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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 • m^{2}___________s^{2}1J =
Equation
v^{2} = 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:
PE_{H} + KE_{H} = PE_{A} + KE_{A} = PE_{B} + KE_{B} = PE_{C} +KE_{C}
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/s^{2} • 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 mv^{2} = ½ * 100 kg * v^{2} (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 mv^{2}
19,600 J = ½ * 100 * v^{2} = 50 v^{2}
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 mv^{2}
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) v^{2} = 2g(H-y)
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