ME 413
System Dynamics
&
Control
Lecture#1
Course Instructor:
Dr. Tajammal Imran.
UHB Main Campus Room #2307,
tajammal@uhb.edu.sa
University of Hafr Al Batin
College of Engineering
Department of Mechanical Engineering
Introduction
to
System Dynamics
and
Control
Lecture#1
Overview to System Dynamics and Control
• Course introduction
• What is expected? Course objectives & outcomes!
• University Rule (& DN Rule)
• Course Rule
• Grading and assessment policy with Exam Schedule
• Lecture duration: [2 Session/Week]
• Week wise syllabus
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Course Objectives (what I should do in this course)
1. To demonstrate the basic modeling methodologies for
dynamic systems.
2. To explain the methods for analyzing dynamic responses.
3. To clarify the classical control techniques using basic
control actions.
4. To deliver the techniques for analyzing systems’ stability.
5. To expose to experimental laboratory applications of
control to various dynamic systems.
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Course outcomes (what student should be able to do)
1. Demonstrate knowledge of the fundamental assumptions related
to the derivation of simple dynamic models for basic engineering
systems.
2. Demonstrate ability to identify dynamic characteristics: natural
frequency, damping, time constant, settling time, etc. of simple
dynamic systems.
3. Demonstrate ability to analyze systems’ dynamic responses, in
both time and frequency domains.
4. Demonstrate knowledge of the basic characteristics,
representations, and utilization of the P, PD, and PID controllers.
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Course Overview
Course code
ME 413
Title
System Dynamics and Control
Lecture
Credit Hour
2-3-3
Lab
Location: Room 0313
Location: Room 0313/0508
Time: 01.00 noon to 01.50 pm (Monday)
10.00 am to 11.40 am (Tuesday)
Time: 02.00 pm to 05.20 pm (Tuesday)
Number of unexcused absences
Warning I Warning II
Lecture of course with 30
lectures /semester
Laboratory session
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Total absences (excused* & unexcused)
DN
DN
2
4
8
10
1
2
3
4
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UHB & Course Rules
• The following should be noted
➢Students must bring textbook, notebook, calculator and pen to the class
➢Attendance in the classes will be taken within five minutes of the
beginning of the class.
i) Any student who arrives class within 5 minutes from the start of class will be
marked as late.
ii)If the student is marked late 3 times, then this is equivalent to 1 unexcused
absence.
iii)Student who arrives after 5 minutes is considered absent with no excuse.
*Note:
Officially authorized excuse of absences must be obtained from Deanship of
Student Affairs and presented to the instructor not later than two days
following the resumption of class attendance.
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Course Description
•
•
•
•
•
•
•
•
•
Dynamics of mechanical, fluid, electrical and thermal systems.
Equations of motion.
Dynamic response of elementary systems.
Transfer functions and pole-zero diagrams.
Simulation of dynamics of complex systems.
Dynamic stability of systems.
Open and closed-loop systems.
Basic control actions.
Laboratory sessions involving use of computers for simulation of
dynamic systems and analysis of control systems.
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Course Rules
• Late Submission of Assignments/Homework/Lab Reports:
•
•
•
•
It is your responsibility to solve the HW as soon as the material is covered in the class.
Homework solution will be posted on blackboard or given in the class.
I will not accept late submission of HW.
All assignments (graded or non-graded) should be submitted in time as per weekly
plan/or as agreed in the class.
• No assignments/reports will be accepted after Two Days from due date.
