Rotational Dynamics
Additional Practice Problems
AP Physics 1
Kuffer
Unit 5 Circular Dynamics and Simple Harmonic Motion
Homework Outline
Due
Class
Date
#
73
Topic
Book
Section(s)
Rolling Motion
Torque
8.6 & 8.7
9.1
Rigid Objects in Equilibrium
(Statics)
Center of Gravity
Newton’s Second Law for
Rotation
Rotational Work and Energy
9.2
9.3
Angular Momentum
Simple Harmonic Motion & Unit
Circle
Energy of Simple Harmonic
Motion
The Pendulum
9.6
10.2
75
78
81
82
84
87
88
89
9.4
9.5
10.3
10.4
Assignment
Read/recall
Vector Cross Product
Worksheet
FOC: 1 P: 3-5, 7-8, 10
FOC: 3, 6, 8 P: 11-13, 16, 18,
20-21,23-25, 27, 30
FOC: 10, 12, 13
P: 31-33, 35, 38-41, 43-47
FOC: 16 P: 48-49, 51-55, 5758
FOC: 17-18 P: 59, 61, 64-67
FOC: 3-4 P: 15-18, 20-23
FOC: 11, 13 P:27-29, 31-32,
34
FOC: 14 P: 43-45, 48-49
Video(s)
5.1-5.2
5.3-5.7
5.8-5.9
5.10
5.11
5.12-5.13
5.14-5.15
5.16
1
I.
Rotational vs Translational (9.1)
2
3
II.
Torque on Rigid Bodies (9.1)
(9.1)
Direction: The torque is positive when the force tends to produce a counterclockwise
rotation about the axis, and negative when the force tends to produce a clockwise
rotation.
SI Unit of Torque:
Equation 9.1 indicates that forces of the same magnitude can produce different torques,
depending on the value of the lever arm.
In this top view, the hinges of a door appear as a black dot (•) and define the axis of rotation. The line of action and lever
arm are illustrated for a force applied to the door (a) perpendicularly and (b) at an angle. (c) The lever arm is zero because
the line of action passes through the axis of rotation
Note: The lever arm (l) is perpendicular to the Force (line of action)
4
III.
Equilibrium of a Rigid Body (9.2)
A rigid body is in equilibrium if it has zero translational acceleration and zero angular
acceleration. In equilibrium, the sum of the externally applied forces is zero, and the sum
of the externally applied torques is zero:
(4.9a and 4.9b)
(9.2)
The reasoning strategy for analyzing the forces and torques acting on a body in
equilibrium is given below. The first four steps of the strategy are essentially the same as
those outlined in Section 4.11, where only forces are considered. Steps 5 and 6 have been
added to account for any external torques that may be present. Example 3 illustrates how
this reasoning strategy is applied to a diving board.
Reasoning Strategy
Applying the Conditions of Equilibrium to a Rigid Body
1. Select the object to which the equations for equilibrium are to be
applied.
2. Draw a free-body diagram that shows all the external forces acting on
the object.
3. Choose a convenient set of x, y axes and resolve all forces into
components that lie along these axes.
4. Apply the equations that specify the balance of forces at
equilibrium:
and
.
5. Select a convenient axis of rotation. The choice is arbitrary. Identify the
point where each external force acts on the object, and calculate the
torque produced by each force about the chosen axis. Set the sum of the
torques equal to zero:
.
Solve
the
equations
in
Steps
4
and
5 for the desired unknown quantities.
6.
5
Physics Web Quest: Torque
Name ________________________
Click on Torque Simulator on Mr. Neff’s website.
(http://phet.colorado.edu/simulations/sims.php?sim=Torque )
Part I: Torque
1.
2.
3.
4.
Click the tab at the top that says torque
Set the force equal to 1 N.
Click Go let this run for at least 10 seconds
What is the torque on the wheel (include direction).
5. What eventually happens to the lady bug? _______________________________
6. From Newton’s second law, a force will cause an __________________________
7. When considering angular motion, a torque will cause an ___________________
____________________ (consider both torque equations)
8. What must be the centripetal force that keeps the lady bug moving in a circle?
__________________________________________________________________
9. Why does this force eventually fail? ____________________________________
__________________________________________________________________
10. Reset all, and set the force back to 1 N.
11. Observe the acceleration vector as you start. Describe how it changes. _________
__________________________________________________________________
12. Will the acceleration vector ever point directly to the center? ________ Why /
Why not? (the next steps might help you answer this question) _______________
__________________________________________________________________
13. Reset all. Set the force back to 1 N.
14. Hit start, wait about 2 seconds, and set the brake force to 1 N. Hit enter and
observe.
15. Describe the motion of the wheel: ______________________________________
16. What happened to the acceleration vector? __________________________ Why?
__________________________________________________________________
__________________________________________________________________
17. What is the net torque? _________
18. Reset all. Set the Force back to 1 N. Hit Start.
19. After a few seconds, set the brake force equal to 3N and hit enter.
20. Right after you set the break force, calculate the net torque (check with the
graph):
6
21. Eventually the disc stops and the net torque is zero. This is because the breaking
torque changed as you can see in the graph. Why did it change?
Part II: Moment of Inertia
1. Click the Moment of Inertia Tab at the top.
2. Disregard any millimeter units. They should all be meters.
3. To best see the graphs, set the scale of the torque graph to show a range of 20 to
-20.
