Roller Coaster Physics

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.

Page Topic2 General3 The Simple Roller Coaster8 The Most Often Used Calculations11 Getting the Coaster Started15 Weightlessness22 Hills and Dips (and Projectile Motion)42 Loops55 Physiological Effects of Acceleration69 Center of Mass72 Banked Curves

Further Support Materials78 Making the Roller Coaster Simulator81 Making a Hot Wheels Train83 Roller Coaster Activities91 Field Estimation Tips (Handout)93 Formulae and Constants94 How to Take Roller Coaster Measurements in the Park99 Amusement Park Labs

125 Intro to Design: Example133 Simple Coaster Practice139 Roller Coaster Test #1 145 Roller Coaster Test #2150 Answers to the Activities153 Extra Credit Coaster Design Activitiy154 One Final Note of Preparation for Next Year

This book assumes some rudimentary knowledge of physical science. It is a simplified view of what design considerations and science a mechanical/civil design engineer must know when designing a roller coaster.

- 1 - by Tony Wayne

- 2 - by Tony Wayne

A roller coaster is a balance between safety and sensation. Naturally, the ride should be as safe as possible. After all, if the people are injured riding the coaster then there would be fewer repeat riders. Fewer repeat riders means a short life span for the coaster. On the other hand, passengers ride a coaster for the “death defying” thrill. The key to a successful coaster is to give the rider the sensation of speed and acceleration. It all comes down to speed control.

To achieve this, the hills, curves, dips, straight always, braking systems and loops are not randomly designed. They follow some simple rules of physics.

In order to understand what is going on, students must understand the difference between velocity and acceleration.

Velocity describes how quickly an object changes it position. The higher the velocity the quicker an object travels between 2 locations. Phrases like, “...how fast..., how quickly,” are used to describe velocity. Often the word speed is substituted for the word velocity in common usage. However, technically the two are different. Velocity is actually speed with direction. For example, “60 mph, west,” is a velocity. “West” is the direction and “60 mph” is the speed. The units of velocity are in the form of

Velocity = Units of Distance Units of Time

Example = metersecond

milehour

furlongweek

Acceleration describes how quickly an object changes its velocity. Phrases like, “...slow down..., ...speed up..., ...change speed... and change velocity...” are used to describe accelerations. If a student wants an easy way to determine if he is visualizing acceleration or a constant velocity along a straight line he only needs to ask himself one question, “Is the object slowing down or speeding up?” If the answer is “Yes,” then it is accelerating. If the answer is “No” then it is moving with a constant velocity. The units of acceleration are in the form of

Acceleration = Units of Distance = Units of Velocity

(Units of Time)2 Units of Time

Example = meter

second2

mile

hour2furlong

week2meter

second•hour2

milehoursec

( ) metersecsec

( ) fathomminsec

( )

- 3 - by Tony Wayne

- 4 - by Tony Wayne

The simple roller coaster started with Galileo Galilei. Below is a description of how to demonstrate Galileo’s experiment using the “Roller Coaster Simulator.” [This experiment could also be duplicated using a HotWheels™ track and a marble.”

Marble STARTS here

Marble STOPS here

Adjust the ramp to lower angles. The marble keeps rolling to the same height. Galileo concluded from a similar experiment that the marble will keep rolling until it reaches the same height it started from if there were no friction. (With friction, it will keep rolling until it reaches the stop height.)

Marble does not reach “stop height” and continues to move.

This has far-reaching implications. (1) The marble could take any path until it reaches the same height it starts from, assuming no friction. In the previous activity, the marble did not roll to the same height it started from because of friction. But it consistently rolls to the same height. To reflect these implications, the track could be reshaped as shown below.

Marble STARTS here

Marble STOPS here

Marble STARTS here

Marble STOPS here

Marble STARTS here

Marble STOPS here

Marble does not make it around the loop segment.

Marble STARTS here

Marble STOPS here

Marble makes it around the loop because it never reaches the stop height.

- 5 - by Tony Wayne

(2) The ball begins to roll down due to the force of gravity. It stops when all the energy gravity gave the ball is used up. The marble accelerates only while a force acts on it in its direction of motion.

Here is a good exercise to draw on a chalk board.

The gray sections of track are where the coaster car accelerates. (Speeds up or slows down.)

The black sections of track are where the coaster car travels at a constant velocity.

The acceleration can be demonstrated experimentally using the roller coaster simulator or HotWheels™ track. If a long enough section is made horizontal, it can be shown that the average velocities calculated at the beginning and at the end of the horizontal section are equal. Form the track in the shape shown below. Roll a marble or steel bearing down the track. It will accelerate along the drop and move at a constant velocity along the horizontal section and slow down as it climbs up the opposite side. When the marble slows down and speeds up on the hills it is visually obvious. What is not so visually obvious is what happens along the horizontal section of the track. The ball’s constant velocity can be shown mathematically. Divide the horizontal section of the track into 2 sections. Calculate the average velocity of the ball along these two sections. If done accurately, the velocities will be nearly equal. To obtain more accurate results, use fairly long sections of horizontal track. The longer the sections of track, the greater the time measurement. Longer time measurements mean lower percent errors.

- 6 - by Tony Wayne

Marble STARTS here Marble STOPS here

V1avg V2avgForce of gravity changes the speed of the ball. Velocity for the ball does

not change

Force of gravity changes the speed of the ball.

Along the horizontal section of the track, ignoring the minimal effects of friction, there are no forces acting on the ball horizontally. Therefore the ball moves at a constant velocity while no force acts on it. This is Galileo’s law of inertia!!!

- 7 - by Tony Wayne

Here is another example of an illustration of Galileo’s experiment.

RELEASE HEIGHT

RING STAND WITH AN ATTACHED PENDULUM.

RELEASE HEIGHT

RELEASE HEIGHT

AS THIS PEG IS MOVED HIGHER AND LOWER, THE BALL WILL CONTINUE TO REACH THE HEIGHT IT IS DROPPED FROM.

RELEASE HEIGHT

The ball continues until it reaches the starting height.

- 8 - by Tony Wayne

- 9 - by Tony Wayne

A roller coaster is called a coaster because once it starts it coasts through the entire track. No outside forces are required for most coasters. (A few have double or triple lift hills and braking sections.) Roller coasters trade height for velocity and velocity for height. Most all calculations rely on using velocity measurements in one way or another. The first step is being able to calculate the changes in speed.

In an ideal world, mechanical energy is conserved. Frictional forces are ignored in early design stages. (This document does not address the nuances of dealing with frictional forces.) Mechanical energy on a roller coaster comes in two basic forms. Kinetic energy, KE = (1/2)mv2, and potential energy, PE = mgh, due to gravity. Total energy, ET, is conserved and is equal to the sum of kinetic and potential at any single location.

ET = KE + PE (at any single location)

95 m 65 m

8.8 m/s 8.8 m/s

95 m 65 m

8.8 m/s

95 m 65 m

All four of these roller coasters look different. But because friction is ignored all four roller coasters will have the same speed at the bottom of each hill and at the top of the second. The only thing that matters is the height of the locations of interest.

95 m 65 m

8.8 m/s

- 10 - by Tony Wayne

to calculate a change in velocity associated with a change in heightStep 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.

What is the velocity at the bottom of the first hill?

ET(TOP) = ET(BOTTOM)

KE + PE = KE +PE(1/2)mv2 + mgh = (1/2)mv2 + mgh

(1/2)v2 + gh = (1/2)mv2 + mgh The masses cancel out because it is the same coaster at the top and bottom.

(1/2)v2 + gh = (1/2)v2 + gh Substitute the numbers at each location

(1/2)(8.8)2 + 9.8(95) = (1/2)v2 + 9.8(0) The height at the bottom is zero because it is the

lowest point when comparing to the starting height.

77.44 + 931 = (1/2)v2

1008.44 = (1/2)v2

2016.88 = v2

∴ v = 44.9 m/s ...at the bottom the the 1st hill.

What is the velocity at the top of the second hill?

ET(TOP OF 1st HILL) = ET(TOP OF 2nd HILL)


(1/2)v2 + gh = (1/2)mv2 + mgh The masses cancel out because it is the same coaster at the top and bottom.

(1/2)v2 + gh = (1/2)v2 + gh Substitute the numbers at each location

(1/2)(8.8)2 + 9.8(95) = (1/2)v2 + 9.8(65) Notice all the numbers on the left side come from the topof the 1st hill while all the numbers on the right sidecome from the top of the 2nd hill.

77.44 + 931 = (1/2)v2 + 637

371.44 = (1/2)v2

742.88 = v2

∴ v = 27.3 m/s ...at the top the the 2nd hill.

This technique can be used to calculate the velocity anywhere along the coaster.



Something has to be done to get the coaster started. In our previous example energy, power, has to added get the coaster up to 8.8 m/s. This is done by doing work on the coaster. A simplified definition of work would be force times displacement when the force and displacement go in the same direction. [This chapter will not go into all the details of calculating work.] Suffice it to say that when the force acting on the coaster and the displacement of the coaster are in the same direction, work adds energy to the coaster. When the force acting on the coaster and the displacement of the coaster are in opposite directions, work removes energy from the coaster.

FORCE DISPLACEMENT

Energy is added tothe car.

FORCEDISPLACEMENT

Energy is removed from the car.

Work = (Force)(Displacement)W = Fd

Where “work” is measured in joules, J. “Force” is measured in Newtons, N, and “displacement” is measured in meters, m.

What is the velocity of the train after being catapulted into motion?

DISPLACEMENT

95 m 65 m

v = ? m/s

12.5 m

Starts from rest

A catapult system propels the car across a distance of 12.5 m by an 8800 N force.

Train mass is 3.0 X 103 kg

ET(BEGINNING) + Work = ET(TOP OF 1st HILL)

KE + PE +W = KE +PE(1/2)mv2 + mgh + Fd = (1/2)mv2 + mgh

(1/2)3000(0)2 + 3000(9.8)(3000) + 8800(12.5)= (1/2)3000v2 + 3000(9.8)(0) Substitute the numbers at each location


110,000 = (1/2)(3000)v2

73.333 = v2

v = 8.6 m/s ... at the end of the catapult.As an aside you can calculate the acceleration of the rider from kinematics equations.(For the curious the acceleration is 7.0 m/s2)

What is the velocity of the train after being catapulted into motion?

`DISPLACEMENT

95 m 65 m

v = ? m/s

145.5 m

Starts from 5.2 m/s after it leaves the station and engages the lifting mechanism.

A pulley system lifts the train up an incline as shown below with a 19,715.464 N force.

Train mass is 3.0 X 103 kg

ET(BEGINNING) + Work = ET(TOP OF 1st HILL)


(1/2)3000(5.2)2 + 3000(9.8)(0) + 19,715.464(145.5) = (1/2)3000v2 + 3000(9.8)(95) 40560 + 2868600.012 = (1/2)(3000)v2 + 2793000

116160.012 = (1/2)(3000)v2

77.44 = v2

v = 8.8 m/s ... at the end of the catapult.

A roller coaster train of mass 3.0 X 103 kg rolls over a 11.5 m high hill at 8.34 m/s before rolling down into the station. Once in the station, brakes are applied to the train to slow it down to 1.00 m/s in 5.44 m. (a) What braking force slowed the train down?(b) How much time did it take to slow the train down?(c) What was the acceleration of the train in g’s?

ET(HILL) = ET(@ 1 m/s) + WorkKE + PE = KE +PE +W

(1/2)mv2 + mgh = (1/2)mv2 + mgh + Fd(1/2)3000(8.34)2 + 3000(9.8)(11.5) = (1/2)3000(1)2 + 3000(9.8)(0) + F(5.44) Substitute the numbers

at each location

442433.400 = 1500 + 5.44F


F = 81053.9 N ...force to slow down the train

Calculate the velocity as the train enters the station. Use this velocity to calculate the time.ET(HILL) = ET(@ STATION ENTRANCE) No work is done because no force acts between the two

locationsKE + PE = KE +PE

(1/2)mv2 + mgh = (1/2)mv2 + mgh(1/2)3000(8.34)2 + 3000(9.8)(11.5) = (1/2)3000v2 + 3000(9.8)(0) Substitute the numbers

at each location

442433.400 = 1500v2

v = 17.174v = 17.2 m/s ...as the train enters the station.

The time is calculated fromvo = 17.174 m/svf = 1.00 m/s

x = 5.44 mt = ?

xt

= vo + vf 2

5.44t

= 17.174 + 1 2

t = 0.599 sec

Calculate the acceleration in m/s2. Then convert it into g’s.F= ma81053.934 N = (3000 kg)aa = 37.018 m/s2

a = 37.018 m/s2/ 9.80 m/s2 = 3.78 g’s ... Yeow! That’s a big jerk on the passengersinto the restraining harness.



Weight is the pull of gravity. Typical weight units are pounds and newtons. (1 pound ≅ 4.45 newtons). On the moon, gravity pulls with 1/6 the force compared to the Earth. Therefore, a student on the moon weighs 1/6 of what she weighs on the Earth. On the Earth, neglecting air resistance, all objects will speed up at a rate of 9.80 m/s every second they fall. That is a speed increase of about 22 mph for every second an object falls.

Time in the Air Velocity

s mph

0 0

1 22

2 44

3 66

4 88

5 110

There are two ways to experience weightlessness. (1) move far enough away from the planets and sun to where their pull is nearly zero. [Gravity acts over infinite distance. One can never completely escape it.] (2) Fall down at a rate equal to the pull of gravity. In other words, accelerate to the Earth speeding up 22 mph every second in the air. In order for a person to feel weight, a person must sense the reaction force of the ground pushing in the opposite direction of gravity.

REACTION FORCE OF THE GROUND

PULL OF GRAVITY (WEIGHT)

In the absence of the reaction force a person will sink through the ground.Many amusement park rides generate the weightless sensation by accelerating down

at 22 mph every second.

gÕsNeglecting air resistance, if a rock is dropped, it will accelerate down at 9.8 m/s2. This

means it will speed up by 9.8 m/s for every second it falls. If a rock you drop accelerates down at 9.8 m/s2, scientists say the rock is in a “1 g” environment, [1 g = 9.8 m/s2 = 22 mph/s].