• Schedule of Exams for ME 413
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Assessment Plan for the Course
Assessment Policy
%
Quizzes 10
Homework 10
Midterm exam 20
Lab Report 10
Individual Lab Project 20
Final examination 30
Total 100%
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Weekly lecture scheme
Week #
Date
Topic
Section/Ref #
Assessment
Introduction to course and introduction to systems dynamics
1
Lab1: General Introduction and Introduction to MATLAB
Introduction to systems dynamics
2
1-1:1-4
Lab2: General Introduction and Introduction to MATLAB
Laplace transform
2-1, 2-2, 2-3
HW: B-2-2, B-2-6, B2-15
2-4, 2-5
HW: B-2-18, B-2-21,
B-2-24, B-2-25
Quiz1
Mechanical systems
3-1, 3-2, 3-3, 3-4
HW: B-3-9, B-3-12,
B-3-13, B-3-14, B-315, B-3-18, B-3-19, B3-20
Lab5: Mechanical Systems (1):
Harmonic Oscillations - Experiment: The Mass-Spring System
Transfer function approach to modeling dynamic systems
4-1, 4-2, 4-3, 4-4
HW: B-4-1, B-4-3, B4-4, B-4-7, B-4-8, B4-12, Quiz2
3
Lab3: General Introduction and Introduction to MATLAB
Laplace transform
4
Lab4: Laplace Transform
5
6
Exam: M1
Lab6: Mechanical Systems (2):
Translational Systems - Simulation: Modeling of a Suspension Car Sys.
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Weekly lecture scheme
7
Electrical and electromechanical systems
6-1, 6-2, 6-3, 6-4,
HW: B-6-4, B-6-7,
B-6-10,
Lab7: Mechanical Systems (3):
Rotational Systems - Simulation: Modeling of Coupled Disks Assembly
Electrical and electromechanical systems plus Fluid & Thermal systems
6-5, 6-6, 7-1, 7-2
HW: B-6-11, B-6-19,
B-7-2, B-7-3,
Quiz3,
Fluid & Thermal systems
7-3:7-6
HW: B-7-4, B-7-5
Lab9: Electromechanical Systems
Time-domain analysis of dynamic systems
8-1, 8-2
HW: B-8-3, B-8-4
8-3, 8-4, 8-5
HW: B-8-7, B-8-8 B8-9, B-8-14 Quiz4
8
Lab8: Transfer Function Approach Using SIMULINK
9
Exam: M2
10
Lab10: Undamped Dynamic Vibration Absorbers
Time-domain analysis of dynamic systems
11
12
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Lab11: CE110 Servo Trainer (1&2):
- Basic Tests and Transducer Calibration
- Response Calculations and Measurements
Frequency-domain analysis of dynamic systems
9-1, 9-2, 9-3, 9-4, 9-5, HW: B-9-1, B-9-2,
9-6
B-9-3, B-9-6, B-9-7,
B-9-12, Quiz5
Lab12: CE110 Servo Trainer (3&4):
- Proportional Control of Servo Trainer Speed
- Proportional Plus Integral Control of Servo Trainer Speed
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Weekly lecture scheme
Time & Frequency domain analysis and design of control systems
13
14
Lab13: CE105 Coupled Tanks (1&2):
- Basic Tests and Transducer Calibration
- Open and Closed Loop Systems
CE105 Coupled Tanks (3&5):
- PI Controller Tuning: Hand Tuning of a Single Tank
- Measurement of System Response Characteristics
Time & Frequency domain analysis and design of control systems
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10-7, 10-8, 10-9, HW: B-10-11, B10-10, 11-1, 11-2, 10-12, B-10-12, B11-3, 11-4
11-2, B-11-4 Quiz6
Lab14: CE105 Coupled Tanks (6):
- Tuning of PI Controllers for a Single Tank System by Damping Factor and Width
Specifications/ Lab Project Presentations
Time & Frequency domain analysis and design of control systems
11-5, 11-6
15
16
10-1, 10-2, 10-3, HW: B-10-2, B-1010-4, 10-5, 10-6, 3, B-10-4, B-10-9,
HW: B-11-8, B-119
Quiz7
Final Exam (comprehensive)
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Motivation for the Course
Car steering system as a dynamic system
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Motivation for the Course
Washing machine system as a dynamic system
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Motivation for the Course
Washing machine system as a dynamic system
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WHAT IS DYNAMIC SYSTEM?
• A dynamic system is a system that is
constantly changing, like the human
body.
• Dynamic systems tend to become
static or attain a state of equilibrium.
