College of Engineering Education
2nd Floor, B&E Building
Matina Campus, Davao City
Telefax: (082) 296-1084
Phone No.: (082)300-5456/300-0647 Local 133
UNIVERSITY OF MINDANAO
College of Engineering Education
Civil Engineering Program
Physically Distanced but Academically Engaged
Self-Instructional Manual (SIM) for
Self-Directed Learning (SDL)
Course/Subject: CEE 118: Dynamics of Rigid Bodies
Name of Author: Engr. Rey Albert Cabotaje
THIS SIM/SDL MANUAL IS A DRAFT VERSION ONLY; NOT FOR REPRODUCTION AND
DISTRIBUTION OUTSIDE OF ITS INTENDED USE. THIS IS INTENDED ONLY FOR THE
USE OF THE STUDENTS WHO ARE OFFICIALLY ENROLLED IN THE COURSE/SUBJECT.
EXPECT REVISIONS OF THE MANUAL.
Page 1 of 149
College of Engineering Education
2nd Floor, B&E Building
Matina Campus, Davao City
Telefax: (082) 296-1084
Phone No.: (082)300-5456/300-0647 Local 133
Table of Contents
Course Outline ................................................................................................................................................................. 5
Course Outline Policy .................................................................................................................................................. 5
Course Information....................................................................................................................................................... 8
Instruction Proper ........................................................................................................................................................ 8
Big Picture 1 ............................................................................................................................................................ 8
ULO1a.................................................................................................................................................................... 9
Metalanguage ................................................................................................................................... 9
Essential Knowledge ........................................................................................................... 9
Introduction to Dynamics .............................................................................................. 9
Newtonian Mechanics ................................................................................................... 12
Rectilinear Kinematics .............................................................................................. …14
Motion Graphs .............................................................................................................. …17
Constant Acceleration[Equations of Kinematics] .......................................... …30
Free-Falling Bodies..................................................................................................... …33
Let’s Check!.............................................................................................................................38
Let’s Analyze!.........................................................................................................................39
In a Nutshell ...........................................................................................................................41
Course Schedule....................................................................................................................43
ULO1b...................................................................................................................................................43
Metalanguage .....................................................................................................................43
Essential Knowledge .........................................................................................................44
Curvilinear Motion:Projectiles................................................................................... 44
Let’s Check!.............................................................................................................................48
Let’s Analyze!.........................................................................................................................49
In a Nutshell ...........................................................................................................................49
Course Schedule....................................................................................................................50
Big Picture 2 .......................................................................................................................................................... 50
ULO2a.................................................................................................................................................................. 51
Metalanguage ................................................................................................................................. 51
Essential Knowledge .........................................................................................................51
Curvilinear Motion:Normal and Tangential Components .............................. 51
Curvilinear Motion:Cylindrical Components ....................................................... 55
Page 2 of 149
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Cylindrical Coordinates ................................................................................................ 57
Time Derivatives ............................................................................................................. 57
Absolute Dependent Motion Analysis of Two Particles .................................. 59
Let’s Check!.............................................................................................................................64
Let’s Analyze!.........................................................................................................................64
In a Nutshell ...........................................................................................................................65
Course Schedule....................................................................................................................67
ULO2b.................................................................................................................................................................. 67
Metalanguage ................................................................................................................................. 67
Essential Knowledge .........................................................................................................67
Newton’s Second Law of Motion............................................................................... 68
Equations of Motions: Rectangular Coordinates ................................................ 69
Friction ................................................................................................................................ 69
Spring ................................................................................................................................... 69
Course Schedule....................................................................................................................75
Big Picture 3 .......................................................................................................................................................... 75
ULO3a.................................................................................................................................................................. 75
Metalanguage ................................................................................................................................. 75
Essential Knowledge .........................................................................................................75
Work of a Force ................................................................................................................ 76
Work of a Weight............................................................................................................. 77
Work of a Spring Force ................................................................................................. 77
Principle of Work and Energy .................................................................................... 79
Power and Efficiency ..................................................................................................... 85
Conservative Forces and Conservation of Mechanical Energy..................... 90
Principle of Linear Impulse and Momentum ....................................................... 92
Principle of Angular Impulse and Momentum .................................................... 97
ULO3b............................................................................................................................................................... 103
Metalanguage .............................................................................................................................. 103
Essential Knowledge ...................................................................................................... 103
Dynamics of Particle Systems ................................................................................. 103
Translating the Reference Frame .......................................................................... 105
Kinematics of Constrained Motion........................................................................ 107
Impact ............................................................................................................................... 119
Coeffiecient of Restitution, e .................................................................................... 120
Page 3 of 149
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Phone No.: (082)300-5456/300-0647 Local 133
Mass Flow ........................................................................................................................ 122
Let’s Check!.......................................................................................................................... 125
Let’s Analyze!...................................................................................................................... 126
In a Nutshell ........................................................................................................................ 127
Course Schedule................................................................................................................. 127
Big Picture 4 ....................................................................................................................................................... 127
ULO4a............................................................................................................................................................... 127
Metalanguage .............................................................................................................................. 128
Essential Knowledge ...................................................................................................... 128
Introduction ................................................................................................................... 128
Plane Angular Motion ................................................................................................. 129
Rotation about a Fixed Axis ..................................................................................... 129
Kinematics of a Point in the Body.......................................................................... 130
ULO4b............................................................................................................................................................... 133
Metalanguage .............................................................................................................................. 133
Essential Knowledge ...................................................................................................... 133
Introduction ................................................................................................................... 133
Mass Moment of Inertia ............................................................................................. 134
Radius of Gyration ....................................................................................................... 134
Parallel-Axis Theorem................................................................................................ 134
Method of Composite Bodies................................................................................... 135
Let’s Check!.......................................................................................................................... 139
Let’s Analyze!...................................................................................................................... 140
In a Nutshell ........................................................................................................................ 141
Course Schedule................................................................................................................. 142
Page 4 of 149
College of Engineering Education
2nd Floor, B&E Building
Matina Campus, Davao City
Telefax: (082) 296-1084
Phone No.: (082)300-5456/300-0647 Local 133
Course Outline: CEE 118 – Dynamics of Rigid Bodies
Course Coordinator:
Danielyn F. Plazos, RCE
Email:
danielynplazos@umindanao.edu.ph
Student Consultation:
By appointment
Mobile:
0916-573-2111
Phone:
(082) 296-1084 or 300-5456 loc. 133
Effectivity Date:
August 17, 2020
Mode of Delivery:
Blended (On-line with face to face or virtual sessions)
Time Frame:
54 hours
Student Workload:
Expected Self-Directed Learning
Pre-requisite:
CEE 117 (Statics of Rigid Bodies)
Co-requisite:
None
Credit:
3.0 units lecture
Attendance Requirements: A minimum of 95% attendance is required at all
scheduled Virtual or face-to-face sessions
Course Outline Policy
Areas of Concern
Contact and Non-contact Hours
Assessment Task Submission
Details
This 4-unit course self-instructional manual is designed
for blended learning mode of instructional delivery
with scheduled face to face or virtual sessions. The
expected number of hours will be 108 including the
face-to-face or virtual sessions. The face-to-face
sessions shall include the summative assessment tasks
(exams) since this course is crucial in the licensure
examination for civil engineers.
Submission of assessment tasks shall be on 3rd, 5th, 7th
and 9th week of the term. The assessment paper shall
be attached with a cover page indicating the title of the
assessment task (if the task is performance), the name
of the course coordinator, date of submission and name
of the student. The document should be emailed to the
course coordinator. It is also expected that you already
paid your tuition and other fees before the submission
of the assessment task.
If the assessment task is done in real time through the
features in the Blackboard Learning Management
System, the schedule shall be arranged ahead of time by
the course coordinator.
Page 5 of 149
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2nd Floor, B&E Building
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Telefax: (082) 296-1084
Phone No.: (082)300-5456/300-0647 Local 133
Turnitin Submission
(if necessary)
Penalties for Late
Assignments/Assessments
Return of Assignments/
Assessments
Assignment Resubmission
Re-marking of Assessment
Papers and Appeal
Since this course is included in the licensure
examination for civil engineers, you will be required to
take the Multiple-Choice Question exam inside the
University. This should be scheduled ahead of time by
your course coordinator. This is non-negotiable for all
licensure-based programs.
To ensure honesty and authenticity, all assessment
tasks are required to be submitted through Turnitin
with a maximum similarity index of 30% allowed. This
means that if your paper goes beyond 30%, the
students will either opt to redo her/his paper or explain
in writing addressed to the course coordinator the
reasons for the similarity. In addition, if the paper has
reached more than 30% similarity index, the student
may be called for a disciplinary action in accordance
with the University’s OPM on Intellectual and Academic
Honesty.
Please note that academic dishonesty such as cheating
and commissioning other students or people to
complete the task for you have severe punishments
(reprimand, warning, expulsion).
The score for an assessment item submitted after the
designated time on the due date, without an approved
extension of time, will be reduced by 5% of the possible
maximum score for that assessment item for each day
or part day that the assessment item is late.
However, if the late submission of assessment paper
has a valid reason, a letter of explanation should be
submitted and approved by the course coordinator. If
necessary, you will also be required to present/attach
evidences.
Assessment tasks will be returned to you two (2) weeks
after the submission. This will be returned by email or
via Blackboard portal.
For group assessment tasks, the course coordinator will
require some or few of the students for online or virtual
sessions to ask clarificatory questions to validate the
originality of the assessment task submitted and to
ensure that all the group members are involved.
You should request in writing addressed to the course
coordinator his/her intention to resubmit an
assessment task. The resubmission is premised on the
student’s failure to comply with the similarity index and
other reasonable grounds such as academic literacy
standards or other reasonable circumstances e.g.
illness, accidents financial constraints.
You should request in writing addressed to the program
Page 6 of 149
College of Engineering Education
2nd Floor, B&E Building
Matina Campus, Davao City
Telefax: (082) 296-1084
Phone No.: (082)300-5456/300-0647 Local 133
coordinator your intention to appeal or contest the
score given to an assessment task. The letter should
explicitly explain the reasons/points to contest the
grade. The program coordinator shall communicate
with the students on the approval and disapproval of
the
request.
Grading System
If disapproved by the course coordinator, you can
elevate your case to the program head or the dean with
the original letter of request. The final decision will
come from the dean of the college.
All culled from BlackBoard sessions and traditional
contact
Course discussions/exercises – 30%
1st formative assessment – 10%
2nd formative assessment – 10%
3rd formative assessment – 10%
All culled from on-campus/onsite sessions (TBA):
Final exam – 40%
Preferred Referencing Style
Student Communication
Submission of the final grades shall follow the usual
University system and procedures.
Depends on the discipline; if uncertain or inadequate,
use the general practice of the APA 6th Edition.
You are required to create a umindanao email account
which is a requirement to access the BlackBoard
portal.
Then, the course coordinator shall enroll the
students to have access to the materials and resources
of the course. All communication formats: chat,
submission of assessment tasks, requests etc. shall be
through the portal and other university recognized
platforms.
You can also meet the course coordinator in person
through the scheduled face to face sessions to raise
your issues and concerns.
Contact Details of the Dean
Contact Details of the Program
Head
Students with Special Needs
For students who have not created their student email,
please contact the course coordinator or program
head.
Dr. Charlito L. Cañesares
Email: clcanesares@umindanao.edu.ph
Phone: (082) 296-1084 or 300-5456 loc. 133
Engr. Showna Lee T. Sales
Email: ssales@umindanao.edu.ph
Phone: (082) 296-1084 or 300-5456 loc. 133
Students with special needs shall communicate with the
course coordinator about the nature of his or her
Page 7 of 149
College of Engineering Education
2nd Floor, B&E Building
Matina Campus, Davao City
Telefax: (082) 296-1084
Phone No.: (082)300-5456/300-0647 Local 133
CEE Help Desk Contact
CEE Blackboard Administrator
Library Contact
GSTC Contact
special needs. Depending on the nature of the need, the
course coordinator with the approval of the program
coordinator may provide alternative assessment tasks
or extension of the deadline of submission of
assessment tasks. However, the alternative assessment
tasks should still be in the service of achieving the
desired course learning outcomes.
Frida Santa O. Dagatan
cee@umindanao.edu.ph
09562082442
082-2272902
Jetron J. Adtoon
jadtoon@umindanao.edu.ph
09055267834
Brigida E. Bacani
library@umindanao.edu.ph
09513766681
Ronadora E. Deala, RPsy, RPm, RGC, LPT
ronadora_deala@umindanao.edu.ph
09212122846
Silvino P. Josol
gstcmain@umindanao.edu.ph
09060757721
Course Information- see/download course syllabus in the BlackBoard LMS
CF’s Voice:
Hello future engineers! Welcome to this course CEE 118: Dynamics of Rigid Bodies.
After dealing with Statics, you are now about to study the branch of physical science and
subdivision of mechanics that is concerned with the motion of material objects in relation to the
physical factors that affect them force, mass, momentum, energy. Let’s make it a fun journey!
CO:
Upon completion of the course, you are expected to:
CO 1. Explain the principles governing the motion of particles modelled as rigid bodies.
CO 2. Analyze rigid-body kinematics in stationary as well as moving frames of
reference.
CO3. Explain fundamental concepts in analytical dynamics, and the ability to use them
to formulate the equations of motion for rigid bodies and systems.
Let us begin!
Big Picture
Week 1-3: Unit Learning Outcomes-Unit 1 (ULO-1): At the end of the unit, you are expected to
Page 8 of 149
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a. Define basic dynamics concepts an introduction and explain the basic kinematic
quantities of rectilinear motion: such as position, displacement, velocity and
acceleration and describe the analysis of Motion by the Area Method (Motion Graphs)
of a particle.
b. Explain the basic kinematic quantities of curvilinear motion: projectile motion.
Big Picture in Focus: ULO-1a. Define basic dynamics concepts an introduction
and explain the basic kinematic quantities of rectilinear motion: such as
position, displacement, velocity and acceleration and describe the analysis of
Motion by the Area Method (Motion Graphs) of a particle.
Metalanguage
The most essential terms below are defined for you to have a better understanding of
this section in the course:
1. Particle is a mass point; it possesses a mass but has no size.
2. Vector is a quantity having direction as well as magnitude, especially as determining the
position of one point in space relative to another.
3. Acceleration is the rate of change of the velocity of an object with respect to time.
4. Position is the place where we can find it at any constant time. Position is normally defined
in terms of co-ordinates of axes. It is a point in space in which an event occurs.
5. Displacement is defined to be the change in position of an object.
6. Gravitation is the movement, or a tendency to move, toward a center of gravity, as in the
falling of bodies to the earth.
7. Rectilinear Motion is a linear motion in which the direction of the velocity remains
constant and the path is a straight line.
8. Kinematics is a study of the geometry of the motion.
9. Kinetics is a study of the forces that cause the motion.
10. Free Body Diagram are diagrams used to show the relative magnitude and direction of all
forces acting upon an object in a given situation.
11. Speed refers to the magnitude of velocity.
12. Average speed is the total distance traveled divided by the total time.
Essential Knowledge
To perform the ULO for the first week of the course, you need to understand the
following key concepts that will be presented in the succeeding pages. You are also expected to
use other references, books, and other resource material that is available in the university’s
library.
Part 1 : Introduction to Dynamics
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“Dynamics is that branch of mechanics which deals with the motion of bodies under the
action of forces. The study of dynamics in engineering usually follows the study of statics, which
deals with the effects of forces on bodies at rest. Dynamics has two distinct parts: kinematics,
which is the study of motion without reference to the forces, which cause motion, and kinetics,
which relates the action of forces on bodies to their resulting motions. A thorough
comprehension of dynamics will provide one of the most useful and powerful tools for analysis
in engineering.” (Meriam and Kraige, 2012)
Basic Concepts
The basic concepts in mechanics are space, time, mass and force. Among these, space,
time, mass are absolute quantities, which mean that they are independent of each other and
cannot be expressed in terms of other quantities or in simpler terms. Force, on the other hand,
is a derived quantity.
Space is the geometric region occupied by bodies. Position in space is specified by linear or
angular measurements with respect to a geometric reference system.
Time is a measure of the succession of events and is considered an absolute quantity in
Newtonian mechanics.
Mass is the quantitative measure of the inertia or resistance to change in motion of a body. Mass
can also be considered as the amount of matter within a body. Although the mass of a body is
an absolute quantity, its weight can change depending on the gravitational force
(𝑊 = 𝑚𝑔).
Force is the action of one body on another. A force possesses both magnitude and direction,
therefore it is a vector quantity.
A particle is a body of negligible dimensions. Generally, a particle is thought to be an
infinitesimally small element, which possesses all properties of a body. However, when the
dimensions of a body are irrelevant to the description of its motion or the action of forces on it,
a large body may also be treated as a particle. A particle has mass but no shape and dimensions.
The body is considered to be concentrated at a single point which usually will be its mass center.
A rigid body is a body whose changes in shape are negligible compared with the overall
dimensions of the body or with the changes in position of the body as a whole. The shape and
dimensions of a rigid body will remain the same under all conditions of loading and at all times.
Position is where the particle is at any given time. More precisely, we need to specify its
position relative to a convenient reference frame. Earth is often used as a reference frame, and
we often describe the position of an object as it relates to stationary objects in that reference
frame.
For example, a professor’s position could be described in terms of where she is in
relation to the nearby white board in the figure below. In other cases, we use reference frames
that are not stationary but rather are in motion relative to Earth.
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This is the position
of the teacher.
Displacement is defined to be the change in position of an object. It can be defined
mathematically with the following equation:
Displacement: ∆𝑥 = 𝑥𝑓 − 𝑥0
𝑥𝑓 = refers to the value of the final position.
𝑥0 = refers to the value of the initial position.
∆𝑥 = symbol used to represent displacement
Displacement is a vector. This means it has a direction as well as a magnitude and is
represented visually as an arrow that points from the initial position to the final position. For
example, consider the professor that walks relative to the whiteboard in the figure below:
A professor paces left and right while lecturing. The +2.0 displacement of the professor relative to the
whiteboard is represented by an arrow pointing to the right.
The professor’s initial position is 𝑥0 = 1.5 𝑚 and her final position is 𝑥𝑓 = 3.5 𝑚. Thus,
her displacement can be found as follows: ∆𝑥𝑓 = 𝑥𝑓 − 𝑥0 = 3.5 𝑚 − 1.5 𝑚 = +2.0 𝑚 . In this
coordinate system, motion to the right is positive, whereas motion to the left is negative.
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Now consider the passenger that walks relative to the plane:
A passenger moves from his seat to the back of the plane. The −4. 0 m displacement of the passenger relative
to the plane is represented by an arrow toward the rear of the plane.
The airplane passenger’s initial position is 𝑥0 = 6.0 𝑚 and his final position is 𝑥𝑓 =
2.0 𝑚, so his displacement can be found as follows, ∆𝑥 = 𝑥𝑓 − 𝑥0 = 2.0 𝑚 − 6.0 𝑚 =
−4.0 𝑚. His displacement is negative because his motion is toward the rear of the plane, or in
the negative x direction in our coordinate system.
In one-dimensional motion, direction can be specified with a plus or minus sign. When
you begin a problem, you should select which direction is positive—usually that will be to the
right or up, but you are free to select positive as being any direction.
Distance is defined to be the magnitude or size of displacement between two positions. Note
that the distance between two positions is not the same as the distance traveled between them.
Distance traveled is the total length of the path traveled between two positions. Distance
traveled is not a vector. It has no direction and, thus, no negative sign. For example, the distance
the professor walks is 2.0 m. The distance the airplane passenger walks is 4.0 m.
Trajectory / Path is a line or a curve obtained when all the points a body occupies within a
specific time period are joined.
NEWTONIAN MECHANICS
Law I. (Equation of Equilibrium). A particle remains at rest or continues to move with
uniform velocity (along a straight line with a constant speed) if there is no unbalanced force
acting on it.
Law II. (Equation of Motion). The acceleration of a particle is proportional to the resultant
force acting on it and is in the direction of this force.
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∑ 𝐹 = 𝑚𝑎 = 𝑘𝑔 ∙
𝑚
𝑠2
= Newton (N)
Law III. (Principle of Action and Reaction) The forces of action and reaction between
interacting bodies are equal in magnitude, opposite in direction, and collinear.
Units Review
Quantity
Mass
Length
Time
Force
SI Units
Unit
kilogram
meter
second
Newton
SI Units → 1 𝑁 = 𝟏 𝑘𝑔 ∙
𝑚
𝑠2
English Units → 1 𝑙𝑏 = 𝟏 𝑠𝑙𝑢𝑔 ∙
Symbol
kg
m
s
N
English Units
Unit
Symbol
slug
slugs
foot
ft
second
sec
pound
lb.
𝑓𝑡
𝑠𝑒𝑐 2
Newton’s Law of Gravitation, which governs the mutual attraction between bodies, is stated
as:
𝑚1 𝑚2
𝐹1 = 𝐹2 = 𝐺
𝑟2
G = a universal constant called the constant of gravitation
𝐺 = 6.673 𝑥 10−11
𝑚3
𝑘𝑔 ∙ 𝑠 2
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Supplementary Problems
Example ULO1a-1. For the 3600-lb car, determine:
a. its mass in slugs
b. its weight in Newtons
c. its mass in kilograms
Solution:
a. 1 𝑙𝑏 = 𝑠𝑙𝑢𝑔 𝑥
𝑙𝑏
𝑓𝑡 2
𝐹 = 𝑚𝑎 → a = 32.2
𝑙𝑏
𝑓𝑡 2
3600 𝑙𝑏 = 𝑚𝑎𝑠𝑠 𝑥 32.2
𝑙𝑏
𝑓𝑡 2
Ans.
𝒎𝒂𝒔𝒔 = 𝟏𝟏𝟏. 𝟖 𝑠𝑙𝑢𝑔𝑠
b. 1 lb = 4.448 N
4.448 𝑁
3600 𝑙𝑏 ( 1 𝑙𝑏 ) = 16, 012.8 N
c. 𝐹 = 𝑚𝑎 → a = 9.81
𝑚
𝑠2
16, 012.8 N = 𝑚𝑎𝑠𝑠 𝑥 9.81
𝒎𝒂𝒔𝒔 = 𝟏𝟔𝟑𝟐. 𝟐𝟗𝟒 𝑘𝑔
Ans.
𝑚
𝑠2
Ans.
Example ULO1a-2. Determine the force of gravitational attraction between the earth (m =
5.98 x 1024 kg) and a 70-kg physics student if the student is standing at sea level, a distance of
6.38 x 106 m from earth's center.
𝑚3
(5.98 x 1024 kg)(70 𝑘𝑔)
]
𝑘𝑔 ∙ 𝑠 2
(6.38 x 106 𝑚)2
𝑭 = 𝟔𝟖𝟔. 𝟐𝟓𝟓 𝑵
Ans.
𝐹1 = 𝐹2 = [6.673 𝑥 10−11 ∙
Part 2: Rectilinear Kinematics
Rectilinear Kinematics. The kinematics of a particle is characterized by specifying, at a given
instant, the particle's position, velocity, and acceleration.
Let us recall what we have learned about position and displacement from previous ULO1a.
Now let us continue the journey:
Velocity. If the particle moves through a displacement ∆𝑥 during the time interval ∆𝑡, the
average velocity of the particle during this time interval is:
𝑣𝑎𝑣𝑔 =
∆𝑥
∆𝑡
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If we take smaller and smaller values of ∆𝑡 the magnitude of ∆𝑥 becomes smaller and smaller.
∆𝑥
Consequently, the instantaneous velocity is a vector defined as 𝑣 = lim ( ∆𝑡 ) or:
∆𝑡→0
𝑣=
𝑑𝑥
𝑜𝑟 𝑥̇
𝑑𝑡
where: ∆𝑥 = 𝑥𝑓 − 𝑥0
𝑥𝑓 = refers to the value of the final position.
𝑥0 = refers to the value of the initial position.
∆𝑡 = 𝑡𝑓 − 𝑡0
𝑡𝑓 = refers to the value of the final position.
𝑡0 = refers to the value of the initial position.
•
•
If ∆𝑥 is moving to the right, then it is positive.
If ∆𝑥 is moving to the left, then it is negative.
Average Speed. total distance traveled by a particle, 𝑥𝑇 , divided by the elapsed time ∆𝑡;
𝑣𝑎𝑣𝑒 =
𝑥𝑇
∆𝑡
∆𝑥
𝑥𝑇
Average velocity and Average speed
Acceleration. Provided the velocity of the particle is known at two points, the average
acceleration of the particle during the time interval ∆𝑡 is defined as:
𝑎𝑎𝑣𝑒 =
or instantaneous acceleration,
𝑑2𝑥
𝑎=
𝑑𝑡 2
∆𝑣
∆𝑡
𝑜𝑟
𝑥̈
where: ∆𝑣 = 𝑣𝑓 − 𝑣0
𝑣𝑓 = refers to the value of the final position.
𝑣0 = refers to the value of the initial position.
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∆𝑡 = 𝑡𝑓 − 𝑡0
𝑡𝑓 = refers to the value of the final position.
𝑡0 = refers to the value of the initial position.
•
•
•
If ∆𝑣 is negative, then the particle is slowing down or decelerating.
If ∆𝑣 is positive, then the particle is accelerating.
If ∆𝑣 = 0, then the acceleration is zero, the velocity is constant
Supplementary Problems:
Example ULO1a-3. Four objects A, B, C and D move according to the paths shown in the diagram
below. Assume the units of the horizontal scale are given in meters.
What was the displacement of each object?
For object A:
Object A had an initial position of 0 m and a final position of 7 m. The displacement of
object A can be shown with this equation:
∆𝑥𝐴 = 7 𝑚 − 0 𝑚 = +7 𝑚
For object B:
Object B had an initial position of 12 m and a final position of 7 m. The displacement of
object B can be shown with this equation:
∆𝑥𝐵 = 7 𝑚 − 12 𝑚 = −5 𝑚
For object C:
Object C had an initial position of 2 m and a final position of 10 m. The displacement of
object C can be shown with this equation:
∆𝑥𝐶 = 10 𝑚 − 2 𝑚 = +8 𝑚
For object D:
Object D had an initial position of 9 m and a final position of 5 m. The displacement of
object D can be shown with this equation:
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∆𝑥𝐷 = 5 𝑚 − 9 𝑚 = −4 𝑚
Example ULO1a-4. Four objects A, B, C and D move according to the paths shown in the diagram
below. Assume the units of the horizontal scale are given in meters.
What was the total distance of each object?
Solution:
•
•
•
•
Object A travels a total distance of 7 m
Object B travels a total distance of 5 m
Object C travels a total distance of 8 𝑚 + 2 𝑚 + 2 𝑚 = 𝟏𝟐 𝒎
Object D travels at a total distance of 6 𝑚 + 2 𝑚 = 𝟖 𝒎
Example ULO1a-5. An iguana with a poor sense of spatial awareness is walking back and forth
in the desert. First the iguana walks 12 meters to the right in a time of 20 seconds. Then the
iguana runs 16 meters to the left in a time of 8 seconds.
What was the average speed and average velocity of the iguana for the entire trip?
Solution:
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𝐀𝐧𝐬.
