Lecture 3: Dynamic Models
• Dynamics of Mechanical Systems
─ Newton’s 2nd law
o 𝐹 = 𝑚𝑎 (translation)
o 𝑀 = 𝐼𝛼 (rotation)
─ Mass-Spring-Damper Model
o Mass (m)
o Spring (spring force 𝐹𝑠 = 𝑘𝑥)
o Damper (damping force 𝐹𝑑 = 𝑏𝑥)
Spring: stores energy
Damper: dissipates energy
Newton’s 2nd Law: Translational Motion
• Newton’s 2nd law governs the relation between acceleration and force
• Acceleration is proportional to force, and inversely proportional to mass
𝐹𝑛𝑒𝑡 =
𝐹 = 𝑚𝑎
where,
 𝐹𝑛𝑒𝑡 : the vector sum of all forces applied to each body in a
system (N)
 𝑎:
the vector acceleration of each body w.r.t. an inertial
reference frame (m/sec2)
 𝑚:
mass of the body (kg)
Newton’s 2nd Law: Rotational Motion
• Newton’s 2nd law governs the relation between angular acceleration
and moment (torque)
• Angular acceleration is proportional to moment, and inversely
proportional to moment of inertia
𝑀𝑛𝑒𝑡 =
𝑀 = 𝐼𝛼
where,
 𝑀𝑛𝑒𝑡 : the sum of all external moments about the center of mass of a
body in a system, (Nm)
 𝛼:
the angular acceleration of the body w.r.t. an inertial reference
frame (rad/sec2)
 𝐼:
body’s moment of inertia about its center of mass (kgm2)
Moment of Inertia I
• It is a measure of an object’s resistance to changes to its rotation.
Equivalent to mass of an object.
• It should be specified with respect to a chosen axis of rotation.
Moment of Inertia I
• Moment of inertia becomes smaller when mass is concentrated on the
axis of rotation
Moment of Inertia I
Rotation in the
middle of bar
Distributed mass
Lumped mass
L
L
L
m
m
m
Moment of Inertia I
Rotation in the
middle of bar
Distributed mass
Lumped mass
L
L
L
m
m
m
𝐼 = 𝑚𝐿2
1 2
𝐼 = 𝑚𝐿
3
1
𝐼=
𝑚𝐿2
12
Spring Model
Two springs in parallel
Two springs in series
Spring Model
Two springs in parallel
When 𝑘 = 𝑘1 = 𝑘2, 𝑘𝑒𝑞 = 2𝑘
Two springs in series
When 𝑘 = 𝑘1 = 𝑘2, 𝑘𝑒𝑞 = 0.5𝑘
Spring and Damper Model
m
?
?
m
Mass Spring Damper System
• Derive equation of motion
• Transfer function 𝐺 𝑆
• Input: force f
• Output: displacement y
Mass Spring Damper System
• Applying Newton’s 2nd law,
𝑚𝑎 =
𝐹
𝑚𝑦 = −𝑏𝑦 − 𝑘𝑦 + 𝑓
• Taking the Laplace transform
𝑚𝑠 2 + 𝑏𝑠 + 𝑘 𝑌 𝑠 = 𝐹(𝑠)
• Transfer function
𝐺 𝑆 =
𝑌(𝑠)
𝐹(𝑠)
=
1
𝑚𝑠 2 +𝑏𝑠+𝑘
MATLAB Simulation: Mass Spring Dashpot System
• Transfer function 𝐺 𝑆 =
𝑌(𝑠)
𝐹(𝑠)
=
1
𝑚𝑠 2 +𝑏𝑠+𝑘
• 𝑚 = 1, 𝑘 = 1
• Case study
𝑥(𝑡) + 2𝜁𝜔0 𝑥(𝑡) + 𝜔02 𝑥(𝑡) = 𝑢(𝑡)
• 𝑏 = 1 (underdamped  < 1)
• 𝑏 = 2 (critically damped  = 1)
• 𝑏 = 3 (over damped  > 1)
Matlab code:
num = 1
den = [1 b 1]
sys = tf(num, den)
step(sys)
MATLAB Simulation: Mass Spring Dashpot System
