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

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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

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

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.

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