• For example, if a car is assumed to be a
dynamic system, then it requires fuel
to continue moving forward or else it
would come to a stop and become
static.
• Dynamic systems' output depends
upon future and past values.
What quantities are changing constantly in these systems that makes them dynamic systems???????
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WHAT IS SYSTEM DYNAMICS?
• System dynamics deals with the mathematical modeling of
dynamic systems and response analyses of such systems
with a view toward understanding the dynamic nature of
each system and improving the system's performance.
• Think about working with
• automated furnace, hardness/tensile
machine
• Human being as system
• Airplane
• Robot
• Computer
• Instrument, tools and equipment
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WHAT IS CONTROL?
• Dynamic systems require a control system to perform properly
• Control involves writing set of rules that commands what a dynamic system
must do. A good example is a robot arm and the mechanical control
mechanism and conveyor system
A robot arm
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Mechanical system for controlling robot arm
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Mechanical Systems
• Systems that have significant mass, inertia and spring and
energy dissipation (damper) components which are driven by
forces, torques, or by specified displacements are known as
mechanical systems.
• An automobile is a good example of a dynamic mechanical
system.
• It has a dynamic response as it speeds up, slows down, or rounds a
curve in the road.
• The body and the suspension system of the car have a dynamic
response of the position of the vehicle as it goes over a bump.
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Electrical Systems
• Electrical systems include circuits with resistive, capacitive,
or inductive components excited by voltage or current.
• Electronic circuits can include transistors or amplifiers.
• Example are;
1. A television receiver has a dynamic response of the beam that
traces the picture on the screen of the set.
2. The TV tuning circuit, which allows you to select the desired
channel, also has a dynamic response.
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Fluid Systems
• Fluid systems employ orifices, restrictions, control
valves, accumulators (capacitors), long tubes
(inductors), and actuators excited by pressure or
fluid flow.
• Example;
• A city water tower has a dynamic response of the height
of the water as a function of the amount of water
pumped into the tower and the amount being used by
the citizens.
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Thermal Systems
• Thermal systems have components that provide resistance
(conduction, convection or radiation) and a capacitance
(mass a specific heat) when excited by temperature or heat
flow.
• Example;
• A heating system warming a house has a dynamic response as
the temperature rises to meet the set point on the thermostat.
• Placing a pot of water over a burner to boil has a dynamic
response of the temperature.
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Mixed Systems
• Real life dynamic systems use two or more of the engineering
disciplines, with energy conversion between various components.
• Figure given below shows several examples.
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Electro-Mechanical Systems
• Systems employing electromagnetic component that converts a current into
a force generally have a dynamic response.
• Examples are
• a loudspeaker in a stereo system,
• a solenoid actuator,
• and electric motors.
• In a loudspeaker, electrical current from the amplifier is transformed into
movement of the speaker cone and the subsequent air pressure fluctuations
that cause us to hear the amplified sound.
• Fluid-Mechanical Systems?????????
• Thermo-Mechanical Systems???????
• Electro-thermal Systems???????????
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Systems of units
• Absolute system – mass is a primary dimension
• Metric system - International System (S.I.), mks (m, kg, s), cgs (cm, g, s)
• British system
• Gravitational system – force is a primary dimension
• Metric Engineering system
• British Engineering system (BES)
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Basic difference btw
absolute and
gravitational system
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EXAMPLE OF DYNAMIC SYSTEM
Spring-mass-damper system mounted on a cart
• Consider the spring-mass-damper
system mounted on a massless
cart
• u(t) is the displacement of the cart
and is the input to the system.
• The displacement y(t) of the mass
is the output.
• In this system, m denotes the
mass, b denotes the viscousfriction coefficient, and k denotes
the spring constant.
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MODELLING OR MATHEMATICAL MODELLING
• Modeling is simplifying the problem sufficiently and applying the appropriate
fundamental principles.
• The resulting mathematical description is called a mathematical model, or just a model
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MODELLING A DYNAMICAL SYSTEM
• In Dynamical Systems the main objective is to understand
behaviour of states in a system, by using a rule for how the state
changes.
