Physics behind roller coasters

On Thursday Oct.28, students were introduced to the roller coaster project. I think that most of the students in the class, had experienced the breathtaking feeling when riding roller coasters. It is an indescribable emotion when it seems that you can fly without wings.  In order to know how to build it, is necessary to understand how the roller coaster works. Here is where physics takes the lead.

First of all, for those who don't know yet, it is important to mention that a roller coaster doesn't depend on engines in order to move. The two basic principles that are important are potential and kinetic energy. Potential energy is when the energy is stored in a object due to its position. In our case, as the roller coaster's starting point is higher, a greater force can be built in order to push the object down the hill, which will increase the speed at which the object is going down.  As the object goes down the hill the potential energy changes into kinetic energy. Kinetic energy is the energy produced by the motion. As it goes up the hill, it uses kinetic energy that will again change into potential energy at the top of the hill. For the whole motion, the rollercoaster continuously converts the energies in order to go up and down the hills. It is also important to mention about the gravitational forces that act on the roller coaster and create the thrilling feeling in our stomachs.   

Adding Vectors by Components

In order to get to our final result, we have to follow several easy steps in adding Vectors by Components:

1. Determine the positive and negative sides of the axes                  

( it will help in finding the direction of the vectors)

Ex: 

 North and East  = positive axes                                                

South and West = negative axes 

2. Break all the vectors into components: X (horizontal components) and Y(vertical components)

Refer to the diagram given for each vector to correctly use the sign, (+ or -), for each component.

3. Use trigonometry to find X and Y components

 

4. Add all the x- components , to get the total of the x components

5. Add all the y- components , to get the total of the y components

6. Use the "Head to Tail " method to combine the X and Y components.

7. Use The Pythagorean Theory to find the final Resultant

How to derive Equation #4 from a velocity graph

As we did with Equation #3 , it is possible to derive Equation # 4 from a velocity graph. It will be slightly different.

Equation #4 states: d = v2Δt - ½aΔt²

In order to get to our equation we need to follow several steps. First of all we need to find the area of the rectangle OCBD. We can find it using the formula for the area of a triangle which is: (A = lw).While on the graph it will be like this: (d = v2Δt) because we have to multiply v2 with (t2-t1)=Δt

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Secondly we need to find the area of the ACB triangle. The area of a triangle is [A = bh/2]. On our graph is will be rearranged as followed: ½aΔt² since aΔt is isolated from Equation 1, when we multiply  ½Δt with (aΔt) we will get ½aΔt².

 

Lastly in order to get the forth equation we need to subtract the area of the triangle ½aΔt² from the area of the big rectangle v2Δt. 

Our equation will end up looking like this:

d = v2Δt - ½aΔt²

How to derive Equation #3 From a velocity graph

Graph #3 states  :   d= v1Δt + ½aΔt²

While looking the first time at it , it will be difficult to notice any pattern. Actually there is.

I think that equation 3 can be derived from a velocity graph like this: 

First of all as you can see the graph is divided into 2 sections yellow and red.

The same is done with the equation #3.

The yellow part will represent the area o a triangle which is  [A = bh/2],  but on the graph the equation of a triangle would look like this [d = ½(v2 - v1)Δt].  We can simplify this formula because from what we learned before we know that (v2 - v1) = [aΔt] . We end up having  ½aΔt², when we multiply  ½Δt with (aΔt), which is the second part of our equation. 

While the red part is a square. We can find its area based on the following equation: (A = lw).

On the graph this formula can be rearranged like this: (d = v1Δt),because we have to multiply v1 with (t2-t1)

We can conclude that Equation #3 is a combination of the area of a triangle and a rectangle.

When adding the 2 sides together we will get Equation #3 as a result.

Describing the graphs

Graph1

The graph above is a representation of a distance /time graph:

·       Object is at rest at 1m for 1 second, in the positive quadrant, away from the origin

·       Starting from 1 second until 3 seconds, the object is running at a constant speed for 1.5 m [E]

·       During 3 seconds the object is at rest , at 2.7m away from the origin

·       Starting at 6 seconds until 7.5 seconds, the object is walking at a constant speed for 1m [W], toward the origin

·       For 2.5 seconds the object is at rest, at 1.8 m away from the origin

Using the distance/time graph, we can predict the motion on the velocity and acceleration time graphs.

Velocity/time graph:

The movement of the object will start at 0m/s at 1 s. For the next 2 second there will be a horizontal line. Starting at 3 s until 6 seconds the line on the velocity graph will be horizontal at 0m/s. For the next 1.5s there will be a horizontal line in the negative quadrant, because the object is going in the negative direction [w].The graph will end with a horizontal line at 0m/s.

