PH131
Michael Gumuka
Section 15
Chris Ouellette
Group 6 – “ACME”
Anna Shen
Joshua Smith
Physics Team Design
Final Deliverable
December 5, 2003
Introduction:
One important aspect of the field of physics is developing mathematical models to
describe and predict the motion of various objects in the world. For example, models could be
used to describe the rotation of the earth, a collision on the highway, dropping a bomb from an
airplane, or perhaps matchbox cars on a track. For any of these situations, the basic process used
is the same.
In order to scientifically model a system, a scientist’s first task is to identify the system
that he or she wishes to study, and define what it is that they want to model about that system.
Next, the system is studied, and relevant information (such as forces acting on the system) is
described graphically. Scientists then decide which physical laws govern the system, based on
hypotheses. The physical laws are then used to create a theoretical model for the system, which
is compared to the initial hypotheses. Finally, the theoretical solution is tested by comparison to
experimental observations. The theory is then revised and further developed until a model which
satisfactorily predicts experimental observations, and thus accurately models the physical
system, is created. This process usually involves taking more factors into account within the
theoretical model (like friction, for example).
Objectives:
The purpose of this project is to accurately model the motion of a matchbox car on an
arbitrary track. Ultimately, the researchers wish to know the velocity of the matchbox car as a
function of position. Following the construction of this model, the theoretical calculations will be
tested in two ways: first, by comparing experimental velocity data from an arbitrary track to
theoretical predictions, and second, by the success of the predictive model during a series of
challenge sessions.
Experimental Development of Theoretical Model:
In order to predict the motion of a matchbox car on an arbitrary track, a model of the
velocity of the car must be constructed. Velocity is not easily modeled because the effects of
friction, drag, and varying normal forces due to track curvature need to be taken into account.
Instead of attempting to determine the effect of all of these variables in a single step, basic
models are first constructed, and then modified as more effects are considered.
1
Part I: Modeling velocity on a flat track with static friction
The first system studied was a matchbox car moving down a horizontal flat track, with
frictional force acting on it. This model neglects the effects of drag. A free body diagram (with
the relevant forces) is depicted below in Figure 1.
Figure 1
The relationship between velocity and displacement for an object under constant
acceleration can be written as:



v 2  v02  2a  x  x0 
(Eq. 1)
Since the car is traveling on a flat track, using the reference frame given in Figure 1, our
interest is in the x-direction components of Equation 1:
v x2  v02x  2a x x ,
x0 = 0
(Eq. 2)
This can be described graphically as a plot where v2 is plotted against x. A linear best-fit
to experimental data would allow for the determination of the slope. The physical interpretation
of the slope in Equation 2 (2ax) can be described by applying physical laws that describe the
system seen in Figure 1.
In this model only frictional force is considered, while the force due to drag was
neglected. This system can be described using Newton’s Second Law, and a definition of static
frictional force, as shown below:


Newton’s Second Law:  F  ma
Static Frictional Force: f k   k N
2
In the x-direction:
In the y-direction:
F
F
x
 max
y
 ma y
 f k  max
N  mg  0
  k N  max
N  mg
Substituting N and fk back into Equation 2 and solving for ax:
ax  k g
Therefore:
vx2  v02x  2k gx
(Eq. 3)
A linear best-fit of the car’s velocity as a function of position on a flat horizontal track would
yield a slope of -2µkg.
An experiment was designed to test this basic model. A matchbox car with a small flag
attached to the top was released at a high point at the beginning of the track and traveled down
an incline (to gain initial speed), then steadily slowed down as it followed the flat portion of the
track. The basic setup of this experiment is shown below in Figure 2. Photogates were used to
record the velocity of the car along the flat portion of the track. Each gate recorded the time
when the flag attached to the top of the car first activated it, and the time when it was no longer
activated. Using this information, with the knowledge of the width of the flag, the matchbox
car’s velocity at each photogate, along the direction of displacement (x) was determined to be:
v =(Flag width)/(Change in time) since v  x t
Figure 2
With knowledge that drag might be a significant factor in the model, the car was released
from five different heights to get a wide range of velocities on the track. A plot of velocity-
3
squared vs. position is shown below in Plot 1. Five series are shown, represented in different
colors, each which represents seven experimental trials from a single drop height within the
release zone. A linear best fit was done to each data series using Excel.
Velocity Squared vs. Position
Car released from Five Heights
12
Velocity Squared (m2/s2)
10
8
6
4
2
0
0
0.5
1
1.5
2
2.5
3
Position (m)
Plot 1
First to note about Plot 1 is the slope of the trend line for each series. As the height of the
release point increased, the slope of the trend lines became increasingly negative. Calculating µk
from the slope of the best fit line, as represented in Equation 3, for each of the series:
Height 1 µk = .03332 (the lowest height)
Height 2 µk = .03663
Height 3 µk = .03867
Height 4 µk = .04090
Height 5 µk = .04276 (the greatest height)
This data is peculiar for µk because given two surfaces that are in contact, µk should be
constant regardless of other factors, such as velocity.
However, this phenomenon can be
explained by the simplicity of the model: there is an additional force acting on the car, making it
appear as though the frictional force has greater effects at faster velocities, which is actually the
result of drag.
The next section will incorporate the effects of the drag force into the theoretical model.
4
Part II: Incorporating Drag on a Flat Track
Since drag is proportional to the square of velocity, bodies experience greater drag forces
at higher velocities. This explains why the slopes of Plot 1 became increasingly negative when
the car was released from a greater height, when they should have remained constant. This
section incorporates drag into the model developed in Part I, as described in Equation 3.
A Free Body Diagram which accounts for drag force (D) is shown below in Figure 3.
Figure 3
Using Newton’s Second Law, and an equation for drag, the forces in this figure can be related to
the car’s acceleration as shown below:


