Experiment 4
UNIFORMLY ACCELERATED MOTION
Objectives
1.
2.
3.
4.
5.
6.
Measure the position and velocity of an object that is moving with:
a. A constant velocity, and
b. Constant acceleration
Explore the relationships between position, velocity and acceleration,
Analyze the graphs of position and velocity vs. time for a cart’s motion,
Explain how the slope of the velocity graph relates to the acceleration of the object,
Analyze the motion of an object in free fall, and
Measure the acceleration of gravity of an object in free fall
Theory
In this laboratory exercise you will learn the kinematics of an object that travels with (i) uniform velocity
and (ii) uniform acceleration i.e. an object moving with a uniform or constant (i) velocity or (ii)
acceleration. The most common example of an object moving with a constant acceleration is a free falling
object. If the object falls solely under the influence of gravity it is said to be in free fall. In this case the
acceleration of the object is the acceleration due to gravity g. Near the Earth’s surface the acceleration of
gravity is essentially constant with a value of g = 9.81 m/s2 = 981 cm/s2 = 32.2 ft/s2. Air resistance can
affect the acceleration of a freely falling object. However, for relatively heavy objects this effect can be
negligible. In this experiment we will use two methods to measure the acceleration due to gravity. In one
case we will use a motion sensor and in the other case we will use a photo gate timer. The equations and
definitions necessary to make this experiment are the same ones that we presented in experiment 3 of this
course, and some others that we will see next
The purpose of this experiment is to study the relationships between position, velocity, and acceleration
in linear motion when the object moves with (i) a constant velocity (horizontal plane) and (ii) with constant
acceleration (inclined plane)
Horizontal plane
As seen in Figure 4-1, the cart rests on a frictionless horizontal plane. When the cart is given a slight
push, it moves with a constant velocity. Using the motion sensor one can graphically plot the position
versus time, velocity versus time and acceleration versus time of the moving cart
Figure 4-1. On the horizontal plane, with no friction, the cart travels with constant speed
Inclined plane
Figure 4-2 shows a cart moving with a constant acceleration down a frictionless inclined plane. The
value of the acceleration is a = g sin θ. When doing the experiment we will measure the acceleration of
the cart from the slope of the graph of v vs. t. We can determine the value of the angle θ if we know the
height of the left end of the plane, h, and the length of the plane, l. In this case we have θ = sen -1 (h / l).
With this data, and knowing the measured value of a, we can calculate the value of g, the measured value,
and will be able to compare it with 9.81 m/s2 which is its reported value. We must clarify that Figure 4-2
shows an exaggerated inclination, if we compared it with what we will have in the laboratory. Actually h =
1.5 cm approximately
Nicholas J. Pinto and Claudio Guerra‐Vela. Department of Physics and Electronics. University of Puerto Rico at Humacao. Sponsored by the National Science Foundation (NSF) © All rights reserved
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Figure 4-2. The incline makes the cart to move down with a constant acceleration
Free fall
An object in free fall (neglecting air resistance) falls under the influence of gravity, with an acceleration
g. The distance y that the object falls in a time t is given by:
y = yo + vo t +
1 2
gt
2
4-1
where yo is the initial position and vo is the initial velocity. If an object is dropped from rest (vo = 0) and
from the origin (yo = 0), then
y=
1 2
gt
2
4-2
(downward taken as positive). Hence by measuring the time t it takes for an object to fall a distance y, the
acceleration due to gravity can be easily calculated
Figure 4-3. The ball bearing is at a known height and its time of fall will automatically be measured with a photo gate
timer
In this part of the experiment you will use a photo gate timer to measure the time of fall of a steel ball
from a constant height y as seen in Figure 4-3. Take at least ten time measurements. Fill in the data of the
height and times of fall in Table 1 of your lab report. Since the ball is always dropped from rest we can use
Equation 4-2 to find g. Complete Table 1 by calculating t2. Then calculate the value of g for each entry and
obtain the average of all measurements. Compare your average value with 9.81 m/s2
Nicholas J. Pinto and Claudio Guerra‐Vela. Department of Physics and Electronics. University of Puerto Rico at Humacao. Sponsored by the National Science Foundation (NSF) © All rights reserved
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Examples
1. If the angle of inclination θ is 15o in Figure 4-2, what is the acceleration of the cart if the cart is
released from rest at the top of the plane? Assume g = 9.81 m/s2
Solution:
Using information on Figure 4-2:
a = g sinθ = 9.81 sin 15o = 2.54 m/s2
2.
A frightened ostrich moves in a straight line with a velocity described by the velocity-time graph
shown in Figure 4-4. Sketch the acceleration versus time graph
Figure 4-4 Instantaneous velocity of an ostrich as a function of time
Solution:
From the graph in Figure 4-4 we see that at time zero, the speed of the ostrich is zero, however,
after about 0.6 s its speed is 10.0 m/s. Approximately, when t = 1.2 s, the ostrich reaches its
maximum velocity, that is of about 14.3 m/s. This implies that throughout this time the ostrich
accelerated. After that its speed began to diminish, until becoming zero, shortly after passed 3 s.
