WHAT IS NASA PHYSICS?
MODULES
Forces and Motion
Conservation of Momentum & Energy
Temperature and Heat
Fluids
Optics
Electromagnetic Spectrum
Modern Physics
Anticipation Guide 7
Intro to Modern Physics
Blackbody Radiation
The Ultraviolet Catastrophe
The Photoelectric Effect
Bohr's Atom
Spectra
Radioactive Decay
Special Relativity (SR)
Simultaneity
Distance and Time
General Relativity
May the Forces be with You
Modern Physics Notebook
Assessment Problems 7
Useful Things
SITE MAP
Conservation of Momentum & Energy
Conservation of Momentum & Energy
Anticipation Guide
Energy & Conservation Laws
What is Energy?
Kinetic Energy
Potential Energy
Energy Conservation
Roller Coaster Physics
Centripetal Force
Momentum
Angular Momentum
Gravity Assist
Energy Transformation
Notebook
Assessment Problems
A Joule of energy = energy expended in applying a force of one newton through a distance of one meter (1 newton meter or N·m) =
kg • m2___________s21J =
Equation
v2 = 2g(H-y)
2.6
Roller Coaster Physics
We can say that the sum of potential energy plus kinetic energy remains constant at any point in time, again, for a closed system. Let’s take a look at how this applies in one of the most personally exhilarating applications of physics.
Left: Dragon Khan roller-coaster in Universal Port Aventura park in Spain, Wikipedia
A 100 kg roller coaster that does a loop-d-loop must be designed such that the coaster stays on the tracks at all times especially when it is upside down. Using equation 2.4 above, we can write:
PEH + KEH = PEA + KEA = PEB + KEB = PEC +KEC
That is, the sum of the potential and kinetic energy will be the same for all points H, A, B and C on the roller coaster track assuming the ideal case where no energy is lost due to friction. PE can be converted into KE and back, but the sum will always be a constant.
Initially, at point H where y = 20 m above the ground, the roller coaster has no kinetic energy but lots of potential energy. We know it has no kinetic energy because its velocity = 0! Its potential energy can be calculated from Equation 3.3 as:
PE = mgh = 100 kg • 9.8 m/s2 • 20 m = 19,600 J
At the bottom of the roller coaster track (point A), the coaster has no potential energy (h = 0) and lots of kinetic energy. In fact, it has a maximum of kinetic energy and so a maximum velocity. Its kinetic energy can be described as:
KE = ½ mv2 = ½ * 100 kg * v2 (m/s)2
But what is the coaster’s velocity?
Since we know that PE + KE = constant, we know that the KE at the bottom is just equal to the potential energy of the coaster at the top (PE = 19,600 J), and the maximum value of the PE equals the maximum value of KE (assuming there is no friction).
(Left: Rollercoaster © Zechal - Fotolia.com)
PE = 19,600 J = KE = ½ mv2
19,600 J = ½ * 100 * v2 = 50 v2
and so v = (19,600 J/50)½ = 19.8 m/s
For the roller coaster at any point on the track:
ETotal = PE + KE = Potential Energy at maximum height, H.
mgH = mgy + ½ mv2
where y = the height of the roller coaster at that point.
Collecting terms of the last equation gives us a general equation for the velocity of a roller coaster at any point – not very important in the scheme of things, but something you derived and now understand:
velocity squared = 2 x gravity x(maximum height - height at any given point) v2 = 2g(H-y)
© 2013 by Wheeling Jesuit University/Center for Educational Technologies®. 316 Washington Ave., Wheeling, WV 26003-6243. All rights reserved. Privacy Policy and Terms of Use.