• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# Theme Park visit report: The physics of roller coasters.

Extracts from this document...

Introduction

Theme Park Visit Report

On the 11th October 2007, we visited Drayton Manor theme park. We set out to learn more about the physics involved in theme parks. Although there are many aspects of physics present in a theme park I have concentrated my effort on just two aspects in this report. The two aspects I have chosen are power and efficiency.

Before I start calculating power and efficiency I have decided to explain the basic principles of a roller coaster as it helps when explaining the usefulness of knowing power and efficiency how they can be calculated and the equations to calculate them.

Physics of the roller coaster.

The purpose of a roller coasters initial ascent is to build up gravitational potential energy(fig.1). Once you pass over the crest of the initial ascent the built up gravitational potential energy of the cart is transferred to kinetic energy as you descend for the first time(fig.2).  A force of 9.

Middle

To work out energy of a roller coaster for the initial climb we can find the potential energy its gained.

Gravitational potential energy = Mass*Gravitational field strength

*Vertical height

GPE = M*G*H

So Power = M*G*H

Time

How to calculate efficiency

‘Energy conversion efficiency is the ratio between the useful output of an energy conversion machine and the input.’

We are able to find the efficiency of any part of the track although is complex, so to simplify the problem we shall calculated the efficiency of the entire track.

Efficiency = (Useful energy/Total energy)*100

Useful energy

E = (Ke/GPE)*100

Energy out is kinetic and energy in is GPE but GPE equals Ke so if we make energy in and out Ke then we can write the equation as:

E = (V/U)*100                        U= Velocity in

V= Velocity out

To work out V:

V= Δs/Δt

Calculating power and efficiency

Ride name: G-Force

Data known:        m        =        1350Kg

t        =        6.49s

g        =        9.81

U        =        20.25ms-¹

U Δ

Conclusion

>Δs/Δt  V=VΔs/VΔt  V= 8/1.74

V= 4.60ms-¹

Finding Ef

Ef=(V/U)*100  Ef= (4.60/20.25)*100

Ef= 22.72%

Ride name: Apocalypse

Data known:        m        =        500Kg

g        =        9.81

h        =        54m

t        =        15.12s

V Δs        =        51m

VΔt        =        2.72s

Data needed:        GPE=        ?

P=        ?

U        =        ?

V        =        ?

Ef=        ?

Finding GPE

GPE=mgh  GPE = 500*9.81*54

GPE= 264870j

Finding P

P=E/t  P=GPE/t  P= 264870/15.12

P= 17517.86w

Finding U

mgh=0.5mv²  v=(2gh)

v=(2*9.81*54)  v= 32.55ms-²

Finding V

V=Δs/Δt  V=VΔs/VΔt  V= 51/2.72

V= 18.75ms-¹

Finding Ef

Ef=(V/U)*100  Ef= (18.75/32.55)*100

Ef= 57.60%

Bibliography

Physics of the roller coaster:

Fig.1

Fig.2

Fig.3 - sourced form:

http://science.howstuffworks.com/

Gravitational potential energy:

Quote on gravitational potential energy- sourced from:

http://hyperphysics.phy-astr.gsu.edu/

Efficiency:

Quote on energy conversion efficiency - sourced from:

http://en.wikipedia.org/wiki/

Kinetic energy:

Quote on kinetic energy - sourced from:

http://dictionary.reference.com

This student written piece of work is one of many that can be found in our AS and A Level Mechanics & Radioactivity section.

## Found what you're looking for?

• Start learning 29% faster today
• 150,000+ documents available
• Just £6.99 a month

Not the one? Search for your essay title...
• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# Related AS and A Level Mechanics & Radioactivity essays

1.  ## AS OCR B Advancing Physics Coursework - Making Sense of Data

4 star(s)

the results which were established to have less error attached from the very first graph (velocity-distance fallen) when a constant acceleration could be seen. This, as shown by the equation of the line, gives a gradient and value for the acceleration due to gravity as 10.5m s-2.

2.  ## The Physics of Windsurfing

4 star(s)

Tape the shapes to the open ends of the "wing." The flat edge of the shapes should be on the bottom of the wing (see illustration). 10. Taking the 7.5-cm straw handles, one in each hand, draw the fishing line tight and position it so the line is perpendicular to the floor.

1. ## Measure the earth's gravitational field strength.

and m = mass (in kg) Normal Force, Fnorm The normal force is the support force exerted upon an object which is in contact with another stable object. For example, if a book is resting upon a surface, then the surface is exerting an upward force upon the book in order to support the weight of the book.

2. ## Objectives: To determine the center of gravity of a body of irregular shapes

The slope (m) = (3.08-0.35) / (0.73-0) = 3.7397 g = 4?2 / m = g = 4?2 / 3.7397 = 10.56 ms-2(cor. to 4sig fig) Let the slopes of the two good- fit lines be m1 and m2. m1= (0.75 - 3.08)

1. ## Multi-bladed Pumps. Does the number of propellor blades affect the efficiency of a ...

Precautions All apparatus Accident or fire Supervise the experiment at all times and clear away at the end of the session. Store all equipment safely and securely. Boiling water for shaping polypropene propellers Risk of scalding Take care with boiling water, paying attention at all times.

2. ## In this report I will start by exploring the history of the Computerised Tomography ...

Roberts P.D (1990) states, that as they are absorbed or attenuated at different levels, they are able to create a matrix of differing strength. In x-ray machines this matrix is registered on film, whereas in the case of CT the film is replaced by detectors which measure the strength of x-ray.

1. ## Explain how excessive exposure to radiation can cause harm.

Materials with high densities and high atomic numbers are the most effective shielding choice for protection from x and gamma rays. The energy of the photons is reduced by Compton and photoelectric interactions in the shielding material. Thus, substances such as lead, concrete, and steel are very practical shielding materials

2. ## The Physics of an Atomic Bomb

from product decay 10 TOTAL 200 +/- 6 All of the kinetic energy is released to the environment instantly, as are most of the instantaneous gamma rays. The unstable fission products release their decay energies at varying rates, some almost immediately. • Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to 