Roller Coaster Design
(AP) Physics
Abridged Edition
An Abridged Educational Guide
To Roller Coaster Design and Analysis
This resource booklet goes with an final AP physics project.
by Tony Wayne
INTRODUCTION
This booklet will discuss some of the principles involved in the design of a roller coaster. It is intended for the middle or high school teacher. Physics students may find the information helpful as well. Many of the concepts can be applied to topics other than roller coasters. Some sections will use the “Roller Coaster Simulator,” RCS. (See page 78 for instructions on its construction.) The included activities are hands on cookbook type. Each section includes background topics that should have been taught previously.
Calculation Algorithm to calculate a change in velocity associated with a change in height
Step 1 
Identify two locations of interest. One with both a speed and a height and the other location with either speed or height. 
Step 2 
Write an equation setting the total energy at one location equal to the total energy at the other location. 
Step 3 
Solve for the unknown variable. 
Example 1
What is the velocity at the bottom of the first hill?
Solution:
ET_{(TOP)}_{ }= ET_{(BOTTOM)}
KE + PE = KE +PE
(^{1}/2)mv^{2} + mgh = (^{1}/2)mv^{2} + mgh
(^{1}/2)v^{2} + gh = (^{1}/2)v^{2} + gh The masses cancel out because it is the same coaster at the top and bottom.
(^{1}/2)v^{2} + gh = (^{1}/2)v^{2} + gh Substitute the numbers at each location
(^{1}/2)(8.8)^{2} + 9.8(95) = (^{1}/2)v^{2} + 9.8(0) The height at the bottom is zero because it is the lowest point when comparing to the starting height.
38.72 + 931 = (^{1}/2)v^{2
} 969.72 = (^{1}/2)v^{2}
1939.44 = v^{2}
v = 44.04 ^{m}/s at the bottom the the 1st hill.
Example 2
What is the velocity at the top of the second hill?
Solution:
ET_{(TOP OF 1}st_{ HILL)}_{ }= ET_{(TOP OF 2}nd_{ HILL)}
KE + PE = KE +PE
(^{1}/2)mv^{2} + mgh = (^{1}/2)mv^{2} + mgh
(^{1}/2)v^{2} + gh = (^{1}/2)v^{2} + gh The masses cancel out because it is the same coaster at the top and bottom.
(^{1}/2)v^{2} + gh = (^{1}/2)v^{2} + gh Substitute the numbers at each location
(^{1}/2)(8.8)^{2} + 9.8(95) = (^{1}/2)v^{2} + 9.8(65) Notice all the numbers on the left side come from the top of the 1st hill while all the numbers on the right side come from the top of the 2nd hill.
38.72 + 931 = (^{1}/2)v^{2} + 637^{
} 332.72 = (^{1}/2)v^{2}
665.44 = v^{2}
v = 25.80 ^{m}/s at the top the the 2nd hill.
This technique can be used to calculate the velocity anywhere along the coaster.
by Tony Wayne