Newton’s Second Law:  F  ma
Drag: D  kv2
In the x-direction:
In the y-direction:
F
F
x
 max
y
 ma y
 f k  D  max
N  mg  0
  k N  kv2  max
N  mg
Combining relationships from the equations in the x and y direction yields:
  k mg  kv2  max
This equation is slightly misleading because it shows ax as a constant. However, since
drag is a function of velocity, which varies with time, ax is not constant and best represented as
dv/dt, as shown below:
5
 dv 
  k mg  kv2  m 
 dt 
Since we are interested in finding velocity-squared as a function of position, isolating
acceleration and multiplying both sides by dx is the next step.

kv2 
 dv 
dx
 dx    k g 
m 
 dt 

On the left hand side of the equation, dx/dt can be re-written as velocity, leaving v dv.

kv2 
dx
vdv    k g 
m 

Isolating dx, we obtain the following:
v

v0
x
v
dv    dx
kv2
k g 
m
x0
kv2
This integral is solved by a u-substitution where u   k g 
m

 g
m  k
ln
2k 
 k g 

kv2
m
kv02
m


   x  x 
0



To isolate v, the natural log needs to be isolated, and both sides raised to a power of e, yielding:

 k g 


 k g 

kv2
m
kv02
m


2k
  x x 
  e m 0 , then rearranging



2k
kv02   m  x x0 
kv2 


, rearranging more for the final result:
k g 
 k g 
e
m 
m 
m k g  m k g
   x x0 
v 

 v02 e m
, x0 = 0 m
k
 k

2k
2
6
This equation can more easily be represented as:


v 2  v02  A e  Bx  A ,
where A 
m k g
2k
and B 
k
m
(Eq. 4)
With a relationship between velocity and position, incorporating both kinetic-friction and
drag, the coefficient of kinetic-friction and drag can be calculated using the method of least
squares and numerical methods.
Part III: Using Least Squares to Calculate µk and k
The method of least squares can find the “best-fit” for a set of data to a known function.
By finding unknown coefficients in the general form of a function, the best fit can be made to a
given set of data. The method of least squares will be used to find constants A and B in Equation
4.
As the method name indicates, the method of least squares minimizes the sum of the
squares of the difference between the theoretical function value and the data the curve is being fit
to. The sum can be written as:
n
S    yi  y  xi 
2
i 1
In terms of our function, the sum is written as:

