INTERNET AMUSEMENT PARK

PHYSICS PROJECT

TO EXPLORE THE PHYSICS IN AMUSEMENT PARKS

CREATED BY: D. JAMES CHICHESTER 01/98

LINCOLN-WAY HIGH SCHOOL

INTRODUCTION

To introduce students to the use of the internet.

To have students explore different available web sites relating to amusement park physics.

To provide an interactive review of concepts, principles, and equations of physics necessary to successfully complete our Great America Lab.

For students to research various topics relating to amusement park rides.

For students to construct a finished product representing their topic research, similar to a reaction paper.

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TASK

Day 1: Browse sites, collect notes & information.

Day 2: Continue browsing sites, collecting notes & information.

Day 3: Review concepts and equations with sample problems.

Day 4: Continue reviewing concepts & equations with sample problems.

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RESOURCES

The following sight locations are suggestions for neat information about amusement parks, roller coasters, and other stuff. Several locations have links to other sights that you can access withing the location. Investigate, read facts, look at pictures, and go ahead...get excited about this stuff.

SOME COOL COASTER SIGHTS TO TRY

	LIST OF ROLLERCOASTER SIGHTS
	GLENBROOK'S LINK PAGE
	1500 + LINKS
	JOYRIDES
	SIX FLAGS
	CEDAR POINT
	HONORS PHYSICS INTERACTIVE TEXTBOOK
	CONCEPTUAL PHYSICS INTERACTIVE TEXTBOOK

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CONCEPTS PRINCIPLES EQUATIONS

LINEAR MOTION:

Average velocity refers to the total distance traveled per interval of time, units m/sec.

avg velocity = total distance / time

Final velocity, found three ways:

a = (v final - v initial) / time

(v final)^2 = (v intial)^2 + 2*a*d

avg vel = (v final + v initial) / 2

EX: A bus accelerates from rest to a velocity of 20 m/sec in a 8.0 second time interval. Assuming uniform acceleration, find the average velocity of the bus.

avg vel = v final + v initial / 2

avg vel = (20 m/s + 0 m/s) / 2

avg vel = 10 m/s

EX: Find the distance the bus would have traveled in this period of time.

avg vel = total dist / time

10 m/s = (dist) / 8.0 sec

dist = 80 meters

Force:

F = m*a

F (weight) = m*g

Work = F(parallel)*distance

EX: A student estimates that a loaded roller coaster car has a mass of 4000 kg and drops down a hill at an acceleration rate of 4.0 m/s^2. Find the amount of force the coaster has 10.0 meters down the hill, then the force at the end of the hill.

F = (m)(a)

F = (4000kg)(4.0 m/s^2)

F = 16000 N (both 1/2 way down & at the end)

EX: A student estimates the height of the American Eagle hill to be 28.0 meters tall. The base of the hill is 2.0 meters tall. Again they have estimated the mass to be 4000 kg. If it takes 10.0 seconds to raise the coaster to the top of the hill, find the work done in lifting the car to the top of the hill.

W = F(parallel) * distance

W = (4000 kg)(9.8 m/s^2)(26.0 m)

W = 1019200 Nm or Joules

Distance:

avg velocity = distance / time

(v final)^2 = (v initial)^2 + 2*a*d

d = (1/2)a*t^2 + (v initial)(time)

EX: A school bus accelerates uniformly from rest to a velocity of 30 m/sec in a time period of 15 seconds. Using the acceleration value of the bus, find the distance traveled in this interval.

a = (v final - v initial) / (time)

a = (30 m/sec - 0 m/sec) / (15 sec)

a = 2.0 m/s^2

d = 0.5(a)(t^2) + (v initial)(time)

d = 0.5(2.0 m/s^2)(15.0 sec)^2 + (0 m/sec)(15.0 sec)

d = 225 meters

Acceleration:

a = (v final - v initial) / (time)

d = 0.5(a)(t^2) + (v initial)(time)

v final = v initial + (a)(time)

EX: If the school bus that accelerated from rest to 30 m/sec in 15 seconds was to continue at this rate of acceleration for another 15.0 seconds, how fast would they be going?

a = (v final - v initial) / (time)

2.0 m/s^2 = (v final - 0 m/s) /(30 sec)

v final = 60 m/sec

Momentum:

Momentum like energy is conserved. The incoming momentum for a collision is equal to the resulting momentum after a collision. Momentum is due to an objects size(mass) and it's velocity.