4. Set the Moment of Inertia Graph to show a range of 2 kg m2 to – 2 kg m2
5. Set the angular acceleration graph to show 1,000 degrees / s2 to –1000 degrees / s2
6. Calculate the moment of Inertia for the disk with the given information.
7. Hold the mouse over the disk so the mouse finger is pointing anywhere between
the green and pink circles.
8. Hold down the left mouse button. Move your mouse to apply a force.
9. Look at the graph and try to apply a force that creates a torque of 10.
10. Use the ruler to determine the radius at any point between the green and pink
circles. r = ___________m
11. Calculate what the applied force must have been.
12. Calculate the angular acceleration of the disk. Work in SI units, and then convert
to degrees / s2. Compare to the graph to check your answer.
7
13. Predict what will happen to the moment of inertia if you keep the mass of the
platform the same, but you create a hole in the middle (increase inner radius).
__________________________________________________________________
14. Set the inner radius equal to 2. Calculate the moment of inertia for this shape.
Set the disk in motion and check your answer by looking at the moment of inertia
graph.
15. Even when the force on the platform changes, the moment of inertia graph
remains constant. Why? _____________________________________________
__________________________________________________________________
16. Fill in the blanks: When the mass of an object increases, the moment of inertia
________________. When the distance of the mass from the axis of rotation
increases, the moment of inertia ___________________.
Part III
1. Click the Angular Momentum tab at the top.
2. Set the scale of the moment of inertia and angular momentum graphs to show a
range of 2 to -2.
3. Set the angular speed to be 45 degrees / s.
4. What is the SI unit for angular momentum? ______________________________
5. Calculate the angular momentum in SI units (you should have already calculated
the moment of inertia in part II).
6. While the disk is moving, change the inner radius to 2.
7. Observe the graphs.
8. Changing the inner radius automatically changes the angular velocity to 36
degrees / s. Why? (mention moment of inertia and angular momentum in your
answer).
8
IV.
Definition of Center of Gravity (9.3)
(Recall center of mass stuff from last unit? Similar concepts)
The center of gravity of a rigid body is the point at which its weight can be
considered to act when the torque due to the weight is being calculated.
When an object has a symmetrical shape and its weight is distributed
uniformly, the center of gravity lies at its geometrical center. For instance,
Figure 9.9 shows a thin, uniform, horizontal rod of length L attached to a
vertical wall by a hinge. The center of gravity of the rod is located at the
geometrical center. The lever arm for the weight
is
, and the
magnitude of the torque is
. In a similar fashion, the center of gravity
of any symmetrically shaped and uniform object, such as a sphere, disk, cube,
or cylinder, is located at its geometrical center. However, this does not mean
that the center of gravity must lie within the object itself. The center of gravity
of a compact disc recording, for instance, lies at the center of the hole in the
disc and is, therefore, “outside” the object.
Figure 9.9
A thin, uniform, horizontal rod of length L is attached to a vertical wall by a hinge. The center of
gravity of the rod is at its geometrical center.
9
V.
Center of Gravity for a Group of Objects (9.3)
Suppose we have a group of objects, with known weights and centers of gravity, and it is
necessary to know the center of gravity for the group as a whole. As an example, Figure
9.10a shows a group composed of two parts: a horizontal uniform
board
and a uniform box
near the left end of the board. The
center of gravity can be determined by calculating the net torque created by the board and
box about an axis that is picked arbitrarily to be at the right end of the board. Part a of the
figure shows the weights
The net torque is
treating the total weight
and
and their corresponding lever arms
and
.
. It is also possible to calculate the net torque by
as if it were located at the center of gravity and had
the lever arm
, as part b of the drawing indicates:
values for the net torque must be the same, so that
. The two
Figure 9.10
(a) A box rests near the left end of a horizontal board.
(b) The total
weight
acts at the center of gravity of the group.
(c) The group can be balanced by applying an external force
(due to the index finger) at the center of gravity.
This expression can be solved for
, which locates the
center of gravity relative to the axis:
(9.3)
The notation “
” indicates that Equation 9.3 can be extended to
account for any number of weights distributed along a horizontal line.
Figure 9.10c illustrates that the group can be balanced by a single
external force (due to the index finger), if the line of action of the
force passes through the center of gravity, and if the force is equal in
magnitude, but opposite in direction, to the weight of the group.
Example 6 demonstrates how to calculate the center of gravity for the
human arm.
10
11
VI.
Newton’s Second Law for Rotation (9.4)
Toy Plane… The plane's engine produces a net external tangential force
that gives the
plane a tangential acceleration . So…
. The torque produced by this force
is
, where the radius r of the circular path is also the lever arm. As a result, the
torque is
. However, the tangential acceleration is related to the angular
acceleration according to
(from unit 3), where
With this substitution for , the torque becomes
must be expressed in
.
Where I = mr2…
The constant of proportionality is
, which is called the moment of inertia of the
particle. The SI unit for moment of inertia is kg m2. (analogous to linear mass)
If all objects were single particles, it would be just as convenient to use the second law in
the form
as in the form
. The advantage in using
is that it can be
applied to any rigid body rotating about a fixed axis, and not just to a particle.
Newton’s 3rd Law tells us that each mass has the same α, so…
where the expression
is the sum of the external torques,
and
represents the sum of the individual moments
of inertia. The latter quantity is the moment of inertia I of the body:
In this equation, r is the perpendicular radial distance of each particle from the axis of
rotation. Combining Equation 9.6 with Equation 9.5 gives the following result:
12
(9.7)
Requirement:
must be expressed in
.
The version of Newton's second law given in Equation 9.7 applies only for rigid bodies.