S   vi2  v02  Ae  Bxi  A
n

2
i 1
Values for A and B must be determined, which minimize the sum. Following the method of
least squares we obtain the following:
 

n
S
 Bx
  2 vi2  v02  A e
 A  e  Bx  1  0
A i 1




n
S
   2 vi2  v02  Ae  Bxi  A xi e  Bxi A  v02   0
B i 1
(Eq. 5)
(Eq. 6)
Now there are two equations with two unknowns. By solving for A in terms of B in
Equation 6, and substituting back into Equation 5, we will obtain a single equation in terms of B.
We can then solve for B, and thereafter solve for A.
7
Simplifying Equation 6:


n
S
  vi2  v02e  Bxi  Ae Bxi  A xi e  Bxi  0 , separating the sums to isolate A:
B i 1
  Ae
 Bxi
  Ae
 Bxi
n
i 1
n
i 1


n

 A xi e  Bxi   vi2  v02e  Bxi xi e  Bxi  0 , setting the sums equal:
i 1

n


 A xi e  Bxi   vi2  v02e  Bxi xi e  Bxi , solving for A:
i 1
  v

n
A
 v02 e  Bxi xi e  Bxi
2
i
i 1
  e
n
 Bxi
i 1
(Eq. 7)

 1 xi e
 Bxi
Substituting A of Equation 7, into Equation 5:




n
S

2
  vi   v02 

A i 1 




  v
n


 v02 e  Bxi xi e  Bxi 
e  Bxi 

 e  Bxi  1 xi e  Bxi 

2
i
i 1

n
i 1


 v02 e  Bxi xi e  Bxi 
  e  Bxi  1  0

 e  Bxi  1 xi e  Bxi 

  v
n

2
i
i 1

n
i 1



(Eq. 8)
In Excel, the value for B that satisfies Equation 8 can be determined by trial and error. A
can then be determined by plugging the known value for B into Equation 7. With the values for
A and B, µk and k can be calculated, by Equation 4.
Using the velocity and position data acquired on the flat track, values were calculated and
are summarized in Table 1.
A
B
k
µk
7.9547331
0.0555949
0.0010753
0.0227070
Table 1
Part IV: Arbitrary Track Shape
In order to evaluate the predictive value of developed velocity models on an arbitrary
track, experimental velocity data must be acquired, as well as the shape of the corresponding
track. The method of collecting velocity data is the same method used for collecting data on the
flat track in Part I. A diagram of a sample track with relevant co-ordinate systems defined is
below:
8
Figure 4
Since the track is curved, it is necessary to find an equation which represented the car’s
path in the xy-plane, to determine the matchbox car’s theoretical velocity at any point along the
track. This was done using the “digitizer,” which allows an arbitrary track to be represented as a
set of points (x, y). The data collected was fit to a polynomial equation using Matlab. The
resulting representative polynomial graph of the track follows in Plot 2:
Arbitrary Track Shape
y coordinate (m)
1.7
1.5
1.3
1.1
0.9
0.7
0.5
0
0.5
1
1.5
2
2.5
3
3.5
4
x coordinate (m)
Plot 2
The coefficients of the arbitrary track polynomial, as calculated in Matlab, are shown in Table 2
below:
9
Power
of X
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Coefficients
22.11737
-285.864
1761.11
-6365.57
15033.27
-24604.2
28898.38
-24868.8
15856.12
-7515.25
2635.497
-673.922
122.0431
-14.8222
1.082535
-0.03593
Table 2
With the best-fit polynomial, x and y coordinates can be found at any point along the
arbitrary track, and be used in theoretical models needed. Because the best-fit polynomial
equation represents the physical track on which velocity data was collected, direct comparisons
can be made between theoretical velocity predictions and experimental data.
Part V: Testing “Flat-Track” model with known µk and k
With the values of A and B in Equation 4 determined and resultantly k and µk, as was
done in Part III, the accuracy of these values needed to be verified. To do this, the experimental
