Two to three 50-minute periods.

LESSON RATIONALE

Today's modern amusement park and all the rides found there are the results of problem solving by engineers.  Not only must rides be exciting, but they must also fit space and site requirements, as well as meet high safety standards.  Working within those constraints, engineers have used math and science to create new rides that are faster, go higher, and are more thrilling than ever before.

LESSON DESCRIPTION

Through study of a basic roller coaster problem, students will see how the passengers stay in the car and the car stays on the track when both car and passengers are upside down.  The mathematics of the problem will prove that the kinetic energy, speed, and potential energy derived from the long, steep ascent and descent of the car during the first part of the ride all work toward keeping the car in motion and on the up side of the track.  In addition, students will learn why riders don't feel a sensation of falling when the roller coaster car is upside down.

MATERIALS

1. Classroom set of copies of the student reading, and worksheet.
2. Demonstration materials:
   • Hot Wheels or Lego set
   • Bucket of water on a rope or yo-yo or roller coaster video.  A roller coaster video can be obtained by contacting Arrow Dynamics, Inc., PO Box 1386, Clearfield, UT, 801-825-1611.

CONCEPTS

1. Kinetic energy
2. Potential energy
3. Gravity
4. Centrifugal force
5. Centripetal force

LEARNING OBJECTIVES

Students will:

1. See that mathematical equations can relate physical concepts to each other.
2. Understand the relationships between potential energy, kinetic energy, centrifugal force, gravity, and velocity.
3. Determine the first lift of a roller coaster, given the radius of a loop of the roller coaster.
4. Apply mathematical skills to work out real problems by solving for variables within equations and by substituting values for variables within equations.
5. To solve the math equations in this example, students will need to understand equations, constants, variables, unknowns, squares, square roots, and cross multiplication.

SUGGESTED PROCEDURES

1. Introduce the lesson by asking students if they have ever ridden on a roller coaster, the ride in the amusement park that goes in loops that cause them to ride upside down.  How do they feel when they are upside down?  (Students should not be aware of being upside down, other than the fact that they see they are head down.   They should not feel gravity pulling them toward the earth.)
   
   As part of the introduction to this lesson, you could have an engineer come into the classroom to talk about the engineering problems related to the design and construction of an amusement park roller coaster.
   
   Hot Wheels, a toy consisting of a car on a track that can be set up on loops, can also be used.  Allow the students to set up the track in various loops, and have them send the car around at different speeds, noting that as the car goes around the track at a slower speed, it may fall off the loop.
   
   Ask why the car doesn't fall off the track as it goes around the loops at high speed.   Students answers should indicate that they grasp the connection between the initial speed of the car and the car's ability to stay on the track.
2. Distribute the worksheet.  Tell students to complete it as they follow the demonstration of the problem.
3. Distribute the student reading, and demonstrate the concepts:
   • Explain that centrifugal force and centripetal force are the forces that work on a body turning around a center.  Centrifugal force is the tendency of a body in motion to move only in a straight line if it is allowed to do so, which causes a body turning around a center to move away from the center, or outward.  Centripetal force causes a body to move toward the center; it is the force that keeps a body from flying outward.
     
     One way to demonstration the two forces is by having students swing a yo-yo around, allowing them to feel both the centrifugal and centripetal force.  Another is to attach a bucket to a rope, fill the bucket partially (1/2 full) with water, and swing the bucket around in a full circle in the vertical plane, showing that the water stays in the bucket and does not fall out.  Another is to have students use a Hot Wheels set of care and tracks and observe that the cars hold to the tracks as they go around the loops; students need to conclude that the inclination of the track provides the force which keeps the car from flying off (or centripetal force).  Use the demonstration to set off a discussion of the constraints of any problem involving a roller coaster (strength of materials, etc.).
   • Demonstrate gravity by letting marbles roll down the track or off a table.   Point out that the acceleration of gravity is a constant that does not depend on an object's mass or weight.  use the demonstration to set off a discussion of potential energy (the energy when the marble is being held up) and kinetic energy (the energy the marble has as it is falling).

You will be leading the students through the following problem, for the basic design of a roller coaster, using the data in Figure 1:

Figure 1

PROBLEM:

a. DETERMINE THE SPEED OF THE TRAIN:  To keep things simple, let's imagine that the roller coaster train is frictionless. How fast must the train go through a loop with a 15-foot radius in order for the centrifugal force to equal twice the force of gravity?  We want 1g to counteract the normal force of gravity plus 1g to create an artificial gravity outward at the top of the loop.  We need to find v in the following equation:

Centrifugal force (F) =  mv²/R

where:

m = the mass of the train and passengers
v = velocity (speed of the train) in feet per second (ft/s)
R = the radius of the loop

(As noted above, we want the centrifugal force F to equal twice the force due to gravity.  This gives us the equation

F = 2mg

where m, as before, is the mass of the train and its passengers and where g is the acceleration of gravity, which is a constant equal to 32.2 ft/s².

From the two separate equations for F given above, we can conclude that

2mg = mv²/R 

Our first step in solving for v can be to divide both sides of the equation by m.  This eliminates m from the equation altogether and illustrates that velocities as calculated from the gravitational constant are independent of the object's mass.

2g = v²/R 

Multiplying both sides by R, we have

v² = 2gR

To find v, we take the square root of both sides:

v = 2gr

Substituting values for variables or constants, we get

v= 2 x 32.2 x 15

= 988

v = 31.1 ft/s

b. DETERMINE THE HEIGHT OF THE LIFT:  How high must the lift (the original ascent of the roller coaster) be for the train to travel 31.1 ft/s at the top of the loop?  If the roller coaster is frictionless, then the potential energy at the top is converted entirely to kinetic energy at the bottom of the first drop. We use the following formulas:

Potential energy (PE) = mgh
Kinetic energy (KE) = 1/2 mv²

Since the potential energy is completely exchanged for kinetic energy, we can find h by equating the two formulas and solving:

mgh = 1/2 mv²

After the train passes the bottom of the first drop and has traded all of its potential energy for kinetic energy, it begins to climb the loop, and as it climbs, it trades kinetic energy back for potential energy, slowing down as it does so.

When the train reaches the top of the loop it must be going at 31.1 ft/s. Again, if the roller coaster is frictionless, then all of the speed gained as the train dropped through the last 75 feet of the first drop would be lost as it climbs up the loop.   What will be left will be what it gained by dropping on the first part of the first drop (h minus 75 ft).  As we found above,

mgh = 1/2 mv²

We can solve for v, the speed after a drop through a height h, by first dividing both sides by m, to eliminate the unwanted variable:

gh = 1/2 v²

By multiplying both sides by 2, we get

2gh = v²

To find v, we take the square root of both sides:

v = 2gh

The equation now has two unknowns, v and h.  This because we are trying to solve for h and we also do not know the value of v after a drop of height h.  But we do know v for a drop of h - 75 ft: it is 31.1 ft/s, as we determined in the previous group of equations.  Thus, we can substitute the value 31.1 ft/s for v if we also substitute the value h - 75 for h:

v = 2g(h - 75)

Squaring both sides, we have

v² = 2g(h - 75)

Dividing both sides by 2g gives us

v²/2g = h - 75

Now we add 75 to both sides, to get

h = v