Any time an object experiences the pull equal to the force of gravity, it is said to be in a “one g” environment. We live in a 1 g environment. If a rock whose weight on the Earth is 100 lbs was moved to a 2 g environment then it would weigh 200 lbs. In a 9 g environment it would weigh 900 lbs. In a “NEGATIVE 2 g” environment it would take 200 lbs to hold the rock down on the ground. In a “-5 g” environment it would take 500 lbs to hold the rock down to the ground. If the rock were put into a “zero g” environment then it would be weightless. However,


no matter what happens to its weight the rock’s mass would never change. Mass measurement is unaffected by the pull of gravity.

What does it feel like to walk in a 2 g environment? Have students find someone who’s mass is about equal to theirs. Have them give piggyback rides. As they walk around this is what it feels like to be in a 2 g environment. Go outside on the soft ground and have the students step up on something. This is when they will really know what a 2 g environment feels like.

Often engineers will use g’s as a “force factor” unit. The force factor gives a person a way of comparing what forces feel like.

All acceleration can be converted to g’s by dividing the answer, in m/s2, by 9.8 m/s2.

A roller coaster is propelled horizontally by a collection of linear accelerator motors. The mass of the coaster train is 8152 kg. The train starts from rest and reaches a velocity of 26.1 m/s, 55 mph, in 3.00 seconds. The train experiences a constant acceleration. What is the coaster train’s acceleration in g’s?

m = 8152 kgvo = 0 (starts from rest)vf = 26.1 m/st = 3.00 sa = ?

vf = vo + at26.1 = 0 + a(3.00)a = 8.70 m/sin g’s... 8.70 m/s2 / 9.80 m/s2 = 0.89 g’sThis means the rider is being pushed back into his seat by 89% of his weight.

The rider is pushed back into the seat by a force equal to 89 % of his weight.


MATERIALS1 plastic cup2 skinny 4” diameter rubber bands2 50 g masses

BACKGROUNDCut the rubber bands. Tie the ends of each together to make a stretchy string. Tie the weights to the opposite ends of the rubber band. Attach the middle of the rubber band to the inside bottom of the cup. The two masses should be able to hang over the lip of the cup.

RUBBER BAND

50 g mass

CUP

weight (mg)

Normal force exerted by the cup on the weight

Tension in the rubber band

Free body diagram of the forces acting on the left hanging weight.

The masses are in equilibrium with the upward force of the rubber band. The force pulling up of the rubber bands is equal to the force of gravity, [weight of the masses.] Ask the class, “What would happen if the rubber bands pulled with a force greater than the pull of gravity on the masses?” The masses would shoot upward and be pulled into the cup. To show this, pull down on one of the masses and let go.

Now ask, “What would happen if the masses could be magically shielded from the pull of gravity?” With no force stretching the rubber bands, they would sling shot into the cup.

Explain, “We cannot yet shield gravity. But we can momentarily minimize its effects by accelerating the masses, rubber band and cup down at 9.80 m/s2, the acceleration due to gravity. Without saying anymore, stand on a chair. Raise the apparatus with the weights hanging out. Tell the students, “When this cup is dropped everything will speed up equal to the acceleration of gravity. What will you see when this cup is dropped?”

Drop the cup after polling the students. The masses will be pulled into the cup. When everything falls, gravity will not be pulling against the masses when compared to the


rubber band’s pulling force. The masses are said to be weightless. It is the weight of the mass that stretched the rubber band. If the mass is weightless, the rubber band will pull it in.

Continue to the ground

Drop

When everything is falling together, the pull of gravity is no longer experienced by the rubber bands. Therefore, they pull the masses into the cup.


16 INCHES

STEP 1Nail it together.

STEP 2Glue and nail the triangles to the

outside.

STEP 3Attach the nail to the weight and hang the weight

by rubber bands.

STEP 4Insert the balloon in the hole. Inflate it so that it does

not touch the nail. Tie it closed

BALLOON

Drill 1/4” diameter hole here. (The balloon opening will fit through it.)

MATERIALS1 piece of wood 8 ft X 2” X 1” 8 triangles of wood 1/8” thick2 Rubber bands1 20 g mass1 3 inch sharp nail1 balloon

BACKGROUNDMake a frame as shown below.

3 inches

3 inches

The DEMOWith the balloon inflated, hold the frame over a pillow. Hold the frame straight out at

chest height. Ask students to predict what will happen when you release the frame. Guide them to specifics such as where in the fall will the balloon pop. Release the frame. The balloon will pop almost instantaneously. The balloon pops because the weighted pin becomes weightless. The rubber bands are essentially pulling against nothing. This means the rubber bands pull the pin up into the balloon.

Pull of gravity

Pull of the rubber bands

Pull of the rubber bands

When falling, the rubber bands are moving down due to the force of gravity. this motion negates the

Forces acting on the mass when it is at rest.pull of gravity on the hanging mass. The rubber bands can now pull the hanging mass up -because it is weightless in free fall


CUP WITH2 HOLES

MATERIALS1 plastic cup 1 trash can or big bucket1 candle flame large nail1 large nail - water

PREPARATIONHeat up the nail with the candle flame. Be careful not to burn yourself. Poke the hot

nail into opposite sides of the cup at the bottom. This will make a clean hole. Hold your fingers over the two holes and fill the cup half full of water.

THE DEMOStand on a chair and briefly release your fingers from the holes. The water should

stream out the cup’s holes onto the floor. Ask the students, “What will happen when I drop the cup into the trash can?” Listen to all their answers. Drop the cup to see who’s prediction was correct. The water will not flow out of the cup. Water flows out of the cup when the acceleration of the cup is different from the water. When the cup is held the water is allowed to accelerate down at 9.8 m/s2. When the cup falls too, the cup is also accelerating down at 9.8 m/s2. Since there is no difference in their accelerations the water stays in the cup.

CUP WITH2 HOLES

Force holdingthe cup up.

Pull of gravity

CUP WITH2 HOLES

Remove the force holding the cup up by letting go of the cup.

Pull of gravity

.

The water is accelerating down faster than the the cup. (That is because the cup is not accelerating at all.)

The water is not falling down quicker than the cup, so it stays in the cup.



One of the most basic parts of a ride is going from the top of a hill to the bottom. There are two basics ways designs transport riders to the bottom of a hill. The first is called the “Speed Run.”

h

vo

vf

vf = vo2+2gh

vo = velocity at the hill’s top

vf = velocity at the hill’s bottom

h = the height of the drop

g = the acceleration due to gravity (9.80 m/s2)

A speed run is designed to give the rider the feeling of accelerating faster and faster without the feeling of weightlessness. It simulates being in a powerful car with the accelerator held down to the floor. It is a straight piece of track that connects a high point to a low point.

The increase in velocity of the car comes from lost gravitational potential energy being converted into kinetic energy. Next to a horizontal straight piece, the speed run is the easiest piece of the track to design and analyze.

3.2 m/s

v = ?

h = 64 m

Total Energy at the top = Total Energy at the bottomET(TOP) = ET(BOTTOM)KE + PE = KE + PE

(1/2)mv2 + mgh = (1/2)mv2 + 0(1/2)v2 + gh = (1/2)v2 + 0

(1/2)(3.2)2 + (9.80)(64) = (1/2)v2 + 0632.32 = (1/2)v2 + 0

v = 35.56v =36 m/s

When coasting up to a new height the calculations are the same as the example shown above. The shape of the hill does not matter. See the “Intro to Design” section, step 7, for an example of these up hill calculations.


One of the biggest thrills on a roller coaster is the free fall as a rider travels over a hill. The easiest way to experience free fall is to hang from a tall height and drop to the ground. As a person falls he experiences weightlessness. As long as a person travels in the air like a projectile he will feel weightless.

Suppose a ball traveled off a table, horizontally, at 10 m/s. The ball’s path would look like the path shown below.

BALL IN FREE FALL

Now suppose the ball traveled off the table top on a shallow angled ramp. It would look like the one below.

Where the ball would be without

the ramp.

The ball will take longer to descend. The ramp applies an upward force on the ball to change its path and thereby slow its descent. To the ball, this upward force is registered as weight.

Where the ball would be without

the ramp.

For this section of the trip, the ball is weightless.

For this section of the trip, the ball feels weight.


A straight, “speed run,” drop does not match the fall of a rider over a hill.

For this section of the trip, the ball is weightless.

Not the safest of choices if the ball were a roller coaster car full of passengers.

To give riding more of a thrill, the designer needs to design the shape of the hill to match the falling ball.

BALL IN FREE FALL

The hill is the same shape as a projectile in free fall. The roller coaster barely makes contact with the track.

The only problem with curve above is the impact with the floor. To alleviate this problem another curve scoops the balls as they descend. This makes the ride smooth and survivable for the rider.

BALL IN FREE FALL

The speed at the bottom of a free fall drop is calculated the same way as the speed at the bottom of a speed run drop. The only difference is the shape of the hill from the top to the bottom.


A free fall hill shape gives a rider a weightless sensation. To give this weightless sensation over a hill, the hill is designed to have the same shape as the path of a ball being thrown off the top of a hill. Shape is determined by how fast the roller coaster car travels over the hill. The faster the coaster travels over the hill the wider the hill must be. There are two ways to apply projectile motion concepts to design the hill’s shape. The first way is to calculate the coaster’s position as if it drove off a cliff.

The position equation is as follows.

h = the height from the top of the hillx = is the distance away from the center of the hillv = the velocity the roller coaster car travels over the

top of the hill.g = the acceleration due to gravity. 9.80 m/s2 for

answers in meters. 32.15 ft/s2 for answers in feet.

This can be rewritten as

x in meters h in meters

0.00 0

4.52 1

6.39 2

7.82 3

9.04 4

10.10 5

11.07 6

11.95 7

12.78 8

13.55 9

14.29 10

Velocity (v) = 10 m/sAcceleration (g) = 9.8 m/s2

x in meters h in meters

0.00 0

9.04 1

12.78 2

15.65 3

18.07 4

20.20 5

22.13 6

23.90 7

25.56 8

27.11 9

28.57 10

Velocity (v) = 20 m/sAcceleration (g) = 9.8 m/s2

There comes a certain point on the free-fall drop where the track needs to redirect the riders. Otherwise the riders will just plummet into the ground. This point is the transition point from free-fall to controlled acceleration. This point is also the maximum angle of a hill. This angle can be in virtually any range from 35° to 55°.


Tangent line whose slope is 50°

Transition point where the slope of the free-fall hill equals 50°

This section of the track will not follow the equation for free fall.

For the bottom section of the track, the new equation has the desired outcome of changing the direction of the coaster from a downward motion to a purely horizontal motion. The track will need to apply a vertical component of velocity to reduce the coaster’s vertical velocity to zero. The track will also need to increase the horizontal velocity of the coaster to the value determined from energy relationships. The velocity at the bottom of the hill is determined from

(Kinetic Energy) + (Gravitational Potential Energy) = (Kinetic Energy)

Total mechanical energy at the topTotal mechanical energy at the bottom

which is(1/2)m(vT)2 + (mgh) = (1/2)m(vB)2

This simplifies to

vB = (vT)2 + 2gh

where vB is the horizontal velocity at the bottom of the hill. The value for vB will be used in later calculations.

Recall one of the original horizontal equations.

x = xo + (vxo)t +(1/2)(ax)t2

substituting in our expression for “t” yields,

vyo2 + 2(ay)y - vyo

(ay)x = (vxo)( ) vyo2 + 2(ay)y - vyo

(ay)( )2

ax

2+


where vxo is the horizontal velocity of the coaster at the transition angle and vyo is the vertical component of the velocity at the transition angle. “ay” is calculated from

and

vy2 = vyo2 + 2(ay)y

vy2 - vyo2

2y(ay) =

Where “vy” is the final vertical velocity of zero, “vyo” is the vertical component of the velocity at the transition point, and “y” is the distance left to fall from the transition point to the ground. The horizontal velocity is determined from a parameter decided upon by the engineer. The engineer will want to limit the g forces experienced by the rider. This value will be the net g’s felt by the rider. These net g’s are the net acceleration.

ax = anet2 - ay2

∴ anet2 = ax2 + ay2

anetay

ax

and

these values are plugged back into the original equation and x values are calculated as a function of y.


This is the beginning of the Hurler at Paramount’s Kings Dominion in Doswell Virginia. Can you tell which hill is the free fall hill?

(It’s the curved hill in front.)


0

20

60

100

40

80

200

300

400

A person throws 2 balls. The first ball is thrown horizontally at 20 m/s. The second ball is thrown at 40 m/s. Draw as much of each path as possible. Draw the ball’s position every 20 m of vertical flight. Draw a smooth line to show the curve’s shape.

Vertical distance down

(meters)

Horizontal position for the ball with

an initial velocity of 20

m/s



m/s

0 0 0

40

80

120

160

200

240

280

320

360

400


0

20

60

100

40

80

200

300

400

A person throws 2 balls. The first ball is thrown horizontally at 10 m/s. The second ball is thrown at 30 m/s. Draw as much of each path as possible. Draw the ball’s position every 1 second of the flight. Draw a smooth line to show the curve’s shape.

Time (seconds)

Vertical position for the ball with


m/s

Vertical position for the ball with


m/s

0 0 0

1

2

3

4

5

6

7

8

9


0

20

60

100

40

80

200

300

400

A person throws 2 balls. The first ball is thrown horizontally at 20 m/s. The second ball is thrown at 40 m/s. Draw as much of each path as possible. Draw the ball’s position every 20 m of vertical flight. Draw a smooth line to show the curve’s shape.

Velocity = 40 m/s

Velocity = 20 m/s

Horizontal Distance in meters0 100 200 300 400 500 600

Below is a set of answers for 4 velocities.Vertical

distance down (meters)



m/s



m/s



m/s



m/s

0 0 0 0 0

40 28.6 57.1 102.4 136.6

80 40.4 80.8 121.8 162.4

120 49.5 99.0 134.8 179.8

160 57.1 114.3 144.9 193.2

200 63.9 127.8 153.2 204.3

240 70.0 140.0 160.3 213.8

280 75.6 151.2 166.6 222.2

320 80.8 161.6 172.3 229.7

360 85.7 171.4 177.4 236.6

400 90.4 180.7 182.2 242.9


Ruler, marble, horizontal table

To see the effect of speed on arc shape.

Hold the end of the ruler up to three different angles. Repeatedly roll the ball down the ruler until you can draw the path the ball takes for each height. (Make sure the marble does not bounce when it leaves the table.) Write your data down for each trial in the data table below.

L

Length of the ramp the ball rolls down.