data obtained on a flat track from the five heights was manipulated to simulate a single high drop
with a much longer track. Using Excel, the theoretical velocity along the manipulated “longer
track” was calculated and plotted against the experimentally obtained values as can be seen from
the plot below.
10
Flat Track Model Compared with Experimental Data
Modeled Velocities
Experimental Data
10
2
2
Velocity Squared (m /s )
12
8
6
4
2
0
0
2
4
6
8
10
12
14
Postion (m)
Plot 3
As seen from Plot 3, the experimental velocities are predicted fairly well by the
theoretical model. This leads the researchers to believe that the values of k and µk are accurate
for the matchbox car used, and that the model in Equation 4 is a good predictor for velocity as a
function of position on a flat track.
Since the goal of this project is to model the motion of a matchbox car on an arbitrarily
shaped track, the velocity function proposed in Equation 4 should be compared to experimental
data collected on an arbitrary track. Using the arbitrary track shape function described in Part
IV, theoretical velocity values were calculated at a series of positions along the arbitrary track.
Since Equation 4 only accounts for displacement, distance along the arbitrary track (s) was used
instead of x-position (x), which represented displacement in the case of a flat track. Predictably,
since the flat track model does not account for changes in the height of the track or curvature
(and the forces that these features induce), it is a very poor predictor of velocity on an arbitrary
track. This is shown in Plot 4 below:
11
Experimental Velocities on Arbitrary Track
trial 1
Trial 2
Trial 3
Trail 4
Trial 5
Trial 6
Trial 7
Flat Track Model
3
Velocity (m/s)
2.5
2
1.5
1
0.5
0
0
0.5
1
1.5
2
2.5
3
3.5
4
S - Position (m)
Plot 4
Since the flat track model only accounts for displacement, it creates a velocity-position
curve similar to one which could be observed experimentally on a flat track. This is significantly
different than the observed velocities, shown as seven series representing seven trials. However,
this does not mean that the flat-track model is “worthless.” As described in Part III, the flat-track
model allowed the researchers to isolate the values of A and B to determine k and µk, which are
constant regardless of track shape.
Since the flat-track model is insufficient for modeling velocity on an arbitrary track, the
effects of a sloped track will need to be added into the model.
Part VI: Modeling motion on a flat sloped track with Friction and Drag
On a flat track there is no acceleration due to gravity (except for initial acceleration down
the incline which was not modeled). On an inclined plane, however, one component of the
weight force acts in the direction of motion. Our flat track model does not account for this angle
on the surface, and so failed to accurately predict velocities of the car when the height of the car
changed.
12
The following is a Free-Body Diagram for this slope- track theory:
Figure 5
Applying Newton’s Second Law to Figure 5:
In S-direction (direction of displacement):
In N-direction (direction of normal):
F
F
s
 mas
N
 dv 
 D  f k  mg sin   m 
 dt 
 maN
N  mg cos   0
N  mg cos 
 dv 
 kv2   k N  mg sin   m 
 dt 
Where: cos  
dx
 dy
, sin  
ds
ds
Combining S and N directions:
 dv 
 kv2   k mg cos   mg sin   m  , Similar to multiplying by dx when modeling on the flat
 dt 
track; multiplying by ds:
 kv
2
 dv 
  k mg cos   mg sin  ds  m ds , dividing through by m, substituting for sinθ, and
 dt 
cosθ:

kv2
ds   k gdx  gdy  vdv , to solve for v, we integrate:
m
13
y
x
v
 kv2 
ds   k g  dx  g  dy   vdv , trapezoidal rule is needed to integrate the first term.
  
m 
s0 
x0
y0
v0
s
1  kv02   kv2 

  ( s  s0 )
This can be written as: A  
2  m   m 
 A   k g ( x  xo )  g ( y  y 0 ) 


1 2
v  v02 , multiplying through by 2:
2


 2 A  2 k g ( x  xo )  2 g ( y  y0 )  v 2  v02 , substituting back in for A, and moving all v terms
to one side:
v2 
kv2
kv2
( s  s0 )   0 ( s  s0 )  2 k g ( x  x0 )  2 g ( y  y0 )  v02 , factoring out v2:
m
m
kv2
 ks 
v2 
 1  v02  0 s  2 k gx  2 gy , solving for v:
m
 m

v
v02 
kv02
s  2 k gx  2 gy
m
ks
1
m
(Eq. 9)
With a model which accounts for changes in the height of the track, the model should
again be tested against experimental velocity data on an arbitrary track to verify its accuracy in
modeling the matchbox car’s velocity. Plot 5 (below) shows the theoretical velocity of the
matchbox car used on an arbitrary track, plotted with the average of seven trials of experimental
velocity data collected on the same track.
Experimental Velocity in Comparison Sloped
Track Theoretical Models
3
Velocity (m/s)
2.5
2
1.5
1
0.5
0
0
1
2
3
4
S - Displacement (m)
Plot 5
14
The model in Equation 9 predicts the velocity on an arbitrary track much more closely
than the model in Equation 4, but this is not an unexpected result. The model presented in
Equation 4 was a “flat-track” model, not accounting for changes in the track height (which would
cause additional forces on the car). The model developed on an inclined plane accounted for
changes in the track height, and the resulting forces which would act on the car. In turn, the
“inclined plane” model very closely predicts velocity on an arbitrary track. However, this model
makes the assumption that all changes in height of the track are small inclined planes. A more
complex model which accounts for curvature in the track can be developed, more accurately
reflecting the velocity where this model fails.
Part VII: Modeling motion on an Curved Track with Friction and Drag
The first two models predict the velocity of the matchbox car used in the researchers’
experimentation on a flat track, without and then with drag forces. These models only account
for a flat track, as shown in Part IV, whereas the goal model will need to account for an
arbitrarily curved track. The third model accounted for sloped track with friction and drag, but
did not account for track curvature, which creates variation in the normal force. The model
developed below is for a matchbox car traveling down a curved track, accounting for drag and
friction.
The free body diagram for this system is shown in Figure 6.
Figure 6
Using Newton’s Second Law, drag, and friction equations, a model to describe the velocity of
this system can be developed.
15
In S-direction (direction of displacement):
In N-direction (direction of normal):
F
F
s
 mas
N
 maN
 dv 
 D  f k  mg sin   m 
 dt 
v2
N  mg cos   m
r
 dv 
 kv   k N  mg sin   m 
 dt 
v2
N  m  mg cos 
r
2
Where: cos  
dx
 dy 1
f '' ( x )
, sin  
, 
ds
ds r 1  ( f ' ( x)) 2


3/ 2
Combining S and N directions:
v2
 dv 
 kv   k (mg cos   m )  mg sin   m  , Similar to multiplying by dx when modeling
r
 dt 
2
on the flat track; multiplying by ds:


v2
 dv 
2
  kv   k (mg cos   m )  mg sin  ds  m ds , dividing through by m, substituting for
r
 dt 


sinθ, and cosθ:

kv2
v2
ds   k gdx   k
ds  gdy  vdv , to solve for v, we integrate:
m
r
y
x
v
 kv2
v2 
  
  k ds   k g  dx  g  dy   vdv , trapezoidal rule is needed to integrate the first
m
r 
s0 
x0
y0
v0
s
v02   kv2
1  kv02
v 2 


  k   
  k  ( s  s0 )
term. This can be written as follows: A  
2  m
r0   m
r 
 A   k g ( x  xo )  g ( y  y 0 ) 


1 2
v  v02 , multiplying through by 2:
2


 2 A  2 k g ( x  xo )  2 g ( y  y0 )  v 2  v02 , substituting back in for A, and moving all v terms
to one side:
v 2  k
kv2
v2
v2
kv2
( s  s0 ) 
( s  s0 )   0 ( s  s0 )   k 0 ( s  s0 )  2 k g ( x  x0 )  2 g ( y  y0 )  v02
r
m
m
r0
Factoring out v2:
16
kv2
v2
 s ks 
v 2  k