Motion Graphs
Another way of visualizing the motion of an object is to use a graph. A plot of motion graphs are
as follows:
Versus time graphs:
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𝑥, 𝑚
𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛 𝑣𝑠. 𝑡𝑖𝑚𝑒 𝑔𝑟𝑎𝑝ℎ
𝑡, 𝑠
𝑣, 𝑚/𝑠
𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑣𝑠. 𝑡𝑖𝑚𝑒 𝑔𝑟𝑎𝑝ℎ
𝑡, 𝑠
𝑎, 𝑚/𝑠 2
𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑣𝑠. 𝑡𝑖𝑚𝑒 𝑔𝑟𝑎𝑝ℎ
𝑥, 𝑚
𝑡, 𝑠
Note:
•
•
•
Velocity, v is the slope of the position vs. time graph
Acceleration, a is the slope of the velocity vs. time graph,
Change in velocity, ∆𝑣, is the area under acceleration vs. time graph
∆𝑣 = ∫ 𝑎
•
Displacement, ∆𝑥 , is the area under velocity vs. time graph,
∆𝑥 = ∫ 𝑣
•
𝑑𝑡
𝑑𝑡
Hierarchy of graphs (in descending order), if the position vs time is in 2 nd
degree or quadratic, then it follows the velocity vs. time graph becomes 1st
degree or linear, and then ultimately, the acceleration vs time becomes a
horizontal line.
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Versus position graph:
𝑥
Note:
1
2
𝑥
(𝑣1 2 − 𝑣0 2 ) = ∫𝑥 1 𝑎 𝑑𝑥
0
or area under a – x graph
𝑥
Note:
𝑑𝑣
or velocity times slope of v– x graph
𝑎=𝑣 ( )
𝑑𝑠
Supplementary Problems:
Example ULO1a-6. A hungry dolphin is swimming horizontally back and forth looking for
food. The motion of the dolphin is given by the position graph shown below.
Determine the following for the dolphin:
a.
b.
c.
d.
Average velocity between time t = 0 s to t = 6 s.
Average speed between time t = 0 s to t = 6 s.
Instantaneous velocity at time t = 1 s
Instantaneous velocity at time t = 4 s
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𝑥
𝑥
𝑎=𝑣
𝑑𝑣
𝑑𝑥
Solution:
a.
b.
𝑥
∆𝑥
∆𝑡
𝑥𝑡
∆𝑡
0 𝑚−8 𝑚
−8
𝑣𝑎𝑣𝑒 =
= 6 𝑠−0 𝑠 = 6 𝑚/𝑠 or 1.33 m/s to the left
12 𝑚+0 𝑚+4 𝑚
16
𝑣𝑠𝑝𝑑 =
=
= 6 𝑚/𝑠 or 2.67 m/s
Ans.
6 𝑠−0 𝑠
0 𝑚−8 𝑚
−8 𝑚
𝑣𝑖𝑛𝑠𝑡𝑎𝑛𝑡𝑎𝑛𝑒𝑜𝑢𝑠 = 𝑠𝑙𝑜𝑝𝑒 = 2 𝑠−0 𝑠 = 2 𝑠 = −𝟒 𝒎/𝒔
Ans.
𝑥
Ans.
c.
d. Instantaneous speed is the speed at a given moment in time and will be equal to the
magnitude of the slope. Since the slope at t = 4 s is equal to zero, the
instantaneous speed at t = 4 s is also equal to zero.
Ans.
Example ULO1a-7. A bicycle moves along a straight road such that its position is described by
the graph shown in the figure. In this problem, the position is defined as s.
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a. Draw the velocity vs. time graph for time interval: 0 ≤ 𝑡 ≤ 30 𝑠.
i.
Determine the velocity at time 5 𝑠.
ii.
Determine the velocity at time 20 𝑠.
iii.
Determine the displacement at 𝑡 = 10 𝑠.
iv.
Determine the displacement at 𝑡 = 20 𝑠.
b. Draw the acceleration vs. time graph for time interval: 0 ≤ 𝑡 ≤ 30 𝑠.
i.
Determine the acceleration at time 5 𝑠.
ii.
Determine the acceleration at time 20 𝑠.
Solution:
a. Since we are going to plot the velocity vs. time graph, we have to differentiate the
equations of position:
0 ≤ 𝑡 ≤ 10 𝑠
the equation is: 𝒔 = 𝒕𝟐
Thus,
𝑣=
Substitute:
𝑑𝑠
𝑑𝑡
= 2𝑡
𝑓𝑡/𝑠
𝑎𝑡 𝑡 = 0 ; 𝑣 = 0
and so, on and so forth…………
Time, t
velocity, v
0
0
1
2
2
4
3
6
4
8
5
10
6
12
7
14
8
16
9
18
10
20
..
20
..
20
the equation is: 𝒔 = (𝟐𝟎𝒕 − 𝟏𝟎𝟎)
10 ≤ 𝑡 ≤ 30 𝑠
Thus,
𝑣=
Substitute:
𝑎𝑡 𝑡 = 0 ; 𝑣 = 20
𝑑𝑠
𝑑𝑡
= 20
𝑓𝑡/𝑠
and so on and so forth…………
Time, t
velocity, v
10
20
11
20
12
20
13
20
14
20
Page 22 of 149
..
20
..
20
..
20
30
20
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Thus, plot the graph using the value on the table:
i.
Determine the velocity at time 5 𝑠.
From the velocity vs. time graph, we can conclude that the velocity is the value of v
when t = 5.
Type equation here.
𝐴𝑡 𝑡 = 5 𝑠
By similar triangles:
20
𝑣
=
10
5
𝒗 = 𝟏𝟎 𝒇𝒕/𝒔
Ans.
And since we also have the equation for velocity at t = 5, 𝑣 = 2𝑡
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𝑣 = 2(5) = 𝟏𝟎 𝒇𝒕/𝒔
ii.
Ans.
Determine the velocity at time 20 𝑠.
From the velocity vs. time graph, we can conclude that the velocity is the value of v
when t = 5.
Type equation here.
𝐴𝑡 𝑡 = 20 𝑠
𝒗 = 𝟐𝟎 𝒇𝒕/𝒔
Ans.
And since we also have the equation for velocity at t = 2, 𝑣 = 20
Ans.
𝑣 = 𝟐𝟎 𝒇𝒕/𝒔
iii. Determine the displacement at t = 10 s.
The displacement is the area from t = 0 to t = 10 s from the velocity vs. time graph:
1
𝒗 = "𝑎𝑟𝑒𝑎 𝑓𝑟𝑜𝑚 @𝑡 = 0 𝑡𝑜 @𝑡 = 10 𝑠” = 2 (20)(10) = 𝟏𝟎𝟎 𝒇𝒕
or
Page 24 of 149
Ans.
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Using the equation of position t =10 s is in the interval 0≤ 𝑡 ≤ 10: 𝑠 = 𝑡 2
At t = 10 s ; : 𝑠 = 𝑡 2 = 102 = 𝟏𝟎𝟎 𝒇𝒕
Ans.
or
𝑡
10
Using 𝑣 = ∫𝑡 𝑓 𝑣𝑑𝑡 = ∫0 2𝑡 𝑑𝑡
=
0
Ans.
100 𝒇𝒕
b. Draw the acceleration vs. time graph for time interval: 0 ≤ 𝑡 ≤ 30 𝑠.
Solution:
Since we are going to plot the acceleration vs. time graph, we have to differentiate the
equations of velocity:
𝒅𝒔
0 ≤ 𝑡 ≤ 10 𝑠
the equation is: 𝒗 = 𝒅𝒕 = 𝟐𝒕
Thus,
𝒂=
Substitute:
𝒅𝒗 𝒅𝟐 𝒔
=
= 𝟐
𝒅𝒕 𝒅𝒕𝟐
𝒇𝒕/𝒔𝟐
𝑎𝑡 𝑡 = 0 ; 𝑎 = 2
and so, on and so forth…………
time, t
acceleration, a
0
2
1
2
2
2
3
2
4
2
5
2
the equation is 𝑣 =
10 ≤ 𝑡 ≤ 30 𝑠
𝒂=
Substitute:
𝑎𝑡 𝑡 = 0 ; 𝑣 = 20
6
2
𝑑𝑠
𝑑𝑡
7
2
8
2
9
2
10
2
ft/s
= 20
𝒅𝒗 𝒅𝟐 𝒔
=
= 𝟎
𝒅𝒕 𝒅𝒕𝟐
and so on and so forth…………
Time, t
acceleration, a
10
0
11
0
12
0
13
0
14
0
Thus, plot the graph using the value on the table:
Page 25 of 149
..
0
..
0
..
0
..
0
..
0
30
0
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i.
Determine the acceleration at time 5 𝑠.
From the graph, 𝑎𝑡 𝑡 = 5 𝑠; 𝑎 = 𝟐
ii.
𝒇𝒕/𝒔𝟐
Ans.
Determine the acceleration at time 20𝑠.
From the graph, 𝑎𝑡 𝑡 = 20 𝑠; 𝒂 = 𝟎
Ans.
Example ULO1a-8. The position of a particle that moves along the x-axis is defined by:
𝑥 = −3𝑡2 + 12𝑡 − 6 ft, where 𝑡 is in seconds. For the time interval 𝑡 = 0 𝑡𝑜 𝑡 = 3 𝑠,
Do:
a. plot the position, velocity, and acceleration as functions of time;
b. calculate the distance traveled; and
c. determine the displacement of the particle.
Solution:
a. Position vs time graph
Substitute 𝑡 = 0, 𝑡 = 1, 𝑡 = 2 𝑎𝑛𝑑 𝑡 = 3 into the equation:
𝒙 = −𝟑𝒕2 + 𝟏𝟐𝒕 − 𝟔
Time, t
Position, x
0
-6
1
3
Page 26 of 149
2
6
ft
3
3
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The graph will look like this:
velocity vs time graph
𝑣=
𝑑𝑥
𝑑𝑡
= −6𝑡 + 12
Substitute 𝑡 = 0, 𝑡 = 1, 𝑡 = 2 𝑎𝑛𝑑 𝑡 = 3 into the equation:
ft/s
𝒗 = −𝟔𝒕 + 𝟏𝟐
Time, t
Velocity, v
0
12
1
6
2
0
3
-6
The graph will look like this:
acceleration vs time graph
𝑎=
𝑑2 𝑥
= −6
𝑑𝑡 2
Substitute 𝑡 = 0, 𝑡 = 1, 𝑡 = 2 𝑎𝑛𝑑 𝑡 = 3 into the equation:
𝒗 = −𝟔
Time, t
Velocity, v
0
-6
1
-6
Page 27 of 149
ft/s2
2
-6
3
-6
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The graph will look like this:
b. Distance traveled is:
3 + 3 = 𝟏𝟓 ft
𝑥𝑇 = 6 + 3 +
Ans.
c. displacement of the particle.
∆𝑥 = 𝑥𝑓 − 𝑥0
∆𝑥 = 3 − (−6) = 9 ft
Ans.
Example ULO1a-9. The car in the figure starts from rest and travels along a straight
track such that it accelerates at 10 m/s2 for 10 s, and then decelerates at 2 m/s2.
Do:
a.
b.
c.
d.
Plot the velocity vs. time graph
Plot the position vs. time graph
Determine the time t needed to stop the car.
How far has the car traveled?
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Solution:
a. Plot the velocity vs. time graph
• The equation of the acceleration for interval 0 ≤ 𝑡 ≤ 10 is
Since a =
Then
→ 𝒂 = 𝟏𝟎
𝒅𝒗
𝒅𝒕
𝑑𝑣 = 𝑎 𝑑𝑡
∫ 𝑑𝑣 = ∫ 𝑎 𝑑𝑡
𝑣 = ∫𝑎
Thus,
𝑑𝑡 =
𝑡
∫ 10𝑑𝑡
0
𝒗 = 𝟏𝟎𝒕
Substitute the value for 𝑡 in the equation 𝑣 = 10𝑡:
𝐴𝑡 𝑡 = 0 𝑠 ; 𝑣 = 10(0) = 0
𝐴𝑡 𝑡 = 10 𝑠 ; 𝑣 = 10(10) = 100 m/s
•
The equation of the acceleration for interval 10 ≤ 𝑡 ≤ 𝑡 ′ is
𝒅𝒗
Since a = 𝒅𝒕
Then
𝑑𝑣 = 𝑎 𝑑𝑡
∫ 𝑑𝑣 = ∫ 𝑎 𝑑𝑡
𝑣 = ∫𝑎
𝑑𝑡 =
𝑡
∫ −2𝑑𝑡
10
Page 29 of 149
→ 𝒂 = −𝟐
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𝑏𝑢𝑡 𝑎𝑡 𝑡 = 10; 𝑣 = 100 m/s
𝒗 = − 𝟐𝒕 + 𝟐𝟎 + 𝟏𝟎𝟎 = − 𝟐𝒕 + 𝟏𝟐𝟎
𝒎/𝒔
Substitute the value for 𝑡 in the equation 𝒗 = −𝟐𝒕 + 𝟏𝟐𝟎:
𝐴𝑡 𝑡 = 10 𝑠 ; 𝑣 = −𝟐𝒕 + 𝟏𝟐𝟎 = −𝟐(𝟏𝟎) + 𝟏𝟐𝟎 = 𝟏𝟎𝟎 𝒎/𝒔
𝐴𝑡 𝑡 = 𝑡 ′ 𝑠 ; 𝑣 ≈ 0 ∶
𝟎 = −𝟐𝒕 + 𝟏𝟐𝟎
𝒕 = 𝟔𝟎 s
The graph will look like this:
b.
Plot the position vs. time graph
•
The equation of the velocity for interval 0 ≤ 𝑡 ≤ 10 is
Since v =
Then
𝒅𝒙
𝒅𝒕
𝑑𝑥 = 𝑣 𝑑𝑡
∫ 𝑑𝑥 = ∫ 𝑣 𝑑𝑡
𝑥 = ∫𝑣
Thus,
𝑑𝑡 =
𝑡
∫ 10𝑡𝑑𝑡
0
𝒙 = 𝟓𝒕𝟐
Page 30 of 149
→ 𝒗 = 𝟏𝟎𝒕
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Substitute the value for 𝑡 in the equation 𝒙 = 𝟓𝒕𝟐
𝐴𝑡 𝑡 = 0 𝑠 ; 𝑥 = 𝟓𝒕𝟐 = 𝟓(𝟎)𝟐 = 0
𝐴𝑡 𝑡 = 10 𝑠 ; 𝑥 = 𝟓𝒕𝟐 = 𝟓(𝟏𝟎)𝟐 = 𝟓𝟎𝟎 m
•
The equation of the velocity for interval 10 ≤ 𝑡 ≤ 𝑡 ′ is
→ 𝒗 = −𝟐𝒕 + 𝟏𝟐𝟎
𝒅𝒙
Since v = 𝒅𝒕
Then
𝑑𝑥 = 𝑣 𝑑𝑡
∫ 𝑑𝑥 = ∫ 𝑣 𝑑𝑡
𝑡
𝑥 = ∫ 𝑣 𝑑𝑡 = ∫ (−𝟐𝒕 + 𝟏𝟐𝟎)𝑑𝑡
10
𝑥=
−𝒕𝟐 + 𝟏𝟐𝟎𝒕 − 𝟏𝟐𝟎𝟎
So:
𝑥=
Thus,
−𝒕𝟐 + 𝟏𝟐𝟎𝒕 − 𝟏𝟐𝟎𝟎 + 𝟓𝟎𝟎
𝒙 = −𝒕𝟐 + 𝟏𝟐𝟎𝒕 − 𝟔𝟎𝟎
Substitute the value for 𝑡 in the equation 𝒙 = −𝒕𝟐 + 𝟏𝟐𝟎𝒕 − 𝟔𝟎𝟎
𝐴𝑡 𝑡 = 10 𝑠 ; 𝑥 = −(𝒕)𝟐 + 𝟏𝟐𝟎𝒕 − 𝟔𝟎𝟎 = 𝟓𝟎𝟎 𝒎
𝐴𝑡 𝑡 = 60 𝑠 ; 𝑥 = −(𝒕)𝟐 + 𝟏𝟐𝟎𝒕 − 𝟔𝟎𝟎 = 𝟑𝟎𝟎𝟎 𝒎
The graph will look like this:
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c. The car stopped when t = 60 s (time when the velocity becomes 0)
d. At t = 60 s, the distance travelled of the car is 3000 m.
Ans.
Ans.
Part 3: Constant Acceleration[Equations of Kinematics]
𝒂 = 𝒂𝒄
When the acceleration is constant, each of the three kinematic equations 𝑎 =
𝑑𝑥
𝑑𝑡
𝑎𝑛𝑑 𝑎 = 𝑣
𝑑𝑣
𝑑𝑥
can be integrated to obtain formulas that relate 𝑎, 𝑣, 𝑥 and 𝑡.
✓ Velocity as a function of time:
𝑎=
Since 𝑎 = 𝑎𝑐
𝑑𝑣
𝑑𝑡
𝑎𝐶 =
𝑑𝑣
𝑑𝑡
𝑣𝑓
𝑡
∫ 𝑑𝑣 = ∫ 𝑎𝐶 𝑑𝑡
𝑣0
0
𝑣𝑓 − 𝑣0 = 𝑎𝐶 𝑡
Or
𝑣𝑓 = 𝑣0 + 𝑎𝐶 𝑡
✓ Position as a function of time:
𝑣=
𝑑𝑥
𝑑𝑡
𝑑𝑥 = 𝑣𝑑𝑡
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𝑑𝑣
𝑑𝑡
,𝑣 =
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𝑥𝑓
𝑡
∫ 𝑑𝑥 = ∫ (𝑣0 + 𝑎𝐶 𝑡) 𝑑𝑡
𝑥0
0
𝑥𝑓 − 𝑥0 = 𝑣0 𝑡 +
Or
∆𝑥 = 𝑣0 𝑡 +
1
2
1
2
𝑎𝐶 𝑡 2
𝑡 2 𝑎𝐶
✓ Velocity as a function of position:
𝑎𝐶
=𝑣
𝑑𝑣
𝑑𝑥
𝑣𝑑𝑣 = 𝑎𝐶 𝑑𝑥
𝑣𝑓
𝑥𝑓
∫ 𝑣𝑑𝑣 = ∫ 𝑎𝐶 𝑑𝑥
𝑣0
(𝑣𝑓 )
2
2
−
𝑥𝑖
(𝑣𝑖 )2
2
= 𝑎𝐶 (𝑥𝑓 − 𝑥0 )
2
(𝑣𝑓 ) − (𝑣0 )2 = 2𝑎𝐶 (𝑥𝑓 − 𝑥0 )
2
Or
(𝑣𝑓 ) = (𝑣0 )2 + 2𝑎𝐶 ∆𝑥
Example ULO1a-10. Initially, the car travels along a straight road with a speed of 35 m/s. If the
brakes are applied and the speed of the car is reduced to 10 m/s in 15 s, determine the constant
deceleration of the car.
Solution:
Identify the given: 𝑣0 = 35 𝑚/𝑠 ; 𝑣𝑓 = 10 𝑚/𝑠 ; 𝑡 = 15 𝑠
Since the deceleration is constant, we can use this → 𝑣𝑓 = 𝑣0 + 𝑎𝑐 𝑡
𝑎𝑐
𝑎𝑐
=
10 − 35
15
=
𝑣𝑓 − 𝑣0
𝑡
= −𝟏. 𝟔𝟕 𝒎/𝒔𝟐
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Example ULO1a-11. A car starts from rest and with constant acceleration achieves a velocity of
15 m/s when it travels a distance of 200 m. Determine the acceleration of the car and the time
required.
Solution:
Identify the given: 𝑣0 = 0 (𝑠𝑡𝑎𝑟𝑡𝑠 𝑓𝑟𝑜𝑚 𝑟𝑒𝑠𝑡) ; 𝑣𝑓 = 15 𝑚/𝑠 ; ∆𝑥 = 200 𝑚
Since the acceleration is constant, we can use this → 𝑣𝑓 2 = 𝑣0 2 + 2𝑎𝑐 ∆𝑥
152 = 02 + 2𝑎𝑐 (200)
𝑎=
For time required,
152
2(200)
= 𝟎. 𝟓𝟔𝟐𝟓 𝒎/𝒔𝟐
Ans.
𝑣𝑓 = 𝑣0 + 𝑎𝑐 𝑡
15 = 0 + 0.5625(𝑡)
𝑡=
15
0.5625
= 𝟐𝟔. 𝟔𝟕 𝒔
Ans.
Example ULO1a-12. A train starts from rest at a station and travels with a constant acceleration
of 1 m/s2. Determine the velocity of the train when t = 30 s and the distance traveled during this
time.
Solution:
Identify the given: 𝑣0 = 0 (𝑠𝑡𝑎𝑟𝑡𝑠 𝑓𝑟𝑜𝑚 𝑟𝑒𝑠𝑡) ; 𝑎𝑐 = 1 𝑚/𝑠 2 ; 𝑡 = 30 𝑠
Since the acceleration is constant, we can use this →
𝑣𝑓 = 𝑣0 + 𝑎𝑐 𝑡
𝑣𝑓 = 0 + (1)(30)
𝑣𝑓 = 𝟑𝟎 𝒎/𝒔
Ans.
1
For distance traveled:
∆𝑥 = 𝑣0 𝑡 + 𝑎𝑐 𝑡 2
2
1
∆𝑥 = 0 + (1)(30)2
2
∆𝑥 = 𝟒𝟓𝟎 𝒎
Ans.
Example ULO1a-13. A car is traveling at 15 m/s, when the traffic light 50 m ahead turns yellow.
Determine the required constant deceleration of the car and the time needed to stop the car at
the light.
Solution:
Identify the given: 𝑣0 = 15
𝑚
𝑠
; ∆𝑥 = 50 𝑚; 𝑣𝑓 = 0 (𝑡ℎ𝑒 𝑐𝑎𝑟 𝑖𝑠 𝑠𝑡𝑜𝑝𝑝𝑒𝑑)
Since the deceleration is constant, we can use this → 𝑣𝑓 2 = 𝑣0 2 + 2𝑎𝑐 ∆𝑥
𝑣𝑓 2 = 𝑣0 2 + 2𝑎𝑐 ∆𝑥
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0 = 152 + 2𝑎𝑐 (50)
Ans.
𝑎𝑐 = −𝟐. 𝟐𝟓 𝒎/𝒔𝟐
For time needed to stop:
0 = 15 + (−2.25)𝑡
Ans.
𝑡 = 𝟔. 𝟔𝟕 𝒔
Part 4: Free – Falling Bodies
A freely falling object is any object moving freely under the influence of gravity alone, regardless
of its initial motion. Objects thrown upward or downward and those released from rest are all
considered freely falling.
𝒗𝒇 = 𝟎
∆𝒙 = 𝒉
−𝒈
𝒗𝟎 = ?
𝒗𝟎 = 𝟎
∆𝒙 = 𝒉
+𝒈
𝒗𝒇 = ?
✓ Remember that these equations are useful when acceleration is constant.
✓ For free falling bodies use 𝑔 = 9.81 m/𝑠 2 [SI] and use 𝑔 = 32.2 ft/𝑠 2 [English]
Example ULO1a-14. A ball is thrown vertically upward with a speed of 25.0 m/s.
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a.
b.
c.
d.
How high does it rise?
How long does it take to reach its highest point?
How long does the ball take to hit the ground after it reaches its highest point?
What is its velocity when it returns to the level from which it started?
Solution:
For a:
𝐹𝑟𝑜𝑚 𝑔𝑟𝑜𝑢𝑛𝑑 𝑡𝑜 𝑝𝑒𝑎𝑘
Given:
𝑣0 = 25
𝑚
𝑠
;
𝑣𝑓 = 0 ; 𝑎𝑐 = 𝑔 = −9.81 𝑚/𝑠2
2
(𝑣𝑓 ) = (𝑣𝑖 )2 + 2𝑎𝑐 ∆𝑥
∆𝑥 = ℎ
0 = 252 + 2(−9.81)ℎ
𝒉 = 𝟑𝟏. 𝟖𝟔 𝒎
For b:
𝑨𝒏𝒔.
𝑣𝑓 = 0
𝐹𝑟𝑜𝑚 𝑔𝑟𝑜𝑢𝑛𝑑 𝑡𝑜 𝑝𝑒𝑎𝑘
𝑣0 = 25
𝑚
𝑠
;
𝑣𝑓 = 0 ; 𝑎𝑐 = 𝑔 = −9.81 𝑚/𝑠2
𝑣𝑓 = 𝑣0 + 𝑎𝑐 𝑡
0 = 25 + (−9.81)(𝑡)
𝒕 = 𝟐. 𝟓𝟓 𝒔𝒆𝒄𝒔
For c:
𝑨𝒏𝒔.
𝑣0 = 25
𝑚
𝑠
𝐹𝑟𝑜𝑚 𝑝𝑒𝑎𝑘 𝑡𝑜 𝑔𝑟𝑜𝑢𝑛𝑑,
𝐼𝑓 ℎ = 31.86 𝑚 ; 𝑣0 = 0 ; 𝑎𝑐 = 𝑔 = 9.81 𝑚/𝑠
𝑣0 = 0
∆𝑥 = ℎ
ℎ = 𝑣0 𝑡 +
31.86 = 0 +
1
2
𝑔 𝑡2
1
(9.81)𝑡 2
2
𝒕 = 𝟐. 𝟓𝟓 𝒔𝒆𝒄𝒔
For d:
𝑣𝑓 =?
𝑨𝒏𝒔.
𝐹𝑟𝑜𝑚 𝑝𝑒𝑎𝑘 𝑡𝑜 𝑔𝑟𝑜𝑢𝑛𝑑,
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𝐼𝑓 ℎ = 31.86 𝑚 ; 𝑣0 = 0 ; 𝑎𝑐 = 𝑔 = 9.81 𝑚/𝑠 ; 𝑣𝑓 =?
∆𝑥 = ℎ
(𝑣𝑓 )2 = (𝑣0 )2 + 2𝑔ℎ
(𝑣𝑓 )2 = 0 + 2 (9.81)(31.86)
𝒗𝒇 = 𝟐𝟓 𝒎/𝒔
𝑨𝒏𝒔.
Example ULO1a-15. A stone is thrown vertically upward and return to earth in 10 sec. What
was its initial velocity and how high did it go?
Solution:
𝑣𝑓 = 0
𝑣𝑖 = 0
𝑡1 =?
𝑡2 =?
𝑣𝑓 =?
𝑣0 =?
𝐿𝑒𝑡 𝑡1 𝑏𝑒 𝑡ℎ𝑒 𝑡𝑖𝑚𝑒 𝑡𝑜 𝑟𝑒𝑎𝑐ℎ 𝑡ℎ𝑒 𝑝𝑒𝑎𝑘, 𝑎𝑛𝑑 𝑡2 𝑏𝑒 𝑡ℎ𝑒 𝑡𝑖𝑚𝑒 𝑡𝑜 𝑟𝑒𝑎𝑐ℎ 𝑡ℎ𝑒 𝑔𝑟𝑜𝑢𝑛𝑑.
𝑡1 + 𝑡2 = 𝑡𝑜𝑡𝑎𝑙 𝑡𝑖𝑚𝑒 𝑜𝑓 𝑓𝑙𝑖𝑔ℎ𝑡
𝑡1 + 𝑡2 = 10 𝑠𝑒𝑐𝑠
𝑡1 = 𝑡2 = 5 𝑠𝑒𝑐𝑠
For 𝑣0 :
𝑣0 =? ; 𝑣𝑓 = 0 ;
𝑎𝑐 = 𝑔 = −9.81 𝑚/𝑠 2 ; 𝑡 = 5
𝑣𝑓 = 𝑣0 + 𝑔𝑡
0 = 𝑣0 − 9.81 (5)
𝒗𝟎 = 𝟒𝟗. 𝟎𝟓 𝒎/𝒔 𝑨𝒏𝒔.
