Mathematics AA SL

Sample Internal Assessment

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Aim

Research Question

Introduction

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My favorite subject is physics. I find it very interesting. It gives rise to so many questions and finding those answers is what I love the most about the subject.

I am also a fun-loving person. I like playing games and so I love visiting parks. Like every other kid, I have visited parks from childhood numerous numbers of times.

I have the habit of connecting almost everything with physics. My teacher says that it not only clears our concepts but also increase our interest. I totally agree. Physics seem so much fun now. It is not just book, it is everywhere.

Last week when I visited the park with my cousin. My cousin and I love the roller coaster the most and it are always the first thing we ride. She is in grade 6 and they are being taught speed and velocity in physics. She was talking about the topic when a thought crossed my mind.

I looked at the structure of the roller coaster with greater interest. I began wondering what the minimum velocity of the riding cart should be so that it can reach the maximum height of the designed track after a steep depression of the track. Also, I want to find what type of structure should be given to the track after the highest point so that the riders have the maximum thrill.

I went home with the questions in my mind and started looking for answers. I read many writings, research papers to have an idea on this. I read about the working principle of a roller coaster and also came across various laws responsible for determining velocity. But none could give the accurate answers to the questions I had in mind, the minimum velocity and the structure.

This IA is about finding answers to the same.

There are two objectives in this IA. Firstly, the minimum velocity which is required by a roller coaster at the top of a small elevation to reach the highest point of the roller coaster ride just after a depression as mentioned in Figure 1 should be found. Secondly, a model of a roller coaster ride will be prepared based on the guidelines of several authorities provided that, the roller coaster ride will offer maximum thrill. In this IA, I will be designing a part of the track which will offer maximum thrill just after the highest point of the roller coaster ride as mentioned in Figure 1. These are the main objectives of this IA.

To what extent does the shape of track responsible in offering a thrilling experience in a roller coaster ride and also determination of velocity at a certain point of a roller coaster to be able to reach the highest point of the roller coaster ride.

Roller coaster ride is one of the most popular thrilling rides amongst teenagers and adults in this world. The track of the ride is full of ups and downs. The track often reaches a certain height and just a moment after that it reaches down the ground and this offers a thrilling experience to the people.

In my experience in this roller coaster ride in Nicco Park, Kolkata, there was no external force applied to the roller coaster between the points A and C other than the force of gravity which actually enhanced the thrill at this part of the track. Furthermore, the calculation of velocity was so accurate that the velocity of the cart at position C was zero and after the point C, the cart began to slide down due to force of gravity only. Before riding the cart, weight of each individual was measured to set the velocity of the cart at point A which is required to reach the highest point C without application of any external velocity between A and C. In this calculation, the law of conservation of energy of physics is very essential. The Law of Conservation of Energy states that, *“Energy cannot be created nor be destroyed. It can only be converted from one form to another.”* The two forms of energy, i.e., Potential Energy or the energy of a body achieved due to the state of rest and Kinetic Energy or the energy which is achieved by a body due to its state of motion. Their mathematical expressions are shown below:

*Potential Energy = mgh*

Kinetic Energy = \(\frac{1}{2}\) *mv ^{2}*

where, * m *is the mass of the body,

In reference to the last section of the ride, i.e., after point C, the motive is to design a track that will offer maximum thrill abiding by the laws. As we all know, more the speed of descent from the point C, more thrilling the experience becomes for the individual. As there is no initial velocity of the cart at C, the only force that the body gets is gravitational force which increases the speed of the body during the descent. Consequently, maximum speed can be calculated by calculating the time required to cover the entire path of the slide. Minimum the time be, more will be the velocity of descent. In order to find that, the shape of the slide is very essential to determine.

The initial and final positions of roller coaster ride is fixed. From Figure 1, we know that the initial position is C and the let the final position be D. In the Figure 2, we have considered three different types of pathways which can be followed to cover up the distance but the type of pathway which will offer maximum velocity is required to be found.

Thus, the shape of the pathway that should be followed to have minimum time period required to cover the distance between the two points provided no additional force is applied to the body except the gravitational force should be analyzed in this IA.