Momentum (p) = mass * velocity

F * time = mass * change of velocity

(F * t) is referred to as impulse

(mass * change of velocity) is referred to as change of momentum

EX: A bumper car is estimated to have a mass of 70 kg, and a 60 kg student is riding inside the car. If the car is moving at 5.0 m/sec. How much momentum does the car and rider have?

p = m * v

p = (70 kg + 60 kg) * (5.0 m/s)

p = 650 kgm/s

EX: The same car and rider traveling 5.0 m/s now strike a wall and bounce backwards at 4.0 m/sec. If the collision took 0.2 seconds. Find the force the car and rider felt.

F* t = m * v

F * (0.2 sec) = (130 kg)(9.0 m/s)

F = 5850 N

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CIRCULAR MOTION

Period:

Measured in seconds. Time for one complete revolution.

T = (#seconds) / (#revolutions)

Centripetal forces:

Center seeking forces that pull objects from intended straight line tangential paths, due to their interia, into a circular path.

F centripetal = m * centripetal acceleration

centripetal acceleration = (v^2) / radius (uses tangential velocity)

tangential velocity = (2pi*r) / T

EX: A merry go round horse goes around 5 times every 100 seconds. The horse is 5.0 meters from the center of the ride. The rider's mass is 70 kg. Find the rider's linear velocity.

v = (2pi*5m) / (20 seconds)

v = 1.6 m/s

EX: Find the rider's centripetal acceleration.

a centripetal = (1.6m/s)^2 / (5m)

a centripetal = 0.512 m/s^2

EX: Find the rider's centripetal force.

F centripetal = (70 kg)(0.512 m/s^2)

F centripetal = 35.8 Newtons

EX: If the rider moved to a horse 0.5m further out from the center of the circle would that change their centripetal force.

Yes, different radius...different v...different acceleration...same period though

Angular velocity (rotational velocity):

Measured in rotations per second (or minutes). Number of complete rotations per interval of time.

angular velocity = (#revolutions) / (unit of time)

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MEASURING HEIGHTS:

Using your protractor sighting device & length of string.

Look through the straw to the top of the object you are attempting to find the height of. Record the angle measured from the 90 degree position. Walk back in a straight line a distance of 15 meters, measured with your length of string, and again sight the top of the object through the straw. Record the second angle measure. Use the below equation to find the height of the object.

h = (sin angle 1 * sin angle 2) / (sin (angle 1-angle 2)) * walk back length + eye level height

EX: measured a 30 degree angle, walked back 15 m, measured a 21 degree angle,and my eye level height is 1.6 meters

h = (sin30 * sin21) / (sin9) * 15m + 1.6m

h = 18.8 meters (to convert meters to feet multiply by 3.25)

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ACCELEROMETERS:

Using your vertical accelerometer (plastic tube with fishing weight and spring).

Your vertical accelerometer will measure the number of "g's" that you are feeling while riding a roller coasters loop. The reference line represents one g (9.8 m/s^2). The first line down represents 2g's (19.6 m/s^2). The second line down represents 3g's (29.4 m/s^2)

Using your horizontal accelerometer (protractor with washer hanging from string)

Your horizontal accelerometer can be used while accelerating forward, backwards, of around curves. Holding the accelerometer level to the ground below, measure the angle of swing back for the washer. Using the equation a = g * (tan angle) allows us to find the acceleration rate.

EX: While accelerating onto the interstate I find the washer swings back from the reference position of 90 degrees to 82 degrees (or 98 degrees depending upon which way I'm holding it).

a = g * tan (angle)

a = (9.8 m/s^2) * tan 8 degrees

a = 1.38 m/s^2 (which is 1.38/9.8 approximately 0.14g's)

EX: While on the cajun cliffhanger the washer deflected to 25 degrees.

a = (9.8m/s^2) *tan 65 degrees

a = 21 m/s^2

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CONSERVATION OF ENERGY:

Total amount of energy for a closed isolated system always remains the same. What that means is that the stored up potential energy (due to gravity and height) for a ride will be converted into moving kinetic energy. The coaster at the top of a hill has basically all PE, and at the bottom of a hill has basically all KE. The PE was converted into moving KE. (PE + KE = total Mechanical Energy)

PE = m*g*h

KE = 0.5*m*(v^2)

Solving KE for the v: v = (2*g*h)^0.5

EX: A student using their sighting device found the height of the first hill of a coaster to be 32 meters high, and the bottom of the hill is 12 meters high. Find the PE at top and KE at bottom. They estimate the coaster's mass to be 1000kg + another 1000kg for the people inside.

PE change= (2000kg)(9.8m/s^2)(20m)

PE change= 392000 Joules

KE change= 392000 Joules

v change= (2*9.8m/s^2*20m)^0.5

v bottom = 19.8 m/s

Energy is the ability to do work. So the stored up PE has the ability to do Work when released.