The word “rigid” means that the distances , , , etc. that locate each particle
,
, , etc. (see Figure 9.16a) do not change during the rotational motion. In other words, a
rigid body is one that does not change its shape while undergoing an angular acceleration
in response to an applied net external torque.
The form of the second law for rotational motion,
, is similar to the equation for
translational (linear) motion,
, and is valid only in an inertial frame. The
moment of inertia I plays the same role for rotational motion that the mass m does for
translational motion. Thus, I is a measure of the rotational inertia of a body. When using
Equation 9.7, must be expressed in
, because the relation
requires radian measure) was used in the derivation.
(which
It can be seen from Equation 9.6 that the moment of inertia depends on both the mass of
each particle and its distance from the axis of rotation. The farther a particle is from the
axis, the greater is its contribution to the moment of inertia.
13
14
VII. Rotational Work and Energy (9.5)
We know…
To see how this expression can be rewritten using angular variables, consider Figure.
Here a rope is wrapped around a wheel and is under a constant tension F. If the rope is
pulled out a distance s, the wheel rotates through an angle
, where r is the radius
of the wheel and is in radians. Thus,
, and the work done by the tension force in
turning the wheel is
. However,
is the torque applied to the wheel by
the tension, so the rotational work can be written as follows:
Definition of Rotational Work
The rotational work
done by a constant
torque in turning an object through an angle is
Requirement: must be expressed in radians.
SI Unit of Rotational Work: joule (J)
Work–Energy Theorem and KE Application… KE = ½ m v2 … In an analogous manner,
the rotational work done by a net external torque causes the rotational kinetic energy to
change. A rotating body possesses kinetic energy, because its constituent particles are
moving. If the body is rotating with an angular speed , the tangential speed
of a
particle at a distance r from the axis is
If a particle's mass is m, its kinetic energy
is
. The kinetic energy of the entire rotating body, then, is the sum of the
kinetic energies of the particles:
Definition of Rotational Kinetic Energy
The rotational kinetic energy
of a rigid object rotating with an angular speed
a fixed axis and having a moment of inertia I is
about
Requirement: must be expressed in
.
SI Unit of Rotational Kinetic Energy: joule (J)
15
Kinetic energy is one part of an object's total mechanical energy. The total mechanical
energy is the sum of the kinetic and potential energies and obeys the principle of
conservation of mechanical energy (see Section ). Specifically, we need to remember that
translational and rotational motion can occur simultaneously. When a bicycle coasts
down a hill, for instance, its tires are both translating and rotating. An object such as a
rolling bicycle tire has both translational and rotational kinetic energies, so that the total
mechanical energy is
Here m is the mass of the object, is the translational speed of its center of mass, I is its
moment of inertia about an axis through the center of mass, is its angular speed,
and h is the height of the object's center of mass relative to an arbitrary zero level.
Mechanical energy is conserved if
, the net work done by external nonconservative
forces and external torques, is zero. If the total mechanical energy is conserved as an
object moves, its final total mechanical energy
equals its initial total mechanical
energy
.
16
VIII. Angular Momentum (9.6)
In Chapter 7 the linear momentum p of an object is defined as the product of its
mass m and linear velocity that is,
. For rotational motion the analogous concept
is called the angular momentum L. The mathematical form of angular momentum is
analogous to that of linear momentum, with the mass m and the linear velocity being
replaced with their rotational counterparts, the moment of inertia I and the angular
velocity .
Definition of Angular Momentum
The angular momentum L of a body rotating about a fixed axis is the product of the body's moment of
inertia I and its angular velocity with respect to that axis:
(9.10)
Requirement:
must be expressed in
.
SI Unit of Angular Momentum:
Linear momentum is an important concept in physics because the total linear momentum of a system
is conserved when the sum of the average external forces acting on the system is zero. Then, the final
total linear momentum
and the initial total linear momentum
are the same:
. In the
case of angular momentum, a similar line of reasoning indicates that when the sum of the average
external torques is zero, the final and initial angular momenta are the same:
, which is
the principle of conservation of angular momentum.
Principle of Conservation of Angular Momentum
The total angular momentum of a system remains constant (is conserved) if
the net average external torque acting on the system is zero.
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18
19
IX.
Simple Harmonic Motion & Unit Circle (10.2)Elasticity
Simple harmonic motion can be described in terms of displacement, velocity, and acceleration.
Displacement
The Figure above shows the reference circle
and indicates how to determine the
displacement of the shadow on the film. The ball starts on the x axis at
and moves through
the angle in a time t. The circular motion is uniform, so the ball moves with a constant angular
speed (in rad/s), and the angle has a value (in rad) of
. The displacement x of the shadow is
just the projection of the radius A onto the x axis:
&
&
aldfkj
20
Velocity
The reference circle model can also be used to determine the velocity of an object in simple harmonic
motion. The figure below shows the tangential velocity
drawing indicates that the velocity
of the ball on the reference circle. The
of the shadow is just the x component of the vector
;
that is,
, where
. The minus sign is necessary because
points to the left,
in the direction of the negative x axis. Since the tangential speed
is related to the angular
speed by
and since
, it follows that
. Therefore, the velocity in simple
harmonic motion is given by
This velocity is not constant, but varies between maximum and minimum values as time passes. When
the shadow changes direction at either end of the oscillatory motion, the velocity is momentarily zero.
When the shadow passes through the
position, the velocity has a maximum magnitude of
, since the sine of an angle is between
and
:
Both the amplitude A and the angular frequency
Example 3 emphasizes.
determine the maximum velocity, as
21
Acceleration (recall… even with uniform circular mo… there is an accel)
In simple harmonic motion, the velocity is not constant; consequently, there must be an acceleration.