• Given the state variable a dynamical system needs a rule which
defines the dynamics or how the system changes.
• For example, Newton's Laws are incredibly accurate under known
situations and formulate a well-defined dynamical system.
• Position and velocity of a particle are the state variable and Newton's
Laws used as the updating rule.
• In this case, a dynamical system is written as a system of differential
equations.
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DIFFERENTIAL EQUATIONS
• Linear differential equations: Linear differential equations may be classified
as linear, time-invariant differential equations and linear, time varying
differential equations.
• A linear, time-invariant differential equation is an equation in which a
dependent variable and its derivatives appear as linear combinations. An
example of such an equation is:
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TERMS AND DEFINITIONS
• System: combination of elements that guarantee harmonious working relationship.
• Static element: element’s output value depends only on its input value e.g., the current
flowing through a resistor depends only on the present value of the applied voltage (V=IR)
• Dynamic element: element’s output value depends on past input value e.g., the present
position of a bike depends on what its velocity has been from the start (v = u + at)
• Static system: one whose output at any given time depends only on the input at that
time. The output of a static system remains constant if the input does not change.
Independent of time.
• Dynamic system: one whose present output depends on past inputs. In a dynamic
system, the output changes with time if the system is not in a state of equilibrium.
• System dynamics: the study of systems that contain dynamic elements
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CONTINUOUS-TIME AND DISCRETE-TIME SYSTEMS
• Continuous-time systems are systems in
which the signals involved are continuous
in time. These systems may be described by
differential equations. Think about
analogue clock
• Discrete-time systems are systems in which
one or more variables can change only at
discrete instants of time. Think about
digital clock
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SUMMARY AND CONCLUSIONS
• Why knowledge of system dynamics is important
• Types of differential equations
• Procedure for mathematical modeling
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Laplace Transform
Overview to Laplace Transformation
•What is Laplace transformation and why
it is needed?
•Overview of Differential and Algebraic
equations
•Overview of Complex Numbers
•Different input functions
•Application of Laplace transformation
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What is Laplace Transformation?
1. The Laplace transform method is an operational method that can be used
advantageously in solving linear, time-invariant differential equations.
2. Its main advantage is that differentiation of the time function links to
multiplication of the transform by a complex variable “ s ”, and thus the
differential equations in time become algebraic equations in “ s ”.
3. The solution of the differential equation can then be found by using a
Laplace transform table or the partial-fraction expansion technique.
4. Another advantage of the Laplace transform method is that, in solving the
differential equation, the initial conditions are automatically taken care of,
and both “the particular and the complementary” solution can be
obtained simultaneously.
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Why Laplace Transformation?
• Analysis
• Investigation of the performance of a dynamical
system under specified conditions.
• The most crucial step is the mathematical model.
• Design
• Process of finding a system that accomplishes a task.
• Synthesis
• Finding a system which will perform in a specified
way.
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Why Laplace Transform?
•Helps transform model from time-domain (t)
to frequency domain (s)
•It helps reduce complex differential
equations into simpler algebraic equations
•Final solution can be easily converted to
time-domain
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Differential and Algebraic Equations
Algebraic equations:
(linear, quadratic, polynomial, complex)
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Differential equations:
(first, second and multiple order
or partial)
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Complex Algebra
• Review textbook pages from page 8-13 to refresh the following;
• Equality of complex numbers
• Addition of complex numbers or complex with real number
• Subtraction of complex numbers or complex and real number
• Multiplication of complex numbers or complex and real number or j
• Division of complex numbers or complex and real number or j
• Multiplication and division in polar form
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Practice Problem#1
• Given the following variables:
A = 3.4 + 8.5  j
evaluate the followings
1.