Height where the ball starts

Distance to the ball’s landing spot.

Average velocity along the table top ONLY.

H

D

V

(m) (m) (m) (m/s)

Vavg = dist/time

Draw the path the ball takes on the grid on the next page for each ruler height.


RULER Make sure the ball does not bounce when it leaves the table.1 m

The drawn path represents the shape of a free fall roller coaster hill for different coaster car velocities.

1 Which ramp angle gives the ball the greatest speed when it leaves the table?

2 What conclusions can you make about how speed over the top of a hill affects the shape of the hill?


Below are three different starting heights for a hill. For each starting height, rate the hills shape as:

(1) Possible optimum hill. The ball will be in free fall the longest period of time and the ball would softly be caught at the bottom.

(2) Safe hill. The ball stays on the hill but does not remain in free fall for the maximum amount of time.

(3) Unsafe hill. The ball will leave the track and possibly hit the other side of the track.

STARTING HEIGHT #1

STARTING HEIGHT #2

STARTING HEIGHT #3

1

2

3


In this activity the student will visualize the path of a ball as it rolls over different shaped hills. In this activity a ball will be rolled from 3 different heights over three different shaped hills. • Lay out the roller coaster simulator track on the chalk board. • Using a piece of chalk trace the path of the track and mark the starting point of the track.• Roll the ball down the track. If the ball leaves the track, trace its path with chalk. Then roll it

again to see how it compares with the drawn line. • Draw the path the ball takes on the paper. Repeat this process for every hill and every

starting position.

Starting height 1HILL SHAPE “B”

Starting height 1HILL SHAPE “C”

Starting height 1HILL SHAPE “A”


Starting height 2HILL SHAPE “B”

Starting height 2HILL SHAPE “C”

Starting height 2HILL SHAPE “A”


Starting height 3

HILL SHAPE “B”

Starting height 3

HILL SHAPE “C”

Starting height 3

HILL SHAPE “A”


Below are three different starting heights for a hill. For each starting position rate the hills shape as:

(1) Possible optimum hill. The ball will be in free fall the longest period of time and the ball would softly be caught at the bottom.

(2) Safe hill. The ball stays on the hill but does not remain in free fall for the maximum amount of time.

(3) Unsafe hill. The ball will leave the track and possibly hit the other side of the track.

Hill shape C

Starting height 1

Hill shape B

Hill shape A

Starting height 2

Starting height 3

ROLLER COASTER SIMULATOR

1

2

3


Starting height 1

Starting height 2

Starting height 3

Hill shape “A”

Starting height 1

Starting heights 2 & 3

Starting height 2

Starting height 3

Hill shape “B”

Starting height 1

Starting height 2

Starting height 3

Hill shape “C”

All these starting heights will keep the marble on the track. (Starting height 3 may not even make it over the track.)

Starting height 1(Off the track)



Starting height 3


1

2

3

The ball is the highest and will be traveling the fastest. A fast speed means a long “flat” hill.

The ball will get airborne over this hill.




The ball will make it over the hill. The ball will not experience weightlessness because the ball wants to fall quicker than the hill will allow.







The average person on the street has heard of centrifugal force. When asked, they would describe this force as the one pushing an object to the outside of a circle. There is only one problem with this description. There is NO FORCE pushing an object to theoutside.

For a person riding in a car while traveling in a circle, he perceives a force pushing him to the outside of the circle. But what force is physically pushing him? It can not be friction. Frictional forces oppose the direction of motion. It can’t be a “normal force1 .” There is not a surface pushing the rider to the outside. To travel in a circle, a force pointing to the inside of the circle, or curve, is needed. The force pointing to the inside is called the centripetal force.

To understand a source for the misconception of the direction of this force, consider what it feels like when traveling around a corner in the back seat of a car. Everyone who has been in this situation knows that the passenger will slide to the outside of the curve. To understand that there is no force pushing the passenger to the outside, a change of reference frame is needed. Move the point of view from inside the car to a location outside, above, the car.

1 The normal force is the force perpendicular to a surface. The floor is exerting a normal force straight up equal to your weight right now.


CAR

PASSENGER

CURVE

Straight line motion if no force acts on the passenger.

THE CAR ENTERS THE CURVE.

To the passenger, he is sliding to the outside of the curve.

To the passenger, he is sliding farther to the outside of the curve.but notice he is traveling straight.

To the passenger, he is sliding farther to the outside of the curve.

But notice he is traveling straight.

1

2

3

4

5

6

7

8

Force tothe inside

Force tothe inside

FORCE TOTHE INSIDEThis is the

centripetal force.

Many people mistake the reaction force of the rider on the car’s side as a centri fical force.


SUN

EARTH

Gravity supplies the centripetal force that keeps the Earth orbiting the sun.

A BUCKET BEING SWUNG IN A CIRCLE.

The pull of the person’s arm and gravity -when the bucket is upside down- supplies the centripetal force.

A CAR GOING IN A CIRCLE

Friction between the tires and the road supplies the centripetal force. If the road is banked, then gravity will contribute to the centripetal force.


The normal force the track exerts supplies the centripetal force at position #1.

The normal force the track exerts plus the pull of gravity exerts the centripetal force at position #2

POSITION #1

POSITION #2

LOOP

The string exerts the centripetal force on the rock.

Rock twirled on a string overhead in a horizontal circle.


MATERIALS: 6 inches of string, fishing weight (split shot variety), small paper clip

Open the paper clip up like this.

Tie the end of the string to the small end of the paper clip.

Tie a knot at the bottom of the string.

Open the lead shot. Pinch the shot closed.

WASH YOUR HANDS NOW!!PINCH HERE

TO OPEN THE SHOT.

Do not use your teeth to pinch the shot closed. LEAD IN SMALL AMOUNTS IS POISONOUS.

1 2 3

4 5 6

PROCEDURE:

Hold the accelerometer by the paper clip in your hand. Move your hand to the right then to the left. Draw the direction the accelerometer swings. Also indicate the direction of theforce as you move the accelerometer.

Move your hand to the RIGHT.

Move your hand to the LEFT.

Hold the accelerometer in one hand. Hold your hand outstretched and twirl around while watching the accelerometer. Below is the circle your hand makes as viewed from overhead. Draw which way the accelerometer swings. Also label the force’s direction.

You


Move your hand to the RIGHT.

Move your hand to the LEFT.

FORCE FORCE

You

FORCE FORCE


MATERIALS:Bucket of water

PROCEDURE:

Fill the bucket up 1/4 full with water. Swing the bucket in a vertical circle. Swing it fast enough so the water does not come out. Now slow down the swing until the water almost drops out of the bucket.

In order to keep water in the bucket when it is upside down, the water needs to be accelerated,(pulled downwards), faster than what gravity would move the water. The pulling force downward varies with the velocity the bucket moves. By spinning the bucket faster, the pull downwards is greater.

In order to just barely keep water in the bucket when it is upside down, the water needs to be accelerated, (pulled downwards), as fast as what gravity would move it. The pulling force downward varies with the velocity the bucket moves. Thebucket needs to be twirled at a speed so that the pull equals the pull of gravity.

When the water falls out of the bucket, it is because the pull on the water is greater than the pull on the bucket. Gravity is accelerating the water. If swung slow enough the person’s arm is resisting the fall of the bucket. Gravity moves the water out of the bucket. By pulling harder on the bucket, the bucket will be pulled down at an acceleration equal to or greater than what gravity produces. If the bucket is pulled down faster than what gravity pulls, the water stays inside.

PULL ON BUCKET

PULL OF GRAVITY When the pull on the bucket is greater than the

pull of gravity, the water stays in the bucket.

PULL ON BUCKET

PULL OF GRAVITY When the pull on the bucket is equal to the pull of

gravity, the water just barely stays in the bucket.

PULL ON BUCKET

PULL OF GRAVITY

When the pull on the bucket is less than the pull of gravity, the water falls out of the bucket.


This same demo can be duplicated a couple different ways.

Plastic cups filled 1/4 full with water, strong string, wood or cafeteria service tray

Swing in this direction. Start rocking the tray back and forth slowly to build up speed before going in a complete circle.

INDEX FINGER

SAND THE INSIDE FLAT HERE. PLACE THE PENNY HERE TOO.

penny, metal coat hanger

With a file, rub the inside bottom of the coat hanger flat. Balance a penny on the inside of the coat hanger while hanging the hanger on your index finger. Gently rock the hanger back and forth before swinging it all the way around.

The coat hanger supplies the centripetal force to the inside of the circle to keep it going in a circle.

Q. “If I tie a rock onto the end of a string and swing it around my head in a horizontal circle, I feel the rock pulling my arm to the outside. This certainly feels like a centrifical force and not a centripetal force. What is going on here?”

A. The pull the person swinging feels is the pull to the inside against the inertia of the object spinning. Remember, in the absence of a force, a body will travel in a straight line. In order to turn the rock, a force to the inside must be applied. According to Newton’s 3rd law of motion; for every action, there is an equal and opposite reaction.

This is true for anything whose motion changes from a circle to a straight line.


On a well designed roller coaster loop, the riders will not be able to sense when they are traveling upside down. This is done by making sure the force that is exerted on the rider is at least equal to the weight of the rider.

Centripetal force the track applies

Force of gravity

Centripetal force applied to the track depends on the velocity of the car and inversely to the radius. The formula is:

F = Centripetal forcem= mass of the object going in a circlev = Object’s velocityR = Radius of circle of curveac = centripetal acceleration

In order to apply enough centripetal acceleration the roller coaster car has to either be traveling very fast or the radius of the loop has to be made small. Most rides have a tall loop. A tall loop means a big radius. The problem is, as a car goes up, it slows down. The higher it goes, the slower it will be traveling over the top. In order to apply a centripetal force equal to gravity, 1 g, at the top, the car must be traveling extremely fast as the rider enters the loop. On some of the early round loops, the riders actually had their necks broken as a combination of the sudden rise in the loop as they entered at an extremely high rate of speed. As a compromise, the loops today are designed around an irregular shape called a klothoid or spiral of Archimedes. These irregular loops allow a circular figure whose radius changes.


“Klothoid” shaped loop from the Shock Wave at Paramount’s Kings Dominion in Doswell, Virginia.

This is the Loch Ness Monster at Busch Gardens in Williamsburg, Virginia. It has two loops that are designed from the spiral of Archimedes. One loop is easy to identify in the picture. Can you spot the second loop?

RADIUS49 m

RADIUS7 m


For the advanced reader, the formula for the klothoid shape is:

∫0

t

sin(t) dt√t

x = ± A ∫0

t

cos(t) dt√t

y = ± A

Asymptotic points: (± A/2, ± A/2)

The formula for the “Spiral of Archimedes” in polar form is

r = aθwhere “a” describes the magnitude of the spiral and “θ” is the angle through which the spiral is formed. To make a loop, the spiral will have to be mirrored horizontally.

Nothing is perfect in engineering. These designs operate under ideal circumstances. In real life, the curves need to be tweaked into the right shape.

Sometimes it is not necessary to go into all the math to have a little fun with the irregular loop. These loops can be simulated using the combination of semi-circles of different radii.

Can you see the irregular loop in these regular circles?


R1R2R2

The radii can be anything as long as the car will make it around. In this particular drawing the height at the top of the loop from the very bottom is (1/2)R2 + R1.

If the engineer so chose, she could make the radius at the bottom on the way in one value and the bottom radius on the way out a different value.

Do not design a real roller coaster with this method. The transition from different radii would be uncomfortable for the rider and not possible for the roller coaster train.

Other loop possibilities

R1

R1

R2 R2

R3

R3 ≅ 3R1

R2 ≅ 2R1

R1R3

R2



Imagine a passenger riding through a loop on a roller coaster. The passenger’s head is towards the inside of the circle.

ROLLER COASTER

LOOP

Her feet are to the outside of the circle. In order to keep blood in the passenger’s head, a centripetal force needs to be applied to the blood to push it upwards toward the head and the center of the circle. The heart applies the centripetal force on the blood. A passenger can experience many g’s in a loop. Recall that a g is the number of times heavier an object becomes. A 7 g experience means that the passenger feels 7 times heavier. Everything about the passenger becomes 7 times heavier. Her 3 pound brain now weighs 21 pounds. Every ounce of blood becomes 7 times heavier. If the blood feels too heavy the heart cannot apply enough force to push it towards the head. If the brain does not get any blood it will not get the oxygen the blood carries. The passenger will pass out within a second.

You are riding a new untested roller coaster when something goes wrong. As you enter the first big loop, a great pressure pushes you down. You slouch down in the seat from the extra weight. Over the top of the loop the roller coaster car slows down. The extra weight on your legs, lap, and shoulder make it impossible to sense that you are upside down. Out of the loop, over a hill and into another loop. This loop has a smaller radius. The car is traveling much faster now. As the g forces climb up toward 7 g’s, you sink further still in the seat. You can no longer see color. Everything appears in black and white. An instant later, the passenger next to you disappears from view. Your field of vision is shrinking. It now looks like you are seeing things through a pipe. The front corner of the car disappears from view as your peripheral vision disappears. The visual pipe’s diameter is getting smaller and smaller. You sink into the seat further still as the number of g’s climb further. In a flash you see black. You have just “blacked out.” You are unconscious until the number of g’s are reduced and the blood returns to your brain.

Amusement park owners and insurance companies don’t want the previously described situation to occur. It would limit repeat riders and the number of potential consumers who can safely ride the coaster. Most roller coasters keep the g’s felt under 5 g’s on an inside loop or the bottom of a dip after a hill. When a rider travels over a hill at a high rate of speed, he experiences negative g’s. A negative g is the multiple of a person’s weight that is needed to keep a rider in his seat. Negative g’s also force the coaster car to try to come up off the track. Negative g’s are a rider’s heaven and a designer’s nightmare. Negative g’s are avoided as much as possible.

A negative g has a different effect on a rider than a positive g. Both negative and positive g’s can cause a rider to pass out. But negative g’s cause a rider to “red out.” A red out condition occurs when there is too much pressure on the brain caused by too much blood in the head. The extra pressure can cause blood vessels to burst and kill the rider. This is a sure way to limit the number of repeat riders.