For ℎ : 𝑡 = 5
ℎ = 𝑣0 𝑡 +
1
𝑔𝑡 2
2
ℎ = 49.05(5) +
1
(−9.81)(5)2
2
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𝒉 = 𝟏𝟐𝟐. 𝟔𝟑 𝒎
𝑨𝒏𝒔.
Example ULO1a-16. A ball is dropped from the top of a tower 80 ft high at the same instant that
a second ball is thrown upward from the ground with an initial velocity of 40 ft/sec.
a. When do they pass each other?
b. Where do they pass each other?
c. What is the velocity of the second ball as it meets the first ball in the air?
𝐴
ℎ1
𝑡
ℎ=80 ft
𝐵
ℎ2
𝐶
𝑡
Solution:
For a:
𝑇ℎ𝑜𝑢𝑔ℎ𝑡: 𝑆𝑎𝑚𝑒 𝑑𝑢𝑟𝑎𝑡𝑖𝑜𝑛 𝑓𝑜𝑟 𝐴 𝑡𝑜 𝐵 𝑎𝑛𝑑 𝐵 𝑡𝑜 𝐶, 𝑙𝑒𝑡 𝑡𝑖𝑚𝑒 𝑏𝑒 "𝑡".
From A to B:
𝑣𝐴 = 0 ; 𝑔 = 32.2 𝑓𝑡/𝑠 2
ℎ1 = 𝑣𝐴 𝑡 +
ℎ1 =
1
𝑔(𝑡)2
2
1
(32.2)(𝑡)2
2
ℎ1 = 16.1(𝑡)2
→ 𝑒𝑞. 1
From C to B:
𝑓𝑡
; 𝑔 = −32.2 𝑓𝑡/𝑠 2
𝑠
1
𝑣𝐶 𝑡 + 2(-32.2)𝑡 2
1
40𝑡 + 2(-32.2)𝑡 2
2
𝑣𝐶 = 40
ℎ2 =
ℎ2 =
ℎ2 = 40𝑡 − 16. 1𝑡
→ 𝑒𝑞. 2
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From the figure:
Substitute eq. 1 and eq. 2 to eq 3:
For b:
ℎ1 + ℎ2 = 80
→ 𝑒𝑞. 3
2
16.1(𝑡) + 40𝑡 − 16. 1𝑡 2 = 80
𝒕 = 2 secs
Ans:
If the balls will meet when t = 2 secs, they will meet at:
ℎ1 = 16.1(𝑡)2 = 16.1(2)2 = 𝟔𝟒. 𝟒 𝒇𝒕 𝒇𝒓𝒐𝒎 𝑨.
For c:
𝑨𝒏𝒔.
𝑣𝐵(𝑠𝑒𝑐𝑜𝑛𝑑 𝑏𝑎𝑙𝑙) = ?
ℎ2 = 80 − ℎ1 = 80 − 64.4 = 15.6 𝑓𝑡
𝑣𝑐 = 40 𝑓𝑡/𝑠
(𝑣𝐵 )2 = (𝑣𝑐 )2 + 2𝑔ℎ
(𝑣𝐵 )2 = (40 )2 + 2(−9.81)(15.6)
𝑣𝐵 = 35.97 𝑚/𝑠
(of the second ball)
Example ULO1a-17. A stone is dropped from a captive balloon at an elevation of 304.8 m. Two
seconds later another stone is thrown vertically upward from the ground with a velocity of 75.6
m/s. If g = 9.75 m/s2,
a. when will the stones pass each other?
b. where will stones pass each other?
𝑣10 = 0
ℎ1 ?
ℎ = 304.8 𝑚
𝑡
𝑨
𝑣1𝐴 =? ; 𝑣2𝐴 =?
ℎ2 ?
𝑡−2
𝑣20 = 75.6 𝑚/𝑠
Solution:
Let
𝑣10 =
𝑣1𝐴 =
𝑣20 =
𝑣2𝐴 =
be the initial velocity of the first ball
be the velocity of the first ball at point A
be the initial velocity of the second ball
be the velocity of the second ball at point B
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𝑡 = time required for the first ball to reach point A.
𝑡 − 2 = time required for thesecond ball to reach point A.
For the first ball: 𝑣1𝐴 = 0
ℎ1 = 𝑣10 (𝑡) +
1
1
(𝑔)(𝑡)2
2
2
ℎ1 = 0 + (9.75)(𝑡)
2
ℎ1 = 4.875(𝑡)2
For the second ball: 𝑣20 = 75. 6 𝑚/𝑠 at 𝑡 = 𝑡 − 2
ℎ2 = 𝑣20 (𝑡) +
ℎ2 =
ℎ2 =
ℎ2 =
ℎ2 =
ℎ2 =
1
(𝑔)(𝑡)2
2
1
2) + 2 (−9.75)(𝑡
75. 6(𝑡 −
− 2)2
75. 6(𝑡 − 2) − 4.875(𝑡 − 2)2
75. 6𝑡 − 151. 2 − 4.875(𝑡 − 2)2
75. 6𝑡 − 151. 2 − 4.875𝑡 2 + 19.5𝑡 − 19.5
−4.875𝑡 2 + 95.1𝑡 − 170.7
But
ℎ1 + ℎ2 = 304. 8 𝑚
4.875(𝑡)2 − 4.875𝑡 2 + 95.1𝑡 − 170.7 = 304.8
𝑡 = 5 secs
They will pass each other at:
ℎ2 = −4.875𝑡 2 + 85.35𝑡 − 170.7 = −4.875(52 ) + 85.35(5) − 170.7
= 𝟏𝟖𝟐. 𝟗𝟐𝟓 𝒎 𝒂𝒃𝒐𝒗𝒆 𝒕𝒉𝒆 𝒈𝒓𝒐𝒖𝒏𝒅
Ans.
Self-Help! You can also refer to the sources below to help you further
understand the lesson.
Let’s Check!
Activity
Now that you know the basic terms and concepts in the study of dynamics of rigid
bodies, let’s try to check your understanding of these terms. Write the term/s on the space
provided for the statement in each item.
______________________________1. It is a body of negligible dimensions.
______________________________2. Observes motion by considering the forces that cause the motion.
______________________________3. It is a line or a curve obtained when all the points a body occupies
within a specific time period are joined.
______________________________4. Observes motion without considering the forces that cause the
motion.
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______________________________5. It is a body whose changes in shape are negligible compared with
the overall dimensions of the body or with the changes in position of
the body as a whole.
______________________________6. It is the action of one body on another.
______________________________7. It is the quantitative measure of the inertia or resistance to change
in motion of a body.
______________________________8. It is a measure of the succession of events and is considered an
absolute quantity in Newtonian mechanics.
______________________________9. Time rate of change of position coordinates.
______________________________10. It is that branch of mechanics which deals with the motion of
bodies under the action of forces.
______________________________11. The acceleration of a particle is proportional to the resultant force
acting on it and is in the direction of this force.
______________________________12. The acceleration of a particle is proportional to the resultant force
acting on it and is in the direction of this force.
______________________________13. It is defined to be the magnitude or size of displacement between
two positions.
______________________________14. It is defined traveled is the total length of the path traveled
between two positions.
______________________________15. It is a line or a curve obtained when all the points a body occupies
within a specific time period are joined.
Let’s Analyze!
Activity
1. Suppose that two objects attract each other with a gravitational force of 16 units. If the
distance between the two objects is doubled, what is the new force of attraction between the
two objects?
2. Suppose that two objects attract each other with a gravitational force of 16 units. If the
distance between the two objects is reduced in half, then what is the new force of attraction
between the two objects?
3. What is the displacement of the cross-country team if they begin at the school, run 10 miles
and finish back at the school?
4. A cyclist rides 3 km west and then turns around and rides 2 km east. (a) What is her
displacement? (b) What distance does she ride? (c) What is the magnitude of her displacement?
5. Give an example in which there are clear distinctions among distance traveled, displacement,
and magnitude of displacement. Specifically identify each quantity in your example.
6. Under what circumstances does distance traveled equal magnitude of displacement? What
is the only case in which magnitude of displacement and displacement are exactly the same?
7. A particle moves along a horizontal path with a velocity of v = (3t2 - 6t) m/s, where t is the
time in seconds. If it is initially located at the origin 0, determine the distance traveled in 3.5 s,
and the particle's average velocity and average speed during the time interval.
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8. A particle travels along a straight line with an acceleration of 𝑎 = 10 − 0.2𝑠 𝑚/𝑠 2 , where s
is measured in meters. Determine the velocity of the particle when 𝑠 = 10 𝑚 if 𝑣 = 5 𝑚/𝑠 at
s = 0.
9. A baby crawls toward a toy. Its motion is shown on the following graph of x vs t. What is the
average speed of the baby between at 𝑡 = 0 𝑎𝑛𝑑 𝑡 = 4 𝑠?
10. A hungry rabbit sprints toward a vegetable garden in a straight line. Its motion is shown on
the following graph of horizontal position x vs. time t. What is the average velocity of the rabbit
between at 𝑡 = 2 𝑎𝑛𝑑 𝑡 = 12 𝑠?
11. The particle travels along a straight track such that its position is described by position vs.
time graph (s vs. t graph)
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a.
Construct the velocity vs. time graph for the same time interval.
i.
What is the velocity at 𝑡 = 5 𝑠?
ii.
What is the velocity at t = 8 s?
b. Construct the acceleration vs. time graph for the same time interval.
12. A van travels along a straight road with a velocity described by the graph. Construct the
position vs time and acceleration vs time graphs during the same period. Take s = 0 when t
= 0.
13. A car has an initial speed of 25 m/s and a constant deceleration of 3 m/s2. Determine the
velocity of the car when 𝑡 = 4 𝑠. What is the displacement of the car during the
4 𝑠𝑒𝑐𝑜𝑛𝑑𝑠 time interval? How much time is needed to stop the car?
14. A diver jumps from a 3.0 m board with an initial upward velocity of 5.5 m/s. Determine…
a. the time the diver was in the air
b. the maximum height to which she ascended
c. her velocity on impact with the water
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15. A construction worker accidentally drops a brick from a high scaffold.
a. What is the velocity of the brick after 4.0 s?
b. How far does the brick fall during this time?
In a Nutshell!
1. Determine the force of gravitational attraction between the earth (m = 5.98 x 1024 kg) and a
70-kg physics student if the student is in an airplane at 40,000 feet above earth's surface.
This would place the student a distance of 6.39 x 106 m from earth's center.
2. A particle travels along a straight line such that in 2 s it moves from an initial position 𝑥𝐴 =
+0.5 𝑚 to a position 𝑥𝐵 = − 1 .5 𝑚. Then in another 4 s it moves from 𝑥𝐵 𝑡𝑜 𝑥𝐶 = +2.5 𝑚.
Determine the particle's average velocity and average speed during the 6-s time interval.
3. A particle travels along a straight-line path such that in 4 s it moves from an initial position
𝑥𝐴 = −8 𝑚 to a position 𝑥𝐵 = +3 𝑚. Then in another 5 s it moves from 𝑥𝐵 to 𝑥𝐶 = −6 𝑚.
Determine the particle's average velocity and average speed during the 9-s time interval.
4. A particle travels along a straight line with a velocity 𝑣 = 12 − 3𝑡 2 m/s, where t is in seconds.
When 𝑡 = 1 s, the particle is located 10 m to the left of the origin. Determine the acceleration
when 𝑡 = 4 s, the displacement from t = 0 to t = 10 s, and the distance the particle travels
during this time period.
5. The position of a particle along a straight line is given by 𝑠 = ( 1 .5𝑡 3 − 13.5𝑡 2 + 22.5𝑡) ft,
where t is in seconds. Determine the position of the particle when t = 6 s and the total distance
it travels during the 6-s time interval. Hint: Plot the path to determine the total distance
traveled.
6. The velocity vs. time graph of a car while traveling along a road is shown. Draw the position
vs. time and acceleration vs. time graphs for the motion.
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a. What is the displacement at t = 10 sec?
b. What is the acceleration at t =10 sec?
7. The dragster starts from rest and has an acceleration described by the graph. Construct the
velocity vs. time graph for the time interval 0 ≤ 𝑡 ≤ 𝑡 ′ where t’ is the time for the car to come
to rest.
a.
b.
c.
d.
What is the displacement at 𝑡 = 4 𝑠𝑒𝑐?
What is the displacement at 𝑡 = 6 𝑠𝑒𝑐?
What is the velocity at 𝑡 = 4 𝑠𝑒𝑐?
What is the velocity at 𝑡 = 6 𝑠𝑒𝑐?
8. A car starts from rest and moves with a constant acceleration of 1 .5 m/s2 until it achieves a
velocity of 25 m/s. It then travels with constant velocity for 60 seconds. Determine the average
speed and the total distance traveled.
9. A particle moves along a straight line with a velocity 𝑣 = (200𝑠) 𝑚𝑚/𝑠, where s is in
millimeters. Determine the acceleration of the particle at s = 2000 mm. How long does the particle
take to reach this position if s = 500 mm when t = 0?
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10. A particle has an initial speed of 27 m/s. If it experiences a deceleration 𝑜𝑓 𝑎 = −6𝑡 𝑚/𝑠 2 ,
where t is in seconds, determine its velocity, after it has traveled 10 m. How much time does this
take?
11. A parachutist with a camera descends in free fall at a speed of 10 m/s. The parachutist
releases the camera at an altitude of 50 m.
a. How long does it take the camera to reach the ground?
b. What is the velocity of the camera just before it hits the ground?
12. A ball is thrown upward from the ground with an initial speed of 25 m/s; at the same instant,
another ball is dropped from a building 15 m high. After how long will the balls be at the same
height?
13. A ball is shot vertically into the air at a velocity of 193.2 ft per sec. After 4 sec, another ball is
shot vertically into the air. What initial velocity must the second ball have in order to meet
the first ball 386.4 ft from the ground?
Course Schedule:
This section calendars all the activities and exercises including readings and lectures as well as
time for making assignments and doing other requirements in a programmed schedule by days
and weeks, to help the students in SDL pacing, regardless of mode of delivery (OBD or DED):
Activity
Big Picture 1a: Let’s Check Activities!
Big Picture 1a: Let’s Analyze Activities!
Big Picture 1a: In a Nutshell Activities!
Big Picture 1a: Q and A List!
Date
Where to Submit
BB
BB
BB
BB
Big Picture!
Big Picture in Focus: ULO-1b. Explain the basic kinematic quantities of curvilinear
motion: projectile motion.
Metalanguage
The most essential terms below are defined for you to have a better understanding of
this section in the course:
1. Projectile motion is a form of motion where an object moves in parabolic path; the path that
the object follows is called its trajectory.
2. Trajectory is the path of a body as it travels through space.
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3. The time of flight of a projectile motion is the time from when the object is projected to the
time it reaches the surface.
4. The maximum height is reached when 𝑣𝑦 = 0.
5. The range of the motion is fixed by the condition when 𝑦 = 0.
‘
Essential Knowledge
To perform the ULO for the first week of the course, you need to understand the
following key concepts that will be presented in the succeeding pages. You are also expected to
use other references, books, and other resource material that is available in the university’s
library.
Curvilinear Motion: Projectiles
The free-flight motion of a projectile is often studied in terms of its rectangular components. To
illustrate the kinematic analysis, consider a projectile launched at point (𝑥0 , 𝑦0 ), with an initial
velocity of 𝑣0 , having components (𝑣0 )𝑥 and (𝑣0 )𝑦 , When air resistance is neglected, the only
force acting on the projectile is its weight, which causes the projectile to have a constant
downward acceleration of approximately 𝑎𝑐 = g = 9.81 m/s2 or g = 32.2 ft/s2.
Horizontal Motion:
𝑠𝑖𝑛𝑐𝑒 𝑎𝑥 = 0,
𝑣𝑓𝑥 = 𝑣0𝑥 + 𝑎𝑥 𝑡
𝒗𝒇𝒙 = 𝒗𝟎𝒙
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1
(𝑎 )𝑡 2
2 𝑥
𝑥 = 𝑥0 + (𝑣0 )𝑥 𝑡
𝑥 = 𝑥0 + 𝑣0 𝑡 +
∆𝒙 = (𝒗𝟎 )𝒙 𝒕
(𝑣𝑓𝑥 )2 = (𝑣0𝑥 )2 + 2𝑎𝑥 (𝑥 − 𝑥0 )
(𝒗𝒇 )𝒙 = (𝒗𝟎 )𝒙
Vertical Motion:
Since 𝒂 = 𝒂𝒄
𝑣𝑓𝑦 = 𝑣0𝑦 − 𝑎𝑐 𝑡
(𝒗𝒇 ) = (𝒗𝟎 )𝒚 − 𝒈𝒕
𝒚
1
𝑦 = 𝑦0 + 𝑣0 𝑡 − (𝑎𝑐 )𝑡 2
2
1
𝑦 = 𝑦0 + (𝑣0 )𝑦 𝑡 − 𝑔𝑡 2
2
𝟏
∆𝒚 = (𝒗𝟎 )𝒚 𝒕 − 𝒈𝒕𝟐
𝟐
2
2
(𝑣𝑓𝑦 ) = (𝑣0𝑦 ) − 2𝑎𝑐 (𝑦 − 𝑦0 )
(𝒗𝒇𝒚 )𝟐 = (𝒗𝟎𝒚 )𝟐 − 𝟐𝒈∆𝒚
Example ULO1b.1. A sack slides off the ramp, shown in the figure, with a horizontal velocity of
12 m/s. If the height of the ramp is 6 m from the floor, determine the time needed for the sack
to strike the floor and the range R where sacks begin to pile up.
Solution:
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Given: 𝜃 = 0°
𝑣0𝑥 = 𝑣0 𝑐𝑜𝑠𝜃 = 𝑣0 𝑐𝑜𝑠0° = 𝑣0 = 12 𝑚/𝑠
𝑣0𝑦 = 𝑣0 𝑠𝑖𝑛𝜃 = 𝑣0 𝑠𝑖𝑛0° = 0
∆𝑦 = −6 𝑚
∆𝑦 = (𝑣0 )𝑦 𝑡 −
1
1
𝑔𝑡 2
2
2
−6 = − 2 (9.81 )𝑡
𝒕 = 𝟏. 𝟏𝟏 𝒔
𝑅 = 𝑥𝑚𝑎𝑥 = 𝑣0𝑥 𝑡 = 12(1.11) = 𝟏𝟑. 𝟑𝟐 𝒎
Example ULO1b-2.. The chipping machine is designed to eject wood chips at 𝑣0 = 25 𝑓𝑡/𝑠 as
shown in the figure. If the tube is oriented at 30° from the horizontal, determine how high, h,
the chips strike the pile if at this instant they land on the pile 20 ft from the tube.
Solution:
Given:
𝜃 = 30°
𝑣0𝑥 = 𝑣0 𝑐𝑜𝑠𝜃 = 25𝑐𝑜𝑠30° = 𝑣0 = 21.65 𝑓𝑡/𝑠
𝑣0𝑦 = 𝑣0 𝑠𝑖𝑛𝜃 = 25𝑠𝑖𝑛30° = 12.5 𝑓𝑡/𝑠
𝑥𝑚𝑎𝑥 = 20 𝑓𝑡
𝑥𝑚𝑎𝑥 = 𝑣0𝑥 t
20 = 21.65t
𝑡 = 0.92 𝑠𝑒𝑐𝑠
[time of flight]
1
𝑦 = (𝑣0 )𝑦 𝑡 − 2 𝑔𝑡 2
1
𝑦 = 12.5(0.92) − 2 (32.2))(0.92)2
𝑦 = −2. 13 ft
𝑦 + ℎ = −4 − h
−2.13 + (−ℎ) = −4
𝒉 = 𝟏. 𝟖𝟕 𝐟𝐭
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Example ULO1b-3. The track for this racing event was designed so that riders jump off the slope
at 30°, from a height of 1 m. During a race it was observed that the rider shown in remained in
midair for 1.5 s. Determine the speed at which he was traveling off the ramp, the horizontal
distance he travels before striking the ground, and the maximum height he attains. Neglect the
size of the bike and rider.
Solution:
Given:
𝜃 = 30°
𝑣0𝑥 = 𝑣0 𝑐𝑜𝑠𝜃 = 𝑣0 𝑐𝑜𝑠30°
𝑣0𝑦 = 𝑣0 𝑠𝑖𝑛𝜃 = 𝑣0 𝑠𝑖𝑛30°
𝑡 = 1.5 𝑠
1
2
𝑦 = (𝑣0 )𝑦 𝑡 + 𝑔𝑡 2
1
−1 = 𝑣0 𝑠𝑖𝑛30°(1.5) + 2 (−9.81)(1.5)2
𝒗𝟎 = 𝟏𝟑. 𝟑𝟖 𝒎/𝒔
𝑅 = (𝑣0 )𝑥 𝑡 = 𝑣0 𝑐𝑜𝑠30°𝑡 = 13.38𝑐𝑜𝑠30°(1.5)
𝑹 = 𝟏𝟕. 𝟑𝟖 𝒎
(𝑣𝑓𝑦 )2 = (𝑣0𝑦 )2 − 2𝑔𝑦
(𝑣𝐶𝑦 )2 = (𝑣0𝑦 )2 − 2𝑔(ℎ − 1)
0 = (13.38𝑠𝑖𝑛30°)2 − 2(9.81)(ℎ − 1)
𝒉 = 𝟑. 𝟐𝟖 𝒎
𝑚
Example ULO1b-4. The golf ball is hit at A with a speed of 𝑣𝐴 = 40 𝑠 and directed at an angle of
30° with the horizontal as shown. Determine the distance d where the ball strikes the slope at
B.
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Solution:
The x–y coordinate system will be set so that its origin coincides with point A.
𝑚
𝑥 −Motion: (𝑣𝐴 )𝑥 = 40𝑐𝑜𝑠30° = 34.64 𝑠
𝑥𝐴 = 0
5
𝑥𝐵 = 𝑑 ( 2 ) = 0.9806𝑑
√5 +1
𝑥𝐵 = 𝑥𝐴 + (𝑣𝐴 )𝑥 𝑡
0.9806𝑑 = 0 + 34.64𝑡
𝒕 = 𝟎. 𝟎𝟐𝟖𝟑𝟏𝒅
→ 𝒆𝒒. 𝟏
𝑦 −Motion: (𝑣𝐴 )𝑦 = 40𝑠𝑖𝑛30° = 20
𝑦𝐴 = 0
𝑦𝐵 = 𝑑 (√
Thus:
1
√52 +1
𝑚
𝑠
) = 0.1961𝑑
𝑎𝑦 = −𝑔 = −9.81 𝑚/𝑠 2
1
(𝑎𝑦 )𝑡 2
2
1
+ 2 (−9.81)(𝑡 2 )
𝑦𝐵 = 𝑦𝐴 + (𝑣𝐴 )𝑦 𝑡 +
0.1961𝑑 = 0 + 20𝑡
4.905𝒕𝟐 − 𝟐𝟎𝒕 + 𝟎. 𝟏𝟗𝟔𝟏𝒅 = 𝟎
→ 𝒆𝒒. 𝟐
Substituting eq. 1 to eq. 2:
4.905(0.02831𝑑)2 − 20(0.02831𝑑) + 0.1961𝑑 = 0
𝒅 = 𝟗𝟒. 𝟏 𝒎
Self-Help: You can also refer to the sources below to help you further
understand the lesson
* Meriam, J.L. (2016). Engineering Mechanics. 8th Ed., Hoboken, NJ Wiley
*Kuisalaas, J and Pytel, A (2017). Engineering Mechanics: Dynamics. 3rd Ed., Cengage Learning
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*Beer, F (2013). Vector Mechanics for Engineers: Dynamics. 10th Ed., Boston: McGraw Hill
Higher Education.
Let’s Check!
Activity
Now that you know the basic terms and concepts in the study of dynamics of rigid
bodies, let’s try to check your understanding of these terms/ (of this topic):
1. What is the only thing that affects projectiles, no matter which direction it is moving in?
2. If there is no force, there is no___________?
3. Which of the following is NOT considered a projectile motion problem?
a. A football rolling down a hill
b. A golf ball on a tee struck by a driver
c. A baseball thrown from right field to second base
d. A cannon ball shot towards a castle wall
e. All should be solved as projectile motion problems
4. Projectiles follow a ________ path.
5. For an object moving along a trajectory, the horizontal acceleration of an object as its position
changes.
a. is constant b. decreases c. increases
6. For an object moving along a trajectory, the vertical acceleration of an object as its position
changes.
a. is constant b. decreases c. increases
7. Comparing an object that is dropped and an identical object thrown horizontally from the
same height at the same time, we find that the time it takes to hit the ground.
a. is less for the thrown object.
b. for each is the same.
c. is greater for the thrown object.
d. depends on the initial velocity of the objects.
8. In projectile motion, the initial horizontal velocity is the final horizontal velocity.
a. smaller than b. greater than c. the same as
9. For a projectile, the rising and falling times of the object are equal if the launching position is
the landing position.
a. the same height as
b. above
c. below
10. The only force acting on a projectile is _____________?
Let’s Analyze!
1. A stone is thrown vertically upward from the ground with a velocity of 14.72 m per sec. One
second later another stone is thrown vertically upward with a velocity of 96.6 ft per sec 29.44
m per sec. How far above the ground will the stones be at the same level?
2. It is observed that the time for the ball to strike the ground at B is 2.5 s. Determine the speed
𝑣𝐴 and angle 𝑣𝜃𝐴 at which the ball was thrown.
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3. If the motorcycle leaves the ramp traveling at 110 ft/s , determine the height ℎ ramp B must
have so that the motorcycle lands safely.
In a Nutshell!
𝑚
1. The pitching machine is adjusted so that the baseball is launched with a speed of 𝑣𝐴 = 30 𝑠 . If
the ball strikes the ground at 𝐵, determine the two possible angles at which it was launched.
2. The skateboard rider leaves the ramp at A with an initial velocity of 𝑣𝐴 at 30° angle. If he
strikes the ground at B, determine 𝑣𝐴 and the time of flight.
3. A golf ball is struck with a velocity of 80 ft/s as shown. Determine the distance 𝑑 to
where it will land.
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Course Schedule
This section calendars all the activities and exercises including readings and lectures as well as
time for making assignments and doing other requirements in a programmed schedule by days
and weeks, to help the students in SDL pacing, regardless of mode of delivery (OBD or DED):
Activity
Big Picture 1b: Let’s Check Activities!
Big Picture 1b: Let’s Analyze Activities!
Big Picture 1b: In a Nutshell Activities!
Big Picture 1b: Q and A List!
First Exam Schedule
Date
Sept 4, 2020
Where to Submit
BB
BB
BB
BB
BB
Big Picture
Week 4-5: Unit Learning Outcomes-Unit 2 (ULO-2): At the end of the unit, you are expected to
a. Demonstrate knowledge in kinematics-path (normal-tangential) coordinates
kinematics- polar and cylindrical coordinates and;
b. Explain the basic Kinetics: Force- Mass-Acceleration Method
Big Picture in Focus: ULO-2a. Demonstrate knowledge in kinematics-path (normaltangential) coordinates kinematics- polar and cylindrical coordinates.
Metalanguage
The most essential terms below are defined for you to have a better understanding of
this section in the course:
1. Acceleration is the rate of change of the velocity of an object with respect to time.
2. Rectilinear Motion is a linear motion in which the direction of the velocity remains
constant and the path is a straight line.
