the last two years to buy my dream bike but an incident a week before compelled me to reconsider my decision.

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It was Sunday and I was sleeping till late morning when suddenly I was awakened by noises in the street. I rushed to the balcony and found out that a bike met with an accident while turning. There was a man, his wife and a kid. All three were severely injured and was taken to the hospital.

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There was bloodshed on the road and the sight filled me with a dilemma of whether I should buy a bike. After much thinking, I decided to research about all aspects related to roads and ridings.

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I read various sources, various precautions and many more. In the process I came across the fact which I had no idea about and it was that in turnings, one side of the road is always elevated to avoid skidding.

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I wanted to know how it works but could not get convincing resources. Despite going through several research journals and articles on circular motion, banking of road and pseudoforces, my doubts on the topic was not completely cleared; thereby thought of figuring it out myself. This IA is about the same.

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In this IA, I will try and explain how the elevation in turnings help preventing skidding.

The main motive of this IA is the investigate the effect of banking in preventing a car from experiencing a horizontal skid while turning in a road. In association with that, a comparative study will be shown in this IA to establish the requirement of banking at any sharp turn in a road relative to that of a horizontal road without any banking.

To what extent does the angle of banking in a road responsible for prevention of skidding?

Road accidents usually happen when a vehicle skids in the road. The vehicle being persistently out of control due to skidding, often strikes with another vehicle or often strikes with the boundary wall or barrier of any road. One of the most likely places of vehicle skidding is the turning. Vehicles often skid during turning in the road. In case of a sharp turn, vehicles often decrease their speed to avoid skidding which may lead to an accident.

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However, there are several preventive measures that are incorporated while any road is constructed at any turning. Road banking is one of such techniques. It refers to a slight elevation along the side of the road opposite to the turn.

banking of road.svg

From Figure 1, it is evident that, as the road is turning towards right (with respect to the driver of the vehicle), the banking or elevation is done at the left side of the road.

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In this IA, mathematical exploration will be carried on to determine how the banking affects in the skidding of vehicle. Some of the laws of physics that will be required to execute the mathematical exploration are as follows:

<ul>
	<li>Centripetal Force: The force which is responsible for turning mechanism of a body or movement of a body along a circular path is known as Centripetal Force. The expression of centripetal force comprises mass (m) of the body, velocity (v) of the body, and radius (r) of the curvature is given by:</li>
</ul>

\(F_{centripetal} = \frac{mv^2}{r}\)

<ul>
	<li>Frictional Force: It is the resistive force that opposes the external applied force and acts in a direction opposite to that of the applied force. The expression of frictional force between the wheels of the vehicle and the surface of the road comprises (&mu;) which is the coefficient of friction and (N) is the normal reaction force due to weight of the body is given by:</li>
</ul>

ffriction &le; &mu;N

<ul>
	<li>3rd Law of Newton (from Newton&rsquo;s Laws of motion): Every action has an equal and opposite reaction.</li>
	<li>Centrifugal Force: It is a reaction force that appears in presence of centripetal force. Its expression is same as that of Centripetal force but the direction is opposite.</li>
</ul>

In a nutshell, when a vehicle makes a turn, the centripetal force is generated and as a result, the vehicle successfully executes the turning. However, due to generation of centripetal force toward the direction of turning, its reaction force or Centrifugal Force is generated in the opposite direction. This force is responsible for skidding. As a result, whenever a vehicle makes a sharp right turn, it often skids leftward and vice versa. Though skid is not observed in every case of turning. This is because of frictional force that acts between the wheels and road.

In this IA, we will consider three different cases for the study of banking of road. They are:

<ul>
	<li>Road without banking.</li>
	<li>Smooth Road (no friction) with banking.</li>
	<li>Road with banking.</li>
</ul>

Throughout the study, there will be few variables used to denote several physical quantities. Such variables with the respective physical quantities are mentioned as follows:

<ul>
	<li>m = mass of vehicle.</li>
	<li>v = velocity of vehicle.</li>
	<li>g = acceleration due to gravity.</li>
	<li>r = radius of curvature of the turn.</li>
	<li>&mu; = coefficient of friction between road and wheel.</li>
	<li>N = normal reaction force.</li>
	<li>&theta; = angle of banking.</li>
	<li>Cp = centripetal force.</li>
	<li>Cf = centrifugal force.</li>
	<li>F = frictional force.</li>
</ul>

