I am now going to be focusing on the aspect of acceleration, and explaining it in a lot more depth.
The ride can be split into two parts. Freefall and Deceleration. To find out the deceleration once the breaks have been applied, other factors need to be evaluated first.
In order to calculate the deceleration of the ride, we have to calculate other factors such as velocity. Velocity can be obtained through the following calculation:
V2 = u2 + 2as v=final velocity
u = initial velocity
a = acceleration
s= displacement
Acceleration involves calculating a final velocity and an initial velocity. I am not going to be looking at the point when the ride was lifted to the top. I am jus going to be focusing on the vertical drop of 100ft. I am going to try and calculate the deceleration after the drop, when the ride approaches the braking point.
Below are my calculations.
Initial velocity = 0 m s-1
Displacement= 30m – 8m
= 22m
Acceleration = 9.81m s-2
v2 = u2 + 2as
= (0 m s-1)2 + (2×9.81 m s-2×22m)
= 431.64 m s-1
v = √431.64 m s-1
v = 20.8 m s-1
Now that we have both the initial velocity and the final velocity, we can calculate accurate readings for the time. We are going to rearrange the equation shown below in order to obtain our result.
v = u + at v = final velocity
v – u = t u = initial velocity
a a = acceleration
20.8 m s-1 – 0 m s-1 = t
9.81 m s-2
2.1s = t
Now I have fairly accurate readings, I am going to calculate the deceleration of the ride. This is after the vertical drop, when it has approached the braking point.
v2 = u2 + 2as v = 0 m s-1 a=?
u = 20.8 m s-1
s = 8m
Rearranged formula to make ‘a’ the subject of the equation.
a = v2 - u2
2×8m
= 0 m s-1 – 432.64 m s-1
16m
= -27.04 m s-2
This value tells me the rate of deceleration of the ride after it has reached the braking point.
Looking at my calculations, I can see that the acceleration of the ride is –27.04ms-2
In order to gain a clearer insight into the concept of acceleration on the ride, we must study Newton’s Laws of motion. They are needed for dynamics and bodies in motion. They are described below.
- Law of Inertia: - If there is no resultant force acting upon a body, it will remain at rest.
OR
If a body is in motion, it will continue to move at uniform velocity
until acted upon by an external force.
- Second Law: - If a body is acted upon by an external force, it accelerates. The size of
the acceleration is proportional to the resultant force.
- Third Law: - For every action that takes place, there is an equal and opposite reaction.
Newton’s law of inertia is irrelevant in this case as an external force is present. Newton’s second Law of motion applies to this case.
Newton’s second Law of Motion also has a formula which describes the law.
FORCE = MASS × ACCELERATION
If we estimate the mass, we should be able to calculate the force being exerted on the passengers.
Firstly we must estimate the mass of the passengers and the actual ring which is lifted to the top of the tower.
Small child = 30Kg
Female teenager = 50Kg
Male Teenager = 70Kg
Large Adult =90Kg
Small Car = 900kG
Large Car = 1500Kg
As I mentioned earlier, the ride can only seat 12 people. I have chosen to use:
3 female teenagers, 3 male teenagers, 4 adults, and 2 small children.
Mass = 50Kg + 50Kg + 50Kg + 70Kg + 70Kg + 70Kg + 90Kg + 90Kg + 90Kg + 30Kg +30Kg
= 690Kg
I estimate the mass of the tower to be 1000Kg.
Total mass = 1000Kg + 690Kg
=1690Kg
FORCE = 1690Kg × 9.81m s –2
= 16578.9N
This is approximately 16.6kN.
The other aspect which I will be looking at briefly is the changes in energy. When the ride has reached the top of the tower, it has a high level of gravitational potential energy as it is not moving. Kinetic energy is not present as no movement is taking place. When the vertical drop occurs, the level of gravitational potential energy is decreasing, and kinetic energy is increasing. We can clearly see this by the movement that it is taking place.
I am now going to calculate the change in gravitational potential energy using the following formula:
I am going to need to use mass in this formula, so I am going to use the figure which I had previously estimated.
Change in GPE= mgh m = mass (Kg)
= 1690Kg × 9.81m s-2 × 30m g = acceleration due to gravity
= 497367J (497Kj) h = height above the ground
Limitations of my investigation
After collecting my required data for this coursework, I had to take into account that the data was not 100% accurate. The acceleration, we assumed was 9.81. However we know that this is only true when ignoring air resistance. The acceleration would, of course be slightly less than 9.81. This is the main limitation.
Also, the masses I used were merely an estimate as the masses of people who got on the ride varied. However, this may not be a great problem as I am looking at the aspect of acceleration. According to Newton’s theories, all objects accelerate under gravity at the same rate, regardless of their mass.
Strengths of my investigation
The strengths of this coursework were that the distances were accurate as I asked the workers on the ride and I visited the official website to obtain such information.
Use of sources
Throughout this coursework, I used a number of sources in order to help me further my research. They are listed below.
- The official Thorpe Park website
-
(“Detonator at Thorpe Park” was typed into the search engine)
- “Advanced Physics” textbook by Salters Horners
- Letts AS Success Physics Revision Guide
- My Physics teacher (Michael Ishwerwood)