This acceleration can also be determined with the aid of the reference-circle model. As the
figure shows, the ball on the reference circle moves in uniform circular motion, and, therefore, has a
centripetal acceleration
that points toward the center of the circle. The acceleration
of the
shadow is the x component of the centripetal acceleration;
. The minus sign is
needed because the acceleration of the shadow points to the left. Recalling that the centripetal
acceleration is related to the angular speed
by
that
. With this substitution and the fact that
motion becomes
(Equation 8.11) and using
, we find
, the acceleration in simple harmonic
(10.9)
The acceleration, like the velocity, does not have a constant value as time passes. The maximum
magnitude of the acceleration is
(10.10)
Although both the amplitude A and the angular frequency determine the maximum value, the
frequency has a particularly strong effect, because it is squared. Example 5 shows that the acceleration
can be remarkably large in a practical situation.
22
Frequency of Vibration
With the aid of Newton's second law
, it is possible to determine the frequency at
which an object of mass m vibrates on a spring. We assume that the mass of the spring itself is
negligible and that the only force acting on the object in the horizontal direction is due to the spring—
that is, the Hooke's law restoring force. Thus, the net force is
, and Newton's second
law becomes
, where
is the acceleration of the object. The displacement and
acceleration of an oscillating spring are, respectively,
(Equation 10.3)
and
relation
(Equation 10.9). Substituting these expressions for x and
, we find that
into the
which yields
(10.11)
In this expression, the angular frequency must be in radians per second. Larger spring
constants k and smaller masses m result in larger frequencies.
X.
Energy & Simple Harmonic Motion (10.3)
A spring also has potential energy when the spring is stretched or compressed, which we refer to
as elastic potential energy. Because of elastic potential energy, a stretched or compressed spring can
do work on an object that is attached to the spring, as seen in the screen door spring mechanism.
We know… W = ½ k x2… So…
23
Definition of Elastic Potential Energy
The elastic potential energy
is the energy that a spring has by virtue of being stretched or
compressed. For an ideal spring that has a spring constant k and is stretched or compressed by an
amount x relative to its unstrained length, the elastic potential energy is
(10.13)
SI Unit of Elastic Potential Energy: joule (J)
The total mechanical energy E is a familiar idea that we originally defined to be the sum of the
translational kinetic energy and the gravitational potential energy. Then, we included the rotational
kinetic energy. We now expand the total mechanical energy to include the elastic potential energy...
We get…
As Section 6.5 discusses, the total mechanical energy is conserved when external nonconservative
forces (such as friction) do no net work; that is, when
. Then, the final and initial values
ofE are the same:
. The principle of conservation of total mechanical energy is the subject of
the next example.
24
Energy in Simple Harmonic Motion
We can describe an oscillating mass in terms of its position, velocity, and acceleration as
a function of time. We can also describe the system from an energy perspective. In this
experiment, you will measure the position and velocity as a function of time for an
oscillating mass and spring system, and from those data, plot the kinetic and potential
energies of the system.
Energy is present in three forms for the mass and spring system. The mass m, with
velocity v, can have kinetic energy KE
KE  21 mv 2
The spring can hold elastic potential energy, or PEelastic. We calculate PEelastic by using
PE elastic  21 ky 2
where k is the spring constant and y is the extension or compression of the spring
measured from the equilibrium position.
The mass and spring system also has gravitational potential energy (PEgravitational = mgy),
but we do not have to include the gravitational potential energy term if we measure the
spring length from the hanging equilibrium position. We can then concentrate on the
exchange of energy between kinetic energy and elastic potential energy.
If there are no other forces experienced by the system, then the principle of conservation
of energy tells us that the sum KE + PEelastic = 0, which we can test experimentally.
OBJECTIVES

Examine the energies involved in simple harmonic motion.
 Test the principle of conservation of energy.
MATERIALS
computer
Vernier computer interface
Logger Pro
Vernier Motion Detector
wire basket
slotted mass set, 50 g to 300 g in 50 g steps
slotted mass hanger
spring, 1-10 N/m
ring stand
PRELIMINARY QUESTIONS
1. Sketch a graph of the height vs. time for the mass on the spring as it oscillates up and
down through one cycle. Mark on the graph the times where the mass moves the
fastest and therefore has the greatest kinetic energy. Also mark the times when it
moves most slowly and has the least kinetic energy.
2. On your sketch, label the times when the spring has its greatest elastic potential
energy. Then mark the times when it has the least elastic potential energy.
3. From your graph of height vs. time, sketch velocity vs. time.
25
4. Sketch graphs of kinetic energy vs. time and elastic potential energy vs. time.
PROCEDURE
1. Mount the 200 g mass and spring as shown in
Figure 1. Connect the Motion Detector to the
DIG/SONIC 1 channel of the interface. Position
the Motion Detector directly below the
hanging mass, taking care that no extraneous
objects could send echoes back to the
detector. Protect the Motion Detector by
placing a wire basket over the detector. The
mass should be about 60 cm above the
detector when it is at rest. Using amplitudes of
10 cm or less will then keep the mass outside
of the 40 cm minimum distance of the Motion
Detector.
2. Open the file “17a Energy in SHM” from the
Physics with Computers folder.
3. Start the mass moving up and down by lifting
it 10 cm and then releasing it. Take care that
the mass is not swinging from side to side.
Click
to record position and velocity
data. Print your graphs and compare to your
predictions. Comment on any differences.
