2048LLab 2Lab Manual 2 c.pdf1 Lab Manual Irina.docx

2048L/Lab 2/Lab Manual 2 c.pdf
1
Lab Manual
Irina Golub
July 30, 2017

2
PART ONE: Photographic Analysis of a Falling Object
INTRODUCTION
With the great advances that have been made in digital imaging
and analysis, experimental data
is often in the form of photographic images. In this experiment,
you will make displacement
measurements of a tennis ball dropped from a height using
photographic data and your
computer’s mouse positioning system. From two displacement
measurements and the time
between these measurements and one of the five kinematic
equations that describe one-
dimensional motion, the acceleration due to gravity on earth,
“g”, will be estimated. Since you

know what the correct answer should be (9.8 m/s2) you will be
able to calculate the percent
error of your estimate.
Neglecting air resistance, a falling object increases its speed 9.8
meters per second every
second that it falls on earth. This is “g”, the acceleration due to
gravity. Below you will see
snapshots taken of a falling tennis ball at equal intervals of time
(0.1 second between frames).
You can see that the displacement of the tennis ball increases
during each successive time
interval. This is due to the tennis ball speeding up in each
time interval, i.e., the tennis ball is
accelerating.
Read University Physics Volume 1 Chapter # 3: MOTION
ALONG A
STRAIGHT LINE
EQUIPMENT
A PC running MS Internet Explorer web browser. (Other web
browsers may not work for
this experiment.)
OVERVIEW
The photographic data file shows one composite photo made by
splicing six separate images
of a tennis ball dropped straight down. Each of the six separate
images was taken 0.1 second
apart.

Just below the photo in the photographic data file you will see
boxes labelled X and Y with
numbers that change when you move the mouse over the photo.
These numbers are mouse
coordinates in what we will call “mouse units.”
You will record the Y-position of the ball in the first and last
image (i.e., ball image #1
and ball image #6) (the X-direction is not needed as the object
was falling straight down).
Since the numbers you record will be in mouse units and not
meters, only the difference
between these two measurements will be important. You will be
able to convert this
difference from mouse units into meters because there are two
meter sticks joined together
vertically in each photo as a reference.
[If you are unable to see the mouse coordinates in your browser
and are unable to get to
a BCC lab computer you can measure ruler coordinates instead
of mouse coordinates using
a plastic ruler held near (but not touching!) the computer
display. Substitute the phrase ”ruler
coordinates” for ”mouse coordinates” in the procedure and
questions. Take your
measurements in millimeters. This will not be as accurate as
using mouse coordinates which
have a higher resolution.]
NOTE: Since mouse units are relative to the top left side of the
display, it is important to
center the entire photo in your display and not scroll the photo
when you take measurements.
If you cannot see the entire photo on your monitor without
scrolling it, change your display

to 800 x 600 by clicking the “Start” button, then “Control
Pane’, then “Display”, then
“Settings”, and adjusting the Screen Resolution to 800 x 600.
PROCEDURE
1. Determine the conversion factor between mouse units and
meters as follows. In each
3
photo, two meter sticks are joined together vertically. Read the
mouse’s y-coordinate
at the top of the upper meter stick by clicking on that point.
Read the mouse’s y-
coordinate at the bottom end of the lower meter stick by
clicking on that point. The
difference between these measurements represents 2 meters.
For example, if the top of the top meter stick is at y=50 and the
bottom of the bottom meter stick
is at y=650, then 650-50 = 600 mouse units = 2 meters.
Therefore, one (1) mouse unit would equal
2 meters/600 mouse units = 0.0033 meters [This calculation was
just an example of how to perform
the calculation. Do not use 0.0033 meters per mouse unit as the
conversion between mouse units
and millimeters. Measure and perform your own calculation]
2. Open the photographic data file. Measure the y-position (i.e.,
y-coordinate) of the ball
(in mouse units) for the first image and the last (i.e., 6th) image