• Transfer function 𝐺 𝑆 =
𝑌(𝑠)
𝐹(𝑠)
=
1
𝑚𝑠 2 +𝑏𝑠+𝑘
𝑥(𝑡) + 2𝜁𝜔0 𝑥(𝑡) + 𝜔02 𝑥(𝑡) = 𝑢(𝑡)
• 𝑚 = 1, 𝑘 = 1
• Case study
• 𝑏 = 1 (underdamped  < 1)
• 𝑏 = 2 (critically damped  = 1)
• 𝑏 = 3 (over damped  > 1)
Step Response
1.4
underdamped
critically damped
1.2
overdamped
Matlab code:
Amplitude
num = 1
den = [1 b 1]
sys = tf(num, den)
step(sys)
1
0.8
0.6
0.4
0.2
0
0
5
10
Time (seconds)
15
MATLAB Simulink: Mass Spring Dashpot System
• Transfer function 𝐺 𝑆 =
𝑌(𝑠)
𝐹(𝑠)
=
1
𝑚𝑠 2 +𝑏𝑠+𝑘
• 𝑚 = 1, 𝑘 = 1
𝑥(𝑡) + 2𝜁𝜔0 𝑥(𝑡) + 𝜔02 𝑥(𝑡) = 𝑢(𝑡)
• Case study
• 𝑏 = 1 (underdamped  < 1)
• 𝑏 = 2 (critically damped  = 1)
• 𝑏 = 3 (over damped  > 1)
Step Response
1.4
underdamped
critically damped
1.2
overdamped
Amplitude
1
0.8
0.6
0.4
0.2
0
0
5
10
Time (seconds)
15
Mass Spring Damper System
• Automobile suspension system
Problem: Find the transfer function
Mass Spring Damper System
• Automobile suspension system
• The equation of motion for the system
• Taking the Laplace transform
• Transfer function
Cruise Control Model
• Example 2.1
─ Write the equations of motion
─ Find the transfer function
o Input: force u
o Output: velocity
Cruise control model
Free-body diagram
Cruise Control Model
• Example 2.1
─ Applying Newton’s 2nd law
𝑚𝑎 =
𝐹
𝑚𝑥 = −𝑏𝑥 + 𝑢
𝑏
𝑢
𝑥+ 𝑥=
𝑚
𝑚
─ Since v = 𝑥, 𝑣 = 𝑥
𝑏
𝑢
𝑣+ 𝑣=
𝑚
𝑚
─ Transfer function
Free-body diagram
Cruise Control Model
• MATLAB Simulation
− Transfer function
Matlab code:
Free-body diagram
num = 1/m
den = [1 b/m]
sys = tf(num, den)
step(sys)
Step Response
10
9
8
7
Amplitude
6
Parameter values:
5
4
3
2
𝑢 = 500, 𝑚 = 1000𝑘𝑔, 𝑏 = 50𝑁𝑠/𝑚
1
0
0
20
40
60
Time (seconds)
80
100
120
Combined Motion: Rotational and Translational Motion
• Inverted pendulum mounted cart
− Input: force u
− Output: 
• Derive equations of motion
Unstable system
Combined Motion: Rotational and Translational Motion
• Position of the center of gravity
of the pendulum rod
• Rotational motion of pendulum
Free body diagram
Combined Motion: Rotational and Translational Motion
• Horizontal motion of the center
of pendulum
• Vertical motion of the center of
gravity of pendulum
• Horizontal motion of cart
Free body diagram
Combined Motion: Rotational and Translational Motion
• For small 𝜃,
−sin 𝜃 ≈ 𝜃
−cos 𝜃 ≈ 1
Free body diagram
Week 2, Lecture 3: Reading and Practice
Reading for week 2:
- Franklin Textbook Chapter 2, Dynamic Models:
- 2.1: Dynamics of Mechanical Systems
- 2.2: Models of Electric Circuits
- Modern Control Engineering by K. Ogata
- Chapter 3 Mathematical Modeling of Mechanical Systems and
Electrical Systems