➢
➢
&
B = 1.2−3.9  j
Transform A and B into rectangular and polar form
2. A + B
3. A − B
4. A  B
1. A / B
2. j  A
3. j  B
4. A / j
Represent (ii) and (iii) on a complex plane by sketching their plots
Transform (vi) and (vii) from rectangular form to polar form
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Complex Number & Complex Plane presentation
Exercise 1. Convert z = 4.5 + j  3.8 into polar form
Exercise 2. Convert z = 1.5  𝒆 𝒋𝟐𝟔 into rectangular form
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Exercise 2
Convert z = 4.5 + j  3.8 into polar form
For a complex number x + j  y
• |x + j . y|= z = (x2 + y2 )1/2
So, x = 4.5 &
&
tan= y/x
y= 3.8
&
z = (x2 + y2 )1/2 = ( 20.25 + 14.44 )1/2 =5.9
&
tan= 3.8/4.5   = tan-1 (0.844) = 40.2o
So, the rectangular form representation is;
r (cos  + j sin )
5.9 (cos 40.2 + j  sin 40.2)
z = 4.5 + j  3.8 = 5.9 (cos 40.2 + j  sin 40.2)
The above answer is in “degree” mode
So, the polar form representation is;
z = 5.9 𝒆 𝒋𝟒𝟎.𝟐
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Convert z = 1.5 𝒆 𝒋𝟐𝟔 into rectangular form
We know that,
1.5 𝑒 𝑗26 = 1.5 cos 26 +1.5 j  sin 26
Or,
z = z 𝒆 𝒋 = 1.5 𝒆 𝒋𝟐𝟔
or
x = z cos  &
y = z sin 
So,
x = 1.5 cos 𝟐𝟔
&
y = 1.5 sin 26
x= 1.4
&
y = 0.66
x + j  y = 1.4 + j  0.66
Finally, z in rectangular form is;
z = 1.5 𝒆 𝒋𝟐𝟔 = 1.4 + j  0.66
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Complex Conjugate of Complex Number
Complex conjugates are 90 degree apart
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Multiplication & Division by Imaginary
Number “j “ & its Graphical Representation
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Complex Variable & Complex Function
• Complex variable – both real and imaginary components are variables
• Complex function – a function with real and imaginary parts
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Zeros & Poles of a Transfer Functions
• Poles and Zeros of a transfer function are the frequencies for which the value
of the denominator and numerator of transfer function becomes zero
respectively.
• The values of the poles and the zeros of a system determine whether the
system is stable, and how well the system performs
• Zeros ; when F(s) =0; in which numerator is zero
• Poles; when F(s) =; in which denominator is zero (division by zero)
• Simple and multiple poles may exist in a function
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Find Zeros & Poles of the following:
Zeros are
s = -2
&
s = -10
&
Poles are
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s = 0,
s = -1, s = -5, & s = -15
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Definitions & symbols of Laplace Transform
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Definitions & Symbols of Laplace Transform
Inverse Laplace transform
Note:
The Laplace transform of a function is a
complex function of a complex variable.
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Laplace Transforms: Practice
What is the Laplace Transforms of the following?
7
7t
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Laplace Transform of an Exponential Function
Repeat the procedure for positive alpha although – (alpha) indicates an exponential decay
Practice Exercise: Find Laplace transform for f(t) = 4 e-7t
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Laplace Transform of Step Function
Repeat the procedure for unit-step function
Unit step function
Practice Exercise: Find Laplace transform for f(t) = 7
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Laplace Transform of Ramp Function
Recall that integration by part is evaluated as:
Practice problem: Solve for f(t) = A  t2
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Laplace Transform of Sinusoidal Function
Recall:
Now show that
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Laplace Transform of Translated Function
One definition of "to translate" is "to change from one place, state, form, or appearance to
another". When we take a function and pull its rule so that its graph is moved to another spot on
the axis system, yet remains visibly the same graph, we are said to be "translating" the function.