five times. Measure
from the same place on the ball in each photograph (e.g., the top
of the ball)., and
record these values on your data table.
3. Calculate the difference in the y-coordinates (mouse units) of
the ball from the first
image to the last image. For example, if the y-coordinate of the
ball in image #1 was
1369 mouse units and 1748 for the ball in image #2, then the
difference would be 1748-
1369 or 379 mouse units.
4. Multiply the difference you found in step #3 by the
meters/mouse unit conversion
you found in step #2. For example, if you found a difference of
379 mouse units
between image #1 and image #6 and a meters/mouse unit
conversion of 0.0033 in step
#2, the result would be 379 mouse units x 0.0033 meters/mouse
unit = 1.25 meters.
5. Using y=0.5 gt2, solve for g where y is the result you found
in step #4 (in meters) and
t is the time difference between the 1st and 6th photos (0.5
seconds). For example,
using 1.25 meters for y and 0.5 sec for t, g is calculated to be 10
m/s2.
6. Compute the percent error between the value for “g” you
calculated com- pared to the
actual value of “g”.

PART TWO: ESTIMATING REACTION TIME
After you see the rear brake lights on the car in front of you,
how much time does it take for you to step on
your brake pedal? This time interval is one measure of your
reaction time. Reaction time complicates many
time measurements because it introduces a time lag between an
event and the manual recording of the event
(e.g., with a stop watch). Automated measurements employ
sensors that have very short response times.
Read the Free Falling Object with Air Resistance.
1. Neglecting air resistance, an object in free fall moves through
a distance, in meters, given by 0.5gt2.
2. Count the air resistance, an object in free fall moves through
a distance, in meters, given by
�� = �2
��
��
(1 + ��
6��
)
Where k is the constant of proportionality; the numerical value
of k depends upon the shape of the object
being dropped and the density of the atmosphere. Use the
approximal value of the �� = 3.46 ∗ 10−3 ��
��

By measuring the distance an object falls between two events,
the time interval between the events can be
calculated. In this experiment, you will estimate the reaction
time for your hand using one-dimensional
kinematics and free fall.
4
EQUIPMENT
Yardstick with metric rulings (i.e., a meter stick)
The assistance of another person
PROCEDURE
1. Your assistant should hold a meter stick vertically. The top
of your outstretched hand should line up with
the bottom of the meter stick. Your fingers and thumb should
form a “V” around the bottom of the meter
stick. Your hand should neither be too open or too closed for
best results.
2. At a time chosen by your assistant, your assistant will drop
the meter stick through your outstretched hand
and you will safely grab the meter stick.
3. After grabbing the meter stick, record the position of the top
of your hand on the meter stick in centimeters.
4. Perform five trials of the experiment and record your results
into the data table.

5. Determine the weight of the meter stick.
6. Convert the displacements from centimeters to meters.
CALCULATIONS
1. Calculate the average displacement for the five trials.
Calculate the relative error of your measurement of the
displacement.
2. Calculate your average reaction time using the average
displacement. If neglecting air resistance using the
formula:
y=0.5gt2
displacement count the air resistance using the
formula:
�� = �2
��
��
(1 + ��
6��
)
4. Compere your results.
4. Find in the internet the average reaction time of a human.
5. Calculate the percent error between your results and the

theoretical data from the internet.
3. Would you expect the reaction time to move a leg (e.g.,
pressing the brake pedal in response to a visual
stimulus) is greater, the same, or less, than the hand reaction
time measured in this experiment? Why?
INTRODUCTIONEQUIPMENTOVERVIEWPROCEDURE
2048L/Lab 2/free fall theory lab 2c.pdf
Introduction
If you will drop the object from a fixed height and measure the
time t taken to fall between two light
beams which are placed a distance y apart.
a
r
b
1st position
2nd position
Object
T = 0
T = T0
T = T1