In the proceedings of this IA, several concepts of Physics and Mathematics are extensively necessary. Variation calculus and differential calculus will be required to solve the derivations of this IA. Most importantly, one special form of differential equation called Euler Lagrange Equation is essential for this IA. It is a second order partial differential equation whose solution determines the solution for the given function being stationary. According to the Euler Lagrange’s equation, if ** I **is defined as:

\(I = \displaystyle\int f(x,y,y ̇)dx \)

Since, \(\frac{∂y}{∂x}=y ̇\)

Then the function *I *has a stationary value if the following equation is satisfied:

\(\frac{∂f}{∂y}-\frac{d\big(\frac{∂f}{∂y}\big)}{dx}=0\)

There is one more special case in Euler Lagrange’s equation. If and only if, above mentioned function *f*(*x,y,y*) is not explicitly dependent on * x*, then, the equation can be modified as the following:

*f - \(y ̇\frac{∂f}{∂y ̇}=c\)*

where *c *is a constant. This special function is called Beltrami’s Identity.

Pythagorean equation is also important in this IA, which is denoted as:

a^{2} + b^{2} = c^{2}

Where, a, b, and c are three sides of a right-angled triangle.

Additionally, multiple and submultiple angles are also necessary. The formula that will be required in this IA is mentioned below:

2 \(\frac{\theta}{2}=1-cos\ \theta\)

The parametric equation of a cycloid is necessary of this IA. Cycloid is a curve which is traced by a point on a circle when the circle moves along a straight line. The parametric equation of cycloid are as follows:

*x = θ - sin sin θ*

*y = 1 - cos cos θ*

From Figure 1, we have seen that, the height at position A is 10 m and the height at position C is 40 m. The other parametric values at all three positions (A, B and C) are as follows where * h *denotes height,

**At Position A:**

*m _{A} = m kg*(

*v _{A }= v m/s*(

*h _{A }= *10

**At Position B:**

*m _{B} = m kg*

*h _{B }= 0 m*

**At Position C:**

*m _{C} = m kg*

*h _{C }= 40 m*

*v _{C }= 0 m/s*

From the law of conservation of energy, we can say that,

*Total Energy _{A} = Total Energy_{B}*

*=> PE _{A}+KE_{A}=PE_{B}+KE_{B}*

\(=>m_Agh_A+\frac{1}{2}m_Av^2_A=m_Bgh_B+\frac{1}{2}m_Bv^2_B\)

=> mg × 10 + \(\frac{1}{2}mv^2=mg\times0+\frac{1}{2}mv^2_B\)

=> g × 10 + \(\frac{1}{2}\)* v ^{2} = \(\frac{1}{2}v^2_B\)*

=>20g + *v ^{2}* = \(v^2_B\)

Similarly, we can write,

*Total Energy _{B }= Total Energy_{C}*

=> PE_{B} + KE_{B} = PE_{C} + KE_{C}

\(=>m_Bgh_B+\frac{1}{2}m_Bv^2_B=m_cgh_c+\frac{1}{2}m_cv^2_c\)

\(=>mg\times0+\frac{1}{2}m(20g+v^2)=mg\times40+\frac{1}{2}m\times0\)

\(=>\frac{1}{2}m(20g+v^2)=mg\times40\)

\(=>\frac{1}{2}m(20g+v^2)=g\times40\)

*=> 20g + v ^{2 }= 80g*

*=> v ^{2 }*= 80g - 20g

*=> v ^{2}* = 60g

=> *v ^{2 }*= 60 × 9.81

=> *v ^{2 }*= 588.6

=> *v *= \(\sqrt{588.6}\)

=> *v *= 24.26 m/s

So, we can say that the velocity of the roller coaster at Position A is 24.26 m/s which is applicable for all masses of the cart. As the velocity equation does not have any mass component, thus we can say that, velocity of the cart at position A does not depend upon the mass of the individuals riding the cart or the mass of the cart itself.