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Laplace Transform of Pulse Function:
Combine Unit Step Functions
Impulse
Function
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Exercise: Find Laplace transform for f(t) = 8  t  e-2t
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Integration theorem
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Inverse Laplace transform
Find the inverse Laplace transform for the following
𝟏
𝒔
𝑨
𝒔𝟐
𝑨
𝒔𝟑
𝟑
𝒔+𝟒
𝟐
𝒔
𝒔 𝟐 + 𝟏𝟔
𝒔+𝟐
𝒔 + 𝟐 𝟐 + 𝟏𝟔
𝟑
𝒔+𝟒
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Inverse Laplace transformation:
Transformation from s-domain to t-domain
Inverse Laplace transform
• Two ways of doing inverse Laplace transformation
1. Using the Table of Laplace transformation
2. Using partial fraction expansion
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1. Based on Laplace table
2. Based on partial fraction
• Where A and B are polynomials, hence partial fraction must be used
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Partial expansion when F(s) involves distinct poles
If the highest degree of the denominator is higher than numerator, then:
Take LCM of the RHS to eliminate the denominator.
Then substitute s = -1 to obtain a1 or s = -2 to obtain a2.
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• If the order of numerator is higher, then we must factor the polynomials
e.g. {use: conv, deconv in MATLAB}
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Initial and Final Value Theorem
Initial Value Theorem:
To find the value of f(t) at t = 0+ directly
from the Laplace transform of f(t).
Final Value Theorem:
To find the value of f(t) at t =  directly from
the Laplace transform of f(t).
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Initial and Final Value Theorem
• What are initial and final value theorems?
• How, the initial and final values of a function can be, found by using IVT & FVT?
• Example 1 : Initial Value Theorem
Find the initial value for the function f (t) = 2 u (t) + 3 cost u (t)
Solution:
By initial value theorem
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Initial and Final Value Theorem
• How the initial and final values of a function can be, found by using IVT & FVT?
• Example 2 : Final Value Theorem
1
Find the final value for the function F (s) =
𝑠 𝑠+4
Solution: By final value theorem, If f(t) and f'(t) both are
Laplace transformable and s F(s) has no pole in j axis and in
the R.H.P. (Right half Plane) then
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Differential theorem
𝒅𝟒
L[ 𝟒 𝒇(𝒕)]
𝒅𝒕
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ሶ
ሷ
= 𝒔𝟒 𝑭(𝟎) − 𝒔𝟑 𝒇(𝟎) − 𝒔𝟐 𝒇(𝟎)
− 𝒔𝒇(𝟎)
−ሸ
𝒇(𝟎)
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Complex roots
Recall:
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Partial expansion for multiple poles
Find the Laplace transform of :
For partial fraction of multiple poles, recall :
Take LCM of the RHS to eliminate the denominator.
Then substitute s = -1, 0 and 1 to obtained b1, b2 and b3
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Solution:
[Use partial fraction]
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Homework#1
•A-2-1, A-2-4, A-2-5
•B-2-2, B-2-5, B-2-6
•B-2-23, B-2-24, B-2-25
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Solving Linear, Time-Invariant Differential Equations
• Two basic steps:
1. By taking the Laplace transform of each term in the given differential
equation, convert the differential equation into an algebraic equation in s
and obtain the expression for the Laplace transform of the dependent
variable by rearranging the algebraic equation.
2. The time solution of the differential equation is obtained by finding the
inverse Laplace transform of the dependent variable.
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Practice Problems
Recall:
The eq. becomes:
Collect the like terms:
Express as partial equation:
Take inverse of Laplace:
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Practice Problems: Do it yourself
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Addition of Functions (superimposition)
In engineering applications, functions whose values change abruptly at specified values of time t, are very common. One
example is when a voltage is switched on or off in an electrical circuit at a specified value of time t.
By adding ordinates (y-values) of the 3 parts as follows:
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This is Laplace integration by batch over a PERIOD
Always consider the extremes of the impulse/steps
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Practice Problems on Functions
• Determine the Laplace transform of each of the following functions shown below using first principle.
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ME 413
System Dynamics
&
Control
Lecture#3
Course Instructor:
Dr. Tajammal Imran.