y(t)
t = 0
Y = 0
Y = a
Y = a + r
Y = a + r + b
• Figure 1. Note the two types of variables used, the lower case
variables, y, r and t are measured values. For
example: t, corresponds to the time interval, the y measured in
the meter, for the object to travel between the first
and second positions. The upper case quantities, Y and T, are
variables to be determined in the data analysis.
For example: the object has zero velocity at T = 0.
Free Falling Object Without Air Resistance
The equation of motion for a freely falling object is:
m = mg
d 2y
(1.)
dt2
Integration yields:
v(t)= v 0 + gt (2)
where vo is the velocity when the object passes the first
position. Integrating this equation yields:
2
0 2

1
y(t)= v t + gt (3.)
Chaer Falling Object
These equations account for the presence of a gravitational
force only. In practice, air provides a drag
force which can have a significant effect on a falling object.
While y(t) represents the actual distance that the object has
dropped in a time interval t, (see figure 1), in
the actual experiment r is the distance . The distances a and b,
(see figure 1), which are on the order of
a few millimeters, can not be measured directly because they
depend, among other factors, on the
exact position. A rough approximation for both values should be
found during the the experiment and,
as you will see later, very accurate determinations of both a and
b will be made.
Keeping in mind that
y(t)= r(t)+ b (4.)
then g can be determined by linearizing equation 3 and applying
a least square fit to a straight line.
How do you linearize equation 3? Divide it by t; fitting that
equation vs. t results in a linear equation with
slope g/2 and intercept v0.
What is the effect of b, which at this point is not known with
great precision? When we linearize
equation 2 weget:
t t 20

+1 gty (t )= r (t )+ b =v (5.)
If we introduce a small error, δ, in b and then assume that b' = b
+ δ we arriveat:
r (t )+ b′ = r (t )+ b + δ= v
t t 2 t
+ 1 gt + δ0 (6.)
Figure 2.2 shows a plot of equation 6 vs. t for different values
of δ. The asymptotic effect that the δ/t
term produces at t = 0 can clearly be seen.
-2
-4
-6
-8
10
8
6
4
2
0
.6
y/t vs. t for different values of delta

0 0.1 0.2 0.3 0.4 0.5 0
0.1
0.05
0
-0.05
-0.1
• Figure 2. y/t vs. t for different values of δ.
Hence, the "best" value for b can be determined by selecting a
few well chosen values of b' and
calculating the resulting χ2: Only when b' = b, i.e. for δ = 0,
will the data approach a straight line at t=0.
For all other values of b', the data will deviate from a straight
line and the fit will produce a large value of
χ2. Therefore, a minimum χ2 corresponds to the best value of b.
The least square fit automatically also
calculates the best values for g/2 and v0.
Free Falling Object with Air Resistance
For an object falling in air, the equation of motion, 1
becomes:
m
��2��
��2
= �� − ��(��
��
)�� (7.)

where the second term on the right is the drag force and k
specifies the strength of the retarding force.
The value of n depends on the shape of the object and its
velocity. For large velocities, n approaches
unity which corresponds to Stoke’s law of resistance. At small
velocities, like the ones acquired by the
object in this experiment, the resisting force is proportional to
the square of the velocity, in which case it
obeys Newton’s law of resistance:
(8.)
This equation can be integrated to yield the solution:
t
g
y(t )= V
2
(9.)
where Vt is the terminal velocity which is related to k:
mg
kt
V =2 (10.)
If we have a system with Y(T=0) and v0=0 , then equation 9.
can be solved for T:
t −1

g
V
T = cosh (11.)
Since the effect of the drag force is not very transparent in this
equation, the following expansion in Y
may be moreilluminating:
(12.)
The correction due to drag appears in second and higher terms.
Setting Vt=∞ , which is the case when
no retarding media is present, reduces it to the familiar equation
3. (with v0=0 ). Vt is at least on the
order of 1000 cm/sec or larger, we only need to consider the
first two terms in equation 12. Rewriting
this in terms of k we find that T is a function of Y and m, i.e.
the mass of the object:
(13.)
This equation shows that when the mass of the object
approaches infinity, the second term, which
corresponds to the drag force on the object, goes to zero; in that
case, a value for g can be determined
which no longer is affected by air resistance.
m
��2��
��2
= �� − ��(��
��
)2

ln(cosh
��
��
+ ��
��
��
��
)
[��
��
��
2]
�� = 2 ��
��
(1+
��
6��
2+
��2��2
120��
2+…)
�� = 2 ��
��
(1+
��
6��

)
t
g
g − g′
V ≅ v (15.)
for an object dropped from a maximum height ymax is:and
observing that the average velocity v
2
g′y maxv ≅ (16.)
Additional Information
g = 980.616 − 2.5928Cos(2ϕ)+ 0.0069Cos 2(2ϕ)− 3.086×10 −6
H
where ϕ is the latitude and H the elevation above sea level in
centimeters. (Latitude for Minneapolis is
approximately 47 degrees and its elevation is about 700 feet
above sea level.)
A recent measurement by the geophysics department determined
a value of g here at the University as
g = 980.58cm/sec2.
TerminalVelocity
Equation 8. can be also be written as:
where g’ represents the acceleration of an object due to gravity
in a resistive medium and Vt is the

terminal velocity for the particular object. An average value of
g’, i.e. g ′ The corresponding values of
Vt can then be found by solving equation 14 :
��′=g(1- ��
2
��
2 )
(14.)
ü Dropping things from the Empire State Building (for real)
What about throwing things from tall structures in air?
We set up Newton’s Second Law equations for this system as a
differential equation. The key difference is that we are
moving through a fluid that has low density and low viscosity
(think of this as a measure of the syrupy-ness of the
fluid) we need to modify the drag expression:
FG - FD = m g - a v
2 = m
„
„t
v
or
„

„t
v = g -
a
m
v2
Remember, at t=0 the velocity is 0. So the slope of the velocity
is, again, simply g. This looks familiar. As time goes
on the velocity increases so that the slope decreases. The
difference is that the slope changes more rapidly as time
goes on. As the speed increases, the slope decreases more
swiftly. Eventually, at some place close to (D), the slope
reaches 0 and no longer changes significantly.
The slope of the velocity reaches 0 when we reach “terminal
velocity.” For our system described above
this point is vterm =
m
a
g .
We can see that the curve is going to look like the previous
expression for our aqueous exploration but with a steeper
rise sharper turn around:
„v = Ig - a
m
v2M „t
For each tick of the clock the change in the velocity is getting

smaller rapidly as it incrementally adds to the starting
velocity (0) and increments toward the terminal velocity.
Integrating this expression is not as simple as the previous
efforts only because the result doesn’t look like anything we
will find on the reference sheet. As before we will factor
out aêm leaving vterm behind.
„vIvterm2- v2M = am „t
This time we can’t simply transform the variables. Try if you
let u = vterm
2 - v2 you get „u = - 2 v „v. This doesn’t
make the integral look like anything we recognize. We have to
go to a book on integral tables. After scouring the
tables we find the following indefinite integral:Ÿ 1
k2-x2
„x =
tanh-1J x
k
N
k
+ C
We can adapt this fairly quickly to our expression and get the
following result…‡ vHtL0 „vIvterm2- v2M = ‡ t0 am „t ï tanh-
1J vHtLvterm Nvterm = am t
We can look at the dimensions of
a
m
vtermand it is easily to confirm algebraically that this
combination of constants has

units of
[email protected] In other
words: [email protected] = x
So we get for the velocity…
vHtL = vterm tanhIvterm am tM
This is as far we will take this. You can see that the Tanh
function has all the right characteristics as the previous
function only it rises more steeply and bends sharply to
asymptotically arrive at the terminal velocity a bit sooner.
The graph above is scaled so that the two different cases focus
on the different shapes.
free fall theory for lab 2Falling ObjectSlide Number 2Slide
Number 3Slide Number 4free fall theory
Labs/free fall/freefall.html
PHOTOGRAPHIC DATA
Y
X
Labs/free fall/image001.jpg