4. To calculate the spring potential energy, it is
necessary to measure the spring constant k.
Hooke’s law states that the spring force is
proportional to its extension from equilibrium,
or F = –kx. You can apply a known force to
the spring, to be balanced in magnitude by the
spring force, by hanging a range of weights
from the spring. The Motion Detector can
then be used to measure the equilibrium
position. Open the experiment file “17b
Energy in SHM.” Logger Pro is now set up to
plot the applied weight vs. position.
Figure 1
5. Click
to begin data collection. Hang a 50 g mass from the spring and allow
the mass to hang motionless. Click
and enter 0.49, the weight of the mass in
newtons (N). Press ENTER to complete the entry. Now hang 100, 150, 200, 250, and
300 g from the spring, recording the position and entering the weights in N. When
you are done, click
to end data collection.
6. Click on the Linear Fit button, , to fit a straight line to your data. The magnitude of
the slope is the spring constant k in N/m. Record the value in the data table below.
7. Remove the 300 g mass and replace it with a 200 g mass for the following
experiments.
26
8. Open the experiment file “17c Energy in SHM.” In addition to plotting position and
velocity, three new data columns have been set up in this experiment file (kinetic
energy, elastic potential energy, and the sum of these two individual energies). You
may need to modify the calculations for the energies. If necessary, choose Column
Options Kinetic Energy from the Data menu and click on the Column Definition
tab. Substitute the mass of your hanging mass in kilograms for the value 0.20 in the
definition, then click
. Similarly, change the spring constant you determined
above for the value 5.0 in the potential energy column.
9. With the mass hanging from the spring and at rest, click
to zero the Motion
Detector. From now on, all distances will be measured relative to this position. When
the mass moves closer to the detector, the position reported will be negative.
10. Start the mass oscillating in a vertical direction only, with an amplitude of about
10 cm. Click
to gather position, velocity, and energy data.
DATA TABLE
Spring constant
N/m
ANALYSIS
1. Click on the y-axis label of the velocity graph to choose another column for plotting.
Click on More to see all of the columns. Uncheck the velocity column and select the
kinetic energy and potential energy columns. Click
to draw the new plot.
2. Compare your two energy plots to the sketches you made earlier. Be sure you
compare to a single cycle beginning at the same point in the motion as your
predictions. Comment on any differences.
3. If mechanical energy is conserved in this system, how should the sum of the kinetic
and potential energies vary with time? Choose Draw Prediction from the Analyze
menu and draw your prediction of this sum as a function of time.
4. Check your prediction. Click on the y-axis label of the energy graph to choose
another column for plotting. Click on More and select the total energy column in
addition to the other energy columns. Click
to draw the new plot.
5. From the shape of the total energy vs. time plot, what can you conclude about the
conservation of mechanical energy in your mass and spring system?
EXTENSIONS
1. In the introduction, we claimed that the gravitational potential energy could be
ignored if the displacement used in the elastic potential energy was measured from
the hanging equilibrium position. First write the total mechanical energy (kinetic,
gravitational potential, and elastic potential energy) in terms of a coordinate system,
position measured upward and labeled y, whose origin is located at the bottom of the
relaxed spring of constant k (no force applied). Then determine the equilibrium
position s when a mass m is suspended from the spring. This will be the new origin
for a coordinate system with position labeled h. Write a new expression for total
27
energy in terms of h. Show that when the energy is written in terms of h rather than y,
the gravitational potential energy term cancels out.
2. If a non-conservative force such as air resistance becomes important, the graph of
total energy vs. time will change. Predict how the graph would look, then tape an
index card to the bottom of your hanging mass. Take energy data again and compare
to your prediction.
3. The energies involved in a swinging pendulum can be investigated in a similar
manner to a mass on a spring. From the lateral position of the pendulum bob, find the
bob’s gravitational potential energy. Perform the experiment, measuring the
horizontal position of the bob with a Motion Detector.
4. Set up a laboratory cart or a glider on an air track so it oscillates back and forth
horizontally between two springs. Record its position as a function of time with a
Motion Detector. Investigate the conservation of energy in this system. Be sure you
consider the elastic potential energy in both springs.
28
Simple Harmonic Motion
Lots of things vibrate or oscillate. A vibrating tuning fork, a moving child’s playground
swing, and the loudspeaker in a radio are all examples of physical vibrations. There are
also electrical and acoustical vibrations, such as radio signals and the sound you get when
blowing across the top of an open bottle.
One simple system that vibrates is a mass hanging from a spring. The force applied by an
ideal spring is proportional to how much it is stretched or compressed. Given this force
behavior, the up and down motion of the mass is called simple harmonic and the position
can be modeled with
y  A sin 2ft   
In this equation, y is the vertical displacement from the equilibrium position, A is the
amplitude of the motion, f is the frequency of the oscillation, t is the time, and  is a
phase constant. This experiment will clarify each of these terms.
Figure 1
OBJECTIVES

Measure the position and velocity as a function of time for an oscillating mass and
spring system.
 Compare the observed motion of a mass and spring system to a mathematical model
of simple harmonic motion.
 Determine the amplitude, period, and phase constant of the observed simple
harmonic motion.
MATERIALS
computer
Vernier computer interface
Logger Pro
Vernier Motion Detector
200 g and 300 g masses
ring stand, rod, and clamp
spring, with a spring constant of
approximately 10 N/m
twist ties
wire basket
29
PRELIMINARY QUESTIONS
1. Attach the 200 g mass to the spring and hold the free end of the spring in your hand,
so the mass and spring hang down with the mass at rest. Lift the mass about 10 cm
and release. Observe the motion. Sketch a graph of position vs. time for the mass.