Case 1: Skidding in a road without banking

FBD of case 1.svg

From the definitions mentioned in Introduction section, we can write,

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\(C_p = \frac{mv^2}{r}\)

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Also, we can write,

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\(|C_p|=|C_f|=\bigg|\frac{mv^2}{r}\bigg|\)&hellip;&hellip;&hellip;&hellip;(1)

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As there is no banking, from the law of equivalent force, we can write,

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N = mg

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Also,

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f&nbsp; &le;&nbsp; &mu;N

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=&gt;&nbsp;&nbsp;f&nbsp; &le;&nbsp; &mu; &times; mg&hellip;&hellip;&hellip;&hellip;(2)

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From Figure 2, it is observed that to restrict a vehicle from skidding, the frictional force must nullify the centrifugal force as both of them are acting in two opposite directions. However, the expression of frictional force is not an equality. Thus, the condition to resist a vehicle from skidding is expressed as:

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f&nbsp; &ge;&nbsp; Cf

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\(=&gt; μ × mg ≥ \frac{mv^2}{r}\)

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\(=&gt; μ × g ≥ \frac{v^2}{r}\)

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\(=&gt;v≤\sqrt{μgr}\)&hellip;&hellip;&hellip;&hellip;(3)

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From equation (3), we can say that, if there is no banking in a road with a turning, then the velocity of the vehicle should be less than or equal to&nbsp;\(\sqrt{μgr}\) in order to avoid skidding.

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Case 2: Skidding in a smooth road with banking:

FBD of Case 2.svg

From Figure 3, it can be noted that the weight of the body and the normal reaction force of the body are not on the same axis. Thus, in order to solve this problem, resolution of vector should be done.

A section of FBD of case 2.svg

In the triangle shown in figure 4,

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&ang;BAC =&nbsp; &theta;

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&ang;ABC =&nbsp; 90 - &theta;

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&ang;ABD =&nbsp; 90 (by construction)

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&ang;CBD = &theta;

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In triangle ABD,

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\(\frac{BD}{BC} = \, \,cos \, \,cos θ\)

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=&gt; BD = BC cos&nbsp;cos&nbsp;&theta;

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=&gt; BD = mg cos&nbsp;cos&nbsp;&theta;

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Similarly,

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\(\frac{AB}{BC}= \, \,sin \, \,sin θ\)

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=&gt; AB = BC sin&nbsp;sin&nbsp;&theta;

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=&gt; AB = mg sin&nbsp;sin&nbsp;&theta;

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In this case, from figure 3 and 4, we can write,

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N = mg cos&nbsp;cos&nbsp;&theta;

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Therefore, the expression of friction will be:

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f&nbsp; &le;&nbsp;&nbsp;&mu;&nbsp;N

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=&gt; f&nbsp; &le; &mu; &times;&nbsp; mg cos&nbsp;cos&nbsp;&theta;

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However, as the road is smooth, then:

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f &le; 0 &times; mg cos&nbsp;cos&nbsp;&theta;

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f = 0

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Similar to equation (1) of case (1), the expression of centrifugal force will be:

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\(|C_f|=\bigg|\frac{mv^2}{r}\bigg|\)

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From Figure 3 and Figure 4, it is observed that to restrict a vehicle from skidding, the perpendicular component of the weight of the vehicle must nullify the centrifugal force as&nbsp;they are acting in two opposite directions. Thus, the condition to resist a vehicle from skidding is expressed as:

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\(\frac{mv^2}{r} = mg \, \,sin \,\,sin θ\)&nbsp;&hellip;&hellip;&hellip;(4)

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N = mg&nbsp;cos cos&nbsp;&theta;&nbsp;&hellip;&hellip;&hellip;(5)

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From equation (4) and (5), it can be written as:

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\(\frac{\frac{mv^2}{r}}{N}= mg \, \,sin \, \,sin θ\)&nbsp;&hellip;&hellip;&hellip;(4)

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\(=&gt; \frac{\frac{v^2}{r}}{g} = \, \,tan \, \,tan θ\)

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\(=&gt; v = \sqrt{r\ g\ tan \ tan\ \theta}\)&nbsp;&hellip;&hellip;&hellip;&hellip;(6)

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Case 3 - Skidding in a road with banking:

FBD of Case 3.svg

From Figure 5, it can be noted that the weight of the body and the normal reaction force of the body are not on the same axis. Thus, in order to solve this problem, resolution of vector should be done.