 1  v02  0 s   k 0 s  2 k gx  2 gy , solving for v:
r
m
m
r0


v02 
v
kv02
v2
s   k 0 s  2 k gx  2 gy
m
r0
s ks
k

1
r
m
(Eq. 10)
Using this equation, the velocity of the car on an arbitrary track can be determined at any
point, using experimentally determined values of k and µk found in Part III. This model accounts
for friction, drag, and varying normal force due to track curvature.
Part VIII: Testing the Arbitrary track model and comparison of models
The model created in Part VII needs to be tested in comparison to experimental data
collected on an arbitrary track. Using Excel, theoretical velocities of our car were generated for
points along the mathematical representation of the arbitrary track from which we collected
experimental data. The plot comparing theoretical velocity based on Equation 10 (shown in red)
and experimental velocity collected (blue) on the arbitrary track recorded in Part IV follows:
Experimental Velocity in Comparison to Curved
Track Model
3
Velocity (m/s)
2.5
2
1.5
1
0.5
0
0
1
2
3
4
S - Displacement (m)
Plot 6
17
While looking at this model only in comparison to experimental data, the theoretical
velocity generally looks like it is a good fit. One notes, however, there seems to be relatively
large differences in the peaks of the velocities. It would be beneficial to compare the inclinedplane model (light blue) with the curved track model (red). Both of these models are shown in
comparison to arbitrary track velocity data (blue) below.
Experimental Velocity in Comparison to
Theoretical Models
3
Velocity (m/s)
2.5
2
1.5
1
0.5
0
0
1
2
3
4
S - Displacement (m)
Plot 7
Observing Plot 7, it is clear that the curved track model is a better predictor of velocity on
an arbitrary track, than the sloped-track model which does not account for varying normal forces.
With a satisfactory theoretical model developed, the velocity of the matchbox car can
now be predicted on an arbitrary track. This makes it possible to make predictions of the car’s
motion given an arbitrary track: we now solve the following “challenge sessions.”
Challenge Sessions
The following is our solutions for each of the five challenge sessions. These challenges
act as both a test in our ability to develop a solution which fits the given situation and in the
accuracy of our theoretical model. For each challenge we summarize the problem that needs to
be solved, physical principles behind it, a method to obtain a solution, the solution that we
generated, and how the solution compared to the actual result during the challenge session. This
18
section is not meant to provide all data needed to solve the challenge sessions, but provide a
general method for solving each of the problems.
Challenge 1:
The goal of the first challenge was to predict where the car would stop on the given track.
The release point of the car was given, and the zone in which the car would stop needed to be
predicted. The point at which the car would momentarily stop can be calculated by finding
where the velocity changed from a positive value to an undefined value (which means a negative
velocity). The following picture generally describes the car’s path.
Moore
House
1
Starting
Point
2
2
1
P it
3
Predicted
Turning Point Zone
h
Pat
PIT
By implementing our model proposed in Equation 10 in Excel, we calculated the velocity
of the car at points along the given track. Then, by observation, we recorded where the velocity
of the car went to “zero.” This occurred at point (3.3240 m, 1.1366 m). After obtaining this
value, we made sure to check that the velocity of the car did not prematurely go to zero at any
point on the track before the target zone.
During the challenge, the car consistently fell just short of this point (in the yellow zone
just before the red zone).
Challenge 2:
The goal of this challenge was to send the car down through a small valley and up onto a
flat plateau where it should stop as close to the far edge as possible. The closer to the far edge
the flat section, the more points awarded. We needed to determine the car’s release point so that
it would stop at the far edge of the plateau, without rolling off.
19
(x*,y*)
Starting
Point
1
2
3
Zones
To do this, we used Excel to calculate the velocities at points along the track. We then
moved the release point along the track by setting the initial velocity to zero at consecutive
points along the track. By examining the effects of moving the release point on velocity at the
desired stopping point, we moved the release point until the velocity in the targeted region went
to zero. We calculated that the car should be released from the point (.697 m, 1.243 m).
While we checked to make sure that the velocity of the car would not prematurely reach
zero before the target zone with our model, during the challenge session, the car was not able to
make it up onto the plateau. We would attribute this to two issues. First, we were unsure which
part of the car to place at the release point. Second, the window which would allow the car to get
up onto the flat portion of the track without speeding off was small, possibly too small for our
model to accurately predict.
Challenge 3:
The goal of the third challenge was to accurately predict where to release the car in order
to have the car jump over a catapult, placed at the beginning of the plateau, and land within a
target zone.
v0y
v0

v0x
Figure 7
In order to solve this challenge, several steps were required. First, a landing site was
chosen within the target zone. We chose to land within the first half of the target zone, which
20
was where the most points would be awarded. Second, we had to determine where the car would
leave the track, by finding where along the track the normal force went to zero. Then, using
these initial and final positions, kinematics can be used to find the initial velocity needed for the
car to land at the specified landing point/zone, and the resulting release point of the car.
With the landing site chosen, the next step is to find where the car leaves the track. The
free body diagram which best describes the car on this track is Figure 6. The forces which we
are concerned with are those in the N-direction, are summarized here:
F
N
 maN
v2
N  mg cos   m
r
v2
N  m  mg cos 
r
Setting the normal equal to 0, and solving for velocity, we obtain the velocity that the car
must be traveling at a point to leave the track.
v  rg cos  , where cos  
dx
ds
Using Excel and our velocity function (Equation 10) we searched for the point at which
our car’s velocity exceeded the “escape velocity.” However, even when the car was released
from the highest point on the track, the escape velocity was never reached. Thus we attempted an
alternate approach to this problem: we found at what point the normal force was minimized.
Since the normal is a function of velocity, and velocity at the launch point needed to be
determined, we needed to make the assumption that the position where the normal was
minimized (launch point) would not move significantly in order to solve for the velocity at this
point.
After solving for the point at which the normal was minimized, by calculating the normal
at points along the suspected “take-off” range, we developed a model which describes Figure 7.
x-direction:
x  x0  v0 x t 
1 2
at , with no acceleration in the x-direction while car is airborne:
2
21
x  x0  v0 x t , or t 
D
v0 x
y-direction:
y  y0  v0 y t 
1 2
at
2
 D
y  y0  v0 y 
 v0
 x
 1  D
  g
 2  v0