Acceleration/time graph:

The following graph will have only one line that will be at 0 m/s².

Graph2

The graph above is another representation of a distance /time graph, but a different motion:

·       The object started his movement at 3 second away from the origin

·       For the first 2.5 seconds , the object is walking at a constant speed for 1.5 m[W] toward the origin

·       The object is at rest for 1 s at 1.5m away from the origin, in the positive quadrant

·       Starting at 4 second until 5 seconds the object is running at a constant speed for 1m in the negative direction, toward the origin

·       During 2 second the object is at rest at 0.5m away from the origin

·       The last motion that the object does is walking 2.2m for 3 second in the positive direction[E] away from the origin

Prediction of the velocity/time graph:

During the first 3 seconds there will be a horizontal line in the negative quadrant. For the following 2 s the graph will have a straight horizontal line at 0m/s. One more time there will be a horizontal line in the negative quadrant for 1s.  After the velocity graph will have a horizontal line at 0 m/s. The last line that will be represented on this graph will be a horizontal line in the positive quadrant for 3s.

The acceleration graph will have a straight horizontal line at 0m/ s².

Graph3

Graph NR.3, represents a velocity/time graph:

·       For the first 2 seconds the object is at rest

·       During 0.2 s the object is increasing its speed at 1.5 m/s[E] in a positive direction, away from the origin

·       The object is walking at a constant speed for 2.9s at 0.5m/s [E] away from the origin

·       The object is speeding up for 0.1 s at 0.5m/s in the negative direction [W] toward the origin

·       During 1.9 s the object is at rest at 0m/s

·       The object is speeding up at 0.5m/s for 0.1 s [W]

·       For 2.3 s at 0.5 m/s the object is walking at constant speed [W]

The acceleration/time graph of the above velocity graph will look as following:

For the first 2 seconds there is a horizontal line at 0m/ s². During the following 0.1 seconds, there will be a horizontal line in the positive quadrant. After that the line will return at 0m/ s² for a time period of 3 s being horizontal. Again for 0.1 s there will be a horizontal line but now in the negative quadrant . The horizontal line will return back at 0m/ s² during 3.9 s. For another 0.1 s there will be a horizontal line in the negative quadrant and a return back at 0m/ s².

Graph4

This is another graph that shows the movement of an object in a velocity/time graph:

·       At 0.5m/s the object is speeding up for 4 second [E] in the positive direction away from the origin

·       For the next 2 s the object has a constant speed at 0.5m/s

·       During 0.1 s the object is speeding up at 0.m/s [W] passing the origin

·       For 3.9 s at 0.4 m/s the object is walking at the same speed while being in the negative quadrant

·       The object is speeding up for 0.1s at 0.4m/s toward the origin [E]

·       Object at rest for 0.9s

The above graph, can be represented in a acceleration/time graph as following:

There is a positive acceleration , which means that the line will be horizontal in the positive quadrant for 4seconds. For the following 2 seconds the line will return back at 0m/ s². For 0.1 s there will be a negative horizontal line and then it will return at 0m/ s². During another 0.1 seconds the line will be in the positive quadrant because of a positive acceleration. The graph will end being at 0m/ s².

Graph5

Graph 5, position vs. time graph:

·       The object starts it's movement at 0.8m away from the origin

·       For 3.5 s the object is walking 0.8m [E] away from the origin

·       For 3 s the object is at rest at 1.8 m away from the origin

·       During 1.2 m the object runs for 2.7 s [E] in the positive direction away from the origin

Velocity/time graph:

The graph starts with a horizontal line, in the positive quadrant for the first 3.5 seconds. For the next 3 seconds the horizontal line will return at 0m/s. The last line of this graph will be represented with a horizontal line in the positive quadrant for 3.5 seconds.

Acceleration /time graph:

Again, the acceleration/time graph will for the above graphs, will have a straight horizontal line at 0m/ s².

Graph6

This graph is a velocity/time graph:

·       Walk at the same speed at 0.4m/s during 3.2 s [E] in the positive quadrant

·       The object is speeding up, stops and then speeds up again during 0.1 second

·       For 3.3 s the object is walking at the same speed at 0.3 m/s in the negative quadrant

·       The objects speeds up at 0.3m/s during 0.2 s toward the origin

·       The object is at rest for 3.6 second

The acceleration /time graph of the above diagram will look as following:

During the first 3 seconds the line will remain at 0m/ s². After , there will be a horizontal line in the negative quadrant for 0.2 seconds. Again, the line will return at 0m/ s² for 3.5 seconds. The line will move in the positive quadrant for 0.2seconds and return back again at 0m/ s² for the next 3.5 seconds.