There is another way for a rider to experience negative g’s. It is related to the length of the train. The roller coaster track is designed for the dynamics at the center of mass of the coaster train. Negative g’s are experienced by the rider at the back of the train as he travels over a hill. For an empty train, the center of mass is in the middle of the train.Whatever speed is acquired by the center of the train is the speed for the entire train. After the center of a train passes over a hill it begins to gain velocity. As the center speeds up so does the back of the train. This means that the rear of the train will travel over the hill faster than the middle of the train. If the rider travels over the hill faster than the designed velocity of the hill the rear car will be whipped over the hill.

10 m/s

13 m/s

The rear car is traveling over the hill at 3 m/s faster than the center was. If the hill is designed for 10 m/s then the rear train car will have a tendency to leave the tracks. Under-carriage wheels will hold it to the tracks.


SOME ÒgÓ DETERMINATORS:

40 g stress during a quick acceleration in the direction of motion. Death at 40 g’s.

LINEARACCELERATION

20 g stress during a quick acceleration in the direction of motion. Bleeding would occur from ears.

LINEARACCELERATION

INSIDELOOP

8 gblack out

limitA person passes out because of the lack of oxygen in the brain.

OUTSIDELOOP

3 gred out

limit

A person passes out because of too much blood creating too much pressure on the brain.


To calculate the g’s felt, a formula from circular motion will be utilized. Since energy relationships do not utilize time, the circular motion formula used will also not utilize time.

aCENTRIPETAL = vR

2

g’s = aCENTRIPETAL

9.8 ms 2

Where “v” is the velocity of the body and “R” is the radius of the circle traveled. To calculate the velocity a body is traveling, use energy relationships to solve for the kinetic energy and the associated velocity. One more thing. To calculate the g’s felt remember that the g’s felt by the rider is the normal force on the seat of the rider divided by the mass then converted into g’s. As a rider enters a loop he will feel 2 forces.

weight(mg)

Normal Force

Net Force(Centripetal Force)

The real number of interest is the number if g’s felt by the passenger traveling in the vertical circle. The g’s felt are calculated below. ΣFy = m(ac) = (Normal Force) - WeightΣFy = mv2/R = (Normal Force) - mg∴ (Normal Force) = mv2/R + mg

recall that... (Normal Force)/mg = g’s felt by the rider

thus... (Normal Force)/g = mv2/R/mg + mg/mg

∴ at the bottom


ZERO degrees(Horizontal)

θθθ

All angles are measured to the horizontal axis

g’s FELT = g’s - sin(θ)

g’s FELT = g’s + sin(θ)

ABOVE the horizontal

BELOW the horizontal

θ

ZERO degrees(Horizontal)

All angles are measured to the horizontal axis

g’s FELT = g’s - 1

g’s FELT = g’s + 1

ABOVE the horizontal

BELOW the horizontal

SPECIAL CASE (SHORT CUT)

These results can be interpreted easily. As a rider enters the loop, the track has to exert a normal force upwards to supply the necessary centripetal force and acceleration to make the rider travel in a circle. But because the loop is vertical and the rider is at the bottom the normal force not only has to supply the centripetal force but must also overcome the pull of gravity. That’s why 1 g is added in the equation. At the top of the loop, 1 g is subtracted from what is felt because the pull of gravity is helping the normal force exerted by the track instead of needing to be overcome.


10 m30 m/s2

In a roller coaster ride a rider travels as shown to the right. How many g’s will the rider feel at the top of the loop?

SOLUTIONTo calculate the g’s at the top of the loop, you will need to know the velocity of

the rider there. To find velocity, use kinetic energy.

10 m

30 m/s2

ET-TOP = EK + Ug

ET-BOTTOM = EK + Ug

0

ET-TOP = ET-BOTTOM

EK + Ug = EK + 0

(1/2)mv2 + mgh = (1/2)mv2 + 0(1/2)v2 + gh = (1/2)v2

(1/2)v2 + (9.8)17.07106 = (1/2)(30)2

(1/2)v2 + 167.2964 = (1/2)302

282.7036 = (1/2)v2

565.4072 = v2

v = 23.7783 m/s

45°

aCENTRIPETAL = (23.7783 m/s) = 565.4076 = 56.5408 m/s2

10 10

2

g’s = 56.5408 m/s2

9.8 m/s2 = 5.77 g’s

g’s FELT = 5.77-1 = 4.77 g’s felt

The rider will not pass out because 4.77 is less than 8 g’s.


EXAMPLE 2

10 m

30 m/s2

In a roller coaster ride a rider travels as shown to the right. How many g’s will the rider feel at this location of the loop? 45°

SOLUTION

height of the roller coaster car;h = radius + xh = 10 + 10 sin45°h = 10 + 7.07107h = 17.07107 m

10 m

45°x = 10 sin45°10 m

10 m

30 m/s2

ET-TOP = EK + Ug

ET-BOTTOM = EK + Ug

0

ET-TOP = ET-BOTTOM

EK + Ug = EK + 0

(1/2)mv2 + mgh = (1/2)mv2 + 0(1/2)v2 + gh = (1/2)v2

(1/2)v2 + (9.8)17.0711 = (1/2)(30)2

(1/2)v2 + 167.30 = (1/2)302

282.703 = (1/2)v2

565.406 = v2

v = 23.778 m/s

45° =10 sin 45 +10=17.07107 m

aCENTRIPETAL = (23.778 m/s)2 = 565.406 = 56.541 m/s2

10 10

g’s = 56.541 m/s2

9.8 m/s2 = 5.77 g’s

g’s FELT = 5.77 - sin45° = 5.01 g’s felt

The rider will not pass out because 5.01 is less than 8 g’s.


The simple loop is easy enough to calculate. The irregular shaped loop needs a little more work. The velocity as the car enters the loop should be known. First establish the g’s felt at the bottom. Subtract one g to know what the track exerts. Then convert these g’s to m/s2. Now solve for the radius. EXAMPLE

35 m/sR = ?

STEP 1 (I’m randomly choosing 6 g’s as the limit for the rider)Therefore the centripetal acceleration of the track is 6g - 1g = 5g’s.

STEP 2 (convert these g’s to m/s2)

(5g)g

9.80 m/s2( )= 49 m/s2

STEP 3

49 m/s2 = 352

R

R = 25 m

a = v2

R

Now to calculate what the rider feels at the top of the loop.

Decide on the height of the loop. Then decide how many g’s the rider will experience. Use the loop formulae with centripetal acceleration to calculate the radius.

35 m/sR = 25

h = 20 mR = ?

STEP 4 The top of the loop will be at 25 m. (Chosen pretty much at random.)

STEP 5 I’m randomly choosing 6 g’s again as the limit for the rider. It could be almost any number. At the top of the loop add 1 g for the centripetal force. (“Add” because the rider is upside down.)

6g’s + 1 g = 7gSTEP 6 convert g’ to m/s2


(7g)g

9.80 m/s2( ) = 68.6 m/s2

STEP 7

(1/2)(m)(35)2 = (1/2)(m)vo2 +(m)(9.80 m/s2)(20 m)

The m’s divide out.

(1/2)(35)2 = (1/2)vo2 +(9.80 m/s2)(20 m)

vo = 28.86 m/s STEP 8

Calculate the radius at the top

68.6 m/s2 = 28.862

R

R = 12.14 m

a = v2

R

35 m/s

FINAL ANSWER

R = 25h = 20 m

R = 12.14 m

In reality, a person will not pass out the instant he/she reaches 8 g’s. It will take a few seconds of being at 8 g’s for the person to pass out. But for the sake of easy calculations we will assume that the instant 8 g’s is reached a person will pass out. 8 g’s is an average. peaple generally pasout between 6 to 10 g’s. (FYI: The 40 g mark mentioned earlier is instantaneous for death.)


2R

(1/2)R

R

V = ?

h

1 What must the velocity of the car be at the top of the circular loop such that the rider FEELS weightless at the top of the first loop?

2 What must the velocity of the car be at the bottom of the circular loop such that the rider FEELS weightless at the top of the first loop?

3 How many g’s does the rider feel as he enters the circular loop, at the bottom?

4 How fast is the rider traveling when he enters the irregular loop?

5 How many g’s does the rider feel as he enters the irregular loop?

6 How fast is the rider traveling at the top of the irregular loop?

7 How many g’s does the rider feel at the top of the irregular loop?

8 How fast must the car be traveling at the top of the klothoid loop if the rider is to experience 2.00 g’s?

9 How fast would the rider be traveling as she enters the irregular loop?

10 How many g’s does the rider feel as she enters the irregular loop?

11 How many g’s does the rider feel as she enters the circular loop?

12 How many g’s does the rider feel as she passes over the top of the circular loop?

1 15.65 m/s 2 35.00 m/s 3 6 g’s 4 35 m/s5 3.5 g’s 6 16.65 m/s 7 1.00 g’s 8 19.17 m/s9 36.71 m/s 10 3.75 g’s felt 11 6.50 g’s felt

12 0.50 g’s (He feels like he might fall out of his seat.)


The activity on the following page is good for a quick introduction to loop design. It is appropriate for students who lacks the necessary math skills. It could also be used as a quick overview to loop design.

The first page is to be used as a reference. The second page is where the calculations are done on a spreadsheet.


This diagram is to be used in conjunction with the spreadsheet below and the questions on the following page.

Circular loopThe radius is constant

Irregular loopThe radius changes. Usually bigger at the bottom than the top.

The ride starts from rest.

The dropped weight propels the coaster.

The coaster comes to a rest at the top.

Below is the spreadsheet and its formulae for the “Investigating the Loop Using a Spreadsheet” handout. The text on the left hand side is in the “B” column. The “B” column is right justified.

A B C

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

train's mass (kg):train's mass (kg):

weight's mass (kg):weight's mass (kg):

weight's drop height (m):weight's drop height (m):

Train's Acceleration (g's):Train's Acceleration (g's): =(C2/(C3+C2))

Velocity at "A" (m/s):Velocity at "A" (m/s): =SQRT(2*C4*9.8*C3)

Velocity at "B" (m/s):Velocity at "B" (m/s): =C5

Radius of 1st loop (m):Radius of 1st loop (m):

g's felt at "B": =(C6*C6/C7)/9.8+1

Height at "C" (m):Height at "C" (m): =C7*2

Velocity at "C" (m/s):Velocity at "C" (m/s): =SQRT(C6*C6-(2*9.8*C9))

g's felt at "C" =(C10*C10/C7)/9.8-1

Velocity at "D" (m/s):Velocity at "D" (m/s): =C5

Radius at "D" (m):Radius at "D" (m):

g's felt at "D": =(C12*C12/C13)/9.8-1

Radius at "E" (m):Radius at "E" (m):

Height at "E" (m):Height at "E" (m):

Velocity at "E" (m/s):Velocity at "E" (m/s): =SQRT(C6*C6-(2*9.8*C16))

g's felt at "E": =(C17*C17/C15)/9.8+1

Height to "G" (m):Height to "G" (m): =C12*C12/19.6


A 5500 kg coaster train is propelled by a 120,000 kg weight that is dropped 20.0 m to the ground. The first loop has a radius of 25 m.a) How many g’s are felt by the rider as he enters the first loop?b) How fast is the rider traveling as he travels over the top of the first loop?c) How many g’s are felt by the rider as he travels over the top of the first loop?d) Make the radius at the bottom of the irregular loop 25 m. What must the radius at the

top of the second loop be if its height is 42 meters?

A 5500 kg coaster train is propelled by a 91,000 kg weight that is dropped 25.0 m to the ground. a) What must the radius of the first loop be so that a rider feels 2 g’s as she enters the

loop?b) What must the radius of the second loop be so that a rider feels 2 g’s as she enters

the loop?c) How high and what radius must the irregular loop be so that a rider feels the same

g’s at the top and bottom?

Design a roller coaster where the rider feels 2.9 to 3.1 g’s at every acceleration except at the top of the first loop. Enter your numbers at the appropriate locations on the diagram below.

Circular loopThe radius is constant

Irregular loopThe radius changes. Usually bigger at the bottom than the top.

The ride starts from rest.

The dropped weight propels the coaster.

The coaster comes to a rest at the top.

In terms of g’s felt by a rider, what are the benefits of using an irregular loop versus a circular loop?



DEMO Materials: Meter stick, C-clamp

ProcedureHold the meter stick horizontally between two fingers. Slowly slide your hands

together. For a real challenge, close your eyes when sliding your hands together. Your hands will always meet under the center of mass.

ROTATE HANDS

METER STICK

Location of the Center of Mass

Slowly slide hands together

Attach a c-clamp at one end of the meter stick. Redo the demonstration. Your fingers will still meet under the center of mass.

PART 2 of the DEMO

Location of the Center of Mass


The center of mass of a train would be in the center of the train. This is assuming all the riders are of the same mass. The feeling a track is designed for is engineered around the center of mass of a train. A rider gets a different feeling if she is to ride some distance away from the center of mass.

A hill is designed for a specific velocity. The design velocity is chosen such that the rider located at the train’s center of mass will, at most, feel weightless. The hill’s shape determines the design velocity. This shape also dictates a specific velocity at each part of the hill.

These cars, in the front, are traveling faster than the design velocity for the hill’s peak. These riders feel pushed over the hill.

THE CM OF THE TRAIN IS TRAVELING AT JUST

THE RIGHT VELOCITY FOR THIS HILL.

These cars, in the back, are slowing down too quickly. Their riders feel held back as they approach the hill.

The cars in front are traveling too slow for their position on the hill. The riders feel held back. They feel like they are hanging on the hill.



The cars in back are traveling too slow for their position on the hill. The riders feel held back. They feel like they are hanging on the hill.



The cars in front are traveling too slow fortheir position on the hill. The riders feel less of a weightless sensation.

The cars in back are traveling faster than the hill’s design velocity. The riders feel whipped out of their seat prematurely due to the negative g’s. The shoulder or lap harness is holding the rider in place.

This can be demonstrated by using the HotWheels™ train. (Construction of the HotWheels™ train is described on page 81.) Set up a box with a track running horizontally over the top. Slowly roll the train over the hill. As the front of the train begins to pass over the hill it will not speed up until the middle, center of mass of the train, travels over the hill.



A horizontal curve is a curve that does not rise or fall. There are two type of curves, flat curves and banked curves.

A flat curve gives a rider the sensation of being thrown sideways. If the roller coaster car’s velocity is fast enough and the radius small enough, the stresses on the car’s under carriage can be tremendous. For a flat curve the inward net acceleration felt by the rider is calculated from the equation.