3. Kinematics is a study of the geometry of the motion.
4. Kinetics is a study of the forces that cause the motion.
5. Curvilinear Motion is the motion of an object moving in a curved path.
6. Normal Component is the component perpendicular to the curve.
7. Tangential Component is the component tangent to the curve.
8. Radius of Curvature is the radius of the circular arc.
9. Cylindrical Components/Coordinates is the three-dimensional coordinate system that
specifies point positions by the distance from a chosen reference axis, the direction from
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10.
11.
12.
13.
14.
the axis relative to a chosen reference direction, and the distance from a chosen reference
plane perpendicular to the axis.
Time Derivative is the derivative with respect to time.
Free Body Diagram are diagrams used to show the relative magnitude and direction of all
forces acting upon an object in a given situation.
Speed refers to the magnitude of velocity.
Average speed is the total distance traveled divided by the total time.
Pulley is a wheel with a grooved rim around which a cord passes. It acts to change the
direction of a force applied to the cord and is chiefly used (typically in combination) to raise
heavy weights.
Essential Knowledge
Curvilinear Motion: Normal and Tangential Components
When the path along which a particle travels is known, then it is often convenient to describe
the motion using n and t coordinate axes which act normal and tangent to the path, respectively,
and at the instant considered have their origin located at the particle.
Coordinate System
• Provided the path of the particle is known, we can establish a set of n and t coordinates having
a fixed origin, which is coincident with the particle at the instant considered.
• The positive tangent axis acts in the direction of motion and the positive normal axis is
directed toward the path's center of curvature.
Velocity
• The particle's velocity is always tangent to the path.
• The magnitude of velocity is found from the time derivative of the path function. 𝑣 = 𝑥̇
Tangential Acceleration
The tangential component of acceleration is the result of the time rate of change in the magnitude
of velocity. This component acts in the positive 𝑥 direction if the particle's speed is increasing or
in the opposite direction if the speed is decreasing.
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The relations between 𝑎𝑡 , 𝑣, 𝑡 and s are the same as for rectilinear motion, namely,
𝑎𝑡 = 𝑣̇
;
𝑎𝑡 𝑑𝑥 = 𝑣𝑑𝑣
If 𝑎𝑡 is constant, 𝑎𝑡 = (𝑎𝑡 )𝑐 , the above equations, when integrated, yield,
1
(𝑎𝑡 )𝑐 𝑡 2
2
𝑥𝑓 = 𝑥0 + 𝑣0 𝑡 +
or
𝟏
∆𝒙 = 𝒗𝟎 𝒕 + (𝒂𝒕 )𝒄 𝒕𝟐
𝟐
𝒗𝒇 = 𝒗𝟎 + (𝒂𝒕 )𝑪 𝒕
(𝒗𝒇 )𝟐 = (𝒗𝟎 )𝟐 + 𝟐(𝒂𝒕 )𝑪 (∆𝒙)
Normal Acceleration
The normal component of acceleration is the result of the time rate of change in the direction of
the velocity. This component is always directed to the center of the curvature of the path, i.e.,
along the positive n axis.
The magnitude of this component is determined from:
𝑣2
𝑎𝑛 =
𝜌
If the path is expressed as 𝑦 = 𝑓(𝑥), the radius of curvature 𝜌 at any point on the path is
determined by the equation
2
𝜌=
𝑑𝑦
[ 1+ (𝑑𝑥) ]3/2
𝑑2 𝑦
|
𝑑𝑥2
|
Total acceleration of the particle in curvilinear motion:
𝑎 = √(𝑎𝑡 )2 + (𝑎𝑛 )2
Example 2a-1. When the skier reaches point A along the parabolic path in the figure, he has a
speed of 6 𝑚/𝑠 which is increasing at 2 m/𝑠 2 . Determine the direction of his velocity and the
direction and magnitude of his acceleration at this instant. Neglect the size of the skier in the
calculation.
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Solution:
𝑦=
1 2
𝑥
20
𝑑𝑦
1
=
𝑥
𝑑𝑥
10
At 𝑥 = 10 𝑚,
𝑑𝑦
𝑑𝑥
→ 𝐯𝐞𝐥𝐨𝐜𝐢𝐭𝐲 𝐨𝐟 𝐭𝐡𝐞 𝐩𝐚𝐫𝐭𝐢𝐜𝐥𝐞
=1
𝑑𝑦
𝑑𝑥
𝑡𝑎𝑛−1 =
For the direction of the velocity,
𝑡𝑎𝑛−1 = 1
𝜃 = 45°
Thus, the direction of the velocity is at 45° with respect to the horizontal.
From given,
𝑎𝑡 = 2 m/𝑠 2
𝑎𝑛 =
𝑣2
𝜌
For 𝜌 ∶
𝜌=
𝑑𝑦 2 3/2
) ]
𝑑𝑥
𝑑2 𝑦
| 2|
𝑑𝑥
[1+ (
𝑎𝑛 =
𝑣2
𝜌
2
=
1
[1+ ( 𝑥) ]3/2
10
1
| |
10
= 28.28 𝑚
|
𝑥=10 𝑚
𝑚2
=
62 2
36
𝑠
=
m/𝑠 2
28.28 𝑚 28.28
2
36
𝑎𝐴 = √(𝑎𝑡 )2 + (𝑎𝑛 )2 = √22 + (28.28)
∅ = 𝑡𝑎𝑛−1
2
1.273
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= 2.37 𝑚/ 𝑠 2
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∅ = 57. 523°
For the direction of the acceleration with respect to the horizontal:
𝜶
Thus: 𝛼 = 45° + 90° + 57.523° − 180° = 𝟏𝟐. 𝟓𝟐𝟑°
with respect to the horizontal
Example 2a-2. A race car C travels around the horizontal circular track that has a radius of 300
ft, as in the figure shown. If the car increases its speed at a constant rate of 7 ft/s 2 , starting from
rest, determine the time needed for it to reach an acceleration of 8 ft/s 2 . What is its speed at this
instant?
Solution:
Note: The t axis is in the direction of motion, and the positive n axis is directed toward the
center of the circle. This coordinate system is selected since the path is known.
Given:
𝑎 = 8 ft/𝑠 2
𝑎𝑡 = 7 ft/s
𝑣𝑓 = 𝑣
𝑣 = 𝑣0 + (𝑎𝑡 )𝑐 (𝑡)
𝑣 = 7𝑡
𝑎𝑛 =
𝑣2
𝜌
=
72 𝑡 2
300
49
𝑡
300
=
𝑎 = √𝑎𝑡 2 + 𝑎𝑛 2
2
49
8 = √72 + (300 𝑡)
𝒕 = 4.87 s
→ time needed to achieve 8 ft/s2
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The speed at this instant is,
𝑣 = 7𝑡 = 7(4.87) = 𝟑𝟒. 𝟎𝟗 𝒎/𝒔
Remember the velocity will always be tangent to the path, whereas the acceleration will be
directed within the curvature of the path.
Example 2a-3. Starting from rest the motorboat travels around the circular path, 𝜌 = 50 𝑚 at a
speed of 𝑣 = 0.8𝑡 𝑚/𝑠 where t is in seconds. Determine the magnitudes of the boat’s velocity
and acceleration when it has traveled 20 m.
Solution:
The time for which the boat to travel 20 m must be determined first.
𝑑𝑠 = 𝑣𝑑𝑡
𝑡
= ∫0 0.8𝑡𝑑𝑡
𝑡 = 7.071 𝑠
20
∫0 𝑑𝑠
The magnitude of the boat’s velocity is,
𝑣 = 0.8(7.071) = 5.657
𝑚
𝑠
= 𝟓. 𝟔𝟔 𝒎/𝒔
The tangential acceleration is,
𝑎𝑡 = 𝑣̇ = 0.8 m/𝑠 2
To determine the normal acceleration,
𝑎𝑛 =
𝑣2
𝜌
=
5.6572
50
= 0.640 𝑚/𝑠 2
Thus, the magnitude of the acceleration is:
𝑎 = √𝑎𝑡 2 + 𝑎𝑛 2 = √0.82 + 0.6402 = 𝟏. 𝟎𝟐 𝒎/𝒔𝟐
Curvilinear Motion: Cylindrical Components
Sometimes the motion of the particle is constrained on a path that is best described using
cylindrical coordinates. If motion is restricted to the plane, then polar coordinates are used.
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Polar coordinates are a suitable choice for solving problems when data regarding the angular
motion of the radial coordinate r is given to describe the particle's motion. Also, some paths of
motion can conveniently be described in terms of these coordinates. To use polar coordinates,
the origin is established at a fixed point, and the radial line r is directed to the particle. The
transverse coordinate 𝜃 is measured from a fixed reference line to the radial line.
The radial component 𝒗𝒓 is a measure of the rate of increase or decrease in the length
of the radial coordinate while 𝒗𝜽 can be interpreted as the rate of motion along the
𝒅𝜽
circumference of a circle having a radius r. In particular, the term 𝜽= 𝒅𝒕 is called the angular
velocity, since it indicates the time rate of change of the angle 𝜃. Common units used for this
measurement are rad/s.
𝑣𝑟 = 𝑟̇
𝑣𝜃 = 𝑟𝜃̇
Since 𝑣𝑟 and 𝑣𝜃 are mutually perpendicular, the magnitude of velocity or speed is simply
the positive value of:
𝟐
𝒗 = √(𝒓̇ )𝟐 + (𝒓𝜽̇)
and the direction of 𝑣 is of course, tangent to the path.
For acceleration:
𝑎𝑟 = 𝑟̈ − 𝑟𝜃̇ 2
𝑎𝜃 = 𝑟𝜃̈ + 2𝑟̇ 𝜃̇
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𝑑2 𝜃
The term 𝜃̈ = 2 is called the angular acceleration since it measures the change made in
𝑑𝑡
the angular velocity during an instant of time. Units for this measurement are rad/𝑠 2 .
Since 𝑎𝑟 and 𝑎𝜃 are always perpendicular, the magnitude of acceleration is simply the
positive value of:
𝒂 = √(𝒓̈ − 𝒓𝜽̇𝟐 )𝟐 + (𝒓𝜽̈ + 𝟐𝒓̇ 𝜽̇)𝟐
The direction is determined from the vector addition of its two components. In general, 𝑎 will not
be tangent to the path.
Cylindrical Coordinates
If the particle moves along a space curve as shown in the figure, then its location may be specified
by the three cylindrical coordinates, 𝑟, 𝜃, 𝑧. The z coordinate is identical to that used for
rectangular coordinates. Since the unit vector defining its direction, 𝑢𝑧′ is constant, the time
derivatives of this vector are zero, and therefore the position, velocity, and acceleration of the
particle can be written in terms of its cylindrical coordinates as follows:
𝒓𝒑 = √𝒓𝟐 + 𝒛𝟐
𝟐
𝒗 = √𝒓̇ 𝟐 + (𝒓𝜽̇) + (𝒛̇ )𝟐
𝟐
𝒂 = √(𝒓̈ − 𝒓𝜽)̇𝟐 + (𝒓𝜽̈ + 𝟐𝒓̇ 𝜽̇) + (𝒛̈ )𝟐
Time Derivative. The above equations require that we obtain the time derivatives 𝑟̇ , 𝑟,̈ 𝜃̇, and 𝜃̈
in order to evaluate the r and 𝜃 component of 𝑣 and 𝑎. Two types of problems generally occur:
1. If the polar coordinates are specified as time parametric equations, 𝑟 = 𝑟(𝑡) and 𝜃 = 𝜃(t), then
the time derivatives can be found directly.
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2. If the time-parametric equations are not given, then the path 𝑟 = 𝑓(𝜃) must be known. Using
the chain rule of calculus, we can then find the relation between 𝑟̇ and 𝜃̇, and between 𝑟̈ and 𝜃̈.
Example 2a-4. Starting from rest, the boy runs outward in the radial direction from the center of
the platform with a constant acceleration of 0.5 m/𝑠 2 . If the platform is rotating at a constant rate
of 𝜃̇ = 0.2 rad/s, determine the radial and transverse components of the velocity and acceleration
of the boy when 𝑡 = 3 𝑠. Neglect his size.
Solution:
Velocity: When 𝑡 = 3 𝑠, the position of the boy is given by:
(𝑣0 )𝑟 = 0 → [starting from rest]
1
∆𝑥 = (𝑣0 )𝑟 𝑡 + (𝑎𝑐 )𝑟 𝑡 2
2
∆𝑥 =
∆𝑥 =
1
(𝑎 ) 𝑡 2
2 𝑐 𝑟
1
𝑚
(0.5 2 ) (32 𝑠 2 ) = 2.25 𝑚
2
𝑠
In this case, 𝑟 = ∆𝑥 = 2.25
The boy’s final velocity in 𝑟, or the boy’s radial component of velocity:
𝑣𝑟 = (𝑣𝑓 )𝑟 = (𝑣0 )𝑟 + (𝑎𝑐 )𝑟 𝑡
𝑣𝑟 = (𝑣𝑓 )𝑟 = 0 + 0.5(3) = 𝟏. 𝟓𝟎 𝒎/𝒔
The boy’s transverse component of velocity:
𝑣𝜃 = 𝑟𝜃̇ = 2.25(0.2) = 𝟎. 𝟒𝟓𝟎 𝒎/𝒔
Acceleration: When 𝑡 = 3 𝑠, 𝑟 = 2.25 𝑚, 𝑟̇ = 𝑣𝑟 = 1.50
𝑚
,
𝑠
𝑚
𝑟̈ = 0.5 𝑠2 , 𝜃̈ = 0
𝑎𝑟 = 𝑟̈ − 𝑟𝜃̇ 2 = 0.5 − 2.25(0.22 ) = 𝟎. 𝟒𝟏𝟎 𝒎/𝒔𝟐
𝑎𝜃 = 𝑟𝜃̈ + 2𝑟𝜃̇ = 2.25(0) + 2(1.50)(0.2) = 𝟎. 𝟔 𝒎/𝒔𝟐
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Example 2a-5. A car travels along the circular curve of radius r = 300 ft. At an instant shown, its
angular rate of rotation is 𝜃̇ = 0.4 𝑟𝑎𝑑/𝑠, which is increasing at the rate of 𝜃̈ = 0.2 𝑟𝑎𝑑/𝑠 2 .
Determine the magnitudes of the car’s velocity and acceleration at this instant.
Solution:
Velocity: we have,
𝑣𝑟 = 𝑟̇ = 0
𝑣𝜃 = 𝑟𝜃̇ = 300(0.4) = 120 𝑓𝑡/𝑠
Thus, the magnitude of the velocity of the car is:
𝑣 = √(𝑣𝑟 )2 + (𝑣𝜃 )2 = √02 + 1202 = 𝟏𝟐𝟎 𝒇𝒕/𝒔
Acceleration:
𝑎𝑟 = 𝑟̈ − 𝑟𝜃̇ 2 = 0 − 300(0.42 ) = −48.0 𝑓𝑡/𝑠 2
𝑎𝜃 = 𝑟𝜃̈ + 2𝑟̇ 𝜃̇ = 300(0.2) + 2 (0)(0.4) = 60.0 𝑓𝑡/𝑠 2
Thus, the magnitude of the acceleration of the car is:
𝑎 = √(𝑎𝑟 )2 + (𝑎𝜃 )2 = √(−48)2 + (60)2 = 𝟕𝟔. 𝟖 𝒇𝒕/𝒔𝟐
Example 2a-6. The airplane on the amusement park ride moves along a path defined by the
equations 𝑟 = 4 𝑚, 𝜃 = 0.2𝑡 𝑟𝑎𝑑 and 𝑧 = 0.5𝑐𝑜𝑠𝜃 m, where 𝑡 is in seconds. Determine the
cylindrical components of the velocity and acceleration of the airplane when 𝑡 = 6 𝑠.
Solution:
𝑟 = 4𝑚
𝑟̇ = 0
𝑟̈ = 0
𝑧 = 0.5𝑐𝑜𝑠𝜃
𝜃 = 0.2𝑡|𝑡=6 𝑠 = 1.2 𝑟𝑎𝑑
𝜃̇ = 0.2 𝑟𝑎𝑑/𝑠
𝜃̈ = 0
𝑧̇ = −0.5𝑠𝑖𝑛𝜃𝜃̇ |𝜃=1.2 𝑟𝑎𝑑 = −0.0932 𝑚/𝑠
𝑧̈ = −0.5[[𝑐𝑜𝑠𝜃𝜃̇ 2 + 𝑠𝑖𝑛𝜃𝜃̈ ]|𝜃=1.2 𝑟𝑎𝑑 = −0.007247 𝑚/𝑠 2
𝑣𝑟 =
𝑣𝜃 =
𝑣𝑧 =
𝑎𝑟 =
𝑎𝜃 =
𝑎𝑧 =
𝑟̇ = 𝟎
𝑟𝜃̇ = 4(0.2) = 𝟎. 𝟖 𝒎/𝒔
𝑧̇ = −𝟎. 𝟎𝟗𝟑𝟐 𝒎/𝒔
𝑟̈ − 𝑟𝜃̇ 2 = 0 − 4(0.2)2 = −𝟎. 𝟏𝟔 𝒎/𝒔𝟐
𝑟𝜃̈ + 2𝑟̇ 𝜃̇ = 4(0) + 2(0)(0.2) = 𝟎
𝑧̈ = −𝟎. 𝟎𝟎𝟕𝟐𝟓 𝒎/𝒔𝟐
Absolute Dependent Motion Analysis of Two Particles
In some types of problems, the motion of one particle will depend on the corresponding motion
of another particle. This dependency commonly occurs if the particles, here represented by
blocks, are interconnected by inextensible cords which are wrapped around pulleys. For example,
the movement of block A downward along the inclined plane in the figure will cause a
corresponding movement of block B up the other incline. We can show this mathematically by
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first specifying the location of the blocks using position coordinates 𝑠𝐴 and 𝑠𝐵 . If the total cord
length is 𝑙 𝑇 , the two position coordinates are related by the equation:
Differentiating,
𝑑𝑠𝐴
𝑑𝑡
+
𝑑𝑠𝐵
𝑑𝑡
𝑠𝐴 + 𝑠𝐵 = 𝑙 𝑇
=0
since 𝑣 =
𝑣𝐴 + 𝑣𝐵 = 0
𝑑𝑠
𝑑𝑡
𝒗𝑨 = −𝒗𝑩
Differentiating further,
𝑑 2 𝑠𝐴
𝑑𝑡 2
+
𝑑 2 𝑠𝐵
𝑑𝑡 2
=0
𝑎𝐴 + 𝑎𝐵 = 0
𝒂𝑨 = −𝒂𝑩
Relationships vary from one figure to another like this one:
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𝑑2 𝑠
since a = 𝑑𝑡 2
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Differentiating,
2
𝑑𝑠𝐵
𝑑𝑡
+
2𝑠𝐵 + ℎ + 𝑠𝐴 = 𝑙 𝑇
𝑑𝑠𝐴
𝑑𝑡
=0
since 𝑣 =
2𝑣𝐵 + 𝑣𝐴 = 0
𝑑𝑠
𝑑𝑡
𝟐𝒗𝑩 = −𝒗𝑨
Differentiating further,
𝟐𝒂𝑩 = −𝒂𝑨
Example 2a-7. Determine the velocity of block 𝐴 in the figure if block 𝐵 has an upward velocity
of 6 ft/s.
Solution:
Determine the datum for the measurement of length, it is recommended that this be done pulley
to pulley.
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𝑠𝐴 + 3𝑠𝐵 = 𝑙 𝑇
𝑣𝐴 + 3𝑣𝐵 = 0
𝑣𝐴 = −3𝑣𝐵
𝑎𝐴 = −3𝑎𝐵
Thus, the velocity at A must be:
𝑣𝐴 = −3𝑣𝐵 = −3 (−6
𝑓𝑡
) = 18 𝑓𝑡/𝑠
𝑠
or 18 ft/s ↓
Example 2a-8. Determine the velocity of A in the figure if B has an upward velocity of 6 ft/s.
Since there are three lengths involved, two equations must be drawn,
𝑠𝐴 + 2𝑠𝐶 = 𝑙1
𝑑𝑠𝐴
𝑑𝑡
+
2𝑑𝑠𝐶
𝑑𝑡
=0
𝑣𝐴 + 2𝑣𝐶 = 0
−0.5𝑣𝐴 = 𝑣𝐶
→ 𝑒𝑞. 1
𝑠𝐶 + 2(𝑠𝐵 − 𝑠𝐶 ) = 𝑙2
𝑑𝑠𝐶
𝑑𝑡
+
2𝑑𝑠𝐵
𝑑𝑡
−
2𝑑𝑠𝐶
𝑑𝑡
=0
−𝑣𝐶 + 2𝑣𝐵 = 0
𝑣𝐶 = 2𝑣𝐵 → 𝑒𝑞. 2
Subst. 2 to 1
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If 𝑣𝐵 = -6 ft/s,
−0.5𝑣𝐴 = 2𝑣𝐵
−0.5𝑣𝐴 = 2(−6)
𝑣𝐴 = 24 𝑓𝑡/𝑠 or
𝟐𝟒 𝒇𝒕/𝒔 ↓
Example 2a-9. Determine the 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝐵 in the figure below if the end of the cord at A is
pulled down with a velocity of 2 m/s.
Solution:
(𝑠𝐴 − 𝑠𝐶 ) + (𝑠𝐵 − 𝑠𝐶) + 𝑠𝐵 = 𝑙1
𝑑𝑠𝐴 𝑑𝑠𝐶 𝑑𝑠𝐵 𝑑𝑠𝐶 𝑑𝑠𝐵
−
+
−
+
=0
𝑑𝑡
𝑑𝑡
𝑑𝑡
𝑑𝑡
𝑑𝑡
𝑣𝐴 − 𝑣𝐶 + 𝑣𝐵 − 𝑣𝐶 + 𝑣𝐵 = 0
𝒗𝑨 − 𝟐𝒗𝑪 + 𝟐𝒗𝑩 = 𝟎
→ 𝑒𝑞. 1
𝑠𝐶 + 𝑠𝐵 = 𝑙2
𝑑𝑠𝐶 𝑑𝑠𝐵
+
=0
𝑑𝑡
𝑑𝑡
𝒗𝑪 = −𝒗𝑩
→ 𝑒𝑞. 2
Subst. 2 to 1:
𝑣𝐴 − 2𝑣𝐶 + 2𝑣𝐵 = 0
𝑣𝐴 − 2(−𝑣𝐵 ) + 2𝑣𝐵 = 0
𝑣𝐴 + 2𝑣𝐵 + 2𝑣𝐵 = 0
If 𝑣𝐴 = 2 m/s
𝒗𝑨 = −𝟒𝒗𝑩
(2) = −4𝑣𝐵
𝒗𝑩 = -0.5 m/s
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Self-Help: You can also refer to the sources below to help you further
understand the lesson
* Meriam, J.L. (2016). Engineering Mechanics. 8th Ed., Hoboken, NJ Wiley
*Kuisalaas, J and Pytel, A (2017). Engineering Mechanics: Dynamics. 3rd Ed., Cengage Learning
*Beer, F (2013). Vector Mechanics for Engineers: Dynamics. 10th Ed., Boston: McGraw Hill
Higher Education.
Let’s Check!
Activity
Now that you know the basic terms and concepts in the study of dynamics of rigid
bodies, let’s try to check your understanding of these terms. Write the term/s on the space
provided for the statement in each item.
1. Acceleration is the rate of change of the __________ of an object with respect to time.
2. Rectilinear Motion is a linear motion in which the direction of the velocity remains
_______ and the path is a ___________ line.
3. _____________ is a study of the geometry of the motion.
4. Kinetics is a study of the forces that ________ the motion.
5. _________________is the motion of an object moving in a curved path.
6. Normal Component is the component _____________ to the curve.
7. _____________ Component is the component tangent to the curve.
8. Radius of Curvature is the ______________ of the circular arc.
9. ____________________ is the three-dimensional coordinate system that specifies point
positions by the distance from a chosen reference axis, the direction from the axis
relative to a chosen reference direction, and the distance from a chosen reference
plane perpendicular to the axis.
10. Time Derivative is the derivative with respect to _________.
11. ________________ are diagrams used to show the relative magnitude and direction of all
forces acting upon an object in a given situation.
12. Speed refers to the magnitude of _______________.
13. Average speed is the ___________________ divided by the total time.
14. ___________a wheel with a grooved rim around which a cord pass. It acts to change the
direction of a force applied to the cord and is chiefly used (typically in combination) to
raise heavy weights.
15. In Absolute Dependent Motion Analysis of Two Particles, time differentiation of the
velocities yields ________relation between the accelerations.
Let’s Analyze!
Activity
Answer the following problems:
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1. The automobile has a speed of 80 ft/s at point A and an acceleration a having a magnitude
of 10 ft/𝑠 2 , acting in the direction shown. Determine the radius of curvature of the path at
point A and the tangential component of acceleration.
2. The train passes point A with a speed of 30 m/s and begins to decrease its speed at
𝑚
a constant rate of 𝑎𝑡 = −0.25 𝑠2 . Determine the magnitude of the acceleration of the
train when it reaches point B, where 𝑠𝐴𝐵 = 412 𝑚.
3. The car passes point A with a speed of 25 m/s after which its speed is defined by
𝑚
𝑣 = (25 − 0.15𝑠) 𝑠 . Determine the magnitude of the car’s acceleration when it
reaches point B, where s = 51.5 m.
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4. A particle moves along a circular path of radius 300 mm. If its angular velocity is 𝜃̇ =
(2𝑡 2 ) 𝑟𝑎𝑑/𝑠, where t is in seconds, determine the magnitude of the particle’s
acceleration when t = 2 s.
5. A particle moves along a circular path of radius 5 ft. If its position is 𝜃 =
(𝑒 0.5𝑡 ) 𝑟𝑎𝑑/𝑠, where t is in seconds, determine the magnitude of the particle’s
acceleration when 𝜃 = 90°.
6. The position of a particle is described by 𝑟 = (300𝑒 −0.5𝑡 ) 𝑚𝑚 and 𝜃 = (0.3𝑡 2 ) 𝑟𝑎𝑑
,where t is in seconds. Determine the magnitudes of the particle’s velocity and
acceleration at the instant 𝑡 = 1.5 𝑠.
7. The box slides down the helical ramp which is defined by 𝑟 = 0.5 𝑚 , and 𝜃 =
(0.5𝑡 3 ) 𝑟𝑎𝑑, and 𝑧 = (2 − 0.2𝑡 2 ) 𝑚 where t is in seconds. Determine the
magnitudes of the velocity and acceleration of the box at the instant 𝜃 = 2𝜋 𝑟𝑎𝑑.
In a Nutshell
Curvilinear motion along the path can be resolved into rectilinear motion along the x, y, z axes.
The equation of the path is used to relate the motion along each axis.
Projectile Motion
Free-flight motion of a projectile follows a parabolic path. It has a constant velocity in the
horizontal direction, and a constant downward acceleration of 𝒈 = 𝟗. 𝟖𝟏 𝒎/𝒔𝟐 𝒐𝒓 𝟑𝟐. 𝟐 𝒇𝒕/
𝒔𝟐 in the vertical direction. Any two of the three equations for constant acceleration apply in
the vertical direction, and in the horizontal direction only one equation applies.
Curvilinear Motion 𝒏 , 𝒕
If normal and tangential axes are used for the analysis, then 𝑣 is always in the positive t
direction. The acceleration has two components. The tangential component, 𝑎𝑡 , accounts for
the change in the magnitude of the velocity; a slowing down is in the negative t direction, and
a speeding up is in the positive t direction. The normal component 𝑎𝑛 accounts for the change
in the direction of the velocity. This component is always in the positive n direction.
