RollerCoaster-ology Mini-Degree Workbook
The Physics of Amusement Park rides
Solutions for May 27th and June 3rd, 2005
SECTION 1 –THE VORTEX
Measurements:
1. What is the total duration of the ride: 71.4 s
2. What is the total length of the 6-car train: 16.4 m
3. How long does it take for the train to be pushed from the bottom to the top of the highest
hill: 20.2 s
4. How long does it take for the train to travel its own length while being pushed up hill: 6.5 s
5. How long does it take for the train to travel its own length at the top of the first downhill run
(just as it begins its descent down the incline): 1.8 s
6. How long does it take for the train to travel its own length at the bottom of the first downhill
run (just as it leaves the downward incline): 0.8 s
QUALITATIVE OBSERVATIONS and QUESTIONS
‘
FUN-NESS’
1. Where do you think you should sit to have the most Fun on the Vortex –front, middle, or
back of the train? Does it matter? Explain.
2. Where on the ride did you feel you were going the fastest? Why?
3. After passing through the corkscrew section of the Vortex, were you dizzy? Explain.
RIDE ENGINEERING
1. Where was the highest hill on the ride? Why is it there?
2. Observe and describe the wheel assembly, noting how many wheels there are on each car
and how they are positioned. Sketch a diagram if possible.
3. What feature of the wheel assembly is designed to keep the train from leaving the track if it
is traveling too fast as it goes over the top of the hill?
4. What feature of the track design helps keep the train on the track as it goes around a sharp
curve?
At the bottom of the first loop:
RollerCoaster-ology Mini-degree Workbook Solutions 1
QUANTITATIVE CALCULATIONS
ENERGY
1. What was the speed of the train while being pushed uphill?
2.5 m/s or 9.0 km/h
2. What was the speed of the train at the top of the first downhill run?
9.1 m/s or 32.8 km/h
3. What was the speed of the train at the bottom of the first downhill run?
20.5 m/s or 73.8 km/h
4. Use the conservation of energy to determine the speed of the train at the bottom of the first
downhill run. 20.1 m/s or 72.4 km/h (note; accelerometer / altimeter data indicates a
total vertical descent of 20.5 m, significantly more than stated in the appendix)
5. Compare answers 3 and 4. Should they be the same? If they are not, explain why.
6. If each car has a mass of 1200 kg and the coaster is filled with riders whose average mass
is 65 kg, how much total work is being done getting the filled coaster to the top of the
highest hill? Work = Force x distance parallel to Force = 2.94 x 10^5 Joules
7. How much power does the motor have to provide to lift this loaded coaster to the top of the
highest hill? Power = Work / time = 1.45 x 10^4 Watts
FORCES AND ACCELERATIONS
1. A passenger in a car at the top of a loop is nearly weightless. Assuming their car follows a
circular arc of radius 3.5 m, what is the approximate speed of their car?
5.9 m/s or 21.0 km/h (note this is much less than 20 m/s; did you feel weightless? Is
3.5 m the proper value for the radius of curvature?)
2. Consider the ride as experienced by two different people - a passenger at the very front of
the train, and a passenger at the very back of the train. Consider the moments when these
two passengers move past the crest of the first downhill run. Do these two passengers
have the same apparent weight? If not, which one feels lighter? Why? The passenger at
the rear since he / she is moving at a larger speed as they travel over the curved
track.
3. Did you feel lateral forces while on the ride? That is, were you thrown from side to side in
the train car? If so, what forces caused that feeling? If not, why not? If necessary, use a
diagram to help explain. The track is very well engineered. That is, when the train is
moving fast, the track’
sbanking angles match the train’
s instantaneous speed,
virtually eliminating lateral forces.
CHALLENGE QUESTIONS
1. Does the total mass of the train (cars plus passengers) play a role in the speed of the
coaster, the forces and accelerations experienced during the ride, or the duration of the
ride? Explain. If track-wheel friction, and wind resistance, were eliminated, then total
mass would not affect these aspects of the ride.
2. If the acceleration due to gravity was smaller (say by 50%), and all other variables were the
same, would the total duration of the ride be greater, less, or, the same? Explain. If g is
smaller, it would take longer to go down the first big hill, and longer to go up and
down subsequent hills. The total duration of the ride goes up as g goes down.
