Lab 2: Free Fall Motion
Velocity (cm/sec)= distance moved (cm) / time (sec)
Acceleration (cm/sec2) = velocity change (cm/sec) / time (sec)
time
(s)
position
(cm)
distance
(cm)
velocity
(cm/s)
0
0
0
0.01
0.049
0.049
4.9
0.02
0.196
0.147
14.7
0.03
0.441
0.04
0.784
0.05
1.225
0.06
1.764
0.07
2.401
0.08
3.136
0.09
3.969
0.1
4.9
0.11
5.929
0.12
7.056
0.13
8.281
0.14
9.604
0.15
11.025
0.16
12.544
0.17
14.161
0.18
15.876
0.19
17.689
0.2
19.6
0.21
21.609
0.22
23.716
0.23
25.921
0.24
28.224
0.25
30.625
0.26
33.124
0.27
35.721
0.28
38.416
0.29
41.209
0.3
44.1
0.31
47.089
0
Find the acceleration, and % error
gofficial = 980 cm/s2
Question 1:
If an object initially rest is dropped from height of 50 meters,
how long will it take it to hit the ground and what is its final
speed when it hits the ground?
Question 2:
What if the same object is dropped on the moon whose
acceleration is 1/6 that of earth, gmoon = 1.63 m/s2. Answer the
same question.
Time (s)
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.1
0.11
0.12
0.13
0.14
0.15
0.16
0.17
0.18
0.19
0.2
0.21
0.22
0.23
0.24
0.25
0.26
0.27
0.28
0.29
0.3
0.31
Position(cm)
0
0.049
0.196
0.441
0.784
1.225
1.764
2.401
3.136
3.969
4.9
5.929
7.056
8.281
9.604
11.025
12.544
14.161
15.876
17.689
19.6
21.609
23.716
25.921
28.224
30.625
33.124
35.721
38.416
41.209
44.1
47.089
Distance(cm)
0
0.049
0.147
0.245
0.343
0.441
0.539
0.637
0.735
0.833
0.931
1.029
1.127
1.225
1.323
1.421
1.519
1.617
1.715
1.813
1.911
2.009
2.107
2.205
2.303
2.401
2.499
2.597
2.695
2.793
2.891
2.989
Velocity(cm/s)
0
4.9
14.7
24.5
34.3
44.1
53.9
63.7
73.5
83.3
93.1
102.9
112.7
122.5
132.3
142.1
151.9
161.7
171.5
181.3
191.1
200.9
210.7
220.5
230.3
240.1
249.9
259.7
269.5
279.3
289.1
298.9
Velocity
350
300
Velocity (cm/s)
250
200
150
100
50
0
0
0.05
0.1
0.15
0.2
Seconds(s)
0.25
0.3
0.35
Lab II, Problem 5: Velocity and Force
Jenny Smith
July 14, 2011
Physics 1201W, Professor Johnson, TA Min Li
Abstract
The final velocity of a gravity-driven launch cart for a pterosaur model was
determined. The mechanism was tested for four driving masses at similar launch
heights. The final velocity predicted by Newton’s second law and kinematics
was confirmed to within experimental error.
Introduction
A research group is investigating the hypothesis that modern birds are the descendants
of pterosaurs. As part of this investigation, the flight of pterosaurs is being studied via
models. The models are to be launched by a mechanism consisting of a cart accelerated
down a straight, level track by a string. The string runs horizontally, over a pulley, and
then vertically to a hanging mass which is pulled downward by gravity. To ensure the
desired launch parameters for the flight, it is necessary to be able reliably to predict the
final velocity of the cart. This experiment studied the final velocity as a function of the
driving mass and the height of its release.
Prediction
The final velocity can be easily calculated by application of Newton’s second law and
kinematics. Let the mass of the cart be M; the hanging mass, m; and the release
height, h. The force on the system is the downward force of gravity on the hanging
mass, F=mg; using Newton’s second law for the compound system,
(1)
Letting the initial position be 0m and the initial velocity be 0m/s, kinematics then yields
the final velocity:
(2)
The final velocity of the cart is predicted to be that given by Equation 2.
Procedure
A cart was placed on a straight, level, elevated track with a pulley at one end. A string
was tied to the cart and run over a pulley. A mass was tied to the other end and allowed
to hang down over the edge of the track. The string was pulled taught, and the mass was
allowed to fall from rest, pulling the cart with it. Its initial height was less than the
length of string between cart and pulley so that the cart would undergo an initial,
accelerating phase and a final, coasting phase in its motion. This mechanism is depicted
in Figure 1.
Figure 1: The test mechanism used for this experiment. b > h.
The motion of the cart was recorded with a video camera. MotionLab software was
used to plot the position and velocity of the cart with respect to time. The position and
velocity were then fit by eye in two stages, one for the accelerating and one for the
coasting phase. The final velocity was taken to be the parameter in a zeroth-order
monomial fit to the coasting phase of the velocity measurements.