A section of FBD of case 3.svg

In the triangle shown in figure 6,

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&ang;BAC =&nbsp; &theta;

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&ang;ABC = 90 - &theta;

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&ang;ABD = 90 (by construction)

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&ang;CBD = &theta;

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In triangle ABD,

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\(\frac{BD}{BC}= \, \, cos \, \,cos θ\)

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=&gt; BD = BC cos&nbsp;cos&nbsp;&theta;

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=&gt; BD = mg cos cos &theta;

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Similarly,

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\(\frac{AB}{BC} = \, \,sin \, \,sin θ\)

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=&gt;&nbsp;AB = BC sin&nbsp;sin&nbsp;&theta;

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=&gt;&nbsp;AB = mg sin&nbsp;sin&nbsp;&theta;

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In this case, from figure 3 and 4, we can write,

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N = mg cos&nbsp;cos&nbsp;&theta;

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Therefore, the expression of friction will be:

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f&nbsp;&nbsp;&le;&nbsp; &mu;N

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=&gt;&nbsp; f&nbsp;&nbsp;&le;&nbsp; &mu;&nbsp;&times; mg cos&nbsp;cos&nbsp;&theta;

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However, as the road is smooth, then:

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f&nbsp;&nbsp;&le;&nbsp; 0 &times; mg cos&nbsp;cos&nbsp;&theta;

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f&nbsp; = 0

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From Figure 5 and Figure 6, it is observed that to restrict a vehicle from skidding, the perpendicular component of the weight of the vehicle must nullify the centrifugal force as they are acting in two opposite directions. Thus, the condition to resist a vehicle from skidding is expressed as:

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\(\frac{mv^2}{r} ≤ mg \, \,sin \, \,sin θ + f\)

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\(=&gt;\frac{mv^2}{r} ≤  mg \, \,sin \, \,sin θ + μ\,mg \, \,cos \, \,cos θ\)

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\(=&gt;v≤\sqrt{r\ g\ sin\ sin\ \theta +μrg\ cos\ cos\ \theta}\)&nbsp;&hellip;&hellip;&hellip;&hellip;(7)

In this section, a comparative study between the efficiency of Case 1 and Case 2 will be performed.

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For a road without any banking (only friction), the expression of velocity of vehicle that should be maintained in order to obtain no skidding condition:

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\(v≤\sqrt{μgr}\)

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Similarly, that of the road with banking (no friction) is given as:

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\(v≤\sqrt{gr\ tan\ \theta}\)

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Now, we know that, the value of g is constant and equal to 9.81 m/sec. However, the value of r may vary depending upon the turning of the road. In this comparative study, the value of radius will be considered as 10 m because, this is the most commonly used radius of curvature observed while construction of any banking at any turning.

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In case of road without any banking, the friction is responsible to resist the skidding. However, there is a limit of frictional force to resist that. The only physical parameter which can increase or decrease the frictional force is the coefficient of friction of the road. From various sources, it has been seen that, the maximum coefficient of friction for a road is 0.7. As a result, we will observe all the values of coefficient of friction up to the limit of 0.7.

Coefficient of friction of road.svg

On the other hand, only angle of banking is responsible to resist a vehicle from skidding. The maximum value of angle of banking is 90&deg;. So, in this study, the maximum velocity to avoid any skid will be studied up to the angle of banking being 90&deg;.

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Road without banking -

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The velocity to avoid skidding is represented as -

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\(v≤\sqrt{μgr}\)

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Here, g = 9.81 m/ sec and r = 10 m.

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\(v≤\sqrt{μ×9.81×10}\)

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\(=&gt;v≤9.9×\sqrt{μ}\)

Velocity vs Coefficient.svg

Smooth Road with banking -

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The velocity to avoid skidding is represented as:

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\(v≤\sqrt{gr\ tan\ tan\ \theta} \)

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Here, g = 9.81 m/ sec and r = 10 m.

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\(v≤\sqrt{tan\ tan\ \theta\ \times9.81\times10}\)

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\(v≤9.9×\sqrt{tan\ tan\ \theta} \)