UHB Main Campus Room #2307,
tajammal@uhb.edu.sa
University of Hafr Al Batin
College of Engineering
Department of Mechanical Engineering
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Concept of mass and force
• Mass of a body is the quantity of matter in it, which is assumed to be
constant. Mass gives a body its inertia (ability to move or stop)
• Weight is the magnitude of the force that the earth exerts on the body . 1 N is
the force that will give a 1-kg mass an acceleration of 1 m/s2.
• Far out of space where body becomes weightless, what will be the mass of the body?
• Force is the cause which tends to produce a change in motion of a body on
which it acts. Force could be contact or non-contact (field force). Recall
Newton’s 2nd law of motion.
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MECHANICAL ELEMENTS
1. Spring elements
2. Damper elements
3. Inertia elements
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Inertial elements
• This indicates masses and moments of inertia.
• Inertia is the change in force (torque) required to make a unit change
in acceleration (angular acceleration). Recall Newton’s 2nd law of
motion.
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Spring elements: Deformable body
➢ Linear spring is a mechanical element that can be deformed by an external force or torque such that the
deformation is directly proportional to the force or torque applied to the element. It stores energy when
deformed and release same when expand.
➢ Spring constants indicate stiffness; a large value of k corresponds to a hard spring, a small value of k to
a soft spring. Reciprocal is termed compliance
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Obtain equivalent spring constant for each of the following systems:
Force F is distributed on the 2 springs
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Same force F acts on each of the 2 springs
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Torsional spring
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Damper element
A damper is a mechanical element that dissipates energy in the form of heat instead of
storing it.
b is viscous friction constant
Dash pot
Piston
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Oil-filled cylinder
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Linear/Non-linear spring
Hooke’s law for linear displacement
Linear and Nonlinear
friction law (Viscosity)
Characteristic curve for square-law friction.
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Mathematical Modeling of Simple Mechanical Systems
• Rigid body: a body with negligible deformation upon subjected to an external force.
• Newton’s Laws:
• First Law: Total momentum of a mechanical system is constant in the absence of
externa force: m( v – u) = 0
• Second Law: Rate of change of momentum is proportional to applied force. This
is force-acceleration relationship: F = ma. When this force is zero, then we have
the first law. m( v – u ) / t = F
or
ma=F
• Third law: To every action, there must be equal and opposite reaction.
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Newton’s second law
• The most useful and most applied of the three laws because the
other laws are special cases of the second law
• For translational motion where force is applied along the center of
mass:
• For Rotational motion
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Moment of Inertia about the geometric axis
• For a rigid body capable of undergoing rotation:
“dm” is the elemental mass at distance “r” from the axis
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Moment of Inertial about an axis other than the
geometric axis
Moment of inertial about an axis = moment of inertial about
the geometrical axis + moment of inertial about the new axis
Moment of inertia for a point mass
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J = m  r2
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Concepts of Natural & Forced Response with Natural Frequency
• Forced response: Behaviour or response due to a forcing
function or external force
• Natural response: behaviour due to initial condition without
external force
• Transient period: Period between beginning and end of a
response
• Steady state: A state where response change is negligibly
small.
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Natural Response for Translating system (spring-mass)
• Two forces: spring force (k  y upwards) and gravitational
force (mg, downward)
• So, when the spring is pulled downward by acceleration
y’’, then
Newton’s 2nd law is:
ky
mg
Since the spring force k and the gravitational force mg balance, or k = mg
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Free vibration of spring-mass system
Take Laplace transform
Take inverse Laplace transform
Plot the relationship btw x(t) and t,
and compare with damper
Given initial condition:
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Natural Response for rotational system: a rotor mounted in bearings
• This is a rotor rotating on 2 bearings.
• Assuming that the torque on the rotor due to frictional force is :
Then from Newton’s 2nd law:
Take Laplace transform
Take inverse Laplace transform
Time constant T depicts the time at which the exponential becomes -1
In other words, when the time t in seconds is equal to the time constant, the exponential factor is reduced to approximately 36.8% of its initial value, as shown in figure.
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Free Response of mass-spring-damper system
Given m = 0.1 kg, k = 4 N/m, b = 0.4 kg/s
Take Laplace transform
Take inverse Laplace transform
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1. The problem is now dynamic with initial value, and time dependent parameters.