RollerCoaster-ology Mini-degree Workbook 2
SECTION 2 –OCEAN MOTION
Measurements:
1. What is the time for three complete swings, when Ocean Motion is in full swing mode?
23.3 s
2. Based on the actual physical dimensions of the ride as given in the appendix, what do you
estimate is the radius, R, of the ship’
s circular path? Physical measurements, and
photographic analysis, shows the distance from the pivot point to the floor boards is
very close to 15 m (note; data given in the appendix is somewhat erroneous, for
example, the support structure is not 18 m high). We will use 15 m as the value of R.
QUALITATIVE OBSERVATIONS and QUESTIONS
‘
FUN-NESS’
1. When Ocean Motion was in full swing mode, and you were in a downhill forward facing arc,
where did you feel:
a. the strongest push against your back? ____________________________
b. the most pressure against your seat? ____________________________
c. the least pressure against your seat? ____________________________
2. Where did you feel you were going the fastest? _______________________
3. Would the ride be more FUN if you were blindfolded? Explain?
4. Would the ride be more FUN if R were bigger? Explain?
RIDE ENGINEERING
What is the period of one complete swing? 7.77 s
1. Using your measured period, calculate the radius of the shi
p’
sci
r
cul
arpat
h(assuming
Ocean Motion is a simple pendulum), using the SHM equation given in the appendix.
15.0 m
2. Compare your calculated value for R to your measured value. Why the difference? Is
Ocean Motion engineered anything like a simple pendulum? Ocean Motion is essentially
a simple pendulum based on these values of T and R. This is a strange result
however, since the relation between T and R for a simple pendulum given in the
appendix holds only if the pendulum has small amplitude oscillations. Ocean
Motion has huge amplitude oscillations and so should not mimic simple pendulum
behavior. The rides’drive mechanism is obviously at work.
QUANTITATIVE CALCULATIONS
ENERGY
1. Let
’
sassume Ocean Motion’
smassi
sent
i
r
el
ylocated where the mast is attached to the
ship (ie. where the mast attaches to the floor boards). Using a geometrical analysis,
one can show that the maximum height of the centre of mass, h, above its lowest point, is
RollerCoaster-ology Mini-degree Workbook 3
approximately given by; h = R [ 1 - sin(90L/R)], where L is the length of the ship and R is
the radius of its circular path. Although the length of the boat is 13 m, the length of the
boat between the large triangular support struts (which is really L here) is very close
to 11.5 m. Thus, h = 9.4 m
2. Using simple conservation of energy arguments, what is the speed of the ship at the
bottom of the swing, when Ocean Motion is in full swing mode? vmax = (2gh)^0.5 = 13.6
m/s
3. Assuming that the combined mass of the ship and passengers is 3000 kg, what is the
kinetic energy of the ship at the bottom of the swing? Ek = 2.77 x 105 Joules
FORCES AND ACCELERATIONS
1. What is the apparent weight of a 50 kg person at the bottom of the swing, when Ocean
Motion is in full swing mode? W = mg + mv2/R = 1110 Newtons
2. Two 50 kg passengers board the ship, one in the row closest to the mast and one in the
row furthest from the mast. At the top of the swing, when the ship momentarily comes to
rest, which passenger has the smaller apparent weight? Why are their apparent weights
different? The passenger furthest from the mast since the seat of their bench is
almost perpendicular to the ground.
CHALLENGE QUESTIONS
1. Derive the expression, h = R [ 1 - sin(90L/R)] Hints: When in full-swing-mode, and
the boat is at its maximum height, the highest of the two triangular support struts is
essentially parallel to the ground. The length of the support struts, extended to the
floor boards, is essentially equal to R. L is the length of the boat, as measured along
the arc of the boat, between the support struts. The mast is a distance ½ L from the
part of the boat at the end of the highest support strut.
OceanMot
i
onatmaxi
mum ampl
i
t
udei
n‘
f
ul
l
-swi
ng’mode:
2. Is your estimate of vmax reasonable? That is, did you leave something out in your
conservation of energy analysis? Since the drive system keeps the big boat swinging
essentially like a simple pendulum, our estimate of vmax is reasonable. Otherwise,
you would have to consider the moment of inertia and angular speed of the boat.