Let us consider the highest point of the ride in the co-ordinate system be (* 0,h*) and the end point of the ride be (

*T* = \(\displaystyle\int\limits_{c}^{d}{dt\ldots\ldots\ldots\ldots.(1)}\)

*dt *= \(\frac{ds}{v(x,y)}..............(2)\)

Here, * dt *is a small instant of time,

From the Pythagorean Equation, we can write,

*ds* = \(\sqrt{dx^2+dy^2}\)

=> *ds *= \(\sqrt{dx^2(1+\frac{dy^2}{dx^2}})\)

=> *ds *= *dx **\(\sqrt{1+\frac{dy^2}{dx^2}}............(3)\)*

This is because, * ds *is a very small component of distance traversed from the initial position and

Now, from the law of conservation of energy, we can write,

*Potential Energy _{at} _{C }= Kinetic Energy_{D}+Potential Energy_{D}*

*∴ mgh = \(\frac{1}{2}\)mv ^{2 }+ mgy*

*=> mgh - mgy = \(\frac{1}{2}\)* *mv ^{2}*

=>* v* = \(\sqrt{2g(h-y)}...........(4)\)

From equation 1, 2, 3, and 4, we can write,

*T = \(\displaystyle\int\limits_{A}^{B}\)dt*

*=> T = \(\displaystyle\int\limits_{A}^{B}\frac{\sqrt{1+\frac{{\rm dy}^2}{{\rm dx}^2}dx}}{\sqrt{2g(h-y)}}\)*

=> *T* = \(\displaystyle\int\limits_{A}^{B}\frac{\sqrt{1+\dot{y^2}dx}}{\sqrt{2g(h-y)}}\)

For minimum value of T, the function of T must be stationary but the function of which is inside the integral is not explicitly dependent on * x*. Thus, we can use Beltrami’s Identity for solve the function.

\(\frac{\sqrt{1+\dot{y^2}}}{\sqrt{2g(h-y)}}-\ \begin{matrix}.\\y\\\end{matrix}\frac{\partial\frac{\sqrt{{1+y}^{-2}}}{\sqrt{2g(h-y)}}}{\partial y}=c\)

\(\therefore\frac{\partial\frac{\sqrt{1+\dot{y^2}}}{\sqrt{2g(h-y)}}}{\partial\dot{y}}=\frac{\dot{y}}{\sqrt{1+\dot{y^2,}}\sqrt{2g(h-y)}}\)

Therefore, putting the above value in the equation we get,

\(\frac{\sqrt{1+\dot{y^2}}}{\sqrt{2g(h-y)}}-\begin{matrix}.\\\ y\\\end{matrix}\frac{\frac{\dot{y}}{\sqrt{1+\dot{y^2}}\sqrt{2g(h-y)}}}{\partial\dot{y}}=c\)

\(=>\frac{\sqrt{1+\dot{y^2}}}{\sqrt{2g(h-y)}}-\frac{\dot{y^2}}{\sqrt{1+\dot{y^2}}\sqrt{2g(h-y)}}=c\)

\(=>\sqrt{1+\dot{y^2}}\sqrt{2g(h\ -\ y)}[\frac{\sqrt{1+\dot{y^2}}}{2g(h\ -\ y)}\ -\ \)

\(\frac{\dot{y^2}}{\sqrt{1+\dot{y^2}}\sqrt{2g(h\ -\ y)}}]\)

\(=c\sqrt{1+\dot{y^2}}\sqrt{2g(h - y)} \)

\(=>1+\dot{y^2}\ -\ \dot{y^2}=c\sqrt{1+\dot{y^2}}\sqrt{2g(h\ -\ y)}\)

\(=>1=c\sqrt{1+\dot{y^2}}\sqrt{2g(h\ -\ y)}\)

Squaring both sides, we get,

\(2g(h-y)\bigg(1+\dot{y^2}\bigg)c=1\)

\(=>(h-y)\bigg(1+\dot{y^2}\bigg)=\)\(\frac{1}{2gc}\)

\(=>(h-y)\bigg(1+\dot{y^2}\bigg)=c_1\)

\(where,\ \ c_1=\)\(\frac{1}{2gc}\).