2. “T” is negative because it must act opposite to the direction of rotation to stop the
shaft at 120s. So, we want T at  (120).
𝑻
 (t)=  𝟎 − t
𝑱
t=120s, J=6, T=?
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Solution of Problem A-3-4
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Solution of Problem A-3-5
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Sample Problem
• In Figure, the simple pendulum shown consists of
a sphere of mass m suspended by a string of
negligible mass.
• Neglecting the elongation of the string, find a
mathematical model of the pendulum.
• In addition, find the natural frequency of the
system when  is small.
• Assume no friction.
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l*cosθ
l*sinθ
139
• Find the mathematical model and natural frequency of the system
acosθ
Take moment about the fixed point:
F
l*cosθ
y
asinθ
For small θ; cos θ=1, sin θ= θ
l*sinθ
x
l*sinθ
l
l
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How to find k ???????????
How b/2000 =
1. The spring is massless, and it is not connected to any mass so m=0
2. The force F is applied to calculate the spring constant by creating an extension of 0.05m.
3. F is removed to put the spring into vibration, so it a one-time applied force.
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Homework-3
• B-3-13, B-3-14, B-3-15, B-3-18, B-3-20
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WORK, ENERGY AND POWER
• If force is considered a measure of effort,
• then work is a measure of accomplishment and
• energy is the ability to do work, and
• power is work done per unit time.
• Work done W is force x distance moved by the force
• The energy that a body possesses because of its position
is called potential energy, whereas the energy that a
body has as a result of its velocity is called kinetic energy.
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Potential Energy stored in a spring
• Potential energy.
• Energy stored in a spring
• When a translational spring is compressed or expand from
length x1 to x2
• For torsional spring:
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Kinetic Energy stored in an inertial element
• For a translational inertial element of mass m moving with a velocity v,
the K.E. is:
• For torsional inertial element:
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Energy dissipated or work done by a damper
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Power
• This is the rate of doing work
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An energy method for deriving equations of Linear motion
• conservative systems – no frictional loss is involved, so no energy enters or
leaves the system, thus:
• When no work is done:
• With no friction:
• If dx/dt not zero, then:
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𝑑 2
𝑑
𝑑
𝑑
𝑥ሶ =
𝑥ሶ ∗ 𝑥ሶ = 𝑥ሶ
𝑥ሶ + 𝑥ሶ
𝑥ሶ = 2𝑥ሶ 𝑥ሷ
𝑑𝑡
𝑑𝑡
𝑑𝑡
𝑑𝑡
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An energy method for deriving equations of Angular motion
l  cosθ
l*sinθ
l  sinθ
1 2ሶ
𝐾𝑖𝑛𝑒𝑡𝑖𝑐 𝑒𝑛𝑒𝑔𝑦 𝑇 = 𝐽𝜃
2
Potential energy: 𝑈 = 𝑚𝑔ℎ = 𝑚𝑔 𝑙 − 𝑙𝑐𝑜𝑠𝜃 = 𝑚𝑔𝑙(1 − 𝑐𝑜𝑠𝜃)
1
𝑇𝑜𝑡𝑎𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 = 𝑇 + 𝑈 = 𝐽𝜃 2ሶ + 𝑚𝑔𝑙(1 − 𝑐𝑜𝑠𝜃)
2
𝑑
ሶ
𝑇 + 𝑈 = 0 = 𝐽𝜃ሶ 𝜃ሷ + mgl𝜃𝑠𝑖𝑛𝜃
=0
𝑑𝑡
(𝐽𝜃ሷ + mgl 𝑠𝑖𝑛𝜃)𝜃ሶ = 0
𝑔
ሷ𝜃 + 𝜃 = 0
𝑙
Through Energy method
Through Newton’s equation
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The cylinder rotates and translates due to the attached spring, so the K.E. due
to inertial mass:
The potential energy due to the translational spring:
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Refer to textbook: page 97
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End of the chapter
• Questions ????
Quiz#2 ??????????
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