Lab 06 Force and Motion Part II
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Lab 06: Force and Motion Part II
I.
Develop an experimental mathematical model to describe the behavior of a system
a.
Select an IV to test
Hanging mass
b.
Experimental Design Template
Research Question:
How does the acceleration of a system change when hanging mass changes?
Dependent variable (DV):
Acceleration of a system
Independent variable (IV):
Hanging mass
Control variables (CV):
Mass of the system: 0.3044kg, length of the string: 1.04m, starting point: 0.85m
Testable Hypothesis:
There is a positive correlation between the hanging mass and the acceleration of a system.
Prediction:
Picture 1: Experimental setup
c.
Complete the experimental design.
We will be doing 8 trials and will use values of 0.0094kg, 0.0144kg, 0.0194kg, 0.0244kg
d.
Conduct the experiment.
The uncertainty of the measured values for acceleration is ±0.001m/s2 due to the rotary motion sensor’s estimated scale uncertainty, given in this lab.
The uncertainty of the measured values for the length of the string and position of the system is ±0.0005m due to the meter stick’s estimated uncertainty, given in a previous lab.
The uncertainty of the masses is ±0.001kg due to the scale’s estimated uncertainty, given in a previous lab.
Hanging mass
(kg)
Trial 1
(m/s
2
)
Trial 2
(m/s
2
)
Trial 3
(m/s
2
)
Average
(m/s
2
)
0.0094
0.288 ± 5.2*10
-4
0.289 ± 2.8*10
-4
0.290 ± 2.2*10
-4
0.289
0.0114
0.349 ± 3.0*10
-4
0.346 ± 6.2*10
-4
0.349 ± 5.1*10
-4
0.348
0.0134
0.407 ± 3.4*10
-4
0.407 ± 4.2*10
-4
0.404 ± 1.1*10
-3
0.406
0.0154
0.467 ± 5.0*10
-4
0.467 ± 5.3*10
-4
0.466 ± 4.9*10
-4
0.467
0.0174
0.525 ± 2.8*10
-4
0.526 ± 6*10
-4
0.524 ± 6.4*10
-4
0.525
0.0194
0.575 ± 7.1*10
-4
0.574 ± 5.7*10
-4
0.575 ± 5.5*10
-4
0.575
0.0214
0.630 ± 5.3*10
-4
0.630 ± 6.5*10
-4
0.630 ± 6.9*10
-4
0.630
0.0234
0.683 ± 6.8*10
-4
0.684 ± 9.4*10
-4
0.677 ± 2.8*10
-3
0.681
Data table 1: Acceleration vs Hanging mass
e.
Enter collected data into Excel
0.5
0.7
0.9
1.1
1.3
1.5
1.7
1.9
2.1
2.3
2.5
0.25
0.3
0.35
0.4
0.45
0.5
0.55
0.6
0.65
0.7
f(x) = 0.29 x + 0.03
R² = 1
Hanging Force vs Acceleration
Hanging Force (N)
Acceleration (m/s^2)
Graph 1: Acceleration vs Hanging force
(The error bar for acceleration and hanging force is too small to be seen)
f.
Consider the mathematical model provided by Excel
a = 2.8663F + 0.0295
2.8663 (1/kg) and 0.0295 (m/s
2
)
A causal relationship exists between the acceleration (a) and the gravitational force (F) if the mass of the system, length of the string, starting point is held constant, indicating a positive linear function.
II.
Developing a second experimental mathematical model to describe the behavior of the system.
a.
Select a second IV.
Mass of the system
b.
Repeat all steps in Part I.
Experimental Design Template
Research Question:
How does the acceleration of a system change when the mass of the system changes?
Dependent variable (DV):
Acceleration of a system
Independent variable (IV):
Mass of the system
Control variables (CV):
Hanging mass: 0.0234kg, length of the string: 1.04m, starting point: 0.85m
Testable Hypothesis:
There is a negative correlation between the mass of the system and the acceleration of a system.
Prediction:
c.
Complete the experimental design.
8 trials, values of 0.3044kg, 0.3544kg, 0.4044kg, 0.4544kg, 0.5044kg, 0.5544kg, 0.6044kg, 0.6544kg
d.
Conduct the experiment.
The uncertainty of the measured values for acceleration is ±0.001m/s2 due to the rotary motion sensor’s estimated scale uncertainty, given in this lab.
The uncertainty of the measured values for the length of the string and position of the system is ±0.0005m due to the meter stick’s estimated uncertainty, given in a previous lab.
The uncertainty of the masses is ±0.001kg due to the scale’s estimated uncertainty, given in a previous lab.
System mass
(kg)
Trial 1
(m/s
2
)
Trial 2
(m/s
2
)
Trial 3
(m/s
2
)
Average
(m/s
2
)
0.3044
0.687 ± 4.8*10
-4
0.688 ± 5.6*10
-4
0.687 ± 1.3*10
-4
0.687
0.3544
0.596 ± 9.5*10
-4
0.595 ± 1.0*10
-4
0.597 ± 3.4*10
-4
0.595
0.4044
0.528 ± 6.7*10
-4
0.528 ± 5.2*10
-4
0.528 ± 5.6*10
-4
0.528
0.4544
0.473 ± 4.4*10
-4
0.472 ± 4.8*10
-4
0.471 ± 8.1*10
-4
0.472
0.5044
0.428 ± 4.6*10
-4
0.429 ± 4.1*10
-4
0.428 ± 4.7*10
-4
0.428
0.5544
0.391 ± 4.1*10
-4
0.392 ± 2.2*10
-4
0.392 ± 5.8*10
-4
0.392
0.6044
0.361 ± 4.0*10
-4
0.362 ± 4.0*10
-4
0.361 ± 3.3*10
-4
0.361
0.6544
0.335 ± 2.5*10
-4
0.335 ± 3.9*10
-4
0.335 ± 3.1*10
-4
0.335
Data table 2: Acceleration vs System mass
e.