Where “a” is the acceleration felt by the rider to the inside of the circle, “v” is the velocity of the car and “R” is the radius of the curve. This acceleration can be converted to g’s by dividing it by 9.80 m/s2.

A banked curve reduces the rider’s sensation of being thrown sideways by turning the car sideways. The car is tilted. The trick is to tilt the track just the right amount.

The ideal banked curve is one where no outside forces are needed to keep the car on the track. In other words, if the banked curve were covered with ice -no friction- and the coaster did not have a steering mechanism the car would stay on the track. These are the forces acting on the car as the car travels around horizontal banked curves.

No friction

η = Normal force of the track

weight = mgNet Force = mv2r

Coaster car’s free body diagram as viewed from the rear on the curve.

θ

θη cos(θ)

This diagram yields the followingrelationships

Σ Fx = mv2/r = η cos(θ)

Σ Fy = 0 = η sin(θ) - mg

therefore


η = mv2

(R)cos(θ)from Σ Fx ⇒

η = mg sin(θ)

from Σ Fy ⇒

mg sin(θ)

∴ = mv2

(R)cos(θ)

g sin(θ)

= v2

(R)cos(θ)

R = v2 sin(θ)gcos(θ)

R = v2 tan(θ)g

This is for the ideal banked curve where no friction is required to keep the car from sliding to the outside or inside of the curve. On a given curve if the velocity is greater or less than the design velocity then the cars may need a little frictional help to keep them on the track.

If your not comfortable with trigonometry functions, the equations can be rewritten and used as shown below.

The draw back to this method is in measuring the lengths of “x” and “y.”

x

y

Recall that the g’s felt is equal to the normal force divided by mass and then divided by g to convert to from m/s2 to g’s.

η = mg sin(θ)

from Σ Fy ⇒ ... from the above derivation.


η = mg sin(θ)mg

g’s felt ⇒

g’s felt = 1 sin(θ)

Remember this is for the ideal banked curve with no friction.

Too much bank for the car’s velocity. the car could tip to the inside. The undercarriage wheels are holding the car on. The rider fells a force pushing himself down. Friction is needed to keep the car on the track.

At just the right bank for the car’s velocity, the car does not need any type of undercarriage to stay on the track. The rider feels a force pushing his bottom into the seat. This is the optimum position where no friction is needed to keep the car on the track.

Not enough bank for the car’s velocity. The car could tip to the outside. The undercarriage wheels are holding the car on. The rider fells a force pushing himself to the outside of the curve -sideways. Friction is needed to keep the car on the track.

HIGH SPEED BANK MEDIUM SPEED BANK LOW SPEED BANK


θθ

θ

θ

θ

1 What must the curve’s angle be for a roller coaster car to travel around a curve of radius 30 m at 20 m/s?

2 How many g’s are felt by a rider as he travels around the banked curve in the previous problem?

3 A car is to make it around a banked curve. The radius is 15.35 m and the car will travel at 30 m/s. What is the optimum banking angle of the curve?

4 A car is to make it around a banked curve. The radius is 15.35 m and the car will travel at 30 m/s. This roller coaster is on the moon where the acceleration due to gravity is 1.67 m/s2. What is the optimum banking angle of the curve?

5 A rider is to make it around a curve of radius of 24.28 m so that the rider will feel 2.50 g’s. What is the angle of the banked curve?

6 A rider is to make it around a curve of radius of 31.15 m so that the rider will feel 1.64 g’s. How fast must the rider be traveling?

7 A rider is to make it around a curve of radius of 51.15 m so that the rider will feel 4.52 g’s. How fast must the rider be traveling?

10 m

18.43 m/s

R = ? On the top curve

R = 20 mOn the bottom curve

1.5 g’s felt on the top curve

8 What is the banked angle of the bottom curve?

9 How many g’s are felt by the rider along the bottom curve?

10 What is the optimum angle of the top banked curve after spiraling up 10 m?

11 What is the radius of the top curve?

1 53.68° 2 1.69 g’s 3 80.51° 4 88.37°

5 66.42° 6 19.92 m/s (52.69°) 7 47.01 m/s (77.22°) 8 60.01°

9 2.000 g’s 10 48.19° 11 13.11m (11.99 m/s at the top)


This is a demo that mounts on a blackboard that is magnetic. This demo will allow the teacher to demonstrate Galileo’s inertia experiment and how and why the hills on roller coasters are designed the way they are. When it is built, it will be an adjustable rail road type track.Materials:

• 12 feet of vinyl tubing with an outside diameter of 1/4 inch. Some larger hardware stores sell the tubing. (20¢ per foot)

• 18- 11/8 inch diameter magnets from Radio Shack.• 371/2 inches of dowel rod 3/8 inches in diameter .• 1 small tube of contact cement

• 1 tube of “Household” GOOP® or any other adhesive that will glue vinyl. (Epoxy will not

work.)

STEP 1

Cut the dowel rod into 20 pieces 17/8 inches long. Cut a 1/4 deep slit into one end of each dowel rod. The slit should be about 1/16 inch wide. A band saw is a good tool for doing this.

STEP 2

Insert the slit end of each dowel into a magnet.

STEP 3

Cut some wood into 1/4” X 1/4” X 11/2” RECTANGLES. These will be used as spacers. Mark the tubing every 4 inches with a permanent marker. Place the magnets on the board in a straight line 4 inches apart from each other.

4”

STEP 4

Put some contact cement on the top side of each dowel rod. Put some more contact cement on the vinyl tubing at each mark. With another person’s help, lay the tubing against the magnets on top of the dowel. When you have laid out 18 magnets, 6 feet, stop. Cut the tubing. Lay the other piece of tubing 1/4 inch away from the first piece. Use the cut wood as a spacer between the vinyl tubing. It should look like rail road tracks when you are done

SIDE VIEW

OVERHEAD VIEW

STEP 5

Final step. The track’s glue joints are weak. Reinforce them with GOOP™ on the underside. Be sure not to get any glue on top of the tubing. A little piece of glue on top of the tubing will cause the marble to roll off it.


MAGNET

1 7/8 inches

Marble STARTS here

Marble STOPS here

Adjust the ramp to lower angles. The marble keeps rolling to the same height. Galileo concluded from a similar experiment that the ball will keep rolling until it reaches the same height it started from if there were no friction. (With friction, it will keep rolling until it reaches the stop height.)

Marble STARTS here

Marble STOPheight

Adjust the ramp to different loop radii. Notice if the loop’s radius is too big, the marble falls off. Try adjusting the radius until the ball makes it around the loop.

Marble STARTS hereAdjust the ramp to different hill shapes. Drop the marble from the same height each time. Keep adjusting the hill’s shape until the marble no longer leaves the track. This is the desired hill shape for that start height. After all, if the roller coaster left the tracks each time, the company would lose its customers. (Try a lower start height and see how the hill shape would change.)

Marble STARTS here

Adjust the ramp to make a gradual climb on the left side and steep drop on the right. Trace the path the bottom of the ball makes as it leaves the track. Line the track up with the traced curve. The ball will stay on the track. A coaster’s hills are designed to give the rider the weightless feeling they would get if the track were not there.

Marble STOP height

Marble STOP height

Marble STARTS here PRACTICE SPEED ESTIMATIONEach support is 4 inches apart. To estimate the velocity at the bottom of this curve, count an equal number of supports on each side of the dip’s center. Calculate this distance. Time how long the ball takes to travel the distance. Calculate the average speed, (distance/time). The average will equal the exact speed at the center of the dip -but only if (1) the distance measured is equal on both sides of the dip’s center and (2) the region measured is fairly symmetrical in shape.

20 inchesalong track.


Hot Wheels™ is a registered trademark of Mattel, Inc.- 81 - by Tony Wayne

5 -Hot Wheels™ cars. (Use cars with a short front to back wheel base.)24 inches -Kevlar. (This can be found in most fishing goods departments of larger stores with

the fishing line.)30 minute Epoxy

1 Lay the cars upside down about 1/2 inch apart from each other.

1/2 inch apart

2 Lay the Kevlar from front to back on the row of cars. Tape the Kevlar in place with masking tape. Tape to the center of the car.

KEVLAR TAPE

3 Mix and apply the epoxy over the Kevlar and at each car’s rivet. The cars have two rivets on the undercarriage. Let dry for at least 30 minutes. If the epoxy mix is not right, the cars may have to dry longer.

EPOXY HERE

4 To aide in seeing the center of mass of the train of cars, paint the middle car. It is located at the center of mass.

Hot Wheels™ is a registered trademark of Mattel, Inc.



Stand on the edge of a cliff and throw a ball horizontally. It would travel as shown below.10 m/s

ROLLERCOASTER

TRACK

This is the path an object would take if it were weightless.

The roller coaster track is designed to have the same arc shape as a ball that is thrown off a cliff.

10 m/s

10 m/s

Because the two paths are the same, it stands to reason that the roller coaster car would also be weightless as it travels down the hill.

Hold a large piece of plywood at a 60° angle to the floor. Roll a WET tennis ball horizontally across the top of the board. Look at the water trail it left behind. Roll a WET tennis ball FASTER horizontally across the top of the board. Look at the new water trail it left behind.Draw the results below.

SLOW BALL’S PATH FAST BALL’S PATH


Materials: HotWheels™ Track, Hot Wheels™ carRest the top of the track on the edge of the counter. Make the track’s contact point with the ground 40 cm away from the edge on the floor. Using books:

1 . Create a hill that when the car travels over it, the car becomes airborne on the far side. Write down the measurements shown below.

40 cm A

H

Distance A: _________metersDistance H _________meters

2 . Create a hill that when the car travels over it, the car barely becomes airborne on the far side and it still lands on the track on the way down. Write down the measurements shown below.

40 cm A

H



3 . Create a hill that when the car travels over it, the car never becomes airborne on the far side. Write down the measurements shown below.

40 cm A

H


4 . Will the car come off the far side of the track shown below? Why?

40 cm A

10 cm

Distance A: _________meters

Straight piece laid on a meter stick for support.

A B

5 Over which hill will the car most likely come off the track? Why?


Materials: HotWheels™ Track, Hot Wheels™ car, meter stickConstruct a loop in the shape of a “klothoid.” The klothoid shape is like an upside down tear drop.

Klothoid Loop

Vertical distance to the starting height (meters)Vertical loop

height

Vertical loop height is 15 cm

Start the car at various heights. Determine the starting height in which the car just makes it around the loop without falling off the track. This height is ___________m.

Circular Loop

Start the car from the same height as from the previous klothoid loop shape. Vertical loop

height

Vertical loop height is 15 cm

Does the car make it around the circular loop from the same height as before? ______ If the car falls off the track, draw an arrow on the circular loop to indicate where this happens.

What is the minimum height the car needs to be started from to just make it around the circular loop?

________________________m.

Why are loops on most roller coaster rides klothoid shaped instead of circular?


In order to feel weight there must be a “reaction” force.

Force due to the pull of gravity, called weight.

Reaction force of the floor. This reaction force is the same as the weight

If you jump out of a window, you would feel weightless because there is no reaction force pushing up on your feet.

cup (with holes), water, chair, paper towels, trash can

Poke two holes on opposite sides of the bottom of the cup. Stand in a chair. Hold the cup with your fingers over the holes. Fill the cup 1/3 full with water. Hold the cup with your other hand. Briefly remove your fingers from the holes. Observe what happens. Cover the holes again with your fingers. Hold the cup up as high as possible. Drop the cup and observe what happens to the water this time.

Why doesn’t the water come out of the holes when it is dropped?

When on the roller coaster below do you think you are weightless?Start of the ride


Swing the apparatus around with a super ball in the cup.

Swing it just slow enough so the ball does not fall out.

Hold here Hold here

Use either apparatus

Take whatever measurements you need to calculate the centripetal acceleration of the ball in the cup. ____________.

How many g’s is this? _________g’s.


Mark a starting point and walk 10 steps. Walk naturally. Measure this distance. Calculate the distance traveled for each of your steps.

Distance = _________ Step

Using the previous information and “Two Angle Method”, calculate the height of the edge of the school building.

Building edge height ___________.

In the classroom is a HotWheels™ set up. On this set up, measure the velocity of the HotWheelsTM train as it passes over the second hill. The length of the train is the distance used to calculate the average velocity. The time measured is the time for the entire train to pass one point on the center of the hill.Treat this as a normal lab and repeat the process a number of times and take the average of the trials.

HotWheels™ train

Calculate the velocity of the train here.

HotWheels™ track

HotWheels™ train velocity: _____________

The train is heavier than a single car. When the train is in motion, it will cause the track to slide around. Hold the track down firmly to prevent accidents.



CAR LENGTH

START TIMING WHEN THE FRONT OF THE CAR PASSES A POINT. STOP TIMING WHEN THE END OF THE CAR PASSES THE POINT.

TIME THE DISTANCE FROM POINT 1 TO POINT 2.

THE POINTS MUST BE THE SAME DISTANCE FROM THE CENTER. THE CURVE’S SHAPEMUST BE THE SAME

ON BOTH SIDES. ESTIMATE THE DISTANCE TRAVELED BY COUNTING THE STEEL SUPPORTS & ESTIMATING THE DISTANCE BETWEEN THEM.

1 2

EXAMPLE

There are 20 supports between locations 1 and 2. Each support is 1.5 meters apart. It takes 1.5 seconds to travel from location 1 to 2.

Velocity = 20(1.5) = 30 = 20 m/s 1.5 1.5

CAR LENGTH





1 2

EXAMPLE


Velocity = 20(1.5) = 30 = 20 m/s 1.5 1.5

CAR LENGTH





1 2

EXAMPLE


Velocity = 20(1.5) = 30 = 20 m/s 1.5 1.5

CAR LENGTH





1 2

EXAMPLE


Velocity = 20(1.5) = 30 = 20 m/s 1.5 1.5


x = xo + vot + (1/2)at2

v = vo + atv2 = vo2 +2ax

vavg =

F= map = mv

v+vo

2xt

=

EK = (1/2)mv2

Ug = mghUs = (1/2)kx2

W = FddETOTAL = EK + Ug + Us

P = W/tP = Fv

T =

T =

g =9.80 m/s2

g = 32.15 ft/s2

Lg

2π

mk

2π

Estimate this distance and count support structure up for the height and width.