Curvilinear Motion 𝑟, 𝜃
If the path of motion is expressed in polar coordinates, then the velocity and acceleration
components can be related to the time derivatives of 𝑟, 𝜃.
To apply the time-derivative equations, it is necessary to determine 𝑟, 𝑟̇ , 𝑟̈ , 𝜃, ̇ 𝜃̈ the instant
considered. If the path 𝑟 = 𝑓(𝜃)is given, then the chain rule of calculus must be used to obtain
time derivatives.
Once the data are substituted into the equations, then the algebraic sign of the results will
indicate the direction of the components of 𝑣 or 𝑎 along each axis.
Absolute Dependent Motion of Two Particles
The dependent motion of blocks that are suspended from pulleys and cables can be related by
the geometry of the system. This is done by first establishing position coordinates, measured
from a fixed origin to each block. Each coordinate must be directed along the line of motion of
a block.
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Using geometry and/or trigonometry, the coordinates are then related to the cable length in
order to formulate a position coordinate equation.
The first time derivative of this equation gives a relationship between the velocities of the
blocks, and a second time derivative gives the relation between their accelerations.
Activity
Answer the following problems:
1. If the car passes point A with a speed of 20 m/s and begins to increase its speed at a
𝑚
constant rate of 𝑎𝑡 = 0.5 𝑠2 , determine the magnitude of the car’s acceleration when
𝑠 = 100 𝑚.
2. The truck travels along a circular road that has a radius of 50 m at a speed of 4 m/s.
For a short distance when 𝑡 = 0, its speed is then increased by 𝑎𝑡 = 0.4𝑡 𝑚/𝑠 2 ,
where t is in seconds. Determine the speed and the magnitude of the truck’s
acceleration when t = 4 s.
3. A toboggan is traveling down along a curve which can be approximated by the
parabola 𝑦 = 0.01𝑥 2 . Determine the magnitude of its acceleration when it reaches
point A, where its speed is 𝑣𝐴 = 10 𝑚/𝑠, and it is increasing at the rate of (𝑎𝑡 )𝐴 =
𝑚
3 𝑠2 .
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4. A particle moves along an Archimedean spiral 𝑟 = (8𝜃) ft, where 𝜃 is given in
𝑟𝑎𝑑
radians. If 𝜃̇ = 4 𝑠 (constant), determine the radial and transverse components of
the particle’s velocity and acceleration at the instant 𝜃 = 𝜋/2 rad.
Course Schedules
Activity
Big Picture 2a: Let’s Check Activities!
Big Picture 2a: Let’s Analyze Activities!
Big Picture 2a: In a Nutshell Activities!
Big Picture 2a: Q and A List!
Date
Where to Submit
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Big Picture
Big Picture in Focus: ULO-2b. Explain the basic Kinetics: Force- Mass-Acceleration
Method
Metalanguage
The most essential terms below are defined for you to have a better understanding of
this section in the course:
1. Kinetics is a branch of dynamics that is concerned with the relationship between the
motion of bodies and its causes, namely "forces" and "torques".
2. Friction is a force between two surfaces that are sliding, or trying to slide, across each
other.
3. The normal force is the force that surfaces exert to prevent solid objects from passing
through each other.
Essential Knowledge
Newtons’ Second Law of Motion
Kinetics is a branch of dynamics that deals with the relationship between the change in
motion of a body and the forces that cause this change. The basis of for kinetics is Newton's
second law, which states that when an unbalanced force acts on a particle, the particle will
accelerate in the direction of the force with a magnitude that is proportional to the force.
𝑭 = 𝒎𝒂
where:
mass:
acceleration:
Force:
SI System
kg
m/𝑠 2
N
English System
slugs
ft/𝑠 2
lb.
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In the case of a particle located at or near the surface of the earth, the only gravitational
force having any sizable magnitude is that between the earth and the particle. This force is termed
the "weight" and, for our purpose, it will be the only gravitational force considered.
𝑾 = 𝒎𝒈
g = 9.81 m/𝒔𝟐 or
g = 32.2 ft/𝒔𝟐
A. Equations of Motion: Rectangular Coordinates
𝒛
∑ 𝐹𝑥 = 𝑚𝑎𝑥
∑ 𝐹𝑦 = 𝑚𝑎𝑦
𝒚
∑ 𝐹𝑧 = 𝑚𝑎𝑧
𝒙
Procedure of Analysis Steps:
The equations of motion are used to solve problems which require a relationship between the
forces acting on a particle and the accelerated motion they cause.
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a. Select the inertial coordinate system. Most often, rectangular or x, y, z coordinates are
chosen to analyze problems for which the particle has rectilinear motion.
b. Once the coordinated are established, draw the particle's free-body diagram. Drawing
this diagram is very important since it provides a graphical representation that accounts
for all the which act on the particle, and thereby makes it possible to resolve these forces
into their x, y, z components.
c. The direction and sense of the particle's acceleration a should also be established. If the
sense is unknown, for mathematical convenience assume that the sense of each
acceleration component acts in the same direction as its positive inertial coordinate axis.
d. Identify the unknowns in the problem.
Friction. If a moving particle contacts a rough surface, it may be necessary to use the frictional
equation, which relates the frictional and normal forces 𝐹𝑓 and N acting at the surface of contact
by using the coefficient of kinetic friction, 𝜇𝑘 .
𝑭𝒇 = 𝝁𝒌 𝑵
Remember: 𝑭𝒇 always acts o n the free-body diagram such that it opposes the
motion of the particle relative to the surface it contacts. If the particle is on the
verge of relative motion, then the coefficient of static friction, 𝝁𝒔 should be used.
Spring. If the particle is connected to an elastic spring having negligible mass, the spring force
𝐹𝑆 can be related to the deformation of the spring by the equation,
𝑭𝑺 = 𝒌𝒔.
Where: Here k is the spring's stiffness measured as a force per unit length,
s is the stretch or compression defined as the difference between the deformed
length l and the undeformed 𝑙0 , 𝑠 = 𝑙 − 𝑙0 .
Example 2b-1. The 50-kg crate shown in the figure rests on a horizontal surface for which the
coefficient of kinetic friction is 𝜇𝑘 = 0.3. If the crate is subjected to a 400-N towing force as shown,
determine the velocity of the crate in 3 s starting from rest.
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Solution:
a. Free Body Diagram:
𝜇𝑘 =
𝐹
𝑁𝐶
𝐹 = 𝜇𝑘 𝑁𝐶
𝐹 = 0.3𝑁𝐶
Equilibrium Equations:
∑± 𝐹𝑥 = 𝑚𝑎𝑥 ; → +
400cos30° − 0.3𝑁𝐶 = 50𝑎
∑± 𝐹𝑦 = 𝑚𝑎𝑦 ; ↑ +
Since no acceleration in y direction,
∑± 𝐹𝑦 = 0 ;
↑+
eq. 1
𝑁𝐶 − 490.5 + 400𝑠𝑖𝑛30° = 0
𝑁𝐶 = 290.5 N
eq. 2
Substituting 𝑁𝐶 = 290.5 N to equation 1:
400cos30° − 0.3𝑁𝐶 = 50𝑎
400cos30° − 0.3(290.5 ) = 50𝑎
𝑎 = 5.185
𝑚
𝑠2
Kinematics:
Notice that the acceleration is constant, since the applied force P is constant, since the
applied force P is constant. Since the initial velocity is zero, the velocity of the crate in 3 secs is:
𝑣𝑓 = 𝑣0 + 𝑎𝑐 𝑡
𝑎𝑐 = 𝑎 = 5.185
𝑚
𝑠2
𝑣𝑓 = 0 + 5.185(3) = 𝟏𝟓. 𝟓𝟓𝟓
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𝒔𝟐
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Example 2b-2. The man pushes on the 60-lb crate with a force F. The force always directed down
at 30° from the horizontal as shown, and its magnitude is increased until the crate begins to slide.
Determine the crate’s initial acceleration of the coefficient of static friction is 𝜇𝑠 = 0.6 and the
coefficient of kinetic friction is 𝜇𝑘 = 0.3.
Solution:
60 lb
F
Step 1. Free-Body Diagram:
30°
For static:
𝑭𝒇 = 𝟎. 𝟔 N
N
∑± 𝐹𝑥 = 0;
∑ 𝐹𝑦 = 0;
Eq. 1
𝐹𝑐𝑜𝑠30° − 0.6 𝑁 = 0
𝑁 − 60 − 𝐹𝑠𝑖𝑛30° = 0
±
Eq. 2
−𝐹𝑠𝑖𝑛30° + 𝑁 = 60
𝑁 = 91.80 𝑙𝑏
Step 2. Free-Body Diagram:
F =63.6
lb
30°
and
𝐹 = 63.6 𝑙𝑏
60 lb
For kinetic:
𝑭𝒇 = 𝟎. 𝟑(𝟗𝟏. 𝟖𝟎) = 𝟐𝟕. 𝟓𝟒𝒍𝒃
N
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∑ 𝐹𝑥 = 𝑚𝑎𝑥 ;
63.60𝑐𝑜𝑠30° − 27.54 = (
±
60
)𝑎
32.2
𝒂 = 𝟏𝟒. 𝟖 𝒇𝒕/𝒔𝟐
Example 2b-3. The motor winds in the cable with a constant acceleration, such that the 20-kg
crate moves a distance s = 6 m in 3 s, starting from rest. Determine the tension developed in the
cable. The coefficient of kinetic friction between the crate and the plane is 𝜇𝑘 = 0.3.
Solution:
𝒚’
Rotate the y’ – x’ axis:
𝑭𝒇
𝒙’
𝑾 = mg = 20 kg (9.81m/𝒔𝟐 )=196. 2 N
Since there is no movement in y’ axis:
∑ 𝐹𝑦′ = 0;
𝑁 − 𝑊𝑐𝑜𝑠30° = 0
±
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𝑁 = (196.2)𝑐𝑜𝑠30° = 169.91 N
∑ 𝐹𝑥′ = 𝑚𝑎𝑥′ ; → +
Eq. 1
𝑇 − 𝐹𝑓 − 𝑊𝑠𝑖𝑛30° = 𝑚𝑎𝑥′
±
𝑇 − 0.3(169.91) − (196.2)𝑠𝑖𝑛30° = 20(𝑎𝑥 ′ )
𝑇 − 20(𝑎𝑥 ′ ) = 149.073 𝑁
Eq. 2
IF 𝑠 = 6 𝑚 , 𝑡 = 3 𝑠 and 𝑣𝑜 = 0;
𝑣𝑓 = 𝑣0 + 𝑎𝑥 ′ 𝑡
𝑣𝑓 = 3𝑎𝑥 ′
2
(𝑣𝑓 ) = (𝑣0 )2 + 2𝑎𝑥 ′ 𝑠
2
(𝑣𝑓 ) = (0)2 + 2𝑎𝑥 ′ (6)
(3𝑎𝑥 ′ )2 = (0)2 + 2𝑎𝑥 ′ (6)
𝒂𝒙′ = 1.333 m/𝒔𝟐
Subst. 𝒂𝒙′ to Eq. 2
𝑇 − 20(1.333 ) = 149.073 𝑁
𝑻 = 𝟏𝟕𝟓. 𝟕𝟑𝟑 𝑵
Example 2b-4. The 10-lb block A travels to the right at 𝑣𝐴 = 2 𝑓𝑡/𝑠 at the instant shown. If the
coefficient of kinetic friction is 𝜇𝑘 = 0.2 between the surface and A, determine the velocity of A
when it has moved 4 ft. Block B has a weight of 20 lb.
Solution:
Free Body Diagram at A:
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𝑊
𝑎
𝑇
𝐹𝑓
𝑁
∑ 𝐹𝑦 = 0 ;
𝑁 = 𝑊𝐴 = 10 𝑙𝑏
↑±
Since the block at B is heavier, block A will move to the right; all forces to the right is positive:
∑ 𝐹𝑥 = 𝑚𝑎𝑥 ;
−𝑇 + 𝐹𝑓 = 𝑚𝐴 𝑎𝐴
←±
10
) 𝑎𝐴
32.2
−𝑇 + 0.2(10) = (
10
) 𝑎𝐴
32.2
−𝑇 − (
𝑇
Free Body Diagram at B:
Eq. 1
= −2
𝑇
20 𝑙𝑏
20
∑↓± 𝐹𝑦 = 𝑚𝑎𝑦 ;
20 − 2𝑇 = (32.2) 𝑎𝐵
Eq. 2
𝑠𝐴 + 2𝑠𝐵 = 𝑙
Deriving,
𝑣𝐴 + 2𝑣𝐵 = 0
Deriving,
𝑎𝐴 + 2𝑎𝐵 = 0
𝑎𝐴 = −2𝑎𝐵
Solving equations 1, 2 and 3 simultaneously:
10
−𝑇 − (32.2) (−2𝑎𝐵 ) = −2
−𝑇 + 0.62112𝑎𝐵 = −2
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20
−2𝑇 − (32.2) 𝑎𝐵 = −20
𝑇 = 7.333 𝑙𝑏 ; 𝑎𝐵 = 8.587 𝑚/𝑠 2
𝑎𝐴 = −2(8.587) = (−)17. 174 𝑚/𝑠 2
2
(𝑣𝑓 ) = (𝑣𝑂 )2 + 2𝑎𝐴 𝑥
2
(𝑣𝑓 ) = (2)2 + 2(17. 174)(4)
𝒗𝒇 = 11.89 ft/s
Course Schedules
Activity
Big Picture 2b: Let’s Check Activities!
Big Picture 2b: Let’s Analyze Activities!
Big Picture 2b: In a Nutshell Activities!
Big Picture 2b: Q and A List!
2nd Exam
Date
Sept. 18, 2020
Where to Submit
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Big Picture
Week 6-7: Unit Learning Outcomes-Unit 3 (ULO-3): At the end of the unit, you are expected to
a. Demonstrate knowledge in Work-Energy and Impulse Momentum Principles for a
Particle.
b. Explain the Dynamics of Particle Systems.
c. Explain the Planar Kinematics of Rigid Bodies.
Big Picture in Focus: ULO-3a. Demonstrate knowledge in Work-Energy and
Impulse Momentum Principles for a Particle
Metalanguage
The most essential terms below are defined for you to have a better understanding of
this section in the course:
1. The principle of work and kinetic energy (also known as the work-energy theorem)
states that the work done by the sum of all forces acting on a particle equals the change in
the kinetic energy of the particle.
2. The impulse-momentum theorem states that the impulse is equal to this change in
momentum.
3. Power provides a useful basis for choosing the type of motor or machine which is required
to do a certain amount of work in a given time.
4. Mechanical efficiency is a dimensionless number that measures the effectiveness of a
machine in transforming the power input to the device to power output.
Essential Knowledge
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In the force-mass-acceleration method, the equations of particle motion were obtained
directly from Newton’s second law, F=ma. Solution of these equations required two
integrations, the first to obtain the velocity, and the second to obtain the position. Work-energy
and impulse-momentum methods employ integral forms of the equations of motion. If we
integrate both sides of F=ma with respect to position, we obtain the equations used in the workenergy method. Integrating F=ma with respect to time yields the equations of the impulsemomentum method. The integral forms of the equations of motion can be very efficient in the
solution of certain types of problems. Such that,
• The work-energy method is useful in computing the change in speed during a
displacement of the particle.
• The impulse-momentum method is best suited for determining the change in velocity that
occurs over a time interval.
The Work (U) of a Force
𝒅𝑼 = 𝑭 𝒅𝒔 𝒄𝒐𝒔𝜽
𝟏 𝐉𝐨𝐮𝐥𝐞 (𝑱) = 𝟏 𝐍 − 𝐦
Thus, if computing the Work of a Variable Force done from point 1 to point 2,
𝑠2
𝑈1−2 = ∫ 𝐹𝑐𝑜𝑠𝜃 𝑑𝑠
𝑠1
provided that F can be expressed in terms of s, or position.
For computing the Work of a Constant Force done from point 1 to point 2,
𝒔𝟐
𝑼𝟏−𝟐 = 𝑭𝒄 𝒄𝒐𝒔𝜽 ∫ 𝒅𝒔
𝒔𝟏
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or
𝑼𝟏−𝟐 = 𝑭𝒄 𝒄𝒐𝒔𝜽 (𝒔𝟐 − 𝒔𝟏 )
Work of a Weight
𝒚𝟐
𝑼𝟏−𝟐 = ∫ −𝑾𝒅𝒚
𝒚𝟏
Or
𝑼𝟏−𝟐 = −𝑾∆𝒚
Work of a Spring Force
𝒔𝟐
𝒔𝟐
𝑼𝟏−𝟐 = ∫ 𝑭𝒔 𝒅𝒔 = ∫ −𝒌𝒔𝒅𝒔
Or
𝒔𝟏
𝒔𝟏
𝟏
𝟏
𝑼𝟏−𝟐 = − ( 𝒌𝒔𝟐 𝟐 − 𝒌𝒔𝟏 𝟐 )
𝟐
𝟐
Example 3a-1. Apply the work equation to determine the amount of work done by the applied
force in each of the three situations described below:
Situation 1:
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A 100 N force is applied to move a 15
kg object a horizontal distance of 5 meters at
a constant speed.
𝑾 = 𝑭𝒄𝒐𝒔𝜽𝒔 = 100𝑐𝑜𝑠(0)(5 𝑚) = 500 𝑁 − 𝑚 = 𝟓𝟎𝟎 𝑱
Situation 2:
A 100 N force is applied at angle of
30° to the horizontal to move a 15 kg object
at a constant speed for a horizontal distance
of 5 m.
𝑾 = 𝑭𝒄𝒐𝒔𝜽𝒔 = 100𝑐𝑜𝑠(30°)(5 𝑚) = 433.013 𝑁 − 𝑚 = 𝟒𝟑𝟑. 𝟎𝟏𝟑 𝑱
Situation 3:
An upward force is applied to lift a 15
kg object to a height of 5 meters at constant
speed.
Solution:
𝑊 = 15(9.81) =147.15 N
∑ 𝐹𝑦 = 0 ; 𝑁 = 𝑊
N
𝑾 = 𝑭𝒄𝒐𝒔𝜽𝒔 = 𝑁𝑐𝑜𝑠(0°)(5 𝑚) = 735.75 𝑁 − 𝑚 = 𝟒𝟑𝟑. 𝟎𝟏𝟑 𝑱
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Example 3a-2. The 10-kg block shown in the figure rests on the smooth incline. If the spring is
originally stretched at 0.5 m, determine the total work done by all the forces acting on the block
when a horizontal force P = 400 N pushes the block up the plane s = 2 m.
Solution:
•
Work done by force P:
Since this force is constant, the work U is,
𝑈𝑝 = 400 𝑁 (2 𝑚 𝑐𝑜𝑠30°) = 692.8 𝐽
•
Work done by the spring:
1
𝑁
𝑈𝑠 = − [2 (30 𝑚) (2.5 𝑚)2 −
•
1
𝑁
(30 𝑚) (0.5 𝑚)2 ]
2
= −90 𝐽
Work done by the weight:
𝑈𝑤 = − (98.1 𝑁)(2 𝑚 sin 30°) = −98. 1 𝐽
•
Work done by the normal force acting on the block:
not the direction of the motion.
•
Total Work. The work of all the forces when the block is displaced 2 m is therefore:
No work since its direction is
𝑈𝑇 = 692. 8 𝐽 − 90𝐽 − 98.1 𝐽 = 𝟓𝟎𝟓 𝑱
Principle of Work and Energy
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𝑠
∑ ∫𝑠 2 𝐹𝑑𝑠
1
and
𝐹 = 𝑚𝑎
𝑠2
∑ ∫ (𝑚𝑎)𝑑𝑠
But 𝑎 =
𝑠1
𝑑𝑣
𝑣 𝑑𝑠
𝑠2
𝑑𝑣
∑ ∫ (𝑚 (𝑣 )) 𝑑𝑠
𝑑𝑠
𝑠1
So,
𝑣2
∑ ∫ 𝑚𝑑𝑣
𝑣1
𝑠2
𝑣2
∑ ∫ 𝐹𝑑𝑠 = ∑ ∫
𝑠1
𝑚𝑑𝑣
𝑣1
Thus,
∑ 𝐹 (𝑠2 − 𝑠1 ) =
1
1
𝑚(𝑣2 )2 − 𝑚(𝑣1 )2
2
2
Procedure of Analysis:
1. Establish a coordinate system and draw free body diagram to account for all the forces
acting on the body.
2. A force does work when it moves through a displacement in the direction of the force.
3. Work is positive when the force component in the same sense of direction as its
displacement, otherwise it is negative.
4. The work of a weight is the product of the weight magnitude and vertical displacement. It
is positive when the weight moves downwards
Example 3a-3. The 50-kg crate shown in the figure rests on a horizontal surface for which the
coefficient of kinetic friction is 𝜇𝑘 = 0.3. If the crate is subjected to a 400-N towing force as shown,
determine the velocity of the crate in 3 s starting from rest.
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𝑊 = 𝑚𝑔 = 50(9.81) = 490.5 𝑁
Solution:
Free Body Diagram:
𝜇𝑘 =
𝐹𝑓
𝑁𝐶
𝐹𝑓 = 𝜇𝑘 𝑁𝐶
𝐹𝑓 = 0.3𝑁𝐶
∑ 𝐹𝑥 (𝑠2 − 𝑠1 ) =
Equilibrium Equations:
∑ 𝐹𝑦 = 0
; ↑ +
1
1
𝑚(𝑣2 )2 − 𝑚(𝑣1 )2
2
2
𝑁𝐶 + 400𝑠𝑖𝑛30° − 490.5 = 0
𝑁𝐶 = 290. 5 𝑁
Thus,
𝐹𝑓 = 𝜇𝑘 𝑁𝐶 = 0.3(290. 5 ) = 87.15 𝑁
∑ 𝐹𝑥 (𝑠2 − 𝑠1 ) =
1
1
𝑚(𝑣2 )2 − 𝑚(𝑣1 )2
2
2
400𝑐𝑜𝑠30°(𝑠) − 87.15(𝑠) =
2
1
(50)(𝑣𝑓 )
2
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− 2 (50)(𝑣0 )2
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(400𝑐𝑜𝑠30° − 87.15)𝑠 =
2
2
1
(50)(𝑣𝑓 )
2
25(𝑣𝑓 ) = (400𝑐𝑜𝑠30° − 87.15)𝑠
Eq. 1
2
(𝑣𝑓 ) = (𝑣0 )2 + 2𝑎𝑠
2
(𝑣𝑓 ) = 2𝑎𝑠
1
𝑠 = 𝑣0 𝑡 + 𝑎𝑡 2
2
1
𝑠 = 𝑎(3)2 = 4.5𝑎
2
𝑎=
𝑠
4.5
Eq. 2
𝑣𝑓 = 𝑣0 + 𝑎𝑡
𝑠
4.5
𝑣𝑓 = 𝑎(3) = ( ) (3) =
𝑠=
4.5
𝑣
3 𝑓
3
𝑠
4.5
Eq. 3
2
4.5
25(𝑣𝑓 ) = (400𝑐𝑜𝑠30° − 87.15) ( 3 𝑣𝑓 )
𝒗𝒇 = 𝟏𝟓. 𝟓𝟓𝟔 𝒎/𝒔
Example 3a-4. The motor winds in the cable with a constant acceleration, such that the 20-kg
crate moves a distance 𝑠 = 6 𝑚 in 3 s, starting from rest. Determine the tension developed in the
cable. The coefficient of kinetic friction between the crate and the plane is 𝜇𝑘 = 0.3.
Solution:
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𝒚’
Rotate the y’ – x’ axis:
𝑭𝒇
𝒙’
𝑾 = mg = 20 kg (9.81m/𝒔𝟐 )=196. 2 N
Equilibrium Equation:
∑ 𝐹𝑦 ′ = 0 ; ↑ + ;
𝑁 − 𝑊𝑐𝑜𝑠30° = 0
𝑁 = 196.2𝑐𝑜𝑠30° = 169.914 𝑁
𝐹𝑓 = 0.3(169.914) = 50.9742
∑ 𝐹𝑥′ (𝑠2 − 𝑠1 ) =
1
1
𝑚(𝑣2 )2 − 𝑚(𝑣1 )2
2
2
1
1
2
(𝑇 − 𝐹𝑓 − 𝑊𝑠𝑖𝑛30°)𝑠 = 𝑚(𝑣𝑓 ) − 𝑚(𝑣0 )2
2
2
1
1
2
(𝑇 − 50.9742 − 196.2𝑠𝑖𝑛30°)(6) = (20)(𝑣𝑓 ) − (20)(𝑣0 )2
2
2
2
1
(𝑇 − 50.9742 − 196.2𝑠𝑖𝑛30°)(6) = (20)(𝑣𝑓 )
Eq. 1
2
𝑣𝑓 = 𝑣0 + 𝑎𝑡
𝑣𝑓 = 3𝑎
𝑎=
𝑣𝑓 2
𝑣𝑓
3
Eq. 2
𝑣𝑓 2 = 𝑣0 2 + 2𝑎𝑠
= 2𝑎(6) = 12𝑎
𝑣
𝑣𝑓 2 = 12 ( 3𝑓) Eq. 3
𝑣𝑓 = 4 𝑚/𝑠
1
2
(𝑇 − 50.9742 − 196.2𝑠𝑖𝑛30°)(6) = (20)(4)2
𝑻 = 𝟏𝟕𝟓. 𝟕𝟒 𝑵
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Example 3a-5. For a short time, the crane in the figure lifts the 2.50-Mg beam with a force of
𝐹 = (28 + 3𝑠 2 ) kN. Determine the speed of the beam when it has risen 𝑠 = 3 𝑚. Also, how
much time does it take to attain this height starting from rest?
1
1
𝑚(𝑣2 )2 − 𝑚(𝑣1 )2
2
2
∑ 𝐹𝑦 (𝑠2 − 𝑠1 ) =
Calculate:
1 𝑥 106 𝑔
1 𝑘𝑔
) (1000 𝑔)
1 𝑀𝑔
6
𝑚 = 2.5 𝑀𝑔 (
𝑊 = 2.5 𝑀𝑔 (
= 2500 𝑘𝑔
1 𝑥 10 𝑔
1 𝑘𝑔
9.81 𝑁
)(
) = 24, 525 𝑁
)(
1 𝑀𝑔
1000 𝑔
1 𝑘𝑔
1
1
𝑚(𝑣2 )2 − 𝑚(𝑣1 )2
2
2
𝐹𝑠 − 𝑊𝑠 =
𝑠
1
1
2
∫ (28 + 3𝑠 2 )(103 ) − 24, 525𝑠 = 𝑚(𝑣𝑓 ) − 𝑚(𝑣0 )2
2
2
0
1
2
(28𝑠 + 𝑠 3 )(103 ) − 24, 525𝑠 = (2500)(𝑣𝑓 )
2
2
(28𝑠 + 𝑠 3 )(103 ) − 24, 525𝑠 = (1250)(𝑣𝑓 )
1
1
[28(103 )𝑠 + (103 )𝑠 3 − 24,525 𝑠] 2
𝑣𝑓 = (
) = (2.78𝑠 + 0.8𝑠 3 )2
1250
At 𝑠 = 3 𝑚,
𝒗𝒇 = 𝟓. 𝟒𝟕 𝒎/𝒔
𝑣𝑓 =
𝑑𝑠
𝑑𝑡
1
(2.78𝑠 + 0.8𝑠 3 )2 =
𝑑𝑡 = (
𝑑𝑠
)
(2.78𝑠 + 0.8𝑠 3 )1/2
∫ 𝑑𝑡 = ∫ (
3
𝑡= ∫ (
0
𝑑𝑠
𝑑𝑡
𝑑𝑠
)
(2.78𝑠 + 0.8𝑠 3 )1/2
𝑑𝑠
)
(2.78𝑠 + 0.8𝑠 3 )1/2
𝒕 = 𝟏. 𝟕𝟗 𝒔
Example 3a-6. The 3500-lb automobile shown in the figure travels down the 10° inclined road
at a speed of 20 ft/s. If the driver jams on the brakes, causing his wheels to lock, determine how
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far 𝑠 the tires skid on the road. The coefficient of kinetic friction between the wheels and the
road is 𝜇𝑘 = 0.5.