2. Just below the graph of position vs. time, and using the same length time scale, sketch
a graph of velocity vs. time for the mass.
PROCEDURE
1. Attach the spring to a horizontal rod connected to the ring stand and hang the mass
from the spring as shown in Figure 1. Securely fasten the 200 g mass to the spring
and the spring to the rod, using twist ties so the mass cannot fall.
2. Connect the Motion Detector to the DIG/SONIC 1 channel of the interface.
3. Place the Motion Detector at least 75 cm below the mass. Make sure there are no
objects near the path between the detector and mass, such as a table edge. Place the
wire basket over the Motion Detector to protect it.
4. Open the file “15 Simple Harmonic Motion” from the Physics with Computers folder.
5. Make a preliminary run to make sure things are set up correctly. Lift the mass upward
a few centimeters and release. The mass should oscillate along a vertical line only.
Click
to begin data collection.
6. After 10 s, data collection will stop. The position graph should show a clean
sinusoidal curve. If it has flat regions or spikes, reposition the Motion Detector and
try again.
7. Compare the position graph to your sketched prediction in the Preliminary Questions.
How are the graphs similar? How are they different? Also, compare the velocity
graph to your prediction.
8. Measure the equilibrium position of the 200 g mass. Do this by allowing the mass to
hang free and at rest. Click
to begin data collection. After collection stops,
click the statistics button, , to determine the average distance from the detector.
Record this distance (y0) in your data table.
9. Now lift the mass upward about 5 cm and release it. The mass should oscillate along
a vertical line only. Click
to collect data. Examine the graphs. The pattern
you are observing is characteristic of simple harmonic motion.
10. Using the position graph, measure the time interval between maximum positions.
This is the period, T, of the motion. The frequency, f, is the reciprocal of the period, f
= 1/T. Based on your period measurement, calculate the frequency. Record the period
and frequency of this motion in your data table.
11. The amplitude, A, of simple harmonic motion is the maximum distance from the
equilibrium position. Estimate values for the amplitude from your position graph.
Enter the values in your data table. If you drag the mouse from one peak to another
you can read the dx time interval.
12. Repeat Steps 8 – 11 with the same 200 g mass, moving with a larger amplitude than
in the first run.
30
13. Change the mass to 300 g and repeat Steps 8 – 11. Use an amplitude of about 5 cm.
Keep a good run made with this 300 g mass on the screen. You will use it for several
of the Analysis questions.
DATA TABLE
Run
Mass
(g)
y0
(cm)
A
(cm)
T
(s)
f
(Hz)
1
2
3
ANALYSIS
1. View the graphs of the last run on the screen. Compare the position vs. time and the
velocity vs. time graphs. How are they the same? How are they different?
2. Turn on the Examine mode by clicking the Examine button, . Move the mouse
cursor back and forth across the graph to view the data values for the last run on the
screen. Where is the mass when the velocity is zero? Where is the mass when the
velocity is greatest?
3. Does the frequency, f, appear to depend on the amplitude of the motion? Do you have
enough data to draw a firm conclusion?
4. Does the frequency, f, appear to depend on the mass used? Did it change much in
your tests?
5. You can compare your experimental data to the sinusoidal function model using the
Manual Curve Fit feature of Logger Pro. Try it with your 300 g data. The model
equation in the introduction, which is similar to the one in many textbooks, gives the
displacement from equilibrium. However, your Motion Detector reports the distance
from the detector. To compare the model to your data, add the equilibrium distance to
the model; that is, use
y  A sin 2ft     y0
where y0 represents the equilibrium distance.
a. Click once on the position graph to select it.
b. Choose Curve Fit from the Analyze menu.
c. Select Manual as the Fit Type.
d. Select the Sine function from the General Equation list.
e. The Sine equation is of the form y=A*sin(Bt +C) + D. Compare this to the form of
the equation above to match variables; e.g.,  corresponds to C, and 2f
corresponds to B.
f. Adjust the values for A, B and D to reflect your values for A, f and y0. You can
either enter the values directly in the dialog box or you can use the up and down
arrows to adjust the values.
g. The phase parameter  is called the phase constant and is used to adjust the y value
reported by the model at t = 0 so that it matches your data. Since data collection
31
did not necessarily begin when the mass was at the equilibrium position,  is
needed to achieve a good match.
h. The optimum value for  will be between 0 and 2. find a value for  that makes
the model come as close as possible to the data of your 300 g experiment. You
may also want to adjust y0, A, and f to improve the fit. Write down the equation
that best matches your data.
6. Predict what would happen to the plot of the model if you doubled the parameter for
A by sketching both the current model and the new model with doubled A. Now
double the parameter for A in the manual fit dialog box to compare to your prediction.
7. Similarly, predict how the model plot would change if you doubled f, and then check
by modifying the model definition.
8. Click
, and optionally print your graph.
EXTENSIONS
1. Investigate how changing the spring amplitude changes the period of the motion.
Take care not to use too large an amplitude so that the mass does not come closer
than 40 cm to the detector or fall from the spring.
2. How will damping change the data? Tape an index card to the bottom of the mass and
collect additional data. You may want to take data for more than 10 seconds. Does the
model still fit well in this case?
3. Do additional experiments to discover the relationship between the mass and the
period of this motion.
32
XI.