\(=>h+h\dot{y^2}-y-y\dot{y^2}=c_1\)

\(=>\dot{y^2}=\frac{c_1-(h - y)}{h - y}\)

\(=>\dot{y}=\)\(\sqrt{\frac{c_1-(h-y)}{h-y}}\)

\(=>\frac{∂y}{∂x}=\sqrt{\frac{c_1-(h-y)}{h-y}}\)

\(=>∂x=\)\(\sqrt{\frac{h-y}{c_1-(h-y)}}∂y…………(5)\)

\(Let,\ \ y=h\ -\ c_1\frac{\theta}{2}\)

\(∴\frac{dy}{d\theta}=-c_1sin\ sin\frac{\theta}{2}cos\ cos\frac{\theta}{2}\)

\(=>dy=-c_1\ sin \ sin\frac{\theta}{2}cos\ cos\frac{\theta}{2}d\theta\)

From equation (5), we can write,

\(∂x=\sqrt{\frac{h-h+ c_1\frac{\theta}{2}}{c_1-(h-h+ c_1\frac{\theta}{2}}}[-c_1sin\ sin\frac{\theta}{2}cos\ cos\frac{\theta}{2}d\theta]\)

\(=>∂x=\sqrt{\frac{\frac{\theta}{2}}{1-\frac{\theta}{2}}}[-c_1sin\ sin\frac{\theta}{2}cos\ cos\frac{\theta}{2}d\theta]\)

\(=>∂x=\sqrt{\frac{\frac{\theta}{2}}{\frac{\theta}{2}}}[-c_1sin\ sin\frac{\theta}{2}cos\ cos\frac{\theta}{2}d\theta]\)

\(=>∂x=\frac{sin\ sin\frac{\theta}{2}}{cos\ cos\frac{\theta}{2}}[-c_1sin\ sin\frac{\theta}{2}cos\ cos\frac{\theta}{2}d\theta]\)

\(=>∂x=\frac{\theta}{2}d\theta\)

Integrating both sides,

\(\displaystyle\int ∂x=\displaystyle\int\frac{θ}{2} dθ\)

\(=>x=- c_1\displaystyle\int \frac{1-cos \ cos \theta}{2}d\theta\)

\(=>x=k_1(θ-sin\ sin\ \theta)+k_2…………(6)\)

\(since,\ k_1=- \frac{c_1}{2}\)

Putting the value of ** c_{1} **in the equation of

\(y=h+ 2k_1\frac{\theta}{2}\)

\(=>y=h+ 2k_1\frac{1-cos \ cos\ \theta}{2}\)

\(=>y=h+ k_1(1-cos\ cos\ θ) …………(7)\)

\(=>y=40+ k_1(1-cos\ cos\ θ)\)

Equation 6 and 7 are the solution of the nature of the pathway which is actually the parametric equation of a cycloid. So, we can say that, for Time to be minimum, or speed to be maximum, the shape of the pathway must be cycloid in nature.

In this IA, two cases on a roller coaster ride in an amusement park have been extensively studied. Based upon the criterion and other available data, it has been shown that the velocity of the roller coaster cart at position A should be equal to 24.26 m/sec in order to reach and stop at position B without application of any force. Furthermore, it should be noted that the mass of the roller coaster or the mass of individuals who are sitting in the roller coaster cart is not affecting the required velocity of the cart. So, it can be concluded that the velocity of the cart of those two positions A and B will be the same for every round of roller coaster ride.

In addition to this, the model of the track of roller coaster ride after position B to offer maximum thrill to the individuals who are riding the roller coaster has also been developed. This model is based upon a single assumption that individuals will experience maximum thrill if the track is covered at minimum time. Though it simply suggests that minimum distance offers the shortest time to travel, however, it is not the case. This particular problem is mathematically known as Brachistochrone problem. The solution of this problem has given a trajectory or an equation of the path (or roller coaster track) that should be constructed to offer the individuals the maximum thrill or minimum time to travel from position B to the base or ground is a cycloid. The equation of the trajectory is given below:

\(x=k_1(θ-sin \ sin\ θ) +k_2\)

\(where,\ k_1=- \frac{c_1}{2}\)

\(and, c_1=\frac{1}{2gc}\)

and, c is a constant of Belltrami's Identity

\(y=40+ k_1(1-cos\ cos\ θ) \)

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