Table of Contents.htmlPHY2048L PH CALC I LAB ONLINE
622068 - Unit 2 : Lab - Free Fall and Reaction Time
1. Lab 2 - Free Fall and Reaction Time
2. Free Fall Theory
3. Estimating g from Photograpic Analysis of Freefall
1
Lab Manual
Irina Golub
July 30, 2017

2
PART ONE: Photographic Analysis of a Falling Object
INTRODUCTION
With the great advances that have been made in digital imaging
and analysis, experimental data
is often in the form of photographic images. In this experiment,

you will make displacement
measurements of a tennis ball dropped from a height using
photographic data and your
computer’s mouse positioning system. From two displacement
measurements and the time
between these measurements and one of the five kinematic
equations that describe one-
dimensional motion, the acceleration due to gravity on earth,
“g”, will be estimated. Since you
know what the correct answer should be (9.8 m/s2) you will be
able to calculate the percent
error of your estimate.
Neglecting air resistance, a falling object increases its speed 9.8
meters per second every
second that it falls on earth. This is “g”, the acceleration due to
gravity. Below you will see
snapshots taken of a falling tennis ball at equal intervals of time
(0.1 second between frames).
You can see that the displacement of the tennis ball increases
during each successive time
interval. This is due to the tennis ball speeding up in each
time interval, i.e., the tennis ball is
accelerating.
Read University Physics Volume 1 Chapter # 3: MOTION
ALONG A
STRAIGHT LINE
EQUIPMENT
A PC running MS Internet Explorer web browser. (Other web
browsers may not work for
this experiment.)

OVERVIEW
The photographic data file shows one composite photo made by
splicing six separate images
of a tennis ball dropped straight down. Each of the six separate
images was taken 0.1 second
apart.
Just below the photo in the photographic data file you will see
boxes labelled X and Y with
numbers that change when you move the mouse over the photo.
These numbers are mouse
coordinates in what we will call “mouse units.”
You will record the Y-position of the ball in the first and last
image (i.e., ball image #1
and ball image #6) (the X-direction is not needed as the object
was falling straight down).
Since the numbers you record will be in mouse units and not
meters, only the difference
between these two measurements will be important. You will be
able to convert this
difference from mouse units into meters because there are two
meter sticks joined together
vertically in each photo as a reference.
[If you are unable to see the mouse coordinates in your browser
and are unable to get to
a BCC lab computer you can measure ruler coordinates instead
of mouse coordinates using
a plastic ruler held near (but not touching!) the computer
display. Substitute the phrase ”ruler
coordinates” for ”mouse coordinates” in the procedure and
questions. Take your
measurements in millimeters. This will not be as accurate as

using mouse coordinates which
have a higher resolution.]
NOTE: Since mouse units are relative to the top left side of the
display, it is important to
center the entire photo in your display and not scroll the photo
when you take measurements.
If you cannot see the entire photo on your monitor without
scrolling it, change your display
to 800 x 600 by clicking the “Start” button, then “Control
Pane’, then “Display”, then
“Settings”, and adjusting the Screen Resolution to 800 x 600.
PROCEDURE
1. Determine the conversion factor between mouse units and
meters as follows. In each
3
photo, two meter sticks are joined together vertically. Read the
mouse’s y-coordinate
at the top of the upper meter stick by clicking on that point.
Read the mouse’s y-
coordinate at the bottom end of the lower meter stick by
clicking on that point. The
difference between these measurements represents 2 meters.
For example, if the top of the top meter stick is at y=50 and the
bottom of the bottom meter stick
is at y=650, then 650-50 = 600 mouse units = 2 meters.
Therefore, one (1) mouse unit would equal