Work = Force(parallel) * distance

Work needed to lift an object a distance of (d meters) is equal to the PE the object will have at (d meters)
PE top = W need to lift to top

EX: Find the amount of force necessary to lift the 4000 kg car loaded American Eagle car to the top of it's hill. Student's measure a 45degree angle then walk back 15 meters and measure a 32 degree angle swing back of washer, their eye level is 1.6 meters. The Eagle takes 20 seconds to ascend the hill. The bottom hill is basically ground level (height = 0 meters)

h= (sin45 *sin32) /(sin13) *15m + 1.6m

h = 26.6 meters

PE top = (4000kg)(9.8m/s^2)(26.6m)

PE top = 1042720 Joules

Work to lift = 1042720 Joules

1042720 J = Force * 26.6 meters

Force = 39200 Newtons

Power is how quickly you do work, measured in Joules/second or watts.

Power = Work (Joules) / Time (sec)

EX: For the above American Eagle example, find the power to lift the car to the top of the hill.

Power = 1042720 J / 20 seconds

Power =52136 watts

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FREE FALL:

Free fall refers to objects that are falling straight down towards the earth without air resistance. Objects in true free fall accelerate, or pick up speed, at 9.8 m/s every 1 second (9.8 m/s^2). Since they are increasing their velocity every second, that means they are traveling bigger distances every second of their fall.

d = 0.5*a*(t^2) + (v intial * time)

a = (v final - v initial) / (time)

(v final )^2 = (v initial)^2 + 2*a*d

v final = a*t + v initial

EX: A student has determined the height of the carriage of the giant drop to be 65 meters tall with their sighting device. The student figures the carriage free falls for only the first 40meters, because of the braking distance at the bottom of the ride. The student times the free fall and it takes 3.2 seconds. Calculate the data's acceleration rate.

40m = 0.5*(a)*(3.2 sec)^2 + (0m/s)(3.2)

a = 7.8m/s/s

Therefore the riders feel (7.8/9.8 or 0.80g's and only feel 20% of their normal weight....sensation of near weightlessness)

EX: A student drops their calculator from the top of the giant drop ride. How fast will it strike the ground disregarding air resistance.

(v final)^2 = (0 m/s)^2 + 2*(9.8m/s^2)*(65meters)

v final = 35.7 m/sec

EX: The same student drops their calculator from the top of the giant drop ride. This time do NOT disregard air resistance, and find the calculators final velocity. Assume calculator's acceleration rate is that of the carriage.

(v final)^2 = (0 m/s) + 2*(7.8 m/s^2)*(65meters)

v final = 31.8 m/sec

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GREAT AMERICA ADVICE

ATTIRE AND TOOLS:

Dress extremely warm, it is often cold with the wind. Dress dry, it has rained several times and you don't want to be wet and cold all day.
Bring money for lunch and snacks. The park does not allow food or beverage to be brought into the park.
A large ziplock bag is necessary. It will contain your calculator, your pencils, lab book, accelerometers, string, review notes, timer, and rubber band.
The ride operators have always been great about holding bookbags while student's are riding a ride. Your bookbag is great for holding all your belongings (ziplock bag full of your equipment, rain poncho, money, dry socks....)

TEAMWORK:

Make sure you will be able to constructively work with your partners. I do not want any, I blew off my partners because I don't know them, or I don't like them. Enough said.
Your partner's have a wealth of knowledge, just like you. Work the labs out together, use the combined brain power of the whole. DO NOT JUST COPY THEIR WORK! You will only be hurting yourself. If you don't understand something, ask them.

PLANNING:

Decide in advance what rides your lab group is going to do for the day.
Make sure you've talked them over and you understand how to do the labs BEFORE you get to the park. I can't stress how important preparing in advance is for this project.
Decide if you're going to collect data first for the labs, then crunch the calculations. Or do the labs one at a time.
Typically, groups can have their data collected by lunch time. They then check in with their teacher, so we know you're still physically and mentally OK. Grab lunch somewhere warm and finish the lab reports. The rest of the day is riding for fun, of course be thinking of the physics even then.

SOURCES OF HELP:

The park is filled with physics teachers, student's, and other kid's from LWHS that can help.
LWHS physics teachers take turns at the Elm Street School area, waiting for questions.

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CONCLUSION

WHAT DO I NEED TO TURN IN FROM THIS

INTERNET AMUSEMENT PARK PHYSICS PROJECT?

1. Compose a one page double spaced typed webquest summary paper, including pertinent URL addresses. I am looking for you to summarize your reaction to and findings from this webquest project.

2. This summary paper is due on Friday May 11th. The paper is a 25 point assignment.

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