If this structure is 10 ft high, then the right support is 5 x 10ft = 50 ft high.

x = xo + vot + (1/2)at2


vavg =

F= map = mv

v+vo

2xt

=

EK = (1/2)mv2



P = W/tP = Fv

T =

T =

g =9.80 m/s2

g = 32.15 ft/s2

Lg

2π

mk

2π



x = xo + vot + (1/2)at2


vavg =

F= map = mv

v+vo

2xt

=

EK = (1/2)mv2



P = W/tP = Fv

T =

T =

g =9.80 m/s2

g = 32.15 ft/s2

Lg

2π

mk

2π



x = xo + vot + (1/2)at2


vavg =

F= map = mv

v+vo

2xt

=

EK = (1/2)mv2



P = W/tP = Fv

T =

T =

g =9.80 m/s2

g = 32.15 ft/s2

Lg

2π

mk

2π




This is good method to use when you are on flat, level, ground and you know the distance from where you are standing to the edge of the object whose height you are trying to calculate.

h = ?

θ

tan(θ) = hb

b

b = h

tan(θ)

Angle

Washer

String

ProtractorDrinking Straw

Look at the point of interest through the drinking straw. Read the angle.


This is a good method when you do not know the distance between you and the structure. All you will need to do is take two angle measurements of the structure and measure the distance between the measurements as shown in the diagram below.

h = ?

θ φd

a

b

tan(θ) = hb

tan(φ ) = ha

b = h

tan(θ)a =

htan(φ )

d = b - a

d =h

tan(θ)h

tan(φ )

d =1

tan(θ)1

tan(φ )h

d =1

tan(θ)1

tan(φ )h

h = d[tan(θ) - tan(φ )]


Average velocity equals distance over time.The distance measurement comes from the train of cars. While in the loading station, measure the length of the train. This can be done quickly by measureing from the back of one train car to the back of the adjacent train car.

Measure the distance from the back of one train car to the back of an adjacent train car.

Then multiply this distance by the number of cars in the train.

Time how long it takes for the front and the back of the train to pass the same point on the tracks. Pick a point that is in the very top of a hill or the very bottom of a dip.

Vaverage = time to pass one location

length of train


Average velocity equals distance over time.The distance measurement comes from a known length of track. Most tracks are constructed from track cross ties that are some equal distance apart from each other. Either get this distance form the park, measure or estimate it.

Measure or estimate this length. Use these cross tie supports and their distance as a measuring stick on the track.

Time how long it takes for the middle of the train to pass across all 10 sections. You may use any number of track sections so long as there are an equal number on each side and shape of the track is fairly symmetrical about the middle section of the track.

5 sections of track

5 sections of track

Cross tie supports above this dotted line are not symmetrical about the center of the dip.


Vaverage = time to cover this lengthlength of distance travled


You are a secret agent from a competing amusement park. Your task is to take a roller coaster and find out EVERYTHING you can about the roller coaster.

Group Member’s

PART 1 Most roller coasters wrap themselves in a circle. Draw a SCALE drawing of your roller coaster’s first 5 peaks and dips if you could straighten it out. (You do not have to use all the horizontal length of this graphing space -the numbers are just for reference.)

START 1 2 3 4 5 6 7

8 9 10 11 12 13 14 15

16 17 18 19 20 21 22 23


• Label each hillÕs peak as a letter, A - E on your scale drawing. You do not have to use every letter. Note, the top of a loop counts as a hill peak.

• Label each dip as a number, 1 - 5 on your scale drawing. You do not have to use every number. Note, each time the roller coaster is at the bottom of a loop counts as a different dip.

PART 2 Calculate the height at each peak.PEAKS

A B C D E PART 3 Calculate the velocity at each letter and number.

PEAKSA B C D E

DIPS1 2 3 4 5 6

PART 4 Someone in your group needs to ride the roller coaster. Measure the g’s at the first 4 dips and hills. Note which part of the car you are riding in at the time of the measurement.

gÕs Number of rows of seats from the front (middle is best)1 2 3 4

PART 5 Measure the g’s at the first 2 dips and the first 2 peaks after the initial hill from the back, middle and front of the train.For each location indicate how many seats from the front you are riding.

Location BACK gÕs MIDDLE gÕs FRONT gÕs A B 1 2

PART 6 Using the velocity and acceleration for each dip, calculate the radius of curvature for the first 4 dips.

1 2

3 4

PART 7 What is the average velocity for the entire ride? PART 8 Excluding the time it takes for the roller coaster to travel up the first hill, estimate the length of the remaining track, measure the time it takes for the train to travel this length and calculate the average velocity for this section of track.

Length:Time:

Average Velocity:

PART 9 Estimate the height of the first DROP.

PART 10 Use your estimated numbers and any other numbers you need to measure and/or calculate to calculate the height of the first drop using conservation of energy methods.

Height from energy relationships:


You are a secret agent from a competing amusement park. You task is to take a roller coaster and find out EVERYTHING you can about the roller coaster.

Group Member’s

PART 1 Most roller coasters wrap themselves in a circle. Draw a SCALE drawing of your roller coaster’s first 5 peaks and dips if you could straighten it out. (You do not have to use all the horizontal length of this graphing space -the numbers are just for reference.)

START 1 2 3 4 5 6 7

8 9 10 11 12 13 14 15

16 17 18 19 20 21 22 23


• Label each hillÕs peak as a letter, A - E on your scale drawing. You do not have to use every letter. Note, the top of a loop counts as a hill peak.

• Label each dip as a number, 1 - 5 on your scale drawing. You do not have to use every number. Note, each time the roller coaster is at the bottom of the loop counts as a different dip.

PART 2 Calculate the height at each peak.PEAKS

A B C D E PART 3 Calculate the velocity at each letter and number.

PEAKSA B 1 2

PACES METERS

PART 4 How long is the length of the train of roller coaster cars. How many cars make up the train? How many rows of seats are in the train?

PART 5 Someone in your group needs to ride the roller coaster. Measure the g’s at the first 4 dips and hills. Note which part of the car you are riding in at the time of the measurement.

DIP # gÕs Number of rows of seats from the front (middle is best)1 2

PART 6 Measure the g’s at the first 2 dips and the first 2 peaks after the initial hill from the back, middle and front of the train.For each location indicate how many seats from the front you are riding.

Location BACK gÕs MIDDLE gÕs FRONT gÕs A B 1 2

PART 7 What is the average velocity for the entire ride?

PART 8 Excluding the time it takes for the roller coaster to travel up the first hill, estimate the length of the remaining track, measure the time it takes for the train to travel this length and calculate the average velocity for this section of track.

Length:Time:

Average Velocity:

PART 9 Which roller coaster(s) in the park has/have a hill that is NOT parabolic?

Which hill(s) was/were not parabolic? (1ST, 2ND, 3RD, ETC)


The next lab was designed for the park that does not allow accelerometers to be taken on the rides themselves.


1

2

3

4

What is the length of the roller coaster train?

What is the distance between cross tie supports on the ride?

What is the published length of the track?

How many roller coaster cars make up a train of cars?

What is the mass of each coaster car?

TRAIN LENGTH

Distance between supports

1st Hill’s Height?

Velocity at the top of the first hill?

What is the vertical height of the 1st hill that the roller coaster car is lifted up?


How fast is the car traveling over the top of the hill?

How much time does it take for the middle of the train to reach the highest point of the first hill?

What is the velocity of the train at the bottom of the dip after the first hill?

Using the velocity in the previous question calculate the height of this drop.

What is the velocity of the roller coaster at the bottom of the 1st loop?

What is the velocity of the roller coaster at the top of the loop?

Using the information from the previous questions, calculate the vertical height of the loop.

Suppose your loop was designed using simple geometry. Below is an irregular loop shape that is designed from the splicing together of two large circles with a smaller circle. The smaller circle’s radius is half the larger circle’s radius. Using the information from the previous problem calculate the radii of the two circles.

Bigger circle’s radius

Smaller circle’s radius

12

R

R

H

Using the calculated radius information above, calculate the g’s felt by a rider at the bottom of the loop and at the top of the loop.


Using your previous answers do you think it is possible that the loop is designed according to our simple geometry model?

Calculate the velocity of the car as it travels over the 2nd hill.

Calculate the velocity of the car as it travels past the bottom of the next dip.

Calculate the velocity of the car as it travels over the 3rd hill.

Calculate the velocity of the car as it travels past the bottom of the next dip.

Assuming the initial total mechanical energy of the train is ZERO, how much total mechanical energy is gained by raising the train of cars to the top of the first hill.

The train has to lose its total mechanical energy by the time it reaches the end of the ride. Assuming 2/3’s of this initial energy at the top of the hill is lost due to friction during the length of the entire ride, what is the average force of friction opposing the train’s motion as it travels along the entire length of the track? (HINT: Use energy and work)


How much horsepower was used to raise the train up the first hill?

If electricity costs $0.20/(kW•hr), then how much does it cost to raise the train up the first hill?

How many runs does a train make in 1 hour?

How much does it cost to run the train in a 14 hour day?



Your Name:

Your Grader:

Due Dates:• ____________ : Turn in your roller coaster with your answers.• ____________ : Turn in your roller coaster with your answers and the other groups

answers to your roller coaster -No late papers accepted. This counts as a lab grade.

• You may work in groups of 2.

⇒ Your roller coaster must be neatly drawn in pen. It must take up no more than one 8.5 X 11 inch piece of paper.

⇒ Your questions are to be written on a separate sheet of paper.⇒ Your solutions are to be on a third sheet -SHOW ALL WORK OR LOSE 8 POINTS.⇒ Do not give your answers to the group checking your work. After they have finished

checking your work compare answers and solve any discrepancies among yourselves.

Your (group’s) roller coaster’s points depend on if your roller coaster includes:

Work calculated from Fd equation -with at least 1 question about it. (7 pts.)

Work calculated from a graph of force vs. distance-with at least 1 question about it. The graph cannot be a horizontal line. (7 pts.)

A loop-the-loop -with at least 2 questions about it. (7 pts.)

A spring -with at least 1 question about it. (7 pts.)

A hill -with at least 2 question about it. (7 pts.)

At least one of the above questions must be about the g’s felt by the rider. (7 pts.)

The rider experiences g’s no higher than 10 g’s at every hill top and dip. (7 pts.)

Calculated the solutions to someone else's lab. (25% of the lab grade.)Whose lab did you grade?

Percentage of your lab’s questions that were correct as graded by another group.

___________ (20 pts. maximum)

Total number of points checked above:


Your Name:

Your Grader:

Overview:For this project, you will create a poster that diagrams a made up roller coaster. This roller coaster will contain calculations of g’s, velocities, heights, spring constants, etc.

Due Date • No late papers accepted.• This counts as a lab grade.• You may work in groups of 2 or by yourself.

Presentation guidelines:Neatly drawn with a magic marker. straight lines drawn with a straight edge or computersmooth curves drawn with drawing aides or a computerno white outvirtually no smudgesgrid lines drawn every set distance with a ball point penSizebigger than 10” X 13”smaller than 30” X 36”Must be drawn on poster board or mounted to poster boardThe board must have the coaster without any numbers drawn on it.The board must have a plastic sheet cover.

• The cover contains all numbers and answers. These numbers are to be placed at the corresponding locations on the coaster track.

• The plastic cover must be attached only at the top of the poster board.

Calculation guidelines: Every hill, dip and loop, top and bottom.Calculate and label the velocity, centripetal acceleration expressed in m/s2, and the g’s felt by the rider.Every linear force (either from a graph or pure number)Calculate and label the force at the beginning where the force is applied and at the end of where the force is applied. Also calculate the acceleration due to this force expressed in m/s2 and g’s.Every springCalculate and label the spring constant.Calculate and label maximum compression distance of the spring.Calculate and label the velocity of the coaster’s car before it hits the spring.EVERY SINGLE CALCULATION must be NEATLY written on a 1/2 sheet of paper.


Other Maximum positive g’s must be less than 10 g’s to be considered for an “A” grade. Maximum negative g’s must be less than 4 g’s to be considered for an “A” grade. Must contain at least one loop Must contain one spring OR force that does work -whose force is calculated from a graph.

GRADESGrades are based on the neatness of your presentation and the correctness of your work. It is hard to get an “A.” An “A” project follows all the rules above and is creative. It is the exemplary project. It is easier to get a “B.”





HEIGHT (meters)


HEIGHT (meters)


Roller coasters are about speed control and sensations. When designing a roller coaster the designer needs to be able to predict speeds and the riders’ sensations.

A free fall hill is defined by two formulae. The first formula is for the top of the hill. The second formula is for the bottom of the hill. The ending and beginning locations for the top and the bottom of the hill meet at the transition point. The transition point is the location of the maximum angle. The designer usually defines this angle somewhere between 35° and 60°.

Transition point where the hill is at its maximum angle

This portion of the hill is where the rider is in free fall.

It obeys the formula.

x = VxoVyo( ) − Vyo( )2 − 2 ay( )y

ay

−

ax

2

Vyo( ) − Vyo( )2 − 2 ay( )y

ay

2

The curve below the transition point obeys the formula.x =2y vtop( )2

g

The equation for the section of the hill before the transition point is

y = gx2v

2

2

y = the height from the top of the hillx = is the distance away from the center of the hillv = the velocity the roller coaster car travels over the

top of the hill.a = the acceleration due to gravity. 9.80 m/s2 for

answers in meters. 32.15 ft/s2 for answers in feet.


There is a point on the hill where it no longer follows the equations of projectile motion. This is the transition point. The transition point is the location on the hill where its angle is at a maximum. Below the transition point the curve is created following some different rules.

1) To design a hill from top to bottom decide on the initial velocity as the car travels over the hill, the height of the hill, and the maximum inclination angle. The inclination angle will determine how steep the hill will become. Choose a value between 35° and 60° as a “ball park” starting figure. The bigger the maximum inclination angle, the greater the acceleration at the bottom on a given hill. If the acceleration at the bottom of the hill is too great, make the hill higher or reduce the maximum inclination angle. As the free-fall part of the curve is plotted keep track of the angle between successive points. When the angle of the hill matches or just exceeds the maximum angle of inclination of the hill, stop using the free-fall equations.

When doing the actual calculations, use small vertical step intervals, e.g. 1.0 meter increments. Make the top of the hill zero. For each y step down you will get an x value. These are your x, y coordinates for the hill’s shape.

2) Calculate the velocity of the bottom of the hill using energy relationships. [(vTOP)2 = (vBOTTOM2) +2gh]

θ

θ

Vo

Vy

Vx

If energy methods are used to calculate the velocity here, then the answer is Vo in the

diagram below.

