newton’s 2 nd law: translational motion newton’s 2 nd law governs the relation between...
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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
F=ma
where, • F = the vector sum of all forces applied to each body in
a system, newton (N)• a = the vector acceleration of each body w.r.t. an
inertial reference frame (m/sec2)• m = mass of the body (kg)
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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
M=I
where, • M = the sum of all external moments about the center
of mass of a body in a system, (N-m)• = 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
(kg-m2)
I
M
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Mass Spring Dashpot System• Applying Newton’s 2nd law,
• Taking the Laplace transform
• Transfer function
𝑚�̈�=−𝑏 �̇�−𝑘𝑦+ 𝑓f
(𝑚𝑠2+𝑏𝑠+𝑘 )𝑌 (𝑠 )=𝐹 (𝑠)
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MATLAB SimulationMass Spring Dashpot System
• Transfer function
• m=1, k=1
• Case study– b=1 (underdamped <1)– b=2 (critically damped =1)– b=3 (over damped >1)
f
num = 1den = [1 b 1]sys = tf(num, den)step(sys)
40 5 10 15
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Step Response
Time (seconds)
Am
plitu
de
underdamped
critically damped
overdamped
Cruise Control Model
• Example 2.1– Write the equations of motion– Find the transfer function
• Input: force u• Output: velocity
Cruise control model Free-body diagram5
Cruise Control Model
• Example 2.1– Applying Newton’s 2nd law
– Since
- Transfer function
Free-body diagram
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Cruise Control Model
• MATLAB Simulation
– Transfer function– num = 1/m– den = [1 b/m]– Sys = tf(num*u, den)– Step(sys)
Parameter values: u=500, m=1000kg, b=50Ns/m
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0 20 40 60 80 100 1200
1
2
3
4
5
6
7
8
9
10Step Response
Time (seconds)
Am
plitu
de
A Two-Mass System: Automobile Suspension
• Two masses: Car (m2) and Tire (m1)• Problem
– Write equations of motion for the automobile and wheel motion
– Find the transfer function
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A Two-Mass System: Automobile Suspension
• Free body diagram of each body
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Rotational Motion: Pendulum
• Example– Derive equation of motion
• Nonlinear equation• Linear approximation
– Find transfer function
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Moment
• Moment
Pendulum
– Applying Newton’s 2nd law for rotational motion, M=I
– Equation of motion
– Linear equation of motion
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Is it reasonable assumption?
SIMULINK of Pendulum
Linear model
Nonlinear model
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SIMULINKMass Spring Dashpot System
• Dynamic equation of motion
• Draw the block diagram
f
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𝑚�̈�=−𝑏 �̇�−𝑘𝑦+ 𝑓
SIMULINKMass Spring Dashpot System
• Dynamic equation of motion• m=1, k=1• Case study
– b=1 (underdamped <1)– b=2 (critically damped =1)– b=3 (over damped >1)
f
150 5 10 15
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Step Response
Time (seconds)
Am
plitu
de
underdamped
critically damped
overdamped
𝑚�̈�=−𝑏 �̇�−𝑘𝑦+ 𝑓
y
1
m
1
k
f
1
b
1s
Integrator1
1s
Integrator
�̈� �̇�
Rotational Motion: Satellite Attitude Control Model
• Example– Derive the equation of motion– Find transfer function
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Rotational Motion: Satellite Attitude Control Model
• Example– Applying Newton’s 2nd law for
rotational motion, M=I
• Find transfer function