By definition, the instantaneous acceleration is the derivative of velocity with respect to time.
This means that, to sketch the graph of acceleration against time, we must choose some points in
the graph of Figure 4-4, draw a tangent straight line to the curve in each one of those points, and
calculate its slope. See, for example, Figure 4-5, where we have chosen the origin as the first
point to draw the tangent and to calculate its slope. In agreement with the scales of the axes, the
slope of that straight line is 20.0 m/s2. We have identified points A, B, C, D, E, and F as others in
which we can draw up the tangent and calculate its slope. Evidently the outline of the graph of
acceleration against time will be more precise whichever greater is the number of points in which
we draw up the tangents. The result of the graph of acceleration vs. time is in Figure 4-6,
where, in addition, we have identified the points A, B, C, D, E, and F
Nicholas J. Pinto and Claudio Guerra‐Vela. Department of Physics and Electronics. University of Puerto Rico at Humacao. Sponsored by the National Science Foundation (NSF) © All rights reserved
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Figure 4--5 The acceleration in the origin, at time t = 0, is calculated from the slope of the tangent to the curve in that
point
Figure 4-6 The ordinates of points A, B, C, D, E, and F are the values of slopes of the tangents to the curve of Figure 45 in those same points
Nicholas J. Pinto and Claudio Guerra‐Vela. Department of Physics and Electronics. University of Puerto Rico at Humacao. Sponsored by the National Science Foundation (NSF) © All rights reserved
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3.
How far does the runner whose velocity-time graph is shown in the Figure 4-7 below travel in 16 s.
Solution: In this example we need to use Equation 4-1 with g replaced by a. There are four time
intervals A, B, C, and D. In each of these intervals we need to calculate the acceleration of the
runner
Interval A: a = 4 m/s2
Interval B: a = 0
Interval C: a = - 2 m/s2
Interval D: a = 0
Distance covered in interval A:
yo + vo t +
1
2
at 2 = 0 + 0 +
1
2
× 4 × 22 = 8 m
Distance covered in interval B:
yo + vo t +
1
2
at 2 = 0 + 8 × 8 + 0 = 64 m
10
B
8
Velocity (m/s)
C
6
4
D
A
2
0
0
2
4
6
8
10
12
14
16
18
Time (s)
Figure 4-7 A velocity vs. time graph of a runner
Distance covered in interval C:
yo + vo t +
1
2
at = 0 + 8 × 2 −
2
1
2
× 2 × 22 = 12 m
Distance covered in interval D:
yo + vo t +
1
2
at 2 = 0 + 4 × 4 + 0 = 16 m
Hence total distance traveled by the runner in 16 s is 8 + 64 +12 + 16 = 100 m
Nicholas J. Pinto and Claudio Guerra‐Vela. Department of Physics and Electronics. University of Puerto Rico at Humacao. Sponsored by the National Science Foundation (NSF) © All rights reserved
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4.
A student in Caracas did the free fall experiment in his Physics class and recorded the data shown
below. Find the value of the acceleration of gravity in Caracas by plotting an appropriate graph
y (m)
t (s)
0
10
20
30
40
50
60
70
80
90
100
0
1.41
2.00
2.45
2.83
3.16
3.46
3.74
4.00
4.24
4.47
Solution: We first create new table calculating t2 as shown. Then we plot a graph of y versus t2
and find the slope. Multiply the slope by 2 to get the acceleration due to gravity in Caracas.
y (m)
t (s)
t2 (s2)
0
10
20
30
40
50
60
70
80
90
100
0
1.41
2.00
2.45
2.83
3.16
3.46
3.74
4.00
4.24
4.47
0
1.99
4.00
6.00
8.01
9.98
11.97
13.99
16.00
17.98
19.98
120
100
y (cm)
80
60
m = 5.0045 m/s2
40
20
0
0
5
10
15
20
25
t2 (s2)
The acceleration due to gravity is g = 2*m = 2 * 5.0045 = 10 m/s2
Nicholas J. Pinto and Claudio Guerra‐Vela. Department of Physics and Electronics. University of Puerto Rico at Humacao. Sponsored by the National Science Foundation (NSF) © All rights reserved
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Experiment 4. Laboratory Report
Uniformly accelerated motion
Section _______ Laboratory bench number _______
Date: ______________________________________
Students:
________________________________________________________
________________________________________________________
________________________________________________________
________________________________________________________
PART I: Horizontal plane
1.
Place the linear track in a horizontal position as shown in Figure 4-1. Connect the motion sensor to the
end of the plane with the electrical connections made to the interface.
2.
Select the graph to plot the position, velocity, and acceleration of the cart as a function of time when
the cart is given a slight push from rest.
3.
Take a printout of the three graphs and attach them you your report.
4.
What are your conclusions?
Nicholas J. Pinto and Claudio Guerra‐Vela. Department of Physics and Electronics. University of Puerto Rico at Humacao. Sponsored by the National Science Foundation (NSF) © All rights reserved
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PART II: Inclined plane
1.