𝒚′
𝑵
𝒙′
10°
Solution:
Equilibrium of Equation:
∑ 𝐹𝑦 = 0;
𝑁 − 3500𝑐𝑜𝑠10° = 0
𝑵 = 𝟑𝟒𝟒𝟔. 𝟖𝟐𝟕 𝒍𝒃
𝐹𝑓 = 𝜇𝑁 = 0.5(3446.827 ) = 1723. 41 𝑙𝑏
∑ 𝐹𝑥′ (𝑠2 − 𝑠1 ) =
1
1
𝑚(𝑣2 )2 − 𝑚(𝑣1 )2
2
2
3500𝑠𝑖𝑛10°(𝑠) − 1723. 41𝑠 =
1
1
2
𝑚(𝑣𝑓 ) − 𝑚(𝑣0 )2
2
2
1
1 3500
2
𝑚(𝑣𝑓 ) − (
) (20)2
2
2 32.2
3500𝑠𝑖𝑛10°(𝑠) − 1723. 41𝑠 =
𝒔 = 𝟏𝟗. 𝟒𝟖𝟔 𝒎
Power and Efficiency
Power. The term "power" provides a useful basis for choosing the type of motor or machine
which is required to do a certain amount of work in a given time. For example, two pumps may
each be able to empty a reservoir if given enough time; however, the pump having the larger
power will complete the job sooner. The power generated by a machine or engine that performs
an amount of work 𝑑𝑈 within the time interval 𝑑𝑡 is therefore,
𝑃=
𝑑𝑈
𝑑𝑡
If the work 𝑑𝑈 is expressed as 𝑑𝑈 = 𝐹 ∙ 𝑑𝑟, then:
𝑃=
𝑑𝑈
𝐹 ∙ 𝑑𝑟
𝑑𝑟
=
=𝐹 ∙
𝑑𝑡
𝑑𝑡
𝑑𝑡
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or
𝑷 = 𝑭𝒗
1𝑊 =1𝑁−
𝑚
𝑠
𝑙𝑏
1 ℎ𝑝 = 550 𝑓𝑡 − 𝑠
1 ℎ𝑝 = 746 𝑊
Efficiency. The mechanical efficiency of a machine is defined as the ratio of the output of useful
power produced by the machine to the input of power supplied to the machine. Hence,
Efficiency, ∈ =
𝑝𝑜𝑤𝑒𝑟 𝑜𝑢𝑡𝑝𝑢𝑡
𝑝𝑜𝑤𝑒𝑟 𝑖𝑛𝑝𝑢𝑡
or
∈=
𝑒𝑛𝑒𝑟𝑔𝑦 𝑜𝑢𝑡𝑝𝑢𝑡
𝑒𝑛𝑒𝑟𝑔𝑦 𝑖𝑛𝑝𝑢𝑡
Since machines consist of a series of moving parts, frictional forces will always be
developed within the machine, and as a result, extra energy or power is needed to overcome
these forces. Consequently, power output will be less than power input and so the efficiency of a
machine is always less than 1.
Procedure of Analysis
•
•
•
•
First determine the external force F acting on the body which causes the motion. This
force is usually developed by a machine or engine placed either within or external to
the body.
If the body is accelerating, it may be necessary to draw its free-body diagram and apply
the equation of motion ∑ 𝐹 = 𝑚𝑎 to determine F.
Once F and the velocity 𝑣 of the particle where F is applied have been found, the power
is determined by multiplying the force magnitude with the component of velocity
acting in the direction of 𝐹, (𝑃 = 𝐹 · 𝑣 = 𝐹𝑣 𝑐𝑜𝑠𝜃).
In some problems the power may be found by calculating the work done by 𝐹 per unit
of time (𝑃 𝑎𝑣𝑔 = ∆𝑈 / ∆𝑡).
Example 3a-7. The man in the figure pushes on the 50 kg crate with a force of 𝐹 = 150 𝑁.
Determine the power supplied by the man 𝑡 = 4 𝑠. The coefficient of kinetic friction between
the floor and the crate is 𝜇𝑘 = 0.2. Initially the crate is at rest.
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Equilibrium of Equation:
3
∑↑+ 𝐹𝑦 = 𝑚𝑎𝑦 ;
𝑁 − (5) (150) − 50(9.81) = 0
𝑁 = 580. 5
∑→+ 𝐹𝑥 = 𝑚𝑎𝑥 ;
(5) (150) − 0.2(580.5) = 50𝑎
𝑎 = 0.078 𝑚/𝑠 2
4
The velocity of the crate when 𝑡 = 4 𝑠 is therefore,
𝑣𝑓 = 𝑣0 + 𝑎𝑡
𝑣𝑓 = 0 + (0.078)(4) = 0.312 𝑚/𝑠
The power supplied to the crate by the man when 𝑡 = 4 𝑠 is therefore,
𝑃 = 𝐹𝑣
4
𝑃 = (5) (150)𝑁(0.312 𝑚/𝑠) = 𝟑𝟕. 𝟒𝟒 𝑾
Example 3a-8. The 50-lb block rests on the rough surface for which the coefficient of kinetic
friction is 𝜇𝑘 = 0.2 . A force 𝐹 = (40 + 𝑠 2 ) 𝑙𝑏, where s is in ft, acts on the block in the direction
shown. If the spring is originally unstretched (𝑠 = 0)and the block is at rest, determine the
power developed by the force the instant the block has moved 𝑠 = 1.5 𝑓𝑡.
Solution:
Free Body Diagram:
𝑾 = 𝟓𝟎 𝒍𝒃
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𝑭
30°
𝑭𝒔
𝑭𝒇
𝑵
Equilibrium Equation:
∑ 𝐹𝑦 = 0;
𝑁 − 50 − 𝐹𝑐𝑜𝑠30° = 0
𝑁 = 50 + (40 + 𝑠 2 )𝑠𝑖𝑛30° = (70 + 0.5𝑠 2 )
∑ 𝐹𝑥 𝑠 =
1
1
1
𝑚𝑣𝑓 2 − 𝑚𝑣0 2
2
2
1
𝐹𝑠 = 2 𝑘𝑠 2 = 2 (20)𝑠 2 = 10𝑠 2
𝐹𝑓 = 𝜇𝑘 𝑁 = 0.2(70 + 0.5𝑠 2 )
𝑠
1
1
𝑚𝑣𝑓 2 − 𝑚𝑣0 2
2
2
0
1.5
1
50
1
∫ (40 + 𝑠 2 ) 𝑐𝑜𝑠30°𝑑𝑠 − 10𝑠 2 − 0.2(70 + 0.5𝑠 2 ) = (
) 𝑣𝑓 2 − 𝑚𝑣0 2
2 32.2
2
0
∫ 𝐹 𝑐𝑜𝑠30°𝑑𝑠 − 𝐹𝑠 − 𝐹𝑓 =
𝐴𝑡 𝑠 = 1.5 𝑓𝑡,
1.5
1.5
∫ (40 + 𝑠 2 ) 𝑐𝑜𝑠30°𝑑𝑠 − 10(1.5)2 − 0.2 ∫ (70 + 0.5𝑠 2 ) 𝑑𝑠 =
0
0
1
1 50
(
)𝑣 2
2 32.2 𝑓
50
52.936 − 22.5 − 0.2(105.5625) = 2 (32.2) 𝑣𝑓 2
𝒗𝒇 = 3.465 ft/s
𝐴𝑡 𝑠 = 1.5 𝑓𝑡,
𝐹 = (40 + 𝑠 2 ) = (40 + 1.52 ) = 42.25
𝑃 = 𝐹𝑣 = 𝐹𝑐𝑜𝑠30°𝑣 = 42.25𝑐𝑜𝑠30°(3.465) = 126.78 𝑓𝑡 ∙
𝑙𝑏
1 ℎ𝑝
(
)
𝑠 550 𝑓𝑡 ∙ 𝑙𝑏
𝑠
𝑷 = 𝟎. 𝟐𝟑𝟏 𝒉𝒑
Example 3a-9. The crate has a mass of 150 kg and rests on a surface for which the coefficients
of static and kinetic friction are 𝜇𝑠 = 0.3 and 𝜇𝑘 = 0.3, respectively. If the motor M supplies a
cable force of 𝐹 = (8𝑡 2 + 20) N, where (𝑡 𝑖𝑠 𝑖𝑛 𝑠𝑒𝑐𝑜𝑛𝑑𝑠), determine the power output
developed by the motor when 𝑡 = 5 𝑠.
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Solution:
150(9.81) = 1471. 5
𝑇=𝐹
𝐹𝑓 = 0.3𝑁
Equations of Equilibrium: If the crate is on the verge of slipping, 𝐹𝑓 = 𝜇𝑠 𝑁 = 0.3𝑁
∑↑+ 𝐹𝑦 = 0;
𝑁 − 150(9.81) = 0
𝑁 = 1471. 5 N
∑→+ 𝐹𝑥 = 0;
3𝐹 − 𝐹𝑓 = 0
3(8𝑡 2 + 20) − 0.3(1471. 5) = 0
𝑡 = 3. 987 𝑠
Since the crate moves at 𝑡 = 3.987 s later, 𝐹𝑓 = 𝜇𝑘 𝑁 = 0.2𝑁
∑↑+ 𝐹𝑦 = 𝑚𝑎𝑦 ;
𝑁 − 150(9.81) = 0
𝑁 = 1471. 5 N
∑→+ 𝐹𝑥 = 𝑚𝑎𝑥 ;
𝐹𝑓 − 3𝐹 = 150(−𝑎)
0.2(1471. 5) − 3(8𝑡 2 + 20) = 150(−𝑎)
𝑎 = (0.160𝑡 2 − 1.562) m/𝑠 2
Kinematics: Applying 𝑑𝑣 = 𝑎𝑑𝑡, we have:
𝑣
5
∫0 𝑑𝑣 = ∫3.987(0.160𝑡 2 − 1.562)
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𝑣 = 1.7045 𝑚/𝑠
For power:
At 𝑡 = 5 𝑠,
𝐹 = (8𝑡 2 + 20) = (8(5)2 + 20) = 220 𝑁
𝑃 = 𝐹𝑣 = 3(220)(1.7045) = 𝟏𝟏𝟐𝟒. 𝟗𝟕 𝑾
Example 3a-10. The 50-kg crate is hoisted up the 30° incline by the pulley system and motor M.
If the crate starts from rest and, by constant acceleration, attains a speed of 4 m/s after traveling
8 m along the plane, determine the power that must be supplied to the motor at the instant the
crate has moved 8 m. Neglect friction along the plane. The motor has an efficiency of 𝜖 = 0.74.
Solution:
Kinematics:
𝑣𝑓 2 = 𝑣0 2 + 2𝑎𝑠
42 = 0 + 2𝑎(8)
𝑎 = 1 𝑚/𝑠 2
Equation of Motion:
∑ 𝐹𝑥 ′ = 𝑚𝑎𝑥′ ;
𝐹 − 50(9.81)𝑠𝑖𝑛30° = 50(1)
±
𝐹 = 295.25 𝑁
Power:
The power output at the instant when 𝑣 = 4 𝑚/𝑠 can be obtained:
𝑃 = 𝐹𝑣 = 295.25(4) = 1181 𝑊 = 1.181 𝑘𝑊
Power can be computed by:
𝑃𝑖𝑛𝑝𝑢𝑡 =
𝑃𝑜𝑢𝑡𝑝𝑢𝑡
𝜖
=
1.181
0.74
= 𝟏. 𝟔𝟎 𝒌𝑾
Conservative Forces and the Conservation of Mechanical Energy
A force is said to be conservative if its work depends only on the initial and final
positions of its point of application. All the specific forces discussed in the previous article are
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conservative, because in each case we could determine the work without having to specify the
path between the end points.
It is often convenient to describe the effects of conservative forces in terms of their
potential energies. Roughly speaking, potential energy is the capacity of the total energy (the
sum of all forms of energy) remains constant for a closed system. The form of the energy may
change—for example, electrical energy may be converted to mechanical energy—but the total
energy can neither be created nor destroyed.
In mechanics, we restrict our attention to mechanical energy, defined to be the sum of the
potential and kinetic energies. If all forces acting on a particle, body, or closed system of bodies
are conservative, mechanical energy is conserved, a concept known as the principle of
conservation of mechanical energy.
Conservation of mechanical energy
(𝐾. 𝐸 + 𝑃. 𝐸)𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = (𝐾. 𝐸 + 𝑃. 𝐸)𝑓𝑖𝑛𝑎𝑙
or
∑ 𝑻𝟏 + ∑ 𝑽𝟏 = ∑ 𝑻𝟐 + ∑ 𝑽𝟐
Computation of Potential Energy (𝑉)
Computation of Kinetic Energy (𝑇)
=
𝟏
𝒎𝒗𝟐
𝟐
Example 3a-11. The gantry structure is used to test the response of an airplane during a crash.
As shown in the figure, the plane, having a mass of 8 Mg, is hoisted back until 𝜃 = 60°, and then
the pull-back cable AC is released when the plane is at rest. Determine the speed of the plane
just before it crashes into the ground, 𝜃 = 15°. Also, what is the maximum tension developed in
the supporting cable during the motion? Neglect the size of the airplane and the effect of lift
caused by the wings during the motion.
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Solution:
Establish the datum. In this case, the datum has been established at the top of the gantry.
Conservation of Energy.
𝑻𝑨 + 𝑽𝑨 = 𝑻𝑩 + 𝑽𝑩
0 − 8000 𝑘𝑔 (9.81
𝑚
1
𝑚
2
(20𝑐𝑜𝑠60°)
(8000)(𝑣
)
)
=
−
8000
𝑘𝑔
(9.81
) (20𝑐𝑜𝑠15°)
𝐵
𝑠2
2
𝑠2
𝒗𝑩 = 𝟏𝟑. 𝟓𝟐 𝒎/𝒔
For the tension of the cable:
∑ 𝐹𝑛 = 𝑚𝑎𝑛 ;
𝑎𝑛 =
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𝑇 − 8000(9.81)𝑐𝑜𝑠15° 𝑁 = (8 𝑥 106 𝑔) (
(13. 52 𝑚/𝑠 2 )2
1 𝑘𝑔
)(
)
1000 𝑔
20 𝑚
𝑇 = 148, 922 ≈ 𝟏𝟒𝟗 𝒌𝑵
Example 3a-12. The 30-lb block A is placed on top of two nested springs B and C and then
pushed down to the position shown. If it is then released, determine the maximum height h to
which it will rise.
Solution:
Conservation of Energy:
𝑇1 + 𝑉1 = 𝑇2 + 𝑉2
1
1
𝑚𝑣1 + [𝑉𝑔1 + 𝑉𝑒1 ] = 𝑚𝑣2 + [𝑉𝑔2 + 𝑉𝑒2 ]
2
2
1
1
0 + 0 + (2000)(42 ) + (100)(62 ) = 0 + 30ℎ + 0
2
2
𝒉 = 𝟏𝟏𝟑 𝒊𝒏.
Principle of Linear Impulse and Momentum
From kinematics,
∑ 𝐹 = 𝑚𝑎 = 𝑚
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𝑡2
𝑣2
∑ ∫ 𝐹𝑑𝑡 = 𝑚 ∫ 𝑑𝑣
𝑡1
𝑣1
𝒕𝟐
∑ ∫ 𝑭𝒅𝒕 = 𝒎(𝒗𝟐 − 𝒗𝟏 )
𝒕𝟏
Thus:
𝒕𝟐
∑ ∫ 𝑭𝒙 𝒅𝒕 = 𝒎(𝒗𝒙𝟐 − 𝒗𝒙𝟏 )
𝒕𝟏
𝒕𝟐
∑ ∫ 𝑭𝒚 𝒅𝒕 = 𝒎(𝒗𝒚𝟐 − 𝒗𝒚𝟏 )
𝒕𝟏
𝒕𝟐
∑ ∫ 𝑭𝒛 𝒅𝒕 = 𝒎(𝒗𝒛𝟐 − 𝒗𝒛𝟏 )
𝒕𝟏
Procedure of Analysis:
The principle of linear impulse and momentum is used to solve problems involving force,
time and velocity, since these terms are involved in the formulation:
•
•
•
•
Establish the x, y or z coordinate axis and draw the particles free-body diagram in order
to account for all the forces that produce impulses on the particle.
The direction and sense of the particle initial and final velocities should be established.
If a vector is unknown, assume that the sense of its components is in the direction of the
positive inertial coordinate(s).
In accordance with the established coordinate system, apply the principle of linear
impulse and momentum,
𝒕𝟐
∑ ∫ 𝑭𝒅𝒕 = 𝒎(𝒗𝟐 − 𝒗𝟏 )
•
•
•
•
𝒕𝟏
If motion occurs in the x-y plane, the two scalar component equations can be
formulated by either resolving the vector components of F from the free-body diagram,
or by using the data on the impulse and momentum diagrams.
Realize that every force acting on the particle's free-body diagram will create an
impulse, even though some of these forces will do no work.
Forces that are functions of time must be integrated to obtain the impulse.
Graphically, the impulse is equal to the area under the force-time curve.
Example 3a-13. The 100-kg stone shown in the figure is originally at rest on the smooth
horizontal surface. If a towing force of 200 N, acting at an angle of 45°, is applied to the stone for
10 s, determine the final velocity and the normal force which the surface exerts on the stone
during this time interval.
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Solution:
Free – Body Diagram:
𝒕𝟐
∑ ∫ 𝑭𝒙 𝒅𝒕 = 𝒎(𝒗𝟐 − 𝒗𝟏 )
𝒕𝟏
∑ 𝐹𝑥𝑡 = 𝒎(𝒗𝟐 − 𝒗𝟏 ) ;
→+
200𝑐𝑜𝑠45°(10) = (100)(𝒗𝒙𝟐 − 𝟎)
𝒗𝟐 = 𝒗𝒇 = 14.1 m/s
For Normal Force,
𝒕𝟐
∑ ∫ 𝑭𝒚 𝒅𝒕 = 𝒎(𝒗𝟐 − 𝒗𝟏 )
𝒕𝟏
∑ 𝐹𝑦𝑡 = 𝒎(𝒗𝒚𝟐 − 𝒗𝟏 ) ;
↑+
−981 + 𝑁𝐶 + 200𝑠𝑖𝑛45° = 100 (𝟎 − 𝟎)
(−981 + 𝑁𝐶 + 200sin45°)(10) = 0
𝑵𝑪 = 𝟖𝟑𝟗. 𝟓𝟖 N
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Example 3a-14. The 50-kg crate shown in the figure rests on a horizontal surface for which the
coefficient of kinetic friction is 𝜇𝑘 = 0.3. If the crate is subjected to a 400-N towing force as shown,
determine the velocity of the crate in 3 s starting from rest.
Solution:
Free Body Diagram:
𝜇𝑘 =
𝐹
𝑁𝐶
𝐹 = 𝜇𝑘 𝑁𝐶
𝐹 = 0.3𝑁𝐶
𝒕𝟐
∑ ∫ 𝑭𝒚 𝒅𝒕 = 𝒎(𝒗𝟐 − 𝒗𝟏 )
𝒕𝟏
∑↑+ 𝐹𝑦𝑡 = 𝒎(𝒗𝒚𝟐 − 𝒗𝒚𝟏 );
At 𝑡 = 3 𝑠,
(𝑁 + 400𝑠𝑖𝑛30° − 490.5)𝑡 = 0
(𝑁 + 400𝑠𝑖𝑛30° − 490.5)(3) = 0
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𝑁 = 290.5
∑→+ 𝐹𝑥𝑡 = 𝒎(𝒗𝒙𝟐 − 𝒗𝒙𝟏 );
(400𝑐𝑜𝑠30° − 0.3𝑁)𝑡 = 50(𝒗𝟐 − 𝟎)
At 𝑡 = 3 𝑠 and 𝑁 = 290.5
(400𝑐𝑜𝑠30° − 0.3(290.5))(3) = 50(𝒗𝟐 − 𝟎)
𝒗𝟐 = 𝒗𝒇 = 15.556 m/s
Example 3a-15. The motor winds in the cable with a constant acceleration, such that the 20-kg
crate moves a distance 𝑠 = 6 𝑚 in 3 s, starting from rest. Determine the tension developed in the
cable. The coefficient of kinetic friction between the crate and the plane is 𝜇𝑘 = 0.3.
Solution:
𝒚’
Rotate the y’ – x’ axis:
𝑭𝒇
𝒙’
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𝑾 = mg = 20 kg (9.81m/𝒔𝟐 )=196. 2 N
𝒕𝟐
∑ ∫ 𝑭𝒚′ 𝒅𝒕 = 𝒎(𝒗𝟐 − 𝒗𝟏 )
𝒕𝟏
∑↑+ 𝐹𝑦′𝑡 = 𝒎(𝒗𝒚𝟐 − 𝒗𝒚𝟏 );
(𝑁 − 𝑊𝑐𝑜𝑠30°)𝑡 = 0
At 𝑡 = 3 𝑠 𝑎𝑛𝑑 𝑊 = 196.2,
(𝑁 − (196.2)𝑐𝑜𝑠30°)(3) = 0
∑→+ 𝐹𝑥′𝑡 = 𝒎(𝒗𝒙𝟐 − 𝒗𝒙𝟏 );
𝑁 = 169.914
(𝑇 − 0.3𝑁 − 𝑊𝑠𝑖𝑛30°)𝑡 = 𝑚(𝑣2 − 𝑣1 )
At 𝑡 = 3 𝑠 𝑎𝑛𝑑 𝑊 = 196.2, 𝑎𝑛𝑑 𝑁 = 169.914,
(𝑇 − 0.3(169.914) − 196.2𝑠𝑖𝑛30°)(3) = 20(𝑣2 − 0 )
(𝑇 − 0.3(169.914) − 196.2𝑠𝑖𝑛30°)(3) = 20(𝑣𝑓 ) Eq. 1
1
𝑠 = 𝑣0 𝑡 + 𝑎𝑡2
2
1
6 = 𝑎(32 )
2
𝑎 = 1.333 𝑚/𝑠 2
𝑣𝑓 = 𝑣0 + 𝑎𝑡
𝑣𝑓 = 3𝑎 = 3(1.3333) = 4 𝑚/𝑠
Eq. 2
Substitute Eq. 2 to 1,
(𝑇 − 0.3(169.914) − 196.2𝑠𝑖𝑛30°)(3) = 20(4)
𝑻 = 𝟏𝟕𝟓. 𝟕𝟒 N
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Principle of Angular Impulse and Momentum
The angular momentum of a particle about point 0 is defined as the "moment" of the particle's
linear momentum about O. Since this concept is analogous to finding the moment of a force about
a point, the angular momentum, Ho, is sometimes referred to as the moment of momentum.
Scalar Formulation. If a particle moves along a curve lying in the x-y plane, as in the figure, the
angular momentum at any instant can be determined about point O (actually the z axis) by using
a scalar formulation. The magnitude of Ho is
(𝑯𝑶 )𝒁 = (𝒅)(𝒎𝒗)
Vector Formulation. If the particle moves along a space curve, the vector cross product can be
used to determine the angular about O. In this case,
𝐻𝑂 = 𝑟 𝑥 𝑚𝑣
The box shown in the figure has a mass m and travels down the smooth circular ramp such that
when it is at the angle 𝜃 it has a speed 𝑣. Determine its angular momentum about point O at this
instant and the rate of increase in its speed, 𝑎𝑡 .
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Solution:
𝑯𝑶 = 𝒓𝒎𝒗
∑ 𝑀𝑂 = 𝐻̇𝑂̇
𝐶𝑊 +
𝑚𝑔(𝑟𝑠𝑖𝑛𝜃) =
𝑑
(𝑟𝑚𝑣)
𝑑𝑡
Since r and m are constant,
𝑚𝑔(𝑟𝑠𝑖𝑛𝜃) = 𝑟𝑚
𝒅𝒗
= 𝒈𝒔𝒊𝒏𝜽
𝒅𝒕
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𝒕𝟐
∑ ∫ 𝑴𝑶 𝒅𝒕 = (𝑯𝑶 )𝟐 − (𝑯𝑶 )𝟏
𝒕𝟏
Conservation of Angular Momentum
∑(𝑯𝑶 ) = ∑(𝑯𝑶 )
𝟏
𝟐
Procedure for Analysis
When applying the principles of angular impulse and momentum, or the conservation of angular
momentum, it is suggested that the following procedure be used.
•
•
•
Draw the particle's free-body diagram in order to determine any axis about which angular
momentum may be conserved. For this to occur, the moments of all the forces (or
impulses) must either be parallel or pass through the axis so as to create zero moment
throughout the time period 𝑡1 to 𝑡2 .
The direction and sense of the particle's initial and final velocities should also be
established.
An alternative procedure would be to draw the impulse and momentum diagrams for the
particle.
Example 3a-16. The 1.5-Mg car travels along the circular road as shown in the figure. If the
traction force of the wheels on the road is 𝐹 = (150𝑡 2 ) N, where t is in seconds, determine the
speed of the car when 𝑡 = 5 𝑠. The car initially travels with a speed of 5 m/s. Neglect the size of
the car.
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Solution:
The free-body diagram of the car is shown in the figure. If we apply the principle of angular
impulse and momentum about the z axis, then the angular impulse created by the weight, normal
force, and radial frictional force will be eliminated since they act parallel to the axis or pass
through it.
𝑡2
∑ ∫ 𝑀𝑧 𝑑𝑡 = (𝐻𝑧 )2 − (𝐻𝑧 )1
𝑡1
𝑡2
∫ 𝑟𝐹𝑑𝑡 = 𝑟𝑚𝑐 (𝑣𝑐 )2 − 𝑟𝑚𝑐 (𝑣𝑐 )1
𝑡1
5
∫ 100(150𝑡 2 )𝑑𝑡 = 100(1500)(𝑣𝑐 )2 − 100(1500)(5)
0
5000𝑡 3 |50 = 1500(𝑣𝑐 )2 − 750,000
(𝒗𝒄 )𝟐 = 𝟗. 𝟏𝟕 𝒎/𝒔
Example 3a-17. The 0.8-lb ball B, shown in the figure, is attached to a cord which passes through
a hole at A in a smooth table. When the ball is 𝑟1 = 1.75 𝑓𝑡 from the hole, it is rotating around in
a circle such that its speed is 𝑣 1 =4 ft/s. By applying the force F the cord is pulled downward
through the hole with a constant speed 𝑣 𝐶 =6 ft/s.
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Determine:
a.
b.
the speed of the ball at the instant it is 𝑟2 = 0.6 𝑓𝑡 from the hole, and
the amount of work done by F in shortening the radial distance from 𝑟1 𝑡𝑜 𝑟2 . Neglect
the size of the ball.
Solution:
As the ball moves from 𝑟1 𝑡𝑜 𝑟2 , in the figure, the cord force F on the ball always passes through
the z axis, and the weight and NB are parallel to it. Hence the moments, or angular impulses
created by these forces, are all zero about this axis. Therefore, angular momentum is conserved
about the z axis.
