The Physics Behind Roller Coasters

The roller coaster is driven almost entirely by inertial, gravitational and centripetal forces.
The train is moved by gravity and momentum. When a coaster has reached the highest point along its course, normally a tall hill at the beginning of the circuit, gravity is what provides the force that controls the speed of the ride.  Inertia is the reluctance of a body ( for example, a coaster train) to change its direction of motion. A coaster train that is accelerating down a steep hill will resist the change in direction and head up the next hill.

But before the train can gain momentum and take advantage of gravitational forces, it needs an initial push to get up the hill.
A chain loop can be used to move the train up. This can consist of a gear at the bottom attached to a motor that moves the roller coaster up.
A catapult launch can build up a great amount of kinetic energy over a short time.

The potential energy the roller coaster builds going up the hill is released as kinetic energy, the energy of motion that takes the coaster down the hill.

The coaster tracksare important in channeling the force. They control the way the coaster cars fall. If the tracks slope down, gravity pulls the front of the car toward the ground. The coaster therefore accelerates. If the tracks tilt up gravity applies a downward force on the back of the coaster. The coaster decelerates.

An object in motion tends to stay in motion (Newton's first law of motion.) The roller coaster car will maintain a forward velocity even when it is moving up the track. This movement is opposite the force of gravity. When the coaster ascends one of the smaller hills that follows the initial lift hill, its kinetic energy changes back to potential energy. It is back at the top. The course of the track is constantly converting energy from kinetic to potential vice versa again.

In most roller coasters, the hills decrease in height as they progress along the track. Thsi happens because the total energy reservoir built up in the lift hill is gradually lost to friction.There is friction between the train and the track and between the train and the air. When the train reaches the end of the track the energy reservoir is almost completely empty.

Roller coasters have brake systems. Those brake systems however, are not situated on the coaster itself, but on the tracks. There are a series of clamps built into the track of the train. There is a central computer system which operates the hydraulic system and closes the clamps for the train to stop.


To sum it all up , when a coaster is at the highest point of its track, there is a high potential energy This can be referd to as energy of position. When the coaster accelerates down the hill the potential energy changes into kinetic energy (or energy of motion). Each time the coaster goes up another hill, the kinetic energy turns into potential energy again. The cycle continues. Ideally, the total amount of energy would remain the same. However some is lost to friction between the wheels and the rails. There is wind drag along the train, and also friction applied by the brakes. Due to this energy loss, each successive hill along a coaster track needs to be smaller than the previous hill to allow the train to continue along the course.


Safety Features
By the nature of the laws of physics a roller coaster wants to fly off the track. Any accidents are prevented by the wheel design and the track.
The weight of the train is supported by running wheels (the largest wheels.)
On the sides of the coaster train guide wheels are arranged to provide stability during tight turns and to keep the coaster train from rubbing the structure of the ride.
Upstop wheels are smaller wheels. They are on the underside of the rails that keep the train locked to the track and allow the train more airtime because they eliminate the danger of the train leaving the track.
A rollback can occur when a coaster fails to crest a hill. Instead, it begins to move backward. Coaster trains are equipped with anti-rollbacks paired with the chain dogs. A rollback can be caused from too much gravity.




check out these websites for more information
http://tlc.howstuffworks.com/family/roller-coaster3.htm
http://www.essortment.com/hobbies/rollercoasters_sdzq.htm

How To Add Vectors

For scalar quantities there is no question about the final result of adding two quantities. For vector quantities the value depends on the direction and angle between the two vectors. The resultant vector of the addition needs to be specified by a magnitude and direction. It is called total displacement.

The vector direction needs to be shown in relation to North and South.
If you know the angle from North or South to the direction, use trigonometry to solve for vertical and horizontal displacement.
Use sin and cos to find the horizontal and vertical distance.
Record the distance in a table with a positive sign if N or E and a negative sign if S or W.
Add up the total displacement horizontally and the total displacement for each vector. (This is similar to collecting like terms in math.)
Once you have the total horizontal and vertical displacement you can find the length of the distance and direction.
Use Pythagoras' theorem for length and the tan function for direction and angle compared to N or S.

If you know the length of a bunch of vectors and directions in relevance to North and South it is possible to find the total displacement horizontally and vertically. This can be done using cosine and sine function of an angle, and the hypotenuse. When the total horizontal and vertical displacement of all the vectors are found, they need to be added up. This will give us the two legs of the triangle. Knowing the two legs' length will make it possible to fine the hypotenuse using Phythagoras' theorem. The angle of the final displacement, in relevance to North and South can be found by using the tangent function.
It is important to split up vectors in horizontal and vertical displacement when adding the total of the vectors, because that way all of the directions are the same. It is possible to add them together. Otherwise the vectors would be going in different directions, and when added up would give a not accurate result. By splitting up vectors into horizontal and vertical it is also possible to assign a positive and negative direction.

Deriving Equation Four

Equation four can be derived from the velocity time graph by deriving the first two equations and then manipulating equation 2 so that it can be substituted into equation 1. From equation 2, the V1 value is isolated.

from equation 2
a = (V2 - V1) / Δt
a . Δt = V2 - V1

plug this into equation 1
d = 1/2 (V1 + V2) Δt
d = 1/2 (V2 - a . Δt + V2)Δt
d = 1/2 (2V2 - a . Δt) . Δt
d = V2 . Δt - 1/2 . a . Δt²
This equation is only applicable in problems involving constant acceleration, because it was derived from a graph involving constant acceleration.