 x




2
Since v0 x  v0 cos  , and v0 y  v0 sin  :
1  D 

We obtain: y  y0  D tan   g 
2  v0 cos 
2
Using the above equation, we changed the release point in Excel (similar to the method
described in Challenge 2) of the car until the velocity of the car at the point where it leaves the
track (v0), made the equation equal to the desired landing y-position. This release point was
recorded to be the point (.486 m, 1.595 m). During the actual challenge, the car landed about 2/3
the length of D, before the end of the plateau, having cleared the catapult. It was not released
from a high enough position to clear the whole plateau.
Challenge 4:
The goal of this challenge was to successfully negotiate the loop (not fall off), and have
just enough velocity when reaching the end of the track at (x*,y*) to land on the lowest floor,
receiving the highest score. The track setup is shown in Figure 8 below.
v0
v0y

v0x
Figure 8
The approach to solving this challenge was similar to Challenge 3. Using kinematics
from point (x*,y*), where the car would leave the track, we developed equations which could
predict the final height of the car (y), after traveling distance (D), given the velocity of the car at
22
the take-off point (v0), and the angle of take-off (θ). Since the release point of the car was
variable, we could move the release point to control the velocity at the take off point. The
kinematics equations that describe the flight of the car are developed below:
x-direction:
1
x  x*  v0 x t  at 2 , with no acceleration in the x-direction while car is airborne:
2
x  x*  v0 x t , or t 
D
v0 x
y-direction:
1
y  y *  v0 y t  at 2
2
 D
y  y*  v0 y 
 v0
 x
 1  D
  g
 2  v0