The ball's velocity 𝑣2 is resolved into two components. The radial component, 6 ft/s, is known;
however, it produces zero angular momentum about the z axis. Thus,
𝐻1 = 𝐻2
𝑟1 𝑚𝐵 𝑣1 = 𝑟1 𝑚𝐵 𝑣′2
0.8 𝑙𝑏
0.8 𝑙𝑏
1.75 ft (32.2 𝑓𝑡/𝑠2 ) (4 𝑓𝑡/𝑠) = 0.6 𝑓𝑡 (32.2 𝑓𝑡/𝑠2 ) 𝑣′2
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𝑣′2 = 11.67 𝑓𝑡/𝑠
The speed of the ball is thus,
𝑣2 = √(11.67 𝑓𝑡/𝑠)2 + (6 𝑓𝑡/𝑠)2
𝑣2 = 13.1 𝑓𝑡/𝑠
The only force that does work on the ball is F. (The normal force and weight do not move
vertically.) The initial and final kinetic energies of the ball can be determined so that from the
principle of work and energy we have
𝑈𝐹 =
1
0.8 𝑙𝑏
1
0.8 𝑙𝑏
(
) (13.1 𝑓𝑡/𝑠)2 − (
) (4 𝑓𝑡/𝑠)2
2
2 32.2 𝑓𝑡/𝑠
2 32.2 𝑓𝑡/𝑠 2
Big Picture
𝑼𝑭 = 𝟏. 𝟗𝟒 𝒇𝒕 ∙ 𝒍𝒃
Big Picture in Focus: ULO-3b. Explain the Dynamics of Particle Systems.
Metalanguage
The most essential terms below are defined for you to have a better understanding of
this section in the course:
1. Impulse is a term that quantifies the overall effect of a force acting over time.
2. Momentum is the is a measurement of mass in motion how much mass is in how
much motion.
3. Particle System is a particle system is a collection of many many minute particles that
together represent a fuzzy object.
4. Relative Motion is motion as observed from or referred to some material system
constituting a frame of reference.
Essential Knowledge
Dynamics of Particle Systems
Before we proceed to kinetics of particle systems, it is necessary to consider another
topic of kinematics, namely the concept of relative motion. Relative motion provides a
convenient means of describing the kinematic constraints (geometric restrictions on the
motion) that are usually present in a system of particles.
This chapter also introduces two new applications of dynamics: impact of particles and
mass flow. Impact refers to a collision between particles and is characterized by a very short
time of contact and large contact forces. The term “mass flow” is applied to problems where the
mass is continuously entering or leaving the system, as in the flow of fluids through pipes and
rocket propulsion.
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Kinematics of Relative Motion
Our discussion of kinematics has so far been limited to absolute motion, where the
motion of a particle is described in a fixed (inertial) reference frame. In order to emphasize the
fixed nature of the reference frame, the prefix “absolute” is sometimes added to the names of
kinematic variables (e.g., “absolute velocity”). Newton’s second law, and the work-energy and
impulse-momentum principles derived from it, apply only to absolute motion. In other words,
the position, velocity, and acceleration appearing in these kinetic principles must be absolute.
In kinematics, a fixed frame is not always the most convenient reference for describing
the motion of a particle. For example, the natural reference frame for observing the motion of a
raindrop on the window of a moving car is the window (a moving frame), not the road (a fixed
frame). A description of motion that is based on a moving frame of reference, such as the
window, is termed relative.
From the figure above, we see that the absolute and relative position vectors are related
by,
𝒓𝑩 = 𝒓𝑨 + 𝒓𝑩/𝑨
Differentiating with respect to time and introducing the notation,
𝒗𝑩/𝑨 = 𝒓̇ 𝑩/𝑨
yields,
𝒗𝑩 = 𝒗𝑨 + 𝒗𝑩/𝑨
sides,
The vector 𝒗𝑩/𝑨 shown is known as the velocity of B relative to A. Differentiating both
𝒂𝑩 = 𝒂𝑨 + 𝒂𝑩/𝑨
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where
𝒂𝑩/𝑨 = 𝒗̇ 𝑩/𝑨 = 𝒓̈ 𝑩/𝑨
is called the acceleration of B relative to A.
From figure, we see that the vector drawn from A to B is the negative of the vector from
B to A, which leads to the following identities:
𝒓𝑩/𝑨 = −𝒓𝑨/𝑩
;
𝒗𝑩/𝑨 = −𝒗𝑨/𝑩
;
𝒂𝑩/𝑨 = −𝒂𝑨/𝑩
Translating the reference frame
It is often convenient to describe relative motion with respect to a coordinate system that
moves with the reference particle. In the figure, the xyz axes are fixed in an inertial reference
frame, whereas the x’y’z’ axes are attached to (and move with) the reference particle A. In this
chapter, we consider only the special case in which the x’y’z’ axes always remain parallel to the
fixed axes. In other words, the moving axes translate with the reference particle but they do not
rotate. Therefore, the base vectors i. j, and k of the fixed reference frame are also the base
vectors of the translating frame.
In the translating reference frame, the coordinates of particle B are x’, y’ and z’. Hence,
the relative position vector of B in this coordinate system is:
𝒓𝑩/𝑨 = 𝒙′ 𝒊 + 𝒚′ 𝒋 + 𝒛′𝒌
The relative velocity and relative acceleration of B are obtained by differentiating with
respect to time (note that because the base vectors are constant, we have di/dt = dj/dt = dk/dt
= 0.
𝒗𝑩/𝑨 =
𝒂𝑩/𝑨 =
𝑑𝑥′
𝑑𝑦′
𝑑𝑧′
𝒊+
𝒋+
𝒌
𝑑𝑡
𝑑𝑡
𝑑𝑡
𝑑 2 𝑥′
𝑑 2 𝑦′
𝑑 2 𝑧′
𝒊
+
𝒋
+
𝒌
𝑑𝑡 2
𝑑𝑡 2
𝑑𝑡 2
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Equations above show that 𝒗𝑩/𝑨 and 𝒂𝑩/𝑨 can be interpreted as the velocity and acceleration
of particle B as seen by a nonrotating observer who moves with particle A.
Example 3b-1. Two airplanes A and B are flying with constant velocities at the same altitude.
The positions of the planes at time t = 0 are shown in the figure.
Determine: a. the velocity of plane A relative to B;
b. the position vector of A relative to B as a function of time; and
c. the minimum distance between the planes and the time when this occurs.
Solution for a:
From the geometry in the figure, the velocities of the planes are:
𝑣𝐴 = 580 (
40𝒊 + 30𝒋
) = 464𝒊 + 348𝒋
50
𝑘𝑚/ℎ
40𝒊 − 30𝒋
) = 208𝒊 − 156𝒋
50
𝑘𝑚/ℎ
𝑣𝐵 = 260 (
The velocity of A relative to B is,
𝑣𝐴/𝐵 = 𝑣𝐴 − 𝑣𝐵 = (464𝒊 + 348𝒋) − (208𝒊 − 156𝒋)
𝑣𝐴/𝐵 = 256𝒊 + 504𝒋
𝑘𝑚/ℎ
The magnitude and direction of the vector are
𝑣𝐴/𝐵 = √2562 + 5042 = 565.3 𝑘𝑚/ℎ
𝜃 = 𝑡𝑎𝑛−1
256
= 26.93°
504
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Solution for b:
The position vector of A relative to B can be found by integrating the relative velocity:
𝑟𝐴/𝐵 = ∫ 𝑣𝐴/𝐵 𝑑𝑡 = ∫(256𝑖 + 504𝑗)𝑑𝑡 = (256𝑖 + 504𝑗)𝑡 + 𝑟0
where t is in hours and 𝑟0 is a constant of integration. From the initial condition, 𝑟𝐴/𝐵 = −30𝑗
km at 𝑡 = 0, we get 𝑟0 = −30𝑗 km. Therefore, the relative position vector becomes
𝑟𝐴/𝐵 = 256𝑡𝒊 + (504𝑡 − 30)𝒋
𝑘𝑚
Denoting the distance between the planes by s, we have
2
𝑠 2 = |𝑟𝐴/𝐵 | = (256𝑡)2 + (504𝑡 − 30)2
The minimum value of s occurs when d
𝑠2
𝑑𝑡
𝑘𝑚2
=0
2(256)2 𝑡 + 2(504𝑡 − 30)(504) = 0
which yields,
𝑡 = 0.04732 ℎ𝑜𝑢𝑟𝑠 = 𝟏𝟕𝟎. 𝟑 𝒔
𝑠𝑚𝑖𝑛 = √[256(0.04732)]2 + [504(0.04732) − 30]2 = 𝟏𝟑. 𝟓𝟗 𝒌𝒎
𝑠𝑚𝑖𝑛 = ̅̅̅̅
𝐵𝐶 = 30𝑠𝑖𝑛26.93° = 13.59 𝑘𝑚
The time required to reach that position,
𝑡=
̅̅̅̅
𝐴𝐶
𝑣𝐴/𝐵
=
30𝑐𝑜𝑠26.93°
565.3
= 0.0473 hours = 170.3 s
Kinematics of Constrained Motion
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The following terminology is used frequently when describing the kinematics of particle
systems:*
• Kinematic constraints: geometric restrictions imposed on the motion of particles.
• Equations of constraint: mathematical expressions that describe the kinematic constraints on
particles in terms of their position coordinates.
• Kinematically independent coordinates: position coordinates of particles that are not subject
to kinematic constraints.
• Number of degrees of freedom: the number of kinematically independent coordinates that are
required to completely describe the configuration of a system of particles.
Example 3b-2. The figure shows a system consisting of two blocks A and B connected by an
inextensible cable that runs around two pulleys. Determine the kinematic relationships between
the velocities and the accelerations of the blocks.
Solution:
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Kinetics: Force-Mass-Acceleration Method
∑ 𝑭 = ma
Solving equations of motion of individual particles
The equation of motion of the mass center has somewhat limited application, since it tells us
nothing about the movement of individual particles or the values of internal forces. Solving the
equations of motion of the individual particles suffers none of these drawbacks.
As an illustration, consider the system in the figure that consists of blocks A and B of
masses 𝑚𝐴 and 𝑚𝐵 , respectively. The blocks are connected by a cable that passes over the pulley
C. The problem is to determine the force in the cable and the acceleration of each block, assuming
negligible friction.
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Block A:
∑ 𝐹𝑥 = 𝑚𝑎 → + ;
𝑇 = 𝑚𝐴 𝑎
Block B:
∑ 𝐹𝑦 = 𝑚𝑎 ↓ + ;
𝑚𝐵 𝑔 − 𝑇 = 𝑚𝐵 𝑎
which yield,
𝒂=
𝒎𝑩 𝒈
𝒎𝑨 + 𝒎𝑩
𝑻=
𝒎𝑨 𝒎𝑩 𝒈
𝒎𝑨 + 𝒎𝑩
Example 3b-3. The 90-N force in Fig. (a) is applied to the cable that is attached to the 60-N block
A. In the figure, this force is replaced by a 90-N block B. Neglecting the mass of the pulley,
determine the acceleration of A and the tension in the cable for both cases.
Solution:
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Drawing FBD:
𝑇 = 90 𝑁
From the mass-acceleration diagram for block A, where its acceleration 𝑎 is assumed to be
upward.
∑ 𝐹𝑦 = 𝑚𝑎 ↑ +
for which the acceleration is,
𝑎=
𝑔
2
90 − 60 =
60
𝑎
𝑔
= 4.9 𝑚/𝑠 2
The FBDs and MADs of the blocks are shown in the figure. Due to the inextensible cable, the
acceleration of A is equal in magnitude to the acceleration of B, but in the opposite direction. We
assumed the acceleration of A to be upward. Therefore, the equation of block A is:
∑ 𝑭𝒚 = 𝒎𝒂 ↑ +
𝑇 − 60 = (
For block B, we have:
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60
)𝑎
𝑔
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∑ 𝑭𝒚 = 𝒎𝒂 ↑ +
𝑇 − 90 = (
90
)𝑎
𝑔
Solving these two equations simultaneously:
𝑻 = 𝟕𝟐 𝑵 and 𝒂 =
𝒈
𝟓
= 𝟏. 𝟗𝟔 𝒎/𝒔𝟐
Example 3b-4. The figure below shows a system consisting of three blocks connected by an
inextensible cable that runs around four pulleys. The masses of blocks A, B, and C are 60 kg, 80
kg, and 20 kg, respectively. Using the coordinates shown and neglecting the masses of the pulleys,
find the acceleration of each block and the tension T in the cable.
Solution:
We will use the force-mass-acceleration method to derive the equation of motion for each block.
Letting 𝐿 be the length of the cable that runs around the pulleys in the figure, we have:
differentiating,
differentiating,
𝐿 = 2𝑦𝐴 + 2𝑦𝐵 + 𝑦𝐶 + 𝐶1
𝑑𝐿
= 2𝑣𝐴 + 2𝑣𝐵 + 𝑣𝐶 = 0
𝑑𝑡
2𝑎𝐴 + 2𝑎𝐵 + 𝑎𝐶 = 0
Kinetic Analysis
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The figure shows the free-body diagrams of blocks A and B (together with the massless pulleys
to which they are attached), and block C. Note that the tension T is constant throughout the cable.
Also shown are the corresponding mass-acceleration diagrams. Applying Newton’s law
∑ 𝐹𝑦 = 𝑚𝑎 to each block, the equations of motion are:
+↓
60(9.8) − 2𝑇 = 60𝑎𝐴
+↓
80(9.8) − 2𝑇 = 80𝑎𝐵
+↓
20(9.8) − 𝑇 = 20𝑎𝐶
Solving the three equations simultaneously,
𝑇 = 294 𝑁
𝑎𝐴 = 0
𝑎𝐵 = 2. 45 𝑚/𝑠 2
𝑎𝐶 = −4.9 𝑚/𝑠 2
The signs indicate that the acceleration of B is directed downward, whereas the acceleration of C
is upward.
Principle of Impulse and Momentum
Linear momentum
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where 𝑝 = momentum
𝒏
𝒏
𝒑 = ∑ 𝒑𝒊 = ∑ 𝒎𝒊 𝒗𝒊
𝒊=𝟏
𝒊=𝟏
̅
𝒑 = 𝒎𝒗
Force-momentum relationship
∑ 𝐹 = 𝑚𝑎̅ = 𝑚
𝑑𝑣̅
𝑑
(𝑚𝑣̅ )
=
𝑑𝑡
𝑑𝑡
The above the above force-momentum relationship for a closed system of particles becomes,
∑𝑭 =
𝒅𝒑
𝒅𝒕
Impulse-momentum Principle
∑ 𝑭 𝒅𝒕 = 𝒅𝒑
𝒕𝟐
𝒕𝟐
∫ ∑ 𝑭 𝒅𝒕 = ∫ 𝒅𝒑 = 𝒑𝟐 − 𝒑𝟏
𝒕𝟏
𝒕𝟏
where𝒑𝟏 and 𝒑𝟐 denote the momenta of the system at t = 𝒕𝟏 and 𝒕𝟐 , respectively. Because the left
side of equation is the impulse of the external forces, denoted by 𝑳𝟏−𝟐 , we can write
𝑳𝟏−𝟐 = 𝒑𝟐 − 𝒑𝟏 = ∆𝒑
which is the impulse-momentum principle for a system of particles
Conservation of Linear Momentum m for a System of Particles
We see that if the impulse of the external forces is zero, momentum of the system is conserved.
In other words, if 𝑳𝟏−𝟐 = 0, we obtain,
𝒑𝟏 = 𝒑𝟐 or ∆𝒑 = 𝟎
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or
∑ 𝑚𝑖 (𝑣𝑖 )1 = ∑ 𝑚𝑖 (𝑣𝑖 )2
which is the principle of conservation of momentum for a system of particles.
Example 3b-5. The blocks A and B are connected by a cable that runs around two pulleys of
negligible mass, as shown in the figure. The kinetic coefficient of friction between the inclined
plane and block A is 0.4. If the initial velocity of A is 3 m/s down the plane, determine 𝑠𝐴 of block
A (measured from its initial position) when the system comes to rest.
Solution:
Draw the Free-Body Diagram
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Kinematics
𝒔𝑨 + 𝟐𝒔𝑩 = constant
Therefore, the displacement ∆𝑠𝐴 and ∆𝑠𝐵 and the velocities of the two blocks are related by,
∆𝒔𝑨 + 𝟐∆𝒔𝑩 = 0
𝒗𝑨 + 𝟐𝒗𝑩 = 0
Work-Energy Principle
Let 1 and 2 denote the initial and the final (rest) positions of the system. Applying the workenergy principle, to the system, we get:
The solution for the displacement of block A is:
∆𝒔𝑨 = 𝟏. 𝟓𝟖 𝒎
Example 3b-6. Blocks A and B shown in the figure have a mass of 3 kg and 5 kg, respectively. If
the system is released from rest, determine the velocity of block B in 6 s. Neglect the mass of the
pulleys and cord.
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Solution:
Principle of Impulse and Momentum:
Block A:
𝒕𝟐
∑ ∫ 𝑭𝒚 𝒅𝒕 = 𝒎(𝒗𝑨 )𝟐 − 𝒎(𝒗𝑨 )𝟏
𝒕𝟏
𝟑(𝟗. 𝟖𝟏)(𝟔) − 𝟐𝑻𝑩 (𝟔) = 𝟑(𝒗𝑨 )𝟐 − 𝟎
𝟑(𝒗𝑨 )𝟐 + 𝟐𝑻𝑩 (𝟔) = 𝟏𝟕𝟔. 𝟓𝟖
Eq. 1
Block B:
𝒕𝟐
∑ ∫ 𝑭𝒚 𝒅𝒕 = 𝒎(𝒗𝑩 )𝟐 − 𝒎(𝒗𝑩 )𝟏
𝒕𝟏
𝟓(𝟗. 𝟖𝟏)(𝟔) − 𝑻𝑩 (𝟔) = 𝟓(𝒗𝑩 )𝟐
Kinematics:
2𝑠𝐴 + 𝑠𝐵 = 𝑙
Taking the time derivative,
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2𝑣𝐴 = −𝑣𝐵
As indicated by the negative sign, when B moves downward A moves upward. Substituting this
result into Eq. 1 and solving Eqs. 1 and 2 yields,
(𝑣𝐵 )2 = 35.8
𝑚
↓
𝑠
𝑻𝑩 = 𝟏𝟗. 𝟐 N
Example 3b-7. The 15-Mg boxcar A is coasting at 1 .5 m/s on the horizontal track when it
encounters a 12-Mg tank car B coasting at 0.75 m/s toward it as shown in the figure. If the cars
collide and couple together, determine:
a. the speed of both cars just after the coupling, and
b. the average force between them if the coupling takes place in O.8 s.
Solution:
Conservation of Linear Momentum
∑ 𝑚𝑖 (𝑣𝑖 )1 = ∑ 𝑚𝑖 (𝑣𝑖 )2
𝑚𝐴 (𝑣𝐴 )1 + 𝑚𝐵 (𝑣𝐵 )1 = 𝑚𝐴 (𝑣𝐴 )2 + 𝑚𝐵 (𝑣𝐵 )2
Since “cars collide and couple together”, then 𝑣𝐴 = 𝑣𝐵 = 𝑣2
𝑚𝐴 (𝑣𝐴 )1 + 𝑚𝐵 (𝑣𝐵 )1 = 𝑚𝐴 𝑣2 + 𝑚𝐵 𝑣2
15,000(1.5) − 12,000(0.75) = 27,000𝑣2
𝒗𝟐 = 𝟎. 𝟓 𝒎/𝒔
Principle of Impulse and Momentum
∫ 𝐹 𝑑𝑡 = 𝑚𝐴 𝑣2 − 𝑚𝐴 𝑣1
𝐹(0.8) = 15,000(0.5) − 15,000(1.5)
𝑭 = 𝟏𝟖. 𝟖 𝒌𝑵
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Example 3b-8. The bumper cars A and B in the figure each have a mass of 150 kg and are coasting
with the velocities shown before they freely collide head on. If no energy is lost during the
collision, determine their velocities after collision.
Solution:
Conservation of Momentum,
𝑚𝐴 (𝑣𝐴 )1 + 𝑚𝐵 (𝑣𝐵 )1 = 𝑚𝐴 (𝑣𝐴 )2 + 𝑚𝐵 (𝑣𝐵 )2
150(3) + 150(−2) = 150(𝑣𝐴 )2 + 150(𝑣𝐵 )2
(𝑣𝐴 )2 = 1 − (𝑣𝐵 )2
Eq. 1
Conservation of Energy
𝑻𝟏 + 𝑽𝟏 = 𝑻𝟐 + 𝑽𝟐
𝟏
𝟏
𝟏
𝟏
(𝒎𝑨 )(𝒗𝑨 )𝟏 𝟐 + (𝒎𝑩 )(𝒗𝑩 )𝟏 𝟐 + 𝟎 = (𝒎𝑨 )(𝒗𝑨 )𝟐 𝟐 + (𝒎𝑩 )(𝒗𝑩 )𝟐 𝟐 + 𝟎
𝟐
𝟐
𝟐
𝟐
𝟏
𝟏
𝟏
𝟏
(𝟏𝟓𝟎)(𝟑)𝟐 + (𝟏𝟓𝟎)(𝟐)𝟐 = (𝟏𝟓𝟎)(𝒗𝑨 )𝟐 𝟐 + (𝟏𝟓𝟎)(𝒗𝑩 )𝟐 𝟐
𝟐
𝟐
𝟐
𝟐
(𝒗𝑨 )𝟐 𝟐 + (𝒗𝑩 )𝟐 𝟐 = 𝟏𝟑
Eq. 2
Substituting Eq. 1 to Eq. 2,
(1 − (𝑣𝐵 )2 )𝟐 + (𝒗𝑩 )𝟐 𝟐 = 𝟏𝟑
(𝒗𝑩 )𝟐 = 𝟑 𝒎/𝒔 →
(𝑣𝐴 )2 = 1 − (𝑣𝐵 )2 = 1 − 3 = −2
𝑚
𝑠
or 𝟐
𝒎
𝒔
←
Impact
Impact occurs when two bodies collide with each other during a very short period of time,
causing relatively large (impulsive) forces to be exerted between the bodies. The striking of a
hammer on a nail, or a golf club on a ball, are common examples of impact loadings.
In general, there are two types of impact. Central impact occurs when the direction of
motion of the mass centers of the two colliding particles is along a line passing through the mass
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centers of the particles. This line is called the line of impact, which is perpendicular to the plane
of contact, as in the figure shown. When the motion of one or both of the particles make an angle
with the line of impact, as in the figure, the impact is said to be oblique impact.
Coefficient of Restitution, e. This is equal to the ratio of the relative velocity of the particles'
separation just after impact, (𝑣𝐵 )2 − (𝑣𝐴 )2 , to the relative velocity of the particles' approach just
before impact, (𝑣𝐴 )1 − (𝑣𝐵 )1 / By measuring these relative velocities experimentally, it has been
found that e varies appreciably with impact velocity as well as with the size and shape of the
colliding bodies.
In general e has a value between zero and one, and one should be aware of the physical
meaning of these two limits.
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Elastic Impact (𝑒 = 1). If the collision between the two particles is perfectly elastic, the
deformation impulse is equal and opposite to the restitution impulse.
Plastic Impact (𝑒 = 0 ). The impact is said to be inelastic or plastic when e = O. In this case there
is no restitution impulse, so that after collision both particles couple or stick together and move
with a common velocity.
Example 3b-9. The bag A, having a weight of 6 Ib, is released from rest at the position 𝜃= 0°, as
shown in the figure. After falling to 𝜃 = 90°, it strikes an 18-lb box B. If the coefficient of restitution
between the bag and box is e = 0.5, determine the velocities of the bag and box just after impact.
What is the loss of energy during collision?
Conservation of Energy
With the datum at 𝜃 = 0°, we have:
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𝑇0 + 𝑉0 = 𝑇1 + 𝑉1
0+0=
1
6 𝑙𝑏
(
) 𝑣𝐴1 2
2 32.2 𝑓𝑡/𝑠2
− 6(3)
(𝑣𝐴 )1 = 13.90𝑓𝑡/𝑠
Conservation of Momentum
𝑚𝐵 (𝑣𝐵 )1 + 𝑚𝐴 (𝑣𝐴 )1 = 𝑚𝐵 (𝑣𝐵 )2 + 𝑚𝐴 (𝑣𝐴 )2
0+(
6
18
6
) (13.90) = (
) (𝑣𝐵 )2 + (
) (𝑣𝐴 )2
32.2
32.2
32.2
Eq. 1
(𝑣𝐴 )2 = 13.90 − 3(𝑣𝐵 )2
Coefficient of Restitution
𝑒=
0.5 =
(𝑣𝐵 )2 − (𝑣𝐴 )2
(𝑣𝐴 )1 − (𝑣𝐵 )1
(𝑣𝐵 )2 − (𝑣𝐴 )2
13.90 − 0
(𝑣𝐵 )2 − (𝑣𝐴 )2 = 6.950
𝐄𝐪. 𝟐
Solving Eq. 1 and Eq. 2 simultaneously,
(𝑣𝐴 )2 = −1.74
(𝑣𝐵 )2 = 5.21
𝑓𝑡
𝑠
𝑓𝑡
𝑠
or 𝟏. 𝟕𝟒
or 𝟓. 𝟐𝟏
Loss of Energy
∑ 𝑈1−2 = 𝑇2 − 𝑇1
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𝒇𝒕
𝒔
𝒇𝒕
𝒔
←
←
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1 18
1 6
1 6
∑ 𝑈1−2 = [ (
) (5.21) + (
) (1.74)] − [ (
) (13.9)]
2 32.2
2 32.2
2 32.2
∑ 𝑼𝟏−𝟐 = −𝟏𝟎. 𝟏
𝒍𝒃 − 𝒇𝒕
The energy loss occurs due to inelastic deformation during the collision.
Mass Flow
Up to this point we have restricted our study of impulse and momentum principles to a
system of particles contained within a closed volume. In this section, however, we will apply the
principle of impulse and momentum to the steady mass flow of fluid particles entering into and
then out of a control volume. This volume is defined as a region in space where fluid particles can
flow into or out of a region. The size and shape of the control volume is frequently made to coincide
with the solid boundaries and openings of a pipe, turbine, or pump. Provided the flow of the fluid
into the control volume is equal to the flow out, then the flow can be classified as steady flow.
.
𝒅𝒎
= 𝝆𝒗𝑨 = 𝝆𝑸
𝒅𝒕
The term 𝑄 = 𝑣𝐴 measures the volume of fluid flow per unit of time and is referred to as
the discharge or the volumetric flow.
Procedure for Analysis
Problems involving steady flow can be solved using the following procedure:
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•
•
•
•
•
Identify the control volume. If it is moving, a kinematic diagram may be helpful for
determining the entrance and exit velocities of the fluid flowing into and out of its
openings since a relative motion analysis of velocity will be involved.
The measurement of velocities VA and VB must be made by an observer fixed in an
inertial frame of reference.
Once the velocity of the fluid flowing into the control volume is determined, the mass
flow is calculated.
Draw the free-body diagram of the control volume in order to establish the forces that
act on it. These forces will include the support reactions, the weight of all solid parts
and the fluid contained within the control volume, and the static gauge pressure forces
of the fluid on the entrance and exit sections. The gauge pressure is the pressure
measured above atmospheric pressure, and so if an opening is exposed to the
atmosphere, the gauge pressure there will be zero.