Transition point(Angle of maximum inclination)

Tangent line

90°

ay

ax


3) Calculate the remaining distance to the bottom of the hill from the location of the maximum inclination angle.

Transition point where the hill is at its maximum angle

This portion of the hill is where the rider is in free fall.

RemainingHeight

4) Decide on an acceleration that will be used to change the shape of the hill. Say 1.5 to 2.5 g’s (14.7 to 24.5 m/s2). There are two accelerations the coaster car will feel. One vertically will slow its vertical velocity to zero. A second horizontal velocity will increase the car’s velocity to the final velocity. The total acceleration is

anet = ax( )2+ ay( )2anet

ax

ay

θ

θ = tan− 1 ay

ax

Remember,

in the vertical direction. In the horizontal direction g’s felt = ax. To calculate the g’s felt by the rider use one of the g’s felt values when calculating anet.


5) Calculate the vertical component of velocity at the location of maximum inclination angle.Calculate the horizontal component of velocity at the location of maximum inclination angle. These velocities are the initial velocities for the final section of the track.

6) Use the formula

x = VxoVyo( ) − Vyo( )2 − 2 ay( )y

ay

−

ax

2

Vyo( ) − Vyo( )2 − 2 ay( )y

ay

2

to calculate the bottom half of the hill.



This will show the reader the basic steps to designing a roller coaster. The example coaster will not be the best possible design. (I don’t want students to use it as their own in other projects.)

Draw a picture of what the coaster may look like.

BANKED CURVE

Assign some beginning numbers.Numbers like the initial velocity as the coaster train leaves the station. The mass of the coaster train. And/or an initial velocity as it tops the first hill. (Label the pieces for easy identification when analyzing.)

5.24 m/s

BANKED CURVE

22.5 m

8.26 m/s

124.5 m


Begin to calculate everything you can and check to see if it makes sense.For this design start by calculating the force needed to pull the train up the incline and the power to pull it up the incline.

In order to calculate the power force and power to pull it up the incline I’m going to need the train’s mass. So make up a reasonable mass. This coaster is made up of 6 cars. Each car has a maximum mass, with two riders at 100 kg each, of 735 kg; (535 kg car + 100 kg rider + 100 kg rider.) Therefore, the coaster train will have a mass of 4410 kg.

What angle will the first incline be? The designer can choose this number too. 42° is good. The train is going to be pulled up vertically a distance of, (124.5-22.5), 102.0 m.

102.0 m

42°

Incline length?

Incline length = 102/(sin42°)

Incline length = 152.4 m

ET(OUT OF STATION) + Work = ET(TOP OF 1st HILL)


(1/2)4410(5.24)2 + 4410(9.8)(22.5) + F(152.4) = (1/2)4410(8.26)2 + 4410(9.8)(124.5) 60544.008 + 972405 + F(152.4) = 150441.858 + 5380641

F(152.4) = 4498133.85F = 29515.314 N ... is the pulling force

along the incline.

How much time will it take to travel up the incline?

The acceleration of the train is found from (vf)2 = (vo)2 +2ad

(8.26)2 = (5.24)2 + 2(a)da = 0.134 m/s2

vf = vo + at8.26 = 5.24 + 0.134(t)

t = 22.537 sec ...is the time to climb the incline.(Most initial lift times are between 60 and 120 seconds.)

It is beyond the scope of this book to show how to calculate the time for each track element. It was shown here because of its ease of calculation.

Calculate maximum velocity of the ride.- 126 - by Tony Wayne

5.24 m/s

BANKED CURVE

22.5 m

8.86 m/s

124.5 m

Since the 1st drop is the longest, the velocity at the bottom will be the greatest, (location #3). Energy relationships will be used to calculate the velocity.

ET(LOCATION #2) = ET(LOCATION #3)


(1/2)4410(8.26)2 + 4410(9.8)(124.5) = (1/2)4410(v)2 + 4410(9.8)(0) 150441.858 + 5380641 = 2205(v)2

2508.428 = (v)2

v = 50.084v = 50.1 m/s ... At the bottom of the first hill

That’s 112 mi/hr !!!

The loop.For any loop, the designer would like to know the velocity as the rider enters the loop; at the top of the loop; and as the rider leaves the loop. The designer would also like to know the g’s felt by the passengers. This is the location on the ride where riders are most likely to pass out if the g’s are too much. The radius in the loop below is made up.


5.24 m/s

BANKED CURVE

22.5 m

124.5 m

8.86 m/s

R = 31.2 m

The velocity as the rider enters the loop and as the rider leaves the loop is the same as the velocity at the bottom of the first hill. This is because all three locations are at the same height.

The velocity at the top of the loop is not the same as at the bottom. As the coaster travels up the loop it will lose kinetic energy and gain potential energy.

The height of the loop is simply double the radius. h = 2(31.2) = 62.4 m



(1/2)4410(8.26)2 + 4410(9.8)(124.5) = (1/2)4410(v)2 + 4410(9.8)(62.4) 150441.858 + 5380641 = 2205(v)2 + 2696803.2

1285.388 = (v)2

v = 35.852v = 35.9 m/s ... At the bottom of the first hill

That’s 80.3 mi/hr !!! If you are doing this calculation and you get an expression that requires you to calculate the velocity by taking the SQUARE ROOT OF A NEGATIVE NUMBER, then the loop is too tall for the given velocity at the bottom of the loop. The velocity will need to be increased or the height of the loop decreased.

To calculate the g’s felt by the rider, calculate the centripetal acceleration at each location, convert to g’s and either add or subtract a g as necessary.


v = 50.084 m/sr = 31.2 m

ac = v2

r

ac = 50.0842 m/s31.2 m

ac = 80.398 m/s2

ac = 80.398 m/s2

9.80 m/s2

ac = 8.2 g’sac = 8.2 g’s + 1gac = 9.2 g’s ... That is an incredible amount of g’s. Most

coasters do not go above 5 g’s. To be safe the radius at the bottom of the loop needs to be bigger.

v = 35.852 m/sr = 31.2 m

ac = v2

r

ac = 35.8522 m/s31.2 m

ac = 41.198 m/s2

ac = 41.198 m/s2

9.80 m/s2

ac = 4.2 g’sac = 4.2 g’s - 1gac = 3.2 g’s ... That is an acceptable amount. But 3.2 g’s is

rather high for the top of a loop. Most of the time the g’s at the top of a loop are from 1.5 to 2 g’s.

Because of the need for a larger radius as the rider enters the loop above, this coaster might be a good candidate for an irregular loop like the one below.


5.24 m/s

R1 =??

BANKED CURVE

22.5 m

124.5 m

R2 = ??

h = ??

8.86 m/s

“h” can equal the sum of the 2 radii or a number a little bigger or smaller than their sum. This is not a design to be built so it is up to the engineer.

For this new loop, the designer will have to calculate the new velocity at the top of the loop using the new height. The g’s felt will also need to be recalculated using the 2 new radii. This is left as an exercise for the reader.

The banked curve.The banked curve is a horizontal curve on the ground in this diagram. Because it is at the lowest point it’s velocity is equal to that of location #3, where v = 50.084 m/s. For a first try I’ll make the radius 31.2 m.

5.24 m/s

31.2 m

BANKED CURVE

22.5 m

124.5 m

8.86 m/s

R = 31.2 m

The curve will be designed at the optimum angle where no friction or outside lateral forces are needed to keep the car on the track at speed. “At speed” means the velocity of the track design.


tan(θ) = v2

rg

tan(θ) = 50.0842 m/s31.2 m(9.8) m/s2)

tan(θ) = 0.164

θ = 9.3° .... That’s almost a flat turn. It might be more exciting to try to decrease the radius so a greater banking angle will be needed.

The g’s felt are calculated from

g’s felt = 1cos(θ)

1cos (9.3°)

g’s felt =

g’s felt = 1.013 g’s .... This is not much more than normal gravity. These g’s are the g’s felt applied to your seat. Because this curve is rather flat, it would be wise to examine lateral, centripetal, acceleration in g’s.

Lateral g’s are the g’s felt in the horizontal plain of the curve.v = 50.084 m/sr = 31.2 m ... of the curve. This just happens to be the same as the loop’s radius.

ac = v2

r

ac = 50.0842 m/s31.2 m

ac = 80.398 m/s2

ac = 80.398 m/s2

9.80 m/s2

ac = 8.2 g’s ... Do not add or subtract a “g” because the circular motion is horizontal and not in the vertical plain.

... That is a lot of lateral g’s. That’s 8.2 times the rider’s weight pressing him against the side of the coaster car. It is too extreme. Maybe a value around 1 to 2 g’s would be better tolerated by the rider.

The banked curve needs to be redesigned.


The camel back The camel back humps begin at the lowest part of the track and climb to a height of 21.5 m. The calculations to check the velocity at the top of the hump is similar to the one for the drop from location 2 to 3.

5.24 m/s

31.2 m

BANKED CURVE

22.5 m

124.5 m

8.86 m/s

R = 31.2 m

21.5 m


KE + PE = KE +PE(1/2)mv2 + 0 = (1/2)mv2 + mgh

(1/2)v2 + 0 = (1/2)v2 + gh(1/2)(50.084)2 + 0 = (1/2)(v)2 + (9.8)(21.5)

1254.203528 = (1/2)(v)2 + 210.72087.007056 = (v)2

v = 45.6838v = 45.7 m/s ... At the bottom of the first camel

back hill. That’s 102 mi/hr !!!


STANDARD LOOP COASTER

80 m

20m

35 m

10 m/s

800 kg

1 How fast is the roller coaster car traveling at the bottom of the hill?2 How fast is the roller coaster traveling as it enters the loop?3 What is the centripetal acceleration applied by the track at the bottom of the

loop?4 How many g’s does the rider feel at the bottom of the loop?5 How fast is the roller coaster car traveling at the top of the loop?6 What is the centripetal acceleration applied by the track at the top of the loop?7 How many g’s does the rider feel at the top of the loop?8 How fast is the roller coaster car traveling at the top of the 35 m hill?


IRREGULAR LOOP COASTER

80 m

10 m

35 m

10 m/s

800 kg

1 How fast is the roller coaster car traveling at the bottom of the hill?2 How fast is the roller coaster traveling as it enters the loop?3 What is the centripetal acceleration applied by the track at the bottom of the

loop?4 How many g’s does the rider feel at the bottom of the loop?5 How fast is the roller coaster car traveling at the top of the loop?6 What is the centripetal acceleration applied by the track at the top of the loop?7 How many g’s does the rider feel at the top of the loop?8 How fast is the roller coaster car traveling at the top of the 35 m hill?

50 m

40 m

The roller coaster car enters the loop here.


ROLLER COASTER DETECTIVE

NO FRICTION

h1R

20 m

4g’s felt,

28 m/s

800 kg, 3 m/s

h2


A

B

C

2g’s felt, 9 m/s

D

1 What is the height, between “A” and “B?”2 What is the centripetal acceleration at “B?”3 What is the radius of the track at “B?”4 How high is location “C?”5 What is the radius of the track at location “C?”6 What is the velocity of the car at location “D?”

1 ) 39.54 m 2 ) 29.4 m/s2 3 ) 26.67 m 4 ) 35.87 m 5 ) 2.76 m 6 ) 19.80 m/s


h1R

20 m

4g’s felt,

35 m/s

800 kg, 3 m/s

h2


B

C

D

2g’s felt, 12 m/s

E

h = 30

k = 12,000 N/m x = ?

Stop

A motor pulls the car up the hill before turning off at the top.A

At the lowest height

7 m/s

1 What is the height, between “B” and “C?”2 What is the centripetal acceleration at “C?”3 What is the radius of the track at “C?”4 How high is location “D?”5 What is the radius of the track at location “D?”6 How fast is the car traveling at location “E?”7 How far is the spring compressed from its hanging position, x?8 If the car traveled up the first hill in 1.2 minutes, then how much power was used to pull it up the

hill in horsepower?

1 ) 62.04 m 2 ) 29.4 m/s2 3 ) 41.7 m 4 ) 55.15 m 5 ) 7.34 m 6 ) 28.86 m/s 7 ) 5.89 m

8 ) 8.76 hp, 6533 w


ET4 = EK + UG

1. ET1 = ET2 2. 40.84 m/s because it’s height 4. AC in g’s = 83.4 m/s2

EK + UG = EK + UG does not change from #1.

1/2 mv2 + mgh = 1/2 mv2 AC = 8.51 g’s

1/2 v2 + gh = 1/2 v2 3. AC = v2 Add 1 g at the bottom 1/2 102 + 9.8(80) = 1/2 v2 for what the rider feels. 50 + 784 = 1/2 v2 AC = 40.842 8.51 g + 1 g = 9.51 g’s 1668 = v2

v = 40.84 m/s AC = 83.4 m/s2

STANDARD LOOP COASTER

80 m

20m

35 m

10 m/s

800 kg

ET3 = EK + UGET2 = EK + UG

ET1 = EK + UG

R

20

9.8 m/s2

5. ET1 = ET3

EK + UG = EK + UG

1/2 mv2 + mgh = 1/2 mv2 + mgh 1/2 v2 + gh = 1/2 v2 + gh 1/2(10)2 + 9.8(80) = 1/ + 9.8(40) 50 + 784 = 1/2 v2 + 392 442 = 1/2 v2

v = 29.73 m/s

6. AC = v2

AC = 29.732

AC = 44.2 m/s2

R

20

AC in g’s = 44.2 m/s2 9.8 m/s2

7. AC in g’s = 4.5 g’sSubtract 1 g at the top for what the rider feels.4.5 g - 1 g = 5.5 g’s

8. ET1 = ET4

EK + UG = EK +UG

1/2 mv2 + mgh = 1/2 mv2 + mgh 1/2 v2 + gh = 1/2 v2 + gh 1/2 (10)2 + 9.8(80) = 1/2 v2 + 9.8(35) 50 + 784 = 1/2 v2 + 343 491 = 1/2 v2

v = 31.34 m/s

ANSWERS


IRREGULAR LOOP COASTER

80 m

10 m

35 m

10 m/s

800 kg

50 m

40 m


ET1 = EK + UG

ET2 = EK + UG

ET3 = EK + UG

1. ET1 = ET2 2. 40.84 m/s because it’s height 4. AC in g’s = 33.36 m/s2

EK + UG = EK + UG does not change from #1.