2 meters/600 mouse units = 0.0033 meters [This calculation was
just an example of how to perform
the calculation. Do not use 0.0033 meters per mouse unit as the
conversion between mouse units
and millimeters. Measure and perform your own calculation]
2. Open the photographic data file. Measure the y-position (i.e.,
y-coordinate) of the ball
(in mouse units) for the first image and the last (i.e., 6th) image
five times. Measure
from the same place on the ball in each photograph (e.g., the top
of the ball)., and
record these values on your data table.
3. Calculate the difference in the y-coordinates (mouse units) of
the ball from the first
image to the last image. For example, if the y-coordinate of the
ball in image #1 was
1369 mouse units and 1748 for the ball in image #2, then the
difference would be 1748-
1369 or 379 mouse units.
4. Multiply the difference you found in step #3 by the
meters/mouse unit conversion
you found in step #2. For example, if you found a difference of
379 mouse units
between image #1 and image #6 and a meters/mouse unit
conversion of 0.0033 in step
#2, the result would be 379 mouse units x 0.0033 meters/mouse
unit = 1.25 meters.
5. Using y=0.5 gt2, solve for g where y is the result you found
in step #4 (in meters) and
t is the time difference between the 1st and 6th photos (0.5
seconds). For example,
using 1.25 meters for y and 0.5 sec for t, g is calculated to be 10

m/s2.
6. Compute the percent error between the value for “g” you
calculated com- pared to the
actual value of “g”.
PART TWO: ESTIMATING REACTION TIME
After you see the rear brake lights on the car in front of you,
how much time does it take for you to step on
your brake pedal? This time interval is one measure of your
reaction time. Reaction time complicates many
time measurements because it introduces a time lag between an
event and the manual recording of the event
(e.g., with a stop watch). Automated measurements employ
sensors that have very short response times.
Read the Free Falling Object with Air Resistance.
1. Neglecting air resistance, an object in free fall moves through
a distance, in meters, given by 0.5gt2.
2. Count the air resistance, an object in free fall moves through
a distance, in meters, given by
�� = �2
��
��
(1 + ��
6��
)

Where k is the constant of proportionality; the numerical value
of k depends upon the shape of the object
being dropped and the density of the atmosphere. Use the
approximal value of the �� = 3.46 ∗ 10−3 ��
��
By measuring the distance an object falls between two events,
the time interval between the events can be
calculated. In this experiment, you will estimate the reaction
time for your hand using one-dimensional
kinematics and free fall.
4
EQUIPMENT
Yardstick with metric rulings (i.e., a meter stick)
The assistance of another person
PROCEDURE
1. Your assistant should hold a meter stick vertically. The top
of your outstretched hand should line up with
the bottom of the meter stick. Your fingers and thumb should
form a “V” around the bottom of the meter
stick. Your hand should neither be too open or too closed for
best results.

2. At a time chosen by your assistant, your assistant will drop
the meter stick through your outstretched hand
and you will safely grab the meter stick.
3. After grabbing the meter stick, record the position of the top
of your hand on the meter stick in centimeters.
4. Perform five trials of the experiment and record your results
into the data table.
5. Determine the weight of the meter stick.
6. Convert the displacements from centimeters to meters.
CALCULATIONS
1. Calculate the average displacement for the five trials.
Calculate the relative error of your measurement of the
displacement.
displacement. If neglecting air resistance using the
formula:
y=0.5gt2
displacement count the air resistance using the
formula:
�� = �2
��
��
(1 + ��
6��

)
4. Compere your results.
4. Find in the internet the average reaction time of a human.
5. Calculate the percent error between your results and the
theoretical data from the internet.
3. Would you expect the reaction time to move a leg (e.g.,
pressing the brake pedal in response to a visual
stimulus) is greater, the same, or less, than the hand reaction
time measured in this experiment? Why?
INTRODUCTIONEQUIPMENTOVERVIEWPROCEDURE

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