RollerCoaster-ology Mini-degree Workbook 4
SECTION 3 - BIG ELI
Measurements:
1. What is the time for three complete revolutions? 38.9 s
2. Each passenger car hangs from a supporting bar. What is the distance from the point of
suspension to the seat? 20 cm
3. When Big Eli comes to a stop while you are sitting in a passenger car, and you begin to
swing back and forth like a pendulum, what is the period of this swinging motion? (note;
you could also measure this by watching other people on the ride while you are standing
on the ground) ~1 sec
4. While Big Eli is in steady motion at operating speed, the seat you occupy hangs vertically
when you are at the highest andl
owestpoi
nt
s(
12o’
cl
ockand6o’
clock). At what angle
awayf
r
om v
er
t
i
cali
sy
ourseatat3o’
cl
ock? A careful photographic analysis shows this
angle to be near 8.5 degrees –tough to measure however!
5. Using your home-made vertical accelerometer, when Big Eli is stationary, what is the
acceleration you experience? St
udent
’
swi
l
lgetsomeval
uedepending on the design
of their accelerometer. This value would be the acceleration due to gravity, ag .
6. Using your home-made vertical accelerometer, when Big Eli is at operating speed, what is
the acceleration;
a. at top of t
her
i
de(
12o’
cl
ock)
less than ag
b. at bottom of t
her
i
de(
6o’
cl
ock) greater than ag
c. going up (
3o’
cl
ock)
ag
d. coming down (
9o’
cl
ock)
ag
QUALITATIVE OBSERVATIONS and QUESTIONS
‘
FUN-NESS’
1. Why does Big Eli rotate in the direction it does? Would it be more FUN if it turned the
other way? Why, or why not?
2. Would the ride be more FUN if Big Eli turned a bit faster? A bit slower? A lot faster?
Explain?
3. Where on the ride did you feel you were going the fastest? Why?
RIDE ENGINEERING
1. For safety, each passenger car comes with a bar that riders put across their laps. Why are
seatbelts not required?
2. When riders board their car, they jostle around for a few moments, and then come to rest
i
nt
hei
rv
er
t
i
calor
i
ent
at
i
on.Whenmov
i
ngatBi
gEl
i
’
soper
at
i
ngspeed,t
hecari
sonl
y
v
er
t
i
calat12o’
cl
ockand6o’
cl
ockbecauset
hecar
sar
ef
r
eet
oswi
ngbackandf
or
t
h.
Would the ride be safer if the cars were mechanically constrained to hang vertically as Big
Eli moves round and round? Explain?
RollerCoaster-ology Mini-degree Workbook 5
QUANTITATIVE CALCULATIONS
ENERGY
1. Circumference of Big Eli x (15 m) = 47.1 m
2. Operating speed of Big Eli circumference / period = 47.1 m / (38.9 s / 3) = 3.63 m/s
3. With every car filled with 2 people, each with an average mass of 60 kg, and with each car
having a mass of 150 kg, what is the net change in gravitational potential energy of Big Eli
after 1 complete turn? zero Joules
4. What is the total kinetic energy of riders and cars when Big Eli is turning at operating speed
(use the above mass values)? 16 chairs x ½ (270 kg (3.63 m/s)2) = 2.8 x 104 Joules
FORCES AND ACCELERATIONS
1. Youshoul
dhav
ef
oundt
hatt
heaccel
er
at
i
onsy
oumeasur
edat6o’
cl
ockand12o’
cl
ock,
while you were moving atoper
at
i
ngspeed(
l
et
’
scal
lt
hem a6 and a12), were different than
the acceleration due to gravity measured when you were stationary (
l
et
’
scal
li
tag). From
your measured values, calculate the differences a6 –ag and ag –a12. Should these
differences be the same? If they are not, explain why. St
udent
’
swould probably find
that the two differences; a6 –ag and ag –a12 , are the same to within the accuracy of
their hand made accelerometers. If the chairs had been suspended from a point
above the heads of the passengers, instead of just slightly above their hips, then the
values of R12 and R6 would have been significantly different. This would then make
a6 farther away from ag than ag is from a12 , and the differences; a6 –ag and ag –a12 ,
would be noticeably different according to their accelerometers.