Data
m (g) h (m) experimental vf (m∕s) theoretical vf (m∕s)
+0.20
+0.04
50.0 0.46
1.25
1.24
-0.35
-0.03
70.0 0.41
1.35
+0.25
1.34
+0.04
100.0 0.47
1.65
150.0 0.48
1.90
-0.15
+0.15
-0.30
+0.25
-0.20
1.64
1.94
-0.03
+0.04
-0.04
+0.04
-0.04
Table 1: The masses m, release heights h, and experimental and theoretical values of
the cart’s final velocity vf. The error in all of the masses is 0.3g. The error in all of the
release heights is 0.02m.
Analysis
The experimental measurements and theoretical predictions of the final velocity are
given in Table 1 in the Data section. The errors in the experimental measurements were
calculated by the parameters in a minimum zeroth-order monomial greater than and a
maximum zeroth-order monomial less than all measured data points in the coasting
phase. The errors in the theoretical predictions were calculated by the “worst case”
propagation of error technique, using the errors quoted above and g=9.80±0.02m∕s. The
entire error intervals of the theoretical predictions are within the respective error
intervals of the experimental measurements, so this experiment did confirm the
predictions to within error.
There are numerous sources of error not addressed in this experiment. One is the
compound effect of friction and drag retarding the motion of the cart and pulley. This
was unquantified. Another was the acceleration produced by any deviation from the
horizontal of the track; this was determined to be zero within the static friction in the
axles of the cart in that it was incapable of accelerating the cart from rest. Another was
the distortion in the videos due to the camera. Another was the masses of the string and
the pulley, which were assumed to be massless in the theoretical calculation. All of
these, and any other sources of error not mentioned here, are believed to be insignificant
in comparison to the random error in that the error interval is much wider than the
difference between the theoretical and experimental values of the final velocity.
Conclusion
A proposed launching mechanism for a pterosaur model to be used in the investigation
of the ancestry of modern birds was modeled by a cart on a track. The cart was
accelerated from rest by each of four hanging masses pulled by gravity via a string run
over a pulley. The final velocity of the cart after the hanging mass reached the ground
was measured and compared to the predictions of kinematics and Newton’s second
law. The predictions were confirmed in that they and the measurements both lay within
one another’s error intervals. The research group is justified in using Equation 2 to
predict the final velocity of the cart used to launch the model. However, it must take
into account the mass of the model by using the sum of the model’s and the cart’s masses
rather than just the mass of the cart in its calculations.
=
qv=0
som
an
0 . By motion and equation
ut & at
S=
th
t
늦
2
U=Omls
– 4.9t²
.
It= 3.194 sec. I ans
t
– so a
–
2
By motion 1st equation
V = ut at
a- g earth = 9.8 mis?
V= – 9.8 X 3.194.
V
Va – 31. 30 mis . I downward c-ve).
speed 31.30mis. Ans
nehen the motion is on moon.
So ut & bat?
So = 0 – 5 X 1.63t?
= 7.83 sec ?
&
5
ट
2
–
こ
–
It
grown
)
*
final speed = ahoon t = 1.63*7.83
V= 12.76 mis.
frrrr
AM
.
Lab 2: Free Fall
Purpose: In this lab, we will show that an object under the influence of gravity alone
accelerates at a constant rate. We will also measure this acceleration and compare with
the actual value of 9.8 m/s.
Equipment: Demo Sparking Tape Apparatus, Sparked Tapes, Masking tape,
Two-Meter stick
Theory: An object which is under the influence of gravity alone, is said to be in Free
Fall. Its acceleration under free fall is equal to 9.8 m/s? or 32 ft/s and is called “g”.
This means that every second, its speed increases by 9.8 m/s if it is dropped from a
certain height. If its initial speed is zero, then its speed at t= 1 second will be v=9.8 m/s
and at t=2 seconds will be v = 19.6 m/s and so on. The equation for this is v=gt=9.8t.
Procedure: The Sparking Tape Apparatus is a device which is used to drop an object
and at the same time to create sparks every 1/60 seconds across the object onto a
Sparking Tape. During the first 1/60 of a second, the object will not travel too much so
the 1″ dot will be close to the Original dot when the object was at rest. As the object
picks up speed, the dots will get successively farther apart. Ifs – distance traveled from
the original dot, v = velocity
or speed, a = acceleration, then we can use the following
equation Vavg Distance Traveled/Time interval = As/t = As/(1/60) = 60As to calculate
the Average velocity or speed in each successive time interval
Then, using the equation Aavg=Change of Velocity/Time interval = Av/t = 60Av will
give us the Average acceleration in each successive time interval. This Average
acceleration should be equal to 9.8 because the acceleration should be constant in each
time interval. If we average all these accelerations and compare with the actual value, we
can achieve a % error
AVavg(m/s)
Aarz (m/s)
Example:
So=0 meters
Si=2.5 cm = .025 meters
S2 = 5.6 cm = .056 m
S3 = 9.6 cm = .096 m
S4 = 14.5 cm = .145 m
Dot Number S (meters) AS (m) Vavg(m/s)
60AS
0
0
1
025
025
1.50
2
056
4
031
1.86
3
.096
.040
2.40
4
.145
.049
2.94
Gexp = (21.6 + 32.4 + 32.4)/3 = 86.4/3 = 28,8 m/s?
60AVR
36
54
54
21.6
32.4
32.4
% error = 100% x 128.8 – 9.8/9.8 = 193.9 % error (Wow, that’s off!!!!!!!!!!!)
Purchase answer to see full
attachment