Enter collected data into Excel
0.3
0.35
0.4
0.45
0.5
0.55
0.6
0.65
0.7
0.3
0.35
0.4
0.45
0.5
0.55
0.6
0.65
0.7
f(x) = 0.23 x^-0.94
R² = 1
System Mass vs Acceleration
System Mass (kg)
Acceleration (m/s^2)
Graph 2: Acceleration vs System mass
(The error bar for acceleration and system mass are too small to be seen)
f.
Consider the mathematical model provided by Excel
a
=
0.2252
m
−
0.938
0.2252 (N) and -0.938
A causal relationship exists between the acceleration (a) and the system mass (F) if the hanging mass, length of the string, starting point is held constant, indicating a negative power function.
III.
Connecting Experimental Model to Established Scientific Model
a.
Define established scientific model for the acceleration of a system.
a ∝ F
net
, hanging mass vs. acceleration of a system
a
∝
1
m
, mass of a system vs. acceleration of a system
b.
Compare your experimental model with the established scientific model.
a
=
2.8663
F
+
0.0295
, acceleration of system
a
=
0.2252
m
0.938
, mass of system vs acceleration of system
The relationships represented in our mathematical models and the scientific equation are remarkably similar to the established models for the acceleration of a system in relation to the hanging mass, as the hanging mass
and the acceleration of a system are proportional, while the mass of the system is inversely proportional to the acceleration of the system. The only difference in the models is that the mass value in the equation for the mass of a system vs. acceleration of a system has an exponential value of 0.938, which may be partially attributed to the uncertainty values of the data.
IV.
Newton’s Second Law
a.
Connect your experimental models to the general form of Newton’s Second Law as Σ F
=
m
system
a
, which can be rewritten for motion in one dimension as:
a
x
=
Σ F
m
system
=
F
1
x
m
system
+
F
2
x
m
system
+
F
3
x
m
system
+
…
i.
Compare your group’s mathematical model, which has the form
a
=
C
1
F
app
+
C
2
to the general form of Newton’s Second Law. In the section of your lab records that includes this model, indicate what the constants C
1
and C
2
physically represent. Then, determine what the value for C
1
should be based on your lab set-up and compare it to the value in your model. Knowing that your model describes the behavior of a real system, summarize the conditions of the lab set-up that might cause the values for C
1
and C
2
to be larger or smaller than expected.
Our group’s mathematical model is a = 2.8663F + 0.0295, with C
1
=2.8663 and C
2
=0.0295. This relates to the general form of Newton’s Second Law because C
1
represent 1
m
, and since F is multiplied by C
1
, it is the same as Newton’s second law. C
2
is the acceleration. Hanging mass
Acceleration
Force of gravity on system
System Mass
Fnet on cart in direction of cart's motion
(a)Accelerati
on of System
Experimental relationship: a=2.8663F+0.0295
Fnet increases with larger mhanging
Fnet increases with larger acceleration
Fnet increases with larger Fg
No impact (controlle
d)
Experimental relationship: a=0.2252/(m^(0.938)) Established relationship: a
∝1/m (when F is constant)
Mechanism explained: There is no mechanism to describe here
Established relationship: a
∝
Fnet (when msystem constant
Mechanism explained: There is no mechanism to describe here
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0.7
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coefficient differential equations is the equation for a harmonic oscilator. This equation models
the motion of a mass attached to a spring. The spring is attached to a vertical wall and the
mass is allowed to slide along a horizontal track. We let z denote the displacement of the mass
from its natural resting place (with x > 0 if the spring is stretched and x 0 is the
damping constant, and k> 0 is the spring constant. Newton's law states that the force acting
on the oscillator is equal to mass times acceleration. Therefore the differential equation for the
damped harmonic oscillator is
mx" + bx' + kr = 0.
(1)
k
Lui
Assume the mass m = 1.
(a) Transform Equation (1) into a system of first-order equations.
(b) For which values of k, b does this system have complex eigenvalues? Repeated eigenvalues?
Real and distinct eigenvalues?
(c) Find the general solution of this system in each case.
(d)…
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A mechanical system is presented as
below. There are four simulation
graphs for different values for m, b
and c tested for a step response.
Identify the graph for m=4 kg, b=0.3
N.s/m and k=1 N/m (Hint: you may
need to use matlab simulation to find
it).
m
1.8
1.6
1.4
b
·f(t)
Step Response
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P1.12: An aluminium thin-walled sphere is used as a marker buoy. The sphere has a radius of 65 cm and
a wall thickness of 10 mm. The density of aluminium is 2700 kg/m³. The buoy is placed in the ocean where
the density of the water is 1050 kg/m³. Determine the height H between the top of the buoy and the surface
of the water.
MATLAB Basics
87
H.
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In a linear spring finite element model, as per the finite element model’s sign convention, which one of the following statements is true?
Select one:
a. If Fe1 is positive then the spring is under tension
b. If Fe2 is negative then the spring is under tension
c. If Fe2 is positive then the spring is under compression
d. If Fe1 is positive then the spring is under compression
Clear my choice
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