2. Calculate the centripetal acceleration you experience at 12o’
clock when Big Eli is at
operating speed (l
et
’
scal
lt
hi
scal
cul
at
edv
al
ueA12). The direction of this acceleration is
downward:
A12 = v2/R12 = (3.63 m/s)2 / (7.5 m –0.2 m) = 1.81 m/s2
A12 = 0.184 g’
s
RollerCoaster-ology Mini-degree Workbook 6
Calculate the centripetal acceleration you experience at 6o’
cl
ockwhen Big Eli is at
operating speed (l
et
’
scal
lt
hi
scal
cul
at
edv
al
ueA6). The direction of this acceleration is
upward:
A6 = v2/R6 = (3.63 m/s)2 / (7.5 m + 0.2 m) = 1.71 m/s2
A6 = 0.174 g’
s
3. From your calculated values, what are A6 –ag and ag –A12? How do these differences
compare to what you found from your measured values? Should these differences be the
same? If they are, explain why. (hint: what value of R did you use in your calculations of
A6 and A12 ?)
A6 –ag = 1.174 g’
s(you feel heavier). This is 17% higher than 1g.
ag –A12 = 0.816 g’
s (you feel lighter). This is 23% lower than 1g.
4. The operating speed of Big Eli can be determined using your measurement of the angle
away from vertical that your seat is, when it is at3o’
cl
ock. If we call this angle , then the
operating speed of Big Eli, v, is related to by the equation, tan = v2 / gR. In this
equation, R is the radius of Big Eli, and g is the acceleration due to the force of gravity.
Using this method, what is the operating speed of Big Eli? v = 3.32 m/s
CHALLENGE QUESTIONS
1. When Big Eli is stationary, the passenger cars swing with a period very much like simple
pendul
ums(
l
et
’
scal
lt
hi
sper
i
odTstationary). Calculate Tstationary assuming the passenger cars
swing like a simple pendulums, and compare this to your measured value. Are they
similar? Why, or why not. Values are close, but different due to friction and wind
Tstationary = 0.9 s (calculated)
Tstationary = 1 s (measured)
2. When Big Eli is rotating at its operating speed, each passenger car is swinging back and
forth like a pendulum, but with a period much different from the stationary case (
l
et
’
scal
l
this period Tmoving). What is the value of Tmoving and why is it so different from Tstationary?
Tmoving = 13 seconds (the rotational period of the ride)
3. Derive the equation, tan = v2 /gR,whi
chhol
dswhenev
eracari
sat3o’
cl
ock(
or9
o’
cl
ock)
.
N

N sin= mv2/R
N cos= mg
eliminating N gives
tan= v2/gR
Se
a
ta
t3o’
c
l
oc
k(angle exaggerated)
mg
RollerCoaster-ology Mini-degree Workbook 7
CALAWAY PARK RIDE DATA
RIDE
DESCRIPTION
Vortex
A train of 6 cars travels on a double looping steel track. The
initial vertical drop is approximately 15 metres. The crest of the
first hill is approximately 19m above the ground.
Ocean Motion
A 13m fibreglass boat that swings from an 18m high structure
which swings to a 90angle to the ground to reach 20m
Big Eli
16 steel 2 seater cars hang on a 15m diameter wheel, spinning at
a total height of 16.8m
Swing Around
24 chairs are suspended on 4.2m wire ropes and swing from a
rotating centre structure
Carousel
36 aluminum horses and 2 benches travel on a 13.5m diameter
wooden structured carousel
Dodgem Bumper
Cars
12 –2 seater fibreglass azzurra bumper cars travel on a 24m
diameter stainless steel floor, D.C. powered for 8 different
speeds
Round Em Up
A 9m diameter rotor where passengers stand in stalls on the
outside hub and are spunat18RPM’
s,t
henr
ai
sedt
oa76
angle.
Paratrooper
10 fibreglass cars are suspended between spokes of a wheel
andosci
l
l
at
et
oandf
r
om ast
hewheelr
ot
at
esat12RPM’
sand
tilts to a 45angle.
Mountain Scrambler
3 gyros with 7 cars each spin while rotating clockwise around the
centre structure. The gyros are raised up to a 45angle.
RollerCoaster-ology Mini-degree Workbook 8