The Pendulum (10.4)
A simple pendulum consists of a particle of mass m, attached to a frictionless pivot P by a cable of
length L and negligible mass. When the particle is pulled away from its equilibrium position by an
angle and released, it swings back and forth as Figure 10.20 shows. By attaching a pen to the bottom
of the swinging particle and moving a strip of paper beneath it at a steady rate, we can record the
position of the particle as time passes. The graphical record reveals a pattern that is similar (but not
identical) to the sinusoidal pattern for simple harmonic motion.
The gravitational force
produces this torque.
(The tension
in the cable creates no torque,
because it points directly at the pivot P and, therefore,
has a zero lever arm.) According to the Torque Equation, the
magnitude of the torque is the product of the
magnitude
of the gravitational force and the lever arm ,
so that
. The minus sign is included
since the torque is a restoring torque; it acts to
reduce the angle [the angle is positive
(counterclockwise), while the torque is negative
(clockwise)]. The lever arm is the perpendicular
distance between the line of action of
and
the pivot P. In Figure, is very nearly equal
to the arc length s of the circular path when the
angle is small (about
or less).
Furthermore, if is expressed in radians, the arc
length and the radius L of the circular path are related,
according to
. It follows, then, that
,
and the gravitational torque is
In the equation above, the term
has a constant value , independent of . For small angles,
then, the torque that restores the pendulum to its vertical equilibrium position is proportional to the
angular displacement . The expression
has the same form as the Hooke's law restoring
force for an ideal spring,
. Therefore, we expect the frequency of the back-and-forth
movement of the pendulum to be given by an equation analogous to
. In place
of the spring constant k, the constant
will appear, and, as usual in rotational motion, in
place of the mass m, the moment of inertia I will appear:
33
The moment of inertia of a particle of mass m, rotating at a radius
by
pendulum
about an axis, is given
. Substituting this expression for I into the above Equation reveals that for a simple
The mass of the particle has been eliminated algebraically from this expression, so only the
length L and the magnitude g of the acceleration due to gravity determine the frequency of a simple
pendulum. If the angle of oscillation is large, the pendulum does not exhibit simple harmonic motion,
and the above Equation does not apply. The above Equation provides the basis for using a pendulum
to keep time.
34
Pendulum Period Lab
A swinging pendulum keeps a very regular beat. It is so regular, in fact, that for many
years the pendulum was the heart of clocks used in astronomical measurements at the
Greenwich Observatory.
There are at least three things you could change about a pendulum that might affect the period
(the time for one complete cycle):

the amplitude of the pendulum swing
 the length of the pendulum, measured from the center of the pendulum bob to the
point of support
 the mass of the pendulum bob
To investigate the pendulum, you need to do a controlled experiment; that is, you need to
make measurements, changing only one variable at a time. Conducting controlled
experiments is a basic principle of scientific investigation.
In this experiment, you will use a simulation (link below) to vary the length of string,
mass of the bob, and amplitude of the pendulum in order to calculate the period of one
complete swing of a pendulum. By conducting a series of controlled experiments with the
pendulum, you can determine how each of these quantities affects the period.
35
http://phet.colorado.edu/en/simulation/pendulum-lab
OBJECTIVES
 Observe the period of a pendulum as a function of amplitude.
 Calculate the period of a pendulum as a function of length.
 Observe the period of a pendulum as a function of bob mass.
 Construct a pendulum with a period of one second
PROCEDURE
Determine the period (T) of the pendulum for 5 different length () combinations of
your choice. Do the same for 5 different mass (m) combinations of your choice. Use the
simulation to construct a pendulum with a period of exactly one second. To do this, vary
one variable at a time and keep track of which ones affect the period and which ones do
not.
ANALYSIS
1. Plot a graph of pendulum period T vs. length . Scale each axis from the origin (0,0).
Does the period appear to depend on length?
2. Plot the pendulum period vs. mass. Scale each axis from the origin (0,0). Does the period
appear to depend on mass? Do you have enough data to answer conclusively?
3. To examine more carefully how the period T depends on the pendulum length , create
2
2
the following two additional graphs of the same data: T vs.  and T vs.  . Of the three
period-length graphs, which is closest to a direct proportion; that is, which plot is most
nearly a straight line that goes through the origin?
4. Using Newton’s laws, we could show that for some pendulums, the period T is related to
the length  and free-fall acceleration g by
T  2
 4 2 

  
, or T 2  
g
 g 
Does one of your graphs support this relationship? Explain. (Hint: Can the term in
parentheses be treated as a constant of proportionality?)
36
ANALYSIS QUESTIONS
1.
2.
3.
4.
What effect, if any, does mass have on the period of the pendulum?
What effect, if any, does amplitude have on the period of the pendulum?
What effect, if any, does length have on the period of the pendulum?
If you set your pendulum up on the top of Mt. Everest, would the period be less than,
the same, or greater than it would be in your lab? Why?
5. If you set up your pendulum aboard an orbiting space station, would the period be less
than, the same, or greater than it would be in your lab? Why?
37
Pendulum Periods
A swinging pendulum keeps a very regular beat. It is so regular, in fact, that for many
years the pendulum was the heart of clocks used in astronomical measurements at the
Greenwich Observatory.
There are at least three things you could change about a pendulum that might affect the period
(the time for one complete cycle):

the amplitude of the pendulum swing
 the length of the pendulum, measured from the center of the pendulum bob to the
point of support
 the mass of the pendulum bob
To investigate the pendulum, you need to do a controlled experiment; that is, you need to
make measurements, changing only one variable at a time. Conducting controlled
experiments is a basic principle of scientific investigation.