1/2 mv2 + mgh = 1/2 mv2 AC = 3.40 g’s

1/2 v2 + gh = 1/2 v2 3. AC = v2 Add 1 g at the bottom 1/2 102 + 9.8(80) = 1/2 v2 for what the rider feels. 50 + 784 = 1/2 v2 AC = 40.842 3.40 g + 1 g = 4.40 g’s 1668 = v2

v = 40.84 m/s AC = 33.36 m/s2

R

50

9.8 m/s2

5. ET1 = ET3

EK + UG = EK + UG

1/2 mv2 + mgh = 1/2 mv2 + mgh 1/2 v2 + gh = 1/2 v2 + gh 1/2(10)2 + 9.8(80) = 1/2v2 + 9.8(40) 50 + 784 = 1/2 v2 + 392 442 = 1/2 v2

v = 29.73 m/s

6. AC = v2

AC = 29.732

AC = 88.39 m/s2

R

10

AC in g’s = 88.39 m/s2 9.8 m/s2

7. AC in g’s = 9.02 g’sSubtract 1 g at the top for what the rider feels.9.02 g - 1 g = 8.02 g’s felt

8. ET1 = ET4

EK + UG = EK +UG

1/2 mv2 + mgh = 1/2 mv2 + mgh 1/2 v2 + gh = 1/2 v2 + gh 1/2 (10)2 + 9.8(80) = 1/2 v2 + 9.8(35) 50 + 784 = 1/2 v2 + 343 491 = 1/2 v2

v = 31.34 m/s

ET4 = EK + UG

ANSWERS


The information on this diagram is to be used for the accompanying questions.

Using the information shown on the diagram on the other page, determine the correct answers to the following questions.

1 How much power was used to pull the coaster car up the first hill? (in watts and horsepower.)

2 How fast is the car traveling at location “A” if it coasted down the back side of the first hill?

3 How many g’s does the rider feel as he enters the bottom of the next element if it’s radius is 60.0 m, at location “B?”

4 How fast is the car traveling when it reaches the banked curve?5 If the banked curve is at a 52.3° angle with the horizontal, then what is the radius of the

banked curve?6 How many g’s does the rider feel in the banked curve?7 How many g’s does a rider feel as he enters the bottom of the loop?8 How fast is the rider traveling at the top of the loop?9 How many g’s does the rider feel at the top of the loop?

10 As the car enters the station at the end of the ride, its brakes are applied. How much power do the brakes exert on the car? (in watts and horsepower.)

Use Energy relationships and power to solvevo = 5.50 m/svf = 4.44 m/sh = 62.5 mt = 45.8 sm =6500 kg

Power = Worktime

= ∆Energytime

∆ET = [KEf + PEf] - [KEo + PEo]∆ET = [(1/2)m(vf)2 + mg(hf)] - [(1/2)m(vo)2 + 0]∆ET = [(1/2)(6500)(4.44)2 + (6500)(9.80)(62.5)] - [(1/2)(6500)(5.50)2]∆ET = 64069.2 + 3981250 - 98312.5∆ET = 3947006.7 JPower = =∆Energy

time3947006.7 J

45.8 s

Power = 86,179.186 watts ...746 w = 1 hpPower = 115.52 hp

Use Energy relationships to solvevo = 4.44 m/s ...at the hill topvf = ???? ...at the hill bottomh = 62.5 + 22.1 = 84.6 m

ET @ the hill top = ET @ the hill bottomKE + PE = KE + PE

(1/2)mv2 + mgh = (1/2)mv2 + 0(1/2)v2 + gh = (1/2)v2 + 0

(1/2)(4.44)2 + (9.80)(84.6) = (1/2)v2

838.94 = (1/2)v2

v = 40.96 m/s

Use circular motion relationships to solvev = 45.5 m/s, r = 60.0 m

ac = V2 = 45.52 = 34.50417 m/s2

r 60.0

ac = 34.50417 m/s2 = 3.52 g’s ...convert ac to g’s 9.80 m/s2

g’s felt at the bottom = ac[in g’s] + 1 gg’s felt at the bottom = 3.52 + 1 gg’s felt at the bottom = 4.52 g’s


Use Energy relationships to solveET @ “B” = ET @ “A”KE + PE = KE + PE

(1/2)mv2 + 0 = (1/2)mv2 + mgh(1/2)v2 = (1/2)v2 + gh

(1/2)(45.5)2 = (1/2)v2+ (9.80)(60.0)447.125 = (1/2)v2

v = 29.904 m/s

Use banked curvesv = 29.90401311 m/s ...from #4θ = 52.3° tan(θ) = v2

(rg) tan(52.3°) = (29.90401311)2 (R)(9.80)

R = 70.52601458 mR = 70.5 m

Use banked curves

g’s felt =1

sin(θ)

g’s felt =1

sin(52.3°)

g’s felt = 1.26

Use circular motion relationships to solvev = 45.5 m/s ...because it is at the same height as location “B.”

r = 15 m

ac = V2 = 45.52 = 138.01667 m/s2

r 15




Use Energy relationships to solveET @ bottom = ET @ top

KE + PE = KE + PE(1/2)mv2 + 0 = (1/2)mv2 + mgh

(1/2)v2 = (1/2)v2 + gh(1/2)(45.5)2 = (1/2)v2+ (9.80)(30.0)

741.125 = (1/2)v2

v = 38.5 m/s

Use circular motion relationships to solvev = 38.5 m/s ...from #8

r = 15 m

ac = V2 = 38.52 = 98.8167 m/s2

r 15


g’s felt at the top = ac[in g’s] - 1 gg’s felt at the top = 10.083 - 1 gg’s felt at the top = 9.08 g’s

Use energy relationships and power to solvevo = 25.05 m/svf = 3.03 m/sd = 10 mt = 45.8 sm =6500 kg

Power = Worktime

= ∆Energytime

∆ET = [KEf + PEf] - [KEo + PEo]∆ET = [(1/2)m(vf)2 + 0] - [(1/2)m(vo)2 + 0]∆ET = (1/2)(6500)(25.05)2 - (1/2)(6500)(5.50)2

∆ET = 2010103.875J

distancetime

= (final velocity) + (initial velocity)2

10t

= 25.03 + 3.032

t = 0.712250712 s


Power = =∆Energytime

2010103.8750.712250712

Power = 2822185.841watts ...746 w = 1 hpPower = 3783.09 hp


Using the information shown on the diagram on the other page, determine the correct answers to the following questions.

1 How much force is used to accelerate the train across the first 42.6 m?2 How many g’s does the rider feel as he is pushed by the initial accelerating force?3 How fast is the car traveling at location “A” if it coasted down the back side of the first

hill?4 At location “B” the incline is 90° of a circle. How many g’s does the rider feel as he

enters the loop at location “B?”5 How fast is the car traveling when it reaches location “C”?6 In the banked turn the rider is traveling 21.6 m/s. What is the optimum angle of the

banked curve?7 How many g’s does the rider feel as he enters the loop?8 What is the height of the loop?9 If the radius at the top of the loop is 7.45 m, then how many g’s does the rider feel at this

location?10 How high does the coaster train coast at the end of the track?

Use energy relationships to solveET(BEGINNING) + WORK = ET(BEGINNING)

0 + Fd = (1/2) mv2

F(42.6) = (1/2) 5634)(38.6)2F(42.6) = 4197217.32

F = 98,526.22817 NF = 98,500 N

Use kinematics to solvevf2 = vo2 + 2ax

38.62 = 0 + 2a(42.6)ac = 17.48779 m/s2

ac = 17.48779 m/s2

9.80 m/s2

convert ac to g’s

Use energy relationships to solveET(AFTER FORCE) = ET (at “A”)

KE + PE = KE + PE (1/2)mv2 + mgh = (1/2)mv2 + 0

(1/2)v2 + gh = (1/2) v2 + 0 (1/2)(36.8)2 + 9.8(12.1) = (1/2)v2

863.56 = (1/2)v2

1727.12 = v2

v = 41.55863 m/sv = 41.6 m/s

v = 45.5 m/s, r = 70.3 m

ac = V2 = 45.52 = 29.44879 m/s2

r 70.3




Use energy relationships and compare locations “B” with “C”ETB = ETC

KE + PE = KE + PE (1/2)mv2 + 0 = (1/2)mv2 + mgh

(1/2)v2 + 0 = (1/2)v2 + gh(1/2)(45.5)2 + 0 = (1/2)v2 + 9.8(61.3)

1035.125 = (1/2)v2 +600.74868.77 = v2

v = 29.4749 m/sv = 29.5 m/s

Use banked curve equationsv = 21.6 m/s, r = 42.3 m tan(θ) = v2

(rg) tan(θ) = 21.62

(42.6)(9.80) θ = 48.117768° θ = 48.1°

Use circular motion equationsv = 45.5 m/s, r = 35.3 mac = V2 = 45.52 = 58.6473 m/s2

r 35.3


g’s felt at the bottom = ac[in g’s] + 1 gg’s felt at the bottom = 5.984419 + 1 gg’s felt at the bottom = 6.98g’s

Use energy relationshipsv(BOTTOM) = 45.5 m/s, V(TOP) = 15.2 m/s

ET(TOP) = ET(BOTTOM)

KE + PE = KE + PE (1/2)mv2 + mgh = (1/2)mv2 + 0

(1/2)v2 + gh = (1/2)v2

(1/2)(15.2)2 + (9.80)h = (1/2)(45.5)2

115.52 + 9.8(h) = 1035.125h =93.8372 mh = 93.8 m


Use circular motion relationshipsv = 15.2 m/s, r = 7.45 mac = V2 = 15.22 = 31.01208 m/s2

r 7.45


g’s felt at the top of a loop = ac[in g’s] + 1 gg’s felt at the top of a loop = 3.164498 + 1 gg’s felt at the top of a loop= 2.16 g’s

Use energy relationships to solve.Velocity at the highest point will be zero, (apogee).

ET(BOTTOM) = ET(TOP)

KE + PE = KE + PE (1/2)mv2 + 0 = 0 + mph

(1/2)v2 = gh (1/2)(45.5)2 + = (9.80)h

h =105.625 mh = 106 m


HILL AND DIPS: ACTIVITY 1

RULER

Shape for a low ramp, slow velocity

Shape for a medium ramp, medium velocity

Shape for a medium ramp, medium velocity

1. The steeper ramp yields the greatest velocity when leaving the table.2. The greater the speed over the hill the flatter, more spread out, the hill must become.

See page 22 in the text for the other parts of the solution.

HILL DESIGN: ACTIVITY 1Draw the results below.

SLOW BALL’S PATH FAST BALL’S PATH

HILL DESIGN: ACTIVITY 21-3 For the hill where the car becomes airborne, the hill width will be shorter than the hill where

the car does not become airborne.4 The car will become airborne over the hill because it will be traveling too fast. And the hill

needs to be the same shape as if there were no track, a parabola.5 The car will become airborne over hill B. The two hills are the same height. The steeper the

drop off on the right side of the hill the slower the car needs to go. This is because hill B is the slow ball’s path (shown in Activity 1.)


LOOP DESIGN: ACTIVITY 1The starting height for the klothoid loop will be lower than the starting height for the circular

loop. The advantage of the klothoid loop’s design is that the roller coaster car does not have to be going as fast to stay on the track when the car is upside down.

FREE FALL (Weightlessness): ACTIVITY 1Why doesn’t the water come out of the holes when it is dropped?

The water comes out of the holes when the cup is standing still and the water is allowed to accelerate down relative to the cup. When the cup is dropped, it is accelerating down at the same rate as the cup. The cup has no reaction force holding it up and the water stays in.

Start of the ride

WEIGHTLESSNESS

Hold here

STAYING SEATEDBlock this out when copying:To calculate the centripetal acceleration you will need to know the radius the cup spins in and the time to go around once. The radius can be calculated while holding the cup and board still. Someone should swing the board around with as consistent a velocity as possible. Time how long it takes to go around 10 times. To find the time to go around once,

Time to go around once = Time to go around 10 times10

to find the centripetal acceleration use the equationa = 4π2R T 2

where “a” is the centripetal acceleration, “R” is the radius and “T” is the time to go around once.To convert the acceleration to g’s divide the acceleration by 9.8. (This is assuming your original measurements are in meters and seconds.)EXAMPLE: A board connected to a rope whose length from the

student’s hand to the board is 0.50m. The board takes 10.77 seconds to go around 10 times.

SOLUTION: The time to go around once is 1.077 seconds. The centripetal acceleration is 17 m/s2. This is 1.73 g’s.

RADIUS


PRACTICING YOUR ÒESTIMATIONSÓMEASURING THE VELOCITY OF A MOVING OBJECT

HotWheels™ trainSTART timing when the FRONT of the train passes a point.

HotWheels™ track

HotWheels™ trainSTOP timing when the BACK of the train passes a point.

HotWheels™ track


Objective: To design and build a simple, to scale, 3-D, roller coaster model.

Design a roller coaster on paper, show the velocities, g’s felt, and heights. The diagram needs to be neat and readable. The model is to be to scale on a piece of card board. The track is made from paper. The supports are made from manila folder or foam core board. The lines on the track represent the cross-ties on the track. Please print them off on a computer or draw them VERY neatly and evenly spaced.

Manila folder or foam core

supports

Hanging coaster like the Alpengeist

Card board base

Foam core base

Foam core sides to give this irregular shaped loop some needed rigidity. The track is laid inside the loop.

ROLLER COASTER PHYSICS Roller Coaster Desgin Activity


This year when you go to the amusement park bring a cam corder and a camera. Video tape the roller coaster rides. Choose a hill where the cars are not moving too fast. Zoom the cam corder on the hill without moving the cam corder. The train will pass in and out of view. Next year show the class the video tape. Have the students use the tape to calculate the velocity of the train at this spot on the hill. It is great real life practice. Below is an example of the view frame for such a taping.

VIEW FRAME OF THE CAM CORDER FOCUSED ON THE ROLLER COASTER’S TRACK ON A HILL.

With the camera, take pictures of loops, hills, curves, etc. Either use slide film or make these pictures into overheads. You can show them to the students before they go so they will be prepared to take the right measurements. You can also use these pictures so students can look at the hills on a roller coaster and rate them in terms of velocity over the top of each.

ROLLER COASTER PHYSICS ONE FINAL NOTE OF PREPARATION


Documents

Roller Coaster Physics