Equation 4 can be proved from the velocity time graph.
The area from the graph to the x axis is equal to the total distance.
If velocity two is more than velocity one, then by calculating V2 . t we find the area of a rectangle that would give the distance traveled at a constant velocity of the faster speed.
However the speed was not always the faster speed. It  started of slow and accelerated from V1 to V2.
Therefore we need to subtract the area of a triangle that is not part of the graph. This area is equivalent to 1/2 of the rectangle formed between the difference of V2 and V1 multiplied by the time taken.

Equation Three

There are five equations that can be derived from a velocity time graph. This applies to motion that has a constant acceleration

The first two equations are
1. V2 = V1 + a .Δt
2. d = 1/2 (V2 + V1) Δt
The following three equations can be derived from the first two by rearranging the variables. This is done by isolating each of the variables in the first equation and plugging them into the second equation.
3. d = V1 . Δt  + 1/2 . a . Δt
4. d = V2 . t - 1/2 . a .Δt
5. V2²  = V1 + 2 . a . d

Equation One
The velocity at a certain point in time is equal to the initial velocity + (the acceleration in seconds squared times the change of time in seconds)
V2 = V1 + a . t
Find the slope of the graph.
The rise is V2 - V1
The run is the change in time

Equation Two
The area (of the trapezoid) calculated in this equation is the change in distance.
the base = time
heights = V2 and V1
d = 1/2 (V2 + V1) Δt
Now that we have equation 1 and equation 2 we can find equation 3.

Equation 1
V2 = V1 + a . Δt
V2 is ready for substitution into Equation 2.

d = 1/2 (V2 + V1) Δt
d = 1/2 ( V1 + a .Δt + V1)
d = 1/2 (2V1 + a .Δt)Δt
d = V1 .Δt + 1/2 . a . Δt²           < Equation 3

This can be proven from the velocity time graph.
The graph can be split into a reactangle and a triangle on top of it.
Distance is equal to the area of the reactangle from where the slope starts, and the triangle on top of it.
the area of a triangle can be found using the formula
d = 1/2 (V2- V1) . Δt
The area of the rectangle can be found using the formula
d = V1 . Δt
Putting the two together gives equation 3

We can find three equations from equation 1 and 2, just like we found equation 3 by
isolating V1 in equation 1
isolating V2 in equation 1
isolating Δt in equation 1
This is how we can get the five fundamental equations in kinematics!

Motion Graphs

This was our first distance over time graph. The distance started at 1 m. The distance increased over the first 3 seconds, it plateaued for 3 seconds, and then it decreased to 1.5 m where it stayed for about 3 seconds.
The velocity was positive and constant for the first part of the graph. From 3 - 6 seconds the velocity was 0, followed by a negative velocity for 1 s, and then 0 velocity.
The acceleration for this graph was 0.

The second graph is also a time vs. distance graph. The motion can be described as start at 3 m from the motion detector. Walk away for 2 s. Stay for a second. Walk away for a second. Present position=1.5 m. Stay at 1.5 m for 2 s. Then walk back to 3 m for 3 seconds.
The velocity is positive for the first 3 seconds. Then velocity becomes 0 because there is not movement. For 1 second again there is a positive velocity followed by a zero velocity. For the last 3 seconds there is a negative velocity because there is movement away from the origin and opposing the original direction of movement.
There is no acceleration in this graph.

Our third graph was time vs velocity. The initial speed was 0m/s for 2 seconds. During seconds 2 - 5 the speed becomes 0.5m/s. The speed goes back to being 0 m/s for another 2 seconds. The speed for the last 3 seconds reverses direction and becomes 0.5m/s (in the opposite direction of the initial speed.)
The distance vs. time graph version of the graph above would have a line on the x-axis for the first 2 seconds, followed by a positive slope for the next 3 seconds, that reaches from 0 to 1.5. There would be a line at 1.5 parallel to the x-axis for 3 seconds, and then a line that has a negative slope and reaches from 1.5m on second 7  to 0 m on second 10.
The acceleration graph would have a 0 slope, and can be graphed on the x-axis.

Graph number four shows velocity vs. time. The velocity starts off as 0 and accelerates in a positive direction for four seconds, until it reaches 0.5 m/s.The velocity stays at 0.5 m/s for 2 seconds. At 6 seconds, the velocity drops down to -0.4m/s and remains like that until the 9th second, when the movement stops, and velocity is 0.
The distance reached from the origin of the graph in the first 6 seconds is 2m. Then over the next 3 seconds the distance decreases by 1.5 m and the position is 0.5 m. The distance remains 0.5 m for the last second.
The acceleration for the first 4 seconds of the graph is 0.125m/s. It is positive. There is no acceleration for the rest of the graph.

The fifth graph is a position vs. time graph. The position starts at 0.8m and over the first 3 seconds the position increases steadily going up to 1.8 m. The distance remains 1.8 m for about 4 seconds. After 4 seconds, it increases steadily up to 2.5 m over a 1 s. time period.
The velocity for this graph starts off as positive at about 0.3 m/s for about 3 seconds. Then the velocity remains 0 for four seconds. This means there is no movement. The velocity becomes 0.5m/s for the last second.
The acceleration for the whole graph is 0. There is no increase or decrease in speed.

The graph above shows a velocity vs. time relationship. At first the velocity is less than 0.5 m/s. This lasts for about 3 seconds. Then the velocity becomes a little over -0.5m/s. The velocity stays at this rate for about 4 seconds. Then at 7 seconds, the velocity becomes 0m/s.

First there is a movement of 1.5 m in 3 seconds. This is followed by another movement of 2 m over 4 seconds, in a contrary direction. After the initial 7 seconds, there is a plateau in distance and no movement for 3 seconds.

There is no acceleration in the above graph at an part.