In this experiment, you will use a Photogate capable of microsecond precision to measure
the period of one complete swing of a pendulum. By conducting a series of controlled
experiments with the pendulum, you can determine how each of these quantities affects
the period.
Figure 1
OBJECTIVES

Measure the period of a pendulum as a function of amplitude.
 Measure the period of a pendulum as a function of length.
 Measure the period of a pendulum as a function of bob mass.
MATERIALS
computer
Vernier computer interface
Logger Pro
Vernier Photogate
protractor
string
2 ring stands and pendulum clamp
masses of 100 g, 200 g, 300 g
meter stick
38
PRELIMINARY QUESTIONS
1. Make a pendulum by tying a 1 m string to a mass. Hold the string in your hand and let
the mass swing. Observing only with your eyes, does the period depend on the length
of the string? Does the period depend on the amplitude of the swing?
2. Try a different mass on your string. Does the period seem to depend on the mass?
PROCEDURE
1. Use the ring stand to hang the 200 g mass from two strings. Attach the strings to a
horizontal rod about 10 cm apart, as shown in Figure 1. This arrangement will let the
mass swing only along a line, and will prevent the mass from striking the Photogate.
The length of the pendulum is the distance from the point on the rod halfway between
the strings to the center of the mass. The pendulum length should be at least 1 m.
2. Attach the Photogate to the second ring stand. Position it so that the mass blocks the
Photogate while hanging straight down. Connect the Photogate to DIG/SONIC 1 on the
interface.
3. Open the file “14 Pendulum Periods” in the Physics with Computers folder. A graph
of period vs. time is displayed.
4. Temporarily move the mass out of the center of the Photogate. Notice the reading in
the status bar of Logger Pro at the bottom of the screen, which shows when the
Photogate is blocked. Block the Photogate with your hand; note that the Photogate is
shown as blocked. Remove your hand, and the display should change to unblocked.
Click
and move your hand through the Photogate repeatedly. After the first
blocking, Logger Pro reports the time interval between every other block as the
period. Verify that this is so.
5. Now you can perform a trial measurement of the period of your pendulum. Pull the
mass to the side about 10º from vertical and release. Click
and measure the
period for five complete swings. Click
. Click the Statistics button, , to
calculate the average period. You will use this technique to measure the period under
a variety of conditions.
Part I Amplitude
6. Determine how the period depends on amplitude. Measure the period for five
different amplitudes. Use a range of amplitudes, from just barely enough to unblock
the Photogate, to about 30º. Each time, measure the amplitude using the protractor so
that the mass with the string is released at a known angle. Repeat Step 5 for each
different amplitude. Record the data in your data table.
Part II Length
7. Use the method you learned above to investigate the effect of changing pendulum
length on the period. Use the 200 g mass and a consistent amplitude of 20º for each
trial. Vary the pendulum length in steps of 10 cm, from 1.0 m to 0.50 m. If you have
room, continue to a longer length (up to 2 m). Repeat Step 5 for each length. Record
the data in the second data table below. Measure the pendulum length from the rod to
the middle of the mass.
39
Part III Mass
8. Use the three masses to determine if the period is affected by changing the mass.
Measure the period of the pendulum constructed with each mass, taking care to keep
the distance from the ring stand rod to the center of the mass the same each time, as
well as keeping the amplitude the same. Repeat Step 5 for each mass, using an
amplitude of about 20°. Record the data in your data table
DATA TABLE
Part I Amplitude
Amplitude
(°)
Average period
(s)
Length
(cm)
Average period
(s)
Mass
(g)
Average period
(s)
Part II Length
Part III Mass
100
200
300
40
ANALYSIS
1. Why is Logger Pro set up to report the time between every other blocking of the
Photogate? Why not the time between every block?
2. Use Logger Pro to plot a graph of pendulum period vs. amplitude in degrees. Scale
each axis from the origin (0,0). Does the period depend on amplitude? Explain.
3. Plot a graph of pendulum period T vs. length . Scale each axis from the origin (0,0).
Does the period appear to depend on length?
4. Plot the pendulum period vs. mass. Scale each axis from the origin (0,0). Does the
period appear to depend on mass? Do you have enough data to answer conclusively?
5. To examine more carefully how the period T depends on the pendulum
length ,
create the following two additional graphs of the same data: T 2 vs.  and T vs.  2. Of
the three period-length graphs, which is closest to a direct proportion; that is, which
plot is most nearly a straight line that goes through the origin?
6. Using Newton’s laws, we could show that for some pendulums, the period T is related
to the length  and free-fall acceleration g by
T  2
 4 2 

  
, or T 2  
g
 g 
Does one of your graphs support this relationship? Explain. (Hint: Can the term in
parentheses be treated as a constant of proportionality?)
EXTENSIONS
1. From your graph of T 2 vs.  determine a value for g.
2. Given what you observed in this experiment, write a set of rules for constructing a
pendulum clock that is reliable under a variety of temperatures.
3. Try a larger range of amplitudes than you used in Part I. If you did not see a change in
period with amplitude before, you should now. Check a college physics textbook for
an expression for the period of a pendulum at large amplitudes and compare to your
own data.
4. Try a different method to study how the period of a pendulum depends on the
amplitude. Start the pendulum swinging with a fairly large amplitude. Start data
collection and allow data collection to continue for several minutes. The amplitude of
the swing will gradually decrease and you can see how much the period changed.
5. Use an air table and air table puck as your pendulum. Tip the air table to a variety of
angles, , and determine the relationship between the period and the angle.
41