 x




2
Since v0 x  v0 cos  , and v0 y  v0 sin  :
1  D 

We obtain: y  y  D tan   g 
2  v0 cos 
2
*
(Eq. 11)
Using Excel, Equation 11 referenced the velocity of the car at point (x*,y*) in the
spreadsheets. Then by moving the release point forward along the track, as in the previous two
challenges, the velocity at (x*,y*) decreased. By observing the output of the function, we
determined the release point which would yield the desired final height (y). Since the loop in the
track posed a threat to our car falling off, we checked to make sure our car could negotiate the
loop, given its release point. To do this, consider the following figure:
Figure 9
23
Using Newton’s Second Law:
F
y
 may
 N  mg  m(a )
 N  mg  m
v2
r
Solving for v: v  rg
This equation gives us the velocity that the car must be traveling at the top of the loop (as
in Figure 9) in order to stay on the track. Entering this equation into Excel, using (1/curvature)
as r (defined in Part VII of developing a model), we found the velocity that the car must be going
at the top of the loop to not fall off exceeded the velocity that was predicted by our velocity
model at this point. Since no points were to be awarded for cars that fell off the track, we decided
to move the car’s release point higher until it had just enough speed to negotiate the loop. At this
point, able to negotiate the loop, by Equation 11 we found it would be impossible to land on the
lowest floor shown in Figure 8.
As such, we chose the final release point to be (.307 m, 1.428 m). During the challenge,
our car was able to successfully navigate the loop, and as predicted, did not land in the lowest
garage, but instead landed in the 4th highest one. This was slightly surprising because we had
predicted that, given our release point, the car would have gone much higher, but some variations
in the end of the physical track could have changed the car’s trajectory (one example of this is
the angle of the end of the track). After our official run was over, we ran the car down the track
a few more times from different starting positions. To our surprise, even when moving the car
half a meter more down the track, it was still able to negotiate the loop, and landed in the 4 th
highest garage/pad.
Challenge 5:
The goal of this challenge session was to travel down a given track, and intercept (in midair) a plummeting object (poodle), which would begin its fall in a given amount of time after our
car passed a photogate. Since a police car would be released when our car passed the photogate,
and would always hit the poodle, we decided that if we could pull just in front of the police car
before it landed on the lower track, then we would hit the poodle before the police car got to it.
24
Since the target was given, our job was to decide where to release our car from to intercept the
poodle. Consider the following diagram of the situation:
Plummeting
Poodle
Local
Police
h
Starting
Point
H
Photogate
Location
(x1, y1)
Landing Spot
(x2, y2)
L
Since we developed a model which predicts velocity as a function of position, we need to
introduce a time component into our calculations. Consider the following equation which
describes our car’s motion:
v
ds
ds
, therefore dt 
dt
v
Using Excel we were able to approximate the time interval dt where the car traveled ds at
velocity v. By setting t0 = 0 s at the location of the first photogate, the time elapsed at any
location could then be estimated by summing the small time intervals dt. Now, knowing the
relationship between the car’s position (s) and the time elapsed (t) we could find the proper
release point of the car so that it would be just ahead of the police car when it landed on the
track.
Using similar methods as described in previous challenge sessions, we moved the car’s
release point forward on the track by setting successive initial velocities to zero, continuing to
observe the time at which it would reach point (2.215 m, 1.382 m). This point was found by
adding the length of our car to (x2, y2), where the police car would land, so that our car would be
clear of the police car as it landed.
We selected our release point to be (.328 m, 1.669 m). Released from this point, our car
would reach point (2.215 m, 1.382 m) momentarily before the police car hit the track at t =
1.144s. This would result in our car intercepting the poodle.
During the challenge sessions, we had not yet prepared this solution to test, so a “bestguesstimate” was used instead. This value was (.291 m, 1.720 m). Our car did beat the police
car during the challenge session, but, unlike the solution proposed above, our car was a large
distance ahead of the police car when it hit the track, making our car fly under the poodle before
25
it dropped. We believe that the solution proposed above is a better solution because it is in
agreement with the results of the challenge session: our solution proposed a release point further
down the track, making the car slightly slower to catch the poodle as it falls.
Challenge Session Evaluation:
In the challenge sessions, we found that our velocity model held up reasonably well to
solve the various problems. However, we did note some common errors in our model in the
challenge session solutions. In the first three challenge sessions we overestimated the velocity
that our car would be traveling, and because of this the car fell short of the desired target
zones/goals. For example, in the first challenge we set our target zone where we thought the car
would stop given our theoretical velocity model. However, the velocity of the car went to zero
before this point, meaning that we had overestimated the velocity.
Not surprisingly, this overestimate of velocity can be seen in the graphs comparing our
theoretical model, which includes track curvature, to experimentally determined velocity values
on the same track (seen in Part VIII). However, we were unable to find a way to account for this
overestimate mathematically when developing the model or challenge session solutions.
One extraneous factor in the challenge sessions which we were also unable to account for
is the different type of track used (yellow track versus red track). While not all of the challenges
used this type of track and we are unsure of how great an effect on the coefficient of friction
different track type might have, this is a variable that deserves mention.
Error Considerations
Our general success in the challenge sessions shows the strength of the velocity model
developed in predicting the velocity of our car along an arbitrary track, but the unexpected
variations in the challenge session results show the error present in our model. Since many of
challenge sessions required low margins of error to get full credit, our model was not able to
accurately enough predict the velocity. As can be seen in Part VIII, both the model for sloped
tracks and model that includes the effects of curvature are generally overestimates of
experimentally determined velocity. This error could be attributed to the values of µk and k,
which have greater corrections on velocity as the velocity of the car gets higher. If the values of
these constants were underestimated, the implementation of the velocity model for our car would
26
be overestimated, especially at higher velocities. In retrospect, better values of µk and k could
have been obtained with a longer flat track, allowing us to release the car from higher points, in
turn allowing us to study the velocity of the car along the flat track for a greater distance and
range of velocities.
Conclusions
This paper has demonstrated the method of creating a mathematical model for a physical
system. Beginning with a simple model that neglected drag on a horizontal track, experimental
data was compared with a basic model constructed on the physical laws that govern the system.
Noticing that this simple model was not enough, drag was added to the model. The new model
was tested against experimental data on an arbitrary track, and not surprisingly, had no predictive
value. We then developed a model that accounted for a sloped track without curvature, and
compared this to experimentally collected data. We also developed a more advanced model that
considered the forces that track curvature induces. Evaluating each of the models we found that
the flat track model accurately fit the experimentally collected data on a flat track. This led us to
believe that our coefficients µk and k were accurate for our car. Using the values of µk and k
calculated on the flat track, we evaluated the predictive value of the sloped track and curved
track models for an arbitrary track. We found that the curved track model which accounts for
variation in the normal force is the best predictive model. During the challenges sessions our
model performed reasonably well, allowing us to develop solutions which were fairly accurate
when tested in the real world..
27