Apply the equations of steady flow, using the appropriate components of velocity and
force shown on the kinematic and free-body diagrams.
Example 3b-10. Determine the components of reaction which the fixed pipe joint at A exerts on
the elbow in the figure, if water flowing through the pipe is subjected to a static gauge pressure
of 100 kPa at A. The discharge at B is 𝑄𝐵 = 0.2 𝑚3 /𝑠. Water has a density 𝜌𝑤 = 1000 kg/𝑚3 , and
the water-filled elbow has a mass of 20 kg and center of mass at G.
Solution:
We will consider the control volume to be the outer surface of the elbow. Using a fixed
inertial coordinate system, the velocity of flow at A and B and the mass flow rate can be
obtained from the equation. Since the density of water is constant 𝑄𝐵 = 𝑄𝐴 = 𝑄. Hence,
𝑑𝑚
= 𝜌𝑤 𝑄 = 1000 𝑘𝑔/𝑚3 (0.2 𝑚3 /𝑠)
𝑑𝑡
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𝑑𝑚
= 200 𝑘𝑔/𝑠
𝑑𝑡
𝑣𝐵 =
𝑣𝐴 =
𝑄
0.2 𝑚3 /𝑠
𝑚
=
= 25.46 ↓
2
𝐴𝐵
𝜋(0.05 𝑚)
𝑠
𝑄
0.2 𝑚3 /𝑠
𝑚
=
= 6.37 →
2
𝐴𝐴
𝜋(0.1 𝑚)
𝑠
𝑭𝑨 = 𝒑𝑨 𝑨𝑨
𝐹𝐴 = 𝑝𝐴 𝐴𝐴 = [100,000 𝑁/𝑚2 ][𝜋 (0.1)2 ] = 3141. 6 𝑁
Equations of Steady Flow
∑ 𝐹𝑥 =
→+
∑ 𝐹𝑦 =
↑+
𝑑𝑚
(𝑣 − 𝑣𝐴𝑥 ) ;
𝑑𝑡 𝐵𝑥
−𝐹𝑥 + 3141. 6 = 200(0 − 6.37)
𝑭𝒙 = 𝟒. 𝟒𝟏 𝒌𝑵
𝑑𝑚
(𝑣 − 𝑣𝐴𝑦 ) ;
𝑑𝑡 𝐵𝑦
−𝐹𝑦 − 20(9.81) = 200(−25.46 − 0)
𝑭𝒚 = 𝟒. 𝟗𝟎 𝒌𝑵
If moments are summed about point O, then 𝐹𝑥, 𝐹𝑦 , and the static pressure 𝐹𝐴 are eliminated, as
well as the moment of momentum of the water entering at A. Hence,
∑ 𝑀𝑂 =
𝐶𝑊+
𝑑𝑚
(𝑑 𝑣 − 𝑑𝑂𝐴 𝑣𝐴 )
𝑑𝑡 𝑂𝐵 𝐵
𝑀𝑂 + 20(9.81)(0.125) = 200[0.3(25.46 − 0)]
𝑴𝑶 = 𝟏. 𝟓𝟎 𝒌𝑵 − 𝒎
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Let’s Check!
Answer the following problems:
1. If the contact surface between the 20-kg block and the ground is smooth, determine the
power of force F when t = 4 s. Initially, the block is at rest.
2. If = 10 𝑠 𝑁, where s is in meters, and the contact surface between the block and the ground
is smooth, determine the power of force F when s = 5 m. Initially, the 20-kg block is at rest.
3. The 2 -Mg car increases its speed uniformly from rest to 25 m/s in 30 s up in the inclined
road. Determine the maximum power that must be supplied by the engine, which operates
with an efficiency of 𝜖 = 0.8. Also, find the average power supplied by the engine.
Let’s Analyze!
Answer the following problems:
1. The fluid transmission of a 30 000-lb truck allows the engine to deliver constant power
to the rear wheels. Determine the distance required for the truck traveling on a level
road to increase its speed from to if 90 hp is delivered to the rear wheels. Neglect drag
and rolling resistance.
2. If the engine of a 1.5-Mg car generates a constant power of 15 kW, determine the speed
of the car after it has traveled a distance of 200 m on a level road starting from rest.
Neglect friction.
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3. The 1.2-Mg mine car is being pulled by the winch M mounted on the car. If the winch
exerts a force of on the cable, where t is in seconds, determine the power output of the
winch when, starting from rest.
In a Nutshell!
Answer the following problems:
1. The diesel engine of a 400-Mg train increases the train's speed uniformly from rest to 10
m/s in 100 s along a horizontal track. Determine the average power developed.
2. Determine the power input for a motor necessary to lift 300 Ib at a constant rate of 5 ft/s.
The efficiency of the motor is E = 0.65.
3. An electric streetcar has a weight of 15 000 Ib and accelerates along a horizontal straight
road from rest so that the power is always 100 hp. Determine how far it must travel to reach
a speed of 40 ft/s.
4. The Milkin Aircraft Co. manufactures a turbojet engine that is placed in a plane having a
weight of 13000 lb. If the engine develops a constant thrust of 5200 Ib, determine the power
output of the plane when it is just ready to take off with a speed of 600 mi/h.
5. The engine of the 3500-lb car is generating a constant power of 50 hp while the car is
traveling up the slope with a constant speed. If the engine is operating with an efficiency of
E = O.S, determine the speed of the car. Neglect drag and rolling resistance.
Course Schedule:
This section calendars all the activities and exercises including readings and lectures as well as
time for making assignments and doing other requirements in a programmed schedule by days
and weeks, to help the students in SDL pacing, regardless of mode of delivery (OBD or DED):
Activity
Big Picture 3a, 3b: Let’s Check Activities!
Big Picture 3a,3b: Let’s Analyze Activities!
Big Picture 3a,3b: In a Nutshell Activities!
Big Picture 3a,3b: Q and A List!
3rd Exam
Date
Oct. 2, 2020
Where to Submit
BB
BB
BB
BB
BB
Big Picture
Week 8-9: Unit Learning Outcomes-Unit 4 (ULO-4): At the end of the unit, you are expected to
a. Demonstrate knowledge in Planar Kinematics of Rigid Bodies
b. Explain the basic kinetic quantities of Rigid Bodies
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Big Picture in Focus: ULO-4a. Demonstrate knowledge in Planar Kinematics of
Rigid Bodies
Metalanguage
The most essential terms below are defined for you to have a better understanding of
this section in the course:
1. Angular Motion is an angular motion is one in which the body moves along a curved path
at a constant angular velocity, as when a runner travels along a circular path or an
automobile rounds a curve.
2. Rotation is a circular movement of an object around a center (or point) of rotation.
3. Translation is a motion of a body on a linear path, without deformation or rotation, i.e. such
that every part of the body moves at the same speed and in the same direction.
Essential Knowledge
Introduction
A body is said to be rigid if the distance between any two points in the body remains
constant. In other words, a rigid body does not deform. Of course, the rigid-body concept is an
idealization, because all bodies deform to some extent when subjected to forces. But if the
deformation is sufficiently small (“small” usually means negligible when compared to the
dimensions of the body), the assumption of rigidity is justified.
This chapter is concerned only with the kinematics of plane motion of rigid bodies. A
body undergoes plane motion if all points in the body remain a constant distance from a fixed
reference plane, called the plane of motion of the body.
In other words, all points in the body move in planes that are parallel to the plane of
motion. There are three categories of plane motion:
Translation is the special case in which the body moves without rotation; that is, any line
in the body remains parallel to its initial position, as shown in the figure below. Because all points
in the body have the same displacement, the motion of one point determines the motion of the
entire body.
Rotation about a fixed axis is the special case in which a line in the body, called the axis of
rotation, is fixed in space. Consequently, each point not on the axis of rotation moves in a circle
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about the axis, as illustrated in the figure (the axis of rotation at O is perpendicular to the plane
of the paper).
General plane motion is the superposition of translation and rotation. The rolling disk in the
figure below is an example of such motion: the disk is translating and rotating simultaneously.
Plane Angular Motion
Angular Displacement
∆𝜽 = 𝜽(𝒕 + ∆𝒕) − 𝜽𝒕
Angular Velocity
∆𝜽
𝒅𝜽
=
= 𝜽̇
∆𝒕→𝟎 ∆𝒕
𝒅𝒕
𝝎 = 𝐥𝐢𝐦
Angular Acceleration
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𝒂=
𝒅𝒘 𝒅𝒘 𝒅𝜽
𝒅𝒘
=
= 𝝎
𝒅𝒕
𝒅𝒕 𝒅𝒕
𝒅𝜽
Rotation about a Fixed Axis
Rotation about a fixed axis is the special case of plane motion in which one line in the
body, called the axis of rotation, is fixed in space. The figure below shows a rigid body that is
rotating about an axis. We let B be a point in the body that is a distance R from the axis. Because
the body is rigid, the path of B is a circle of radius R that lies in a plane perpendicular to the axis
of rotation. The center O of the circle lies on the axis, and the angular position coordinate of he
radial line OB is denoted by θ. Since OB lies in the plane of motion, the angular velocity and
angular acceleration of the body are:
𝝎 = 𝜽̇ and 𝒂 = 𝝎̇ = 𝜽̈
Kinematics of a Point in the Body
𝑣 = 𝑅𝜔
𝑎𝑛 = 𝑅𝜔2 =
𝑣2
𝑅
𝑎𝑡 = 𝑅𝛼
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Solution:
Example 4a-1. The disk rotates about a fixed axis at O. During the period 𝑡 = 0 to 𝑡 = 4 𝑠, the
angular position of the line 𝑂𝐴 in the disk varies as 𝜃(𝑡) = 𝑡 3 − 12𝑡 + 6 rad, where 𝑡 is in
seconds. Determine:
1. the angular velocity and the angular acceleration of the disk at the end of the period;
2. the angular displacement of the disk during the period; and
3. the total angle turned through by the disk during the period.
The angular velocity and angular acceleration of the disk are:
𝜔 = 𝜃̇ = 3𝑡 2 − 12
𝑟𝑎𝑑
𝑠
𝛼 = 𝜔̇ = 6𝑡 rad/𝑠 2
when 𝑡 = 4 𝑠, we have:
𝜔 = 3(4)2 − 12 = 36
𝛼 = 6(4) = 24
𝑟𝑎𝑑
𝑠
𝑟𝑎𝑑
𝑠2
𝐶𝐶𝑊
𝐶𝐶𝑊
The angular positions of the line OA at the beginning and at the end of the period are:
𝜃|𝑡=0 = 6 𝑟𝑎𝑑
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𝜃|𝑡=4 = 43 − 12(4) + 6 = 22 𝑟𝑎𝑑
Therefore, the angular displacement of the disk from 𝑡 = 0 𝑡𝑜 𝑡 = 4 s is:
∆𝜃 = 𝜃|𝑡=4𝑠 − 𝜃|𝑡=0 = 22 − 6 = 𝟏𝟔 𝒓𝒂𝒅 𝐶𝐶𝑊
Note that the direction of rotation of the disk changes 𝜔 = 0, that is when:
3𝑡 2 − 12 = 0;
𝑎𝑡 𝑡 = 2 𝑠
The angular position of OA at that time is:
𝜃|𝑡=2 𝑠 = 23 − 12(2) + 6 = −10 𝑟𝑎𝑑
We conclude that the disk rotates clockwise (𝜔 < 0) between 𝑡 = 0 and 𝑡 = 2 𝑠, its angular
displacement being:
∆𝜃1 = 𝜃|𝑡=2 𝑠 − 𝜃|𝑡=0 = −10 − 6 = −16 𝑟𝑎𝑑
Between 𝑡 = 2 𝑠 and 𝑡 = 4 𝑠, the rotation is counterclockwise (𝜔 > 0 ); the corresponding
angular displacement is:
∆𝜃2 = 𝜃|𝑡=4 𝑠 − 𝜃|𝑡=2 𝑠 = 22 − (−10) = 32 𝑟𝑎𝑑
Therefore, the total angle turned through by the disk from 𝑡 = 0 to 𝑡 = 4 𝑠 is:
|∆𝜃1 | + |∆𝜃2 | = 16 + 32 = 𝟒𝟖 𝒓𝒂𝒅
Example 4a-2. Pulley B is being driven by the motorized pulley A that is rotating at 𝜔𝐴 =
20 𝑟𝑎𝑑/𝑠. At time 𝑡 = 0, the current in the motor is cut off, and friction in the bearings causes
the pulleys to coast to a stop. The angular acceleration of A during the deceleration is α𝐴 =
−2.5𝑡 𝑟𝑎𝑑/𝑠 2 , where 𝑡 is in seconds. Assuming that the drive belt does not slip on the pulleys,
Determine:
a. the angular velocity of 𝐵 as a function of time;
b. the angular displacement of B during the period of coasting; and
c. the acceleration of point C on the straight portion of the belt as a function of time.
Solution:
𝑣 = 𝑅𝐴 𝜔𝐴 = 𝑅𝐵 𝜔𝐵
So that,
𝜔𝐵 =
𝑅𝐴
𝑅𝐵
0.10
) 𝜔𝐴
0.20
𝜔𝐴 = (
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Differentiating with respect to time, we obtain the angular acceleration of pulley
B:
𝛼𝐵 = 0.5𝛼𝐴 = 0.5(−2.5𝑡) = −1.25𝑡 𝑟𝑎𝑑/𝑠 2
Because 𝛼𝐵 =
𝑑𝜔𝐵
, we have 𝑑𝜔𝐵
𝑑𝑡
= 𝛼𝐵 𝑑𝑡 or
𝜔𝐵 = ∫ 𝛼𝐵 𝑑𝑡 = ∫ −1.25𝑡𝑑𝑡 = −0.625𝑡 2 + 𝐶1
The initial condition, 𝜔𝐵 = 20
velocity of pulley B is:
𝑟𝑎𝑑
𝑠
when 𝑡 = 0, yields 𝐶1 = 20 𝑟𝑎𝑑/𝑠. Hence, the angular
𝜔𝐵 = −0.625𝑡 2 + 20 𝑟𝑎𝑑/𝑠
We let 𝜃𝐵 be the angular position of a line in B measured from a fixed reference line.
Recalling that 𝜔𝐵 = 𝑑𝜃𝐵 /𝑑𝑡, we integrate 𝑑𝜃𝐵 = 𝜔𝐵 𝑑𝑡 to obtain:
𝜃𝐵 = ∫ 𝜔𝐵 𝑑𝑡 = ∫(−0.625𝑡 2 + 20)𝑑𝑡 = −0.2083𝑡 3 + 20𝑡 + 𝐶2
Letting 𝜃𝐵 = 0 when 𝑡 = 0, we have 𝐶2 = 0, which gives
rad
𝜃𝐵 = −0.2083𝑡 3 + 20𝑡
The pulley comes to rest when 𝜔𝐵 = −0.625𝑡 2 + 20 = 0, which yields 𝑡 = 5.657 𝑠. The
corresponding angular position of the line in B is:
𝜃𝐵 |𝑡=5.657 𝑠 = −0.2083(5.657)3 + 20(5.657) = 112.0 𝑟𝑎𝑑
Therefore, the angular displacement of pulley B as it coasts to a stop is:
∆𝜃𝐵 = 𝜃𝐵 |𝑡=5.657 𝑠 − 𝜃𝐵 |𝑡=0 = 112 − 0 = 112.0 𝑟𝑎𝑑
Because the direction of rotation does not change, the total angle turned through by pulley
B during the deceleration is also 112.0 rad.
Substituting 𝑅𝐵 = 0.2 𝑚 and 𝜔𝐵 = −0.625𝑡 2 + 20 𝑟𝑎𝑑/𝑠 into equation, the speed of
point C (which is the same for all points on the belt) is:
𝑣𝐶 = 0.2 (−0.625𝑡 2 + 20) = (−0.125𝑡 2 + 4)
𝑚/𝑠
Because the path of point C on the belt is a straight line, the acceleration of C is:
𝑎𝐶 = 𝑣𝑐 ̇ = − 𝟎. 𝟐𝟓𝒕
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Big Picture in Focus: ULO-4b. Explain the basic kinetic quantities of Rigid
Bodies.
Metalanguage
The most essential terms below are defined for you to have a better understanding of
this section in the course:
1. Angular Motion is an angular motion is one in which the body moves along a curved path
at a constant angular velocity, as when a runner travels along a circular path or an
automobile rounds a curve.
2. Rotation is a circular movement of an object around a center (or point) of rotation.
3. Translation is a motion of a body on a linear path, without deformation or rotation, i.e. such
that every part of the body moves at the same speed and in the same direction.
Essential Knowledge
Introduction
Planar Kinetics of Rigid Bodies: Force-Mass-Acceleration Method
This section presents the force-mass-acceleration (FMA) method of kinetic analysis for
rigid bodies in plane motion. The equations of motion, which are the basis of the method, can be
obtained from the results for particle systems in the previous chapter by viewing a rigid body as
a collection of particles where the distances between the particles remain constant. The resulting
equations are known as Euler’s laws of motion. The first law governs the motion of the mass
center of the body; it is identical to the equation of motion of the mass center of a particle system.
The second law, which governs the rotational motion of the body, is the moment–angular
momentum relationship for a system of particles.
When this law is specialized for a rigid body, it gives rise to the concept of mass moment
of inertia, which is the topic of the succeeding sections.
Mass Moment of Inertia, I
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It will be seen shortly that this integral is a measure of the ability of the body to resist a
change in its angular motion about the a-axis, just as the mass of the body is a measure of its
ability to resist a change in its translational motion.
Radius of gyration
The radius of gyration 𝑘𝑎 of the body about the a-axis is defined as,
Although the unit of radius of gyration is length (e.g., feet, meters), it is not a distance that can be
measured physically. Instead, its value can be found only by computation using the equation
above. The radius of gyration allows us to compare the rotational resistances of bodies that have
the same mass.
Parallel – axis theorem
Consider the two parallel axes shown in the figure. The location of the a-axis is arbitrary.
We call the other axis, which passes through the mass center 𝐺 of the body, the central a-axis. *
With d being the distance between the two axes, the parallel-axis theorem states that
𝐼𝑎 = 𝐼𝑎̅ + 𝑚𝑑2
Where:
m = the mass of the body
𝐼𝑎 = is the moment of inertia of the body about the a-axis, and
𝐼𝑎̅ = is its moment of inertia about the central a-axis. Note that if the moment of
inertia about a central axis is known, the parallel-axis theorem can be used to calculate
the moment of inertia about any parallel axis without resorting to integration. Table 17.1
lists the moments of inertia about central axes for a few homogeneous bodies.
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Mass Moments of Inertia of Homogeneous Bodies
Method of Composite Bodies
From Equation: 𝐼𝑎 = ∫𝑉 𝑟 2 𝑑𝑚 , we see that the computation of the mass moment of inertia
requires that an integration be performed over the body. Here we will consider only the method of
composite bodies, a method that follows directly from the property of definite integrals:
The integral of a sum is equal to the sum of the integrals. *
Using this property, it can be shown that if a body is divided into composite parts, the moment of
inertia of the body about a given axis equals the sum of the moments of inertia of its parts about that axis.
The following sample problems illustrate the application of this method.
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Example 4a-3. The assembly in the figure is composed of three homogenous bodies: The 10-kg
cylinder, the 2 kg slender rod, and the 4 kg sphere. For this assembly, calculate 𝐼𝑥 , the mass
moment of inertia and radius of gyration about the central x- axis of the assembly.
Solution:
The mass centers of the cylinder 𝐺1 , the rod 𝐺2 and the sphere 𝐺3 are shown in the figure.
By symmetry, the mass center of the assembly 𝐺 lies on the y – axis, with its coordinate 𝑦̅ to be
determined.
Cylinder. Using the table, the moment of inertia of the cylinder about its own central x-axis is:
(𝐼𝑥̅ )1 =
1
1
(10)[3(0.06)2 + 0.32 ] = 0.084 𝑘𝑔 ∙ 𝑚2
𝑚1 (3𝑅2 + ℎ2 ) =
12
12
Utilizing the parallel-axis theorem, the moment of Inertia of the cylinder about the x-axis becomes
(𝐼𝑥 )1 = (𝐼𝑥̅ )1 + 𝑚1 (𝑑1 )2
(𝐼𝑥 )1 = 0.084 + 10(0.24)2 = 0.66 𝑘𝑔 − 𝑚2
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Slender Rod. Because 𝐺2 coincides with the origin of the xyz-axes, the moment of inertia of the
rod about the x-axis is obtained directly from the table.
(𝐼𝑥 )2 = (𝐼𝑥̅ )2 =
(𝐼𝑥 )2 =
1
𝑚 𝐿2
12
1
(2) (0.36)2 = 0.0216 𝑘𝑔 − 𝑚2
12
Sphere. According to the table, the moment of Inertia of the sphere about its own central x-axis
is
2
2
(𝐼𝑥̅ )3 = (𝑚) (𝑅)2 = (4) (0.09)2 = 0.02196 𝑘𝑔 − 𝑚2
5
5
Using the parallel-axis theorem, the moment of inertia about the x-axis is given by
(𝐼𝑥 )3 = (𝐼𝑥̅ )3 + 𝑚3 𝑑32
(𝐼𝑥 )3 = 0.02196 + 4(0.27)2 = 0.30456 𝑘𝑔 − 𝑚2
Assembly. The moment of inertia of the assembly about an axis equals the sum of the moments of
inertia of its parts about that axis. Therefore, adding the values that were found above, we get
𝐼𝑥 = (𝐼𝑥 )1 + (𝐼𝑥 )2 + (𝐼𝑥 )3 = 0.66 + 0.0216 + 0.30456 = 𝟎. 𝟗𝟖𝟔𝟏𝟔 𝒌𝒈 − 𝒎𝟐
Referring to the figure, the y -coordinate of 𝐺 is
𝑦̅ =
∑𝑖 𝑚𝑖 𝑦𝑖
=
∑𝑖 𝑚𝑖
𝑦̅ =
10(0.24) + 2(0) − 4(0.27)
10 + 2 + 4
1.32
= 0.0825 𝑚
16
Because 𝑦̅ is the distance between the 𝑥 – 𝑎𝑥𝑖𝑠 and the central 𝑥 – 𝑎𝑥𝑖𝑠 of the assembly,
the moment of inertia of the assembly about the latter axis is found from the parallel-axis
theorem:
(𝐼𝑥̅ ) = 𝐼𝑥 − 𝑚(𝑦̅)2 = 0.98616 − 16(0.0825)2
(𝑰𝒙 ) = 𝟎. 𝟖𝟕𝟕 𝒌𝒈 − 𝒎𝟐
The corresponding radius of gyration is:
𝐼𝑥̅
0.877
𝑘̅𝑥 = √ = √
= 𝟎. 𝟐𝟑𝟒 𝒎
𝑚
16
Example 4a-4. The 290 – kg machine part in the figure is made by drilling an off-center, 160 –
mm diameter hole through a homogenous, 400 mm cylinder of length 350 mm. Determine 𝐼𝑧
and 𝑘̅𝑧 .
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Solution:
The machine part in the figure can be considered to be the difference between the
homogeneous cylinders A and B shown in the figures respectively. The mass density ρ of the
machine part is,
𝜌=
𝜋(𝑅𝐴2
𝑚
290
3
2 )ℎ = 𝜋(0.202 − 0.082 )(0.35) = 7849 𝑘𝑔/𝑚
− 𝑅𝐵
Consequently, the masses of cylinders A and B are:
𝑚𝐴 = 𝜌𝜋𝑅𝐴2 ℎ = (7849)𝜋(0.2)2 (0.35) = 345.2 𝑘𝑔
𝑚𝐵 = 𝜌𝜋𝑅𝐵2 ℎ = (7849)𝜋(0.08)2 (0.35) = 55.2 𝑘𝑔
(𝐼𝑧 )𝐴 = (𝐼𝑍̅ )𝐴 =
1
1
𝑚𝐴 𝑅𝐴2 = (345.2)(0.20)2 = 6.904 𝑘𝑔 − 𝑚2
2
2
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(𝐼𝑍̅ )𝐵 =
1
1
𝑚 𝑅2 = (55.2)(0.080)2 = 0.177 𝑘𝑔 − 𝑚2
2 𝐵 𝐵
2
Because the distance between the z-axis and the central z-axis of B is d = 0.11 m, the moment of
inertia of B about the z-axis is found from the parallel-axis theorem:
(𝐼𝑧 )𝐵 = (𝐼𝑧̅ )𝐵 + 𝑚𝐵 𝑑2 = 0.177 + (55.2)(0.11)2 = 0.845 𝑘𝑔 − 𝑚2
𝐼𝑧 = (𝐼𝑧 )𝐴 − (𝐼𝑧 )𝐵 = 6.904 − 0.845 = 𝟔. 𝟎𝟓𝟗 𝒌𝒈 − 𝒎𝟐
The corresponding radius of gyration is,
𝑘̅𝑧 = √
𝐼𝑧̅
5.932
= √
= 𝟎. 𝟏𝟒𝟑𝟎 𝒎
𝑚
290
Let’s Check!
1. The 5O-kg wheel has a radius of gyration about its mass center 0 of ko = 400 mm.
Determine its angular velocity after it has rotated 20 revolutions starting from rest.
2. The uniform 5O-lb slender rod is subjected to a couple moment of M = 100 lb · ft. If the
rod is at rest when 𝜃= 0°, determine its angular velocity when 𝜃 = 90°.
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3.
The uniform 50-kg slender rod is at rest in the position shown when P = 600 N is applied.
Determine the angular velocity of the rod when the rod reaches the vertical position.
Let’s Analyze!
1. If the uniform 30-kg slender rod starts from rest at the position shown, determine its
angular velocity after it has rotated 4 revolutions. The forces remain perpendicular to the
rod.
2. The 20-kg wheel has a radius of gyration about its center 0 of ko = 300 mm. When it is
subjected to a couple moment of M = 50 N · m, it rolls without slipping. Determine the
angular velocity of the wheel after its center 0 has traveled through a distance of So = 20
m, starting from rest.
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3. At a given instant the body of mass m has an angular velocity w and its mass center has a
velocity ve. Show that its kinetic energy can be represented as T = 1Ircw2, where Irc is the
moment of inertia of the body computed about the instantaneous axis of zero velocity,
located a distance 'e/lc from the mass center as shown.
In a Nutshell!
1. A force of P = 20 N is applied to the cable, which causes the 175-kg reel to turn without
slipping on the two rollers A and B of the dispenser. Determine the angular velocity of the
reel after it has rotated two revolutions starting from rest. Neglect the mass of the cable.
Each roller can be considered as an 1 8-kg cylinder, having a radius of 0.1 m. The radius
of gyration of the reel about its center axis is ke = 0.42 m.
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2. The spool of cable, originally at rest, has a mass of 200 kg and a radius of gyration of ke =
325 mm. If the spool rests on two small rollers A and B and a constant horizontal force
of P = 400 N is applied to the end of the cable, determine the angular velocity of the
spool when 8 m of cable has been unwound. Neglect friction and the mass of the rollers
and unwound cable.
Course Schedule:
This section calendars all the activities and exercises including readings and lectures as well as
time for making assignments and doing other requirements in a programmed schedule by days
and weeks, to help the students in SDL pacing, regardless of mode of delivery (OBD or DED):
Activity
Big Picture 4: Let’s Check Activities!
Big Picture 4: Let’s Analyze Activities!
Big Picture 4: In a Nutshell Activities!
Big Picture 4: Q and A List!
Final Exam
Date
Oct. 15-16,
2020
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