Place the linear track in an inclined position as shown in Figure 4-2.
2.
Connect the motion sensor to the end of the plane with the electrical connections made to the interface
3.
Select the graph to plot the velocity of the cart as a function of time when the cart is released from rest
at the top of the plane.
4.
Choose the linear portion of the graph and calculate the acceleration of the cart by using the fitting
program.
5.
Taking a hint from Example 1, calculate the acceleration due to gravity.
6.
Attach the graph and your calculations below.
Nicholas J. Pinto and Claudio Guerra‐Vela. Department of Physics and Electronics. University of Puerto Rico at Humacao. Sponsored by the National Science Foundation (NSF) © All rights reserved
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PART III. Free fall
1.
Connect the apparatus as shown in Figure 4-3.
2.
For a constant height, find the time of ten falls and complete the Table below.
3.
Calculate the acceleration due to gravity in Humacao from your data.
4.
Attach your graph to the laboratory report. Hint: See Exercise 4.
Constant height, y = _________________
Trial
1
t (s)
Table 1
t2 (s2)
g = 2y/t2
2
3
4
5
6
7
8
9
10
gave = ______________m/s2
Δ% =
gave − g reported
g reported
× 100 = ______
Remember that greported = 9.81 m/s2
Questions:
1.
In Part III, what should be the shape of the graph of distance of fall (y) versus time (t)?
Nicholas J. Pinto and Claudio Guerra‐Vela. Department of Physics and Electronics. University of Puerto Rico at Humacao. Sponsored by the National Science Foundation (NSF) © All rights reserved
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Questions on Experiment 4
Uniformly accelerated motion
This questionnaire has some typical questions on experiment 4. All students who are taking the
laboratory course of University Physics I must be able to correctly answer it before trying to make the
experiment.
10
B
Velocity (m/s)
8
C
6
4
D
A
2
0
0
2
4
6
8
10
12
14
16
18
Time (s)
Figure 0-8 Graph of velocity vs. time of a runner
NOTE: Figure 4-8 above represents a velocity-time graph of a runner. Questions 1, 2, and 3 next refer to
this graph
1.
How far does the runner travel in the time interval 0 < t < 4 s?
a. 2 m
b. 8 m
c. 6 m
d. 24 m
e. 16 m
2.
What distance does the runner cover in the time interval 10 s < t < 12 s?
a. 12 m
b. 10 m
c. 2 m
d. 40 m
e. 6 m
3.
How far does the runner run in 16 s?
a. 8 m
b. 64 m
c. 100 m
d. 16m
e. 12 m
Nicholas J. Pinto and Claudio Guerra‐Vela. Department of Physics and Electronics. University of Puerto Rico at Humacao. Sponsored by the National Science Foundation (NSF) © All rights reserved
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4.
The position of a particle with respect to time is given by the following expression
x (t ) = 20t − 5t 3
where x is in meters and t is in seconds. When is the particles’ velocity zero?
a. t = 1.25 s
b. t = 4.73 s
c. t = 1.15 s
d. t = 2.00 s
e. t = 0.50 s
5.
When is the particles’ acceleration zero?
a. t = 0 s
b. t = 1.15 s
c. t = 2.00 s
d. t = 0.50 s
e. t = 1.25 s
6.
If the cart is released from rest at the top of the inclined plane in Figure 4-9 below and the acceleration
of the cart is measured to be 3.2 m/s2, what is the inclination of the plane? Assume g = 9.8 m/s2
Figure 0-9 An inclined plane
a.
b.
c.
d.
e.
o
41
19o
73o
15o
10o
7.
If the angle of inclination θ is 10o in Figure 4-9 above, what is the acceleration of the cart, if the cart is
released from rest at the top of the plane? Assume g = 9.8 m/s2.
a. 3.20 m/s2
b. 2.54 m/s2
c. 1.70 m/s2
d. -5.33 m/s2
e. 9.65 m/s2
8.
A student throws a stone vertically downward with an initial speed of 12 m/s from the roof of a
building 30 m above the ground. Ignore air resistance. How long does it take to reach the ground?
a. 3.06 s
b. 1.75 s
c. 2.5 s
d. 3.98 s
e. 1.54 s
Nicholas J. Pinto and Claudio Guerra‐Vela. Department of Physics and Electronics. University of Puerto Rico at Humacao. Sponsored by the National Science Foundation (NSF) © All rights reserved
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9.
What is the speed of the stone at impact?
a. 12 m/s
b. 360 m/s
c. 2.5 m/s
d. 27.1 m/s
e. 0.4 m/s
10. A stone is dropped from rest into a deep well. The sound of the splash is heard 4 seconds later. How
deep is the well? Assume g = 9.8 m/s2.
a. 39.2 m
b. 156.8 m
c. 78.4 m
d. Insufficient information, need the mass of the stone to find the depth.
e. 100 m
Nicholas J. Pinto and Claudio Guerra‐Vela. Department of Physics and Electronics. University of Puerto Rico at Humacao. Sponsored by the National Science Foundation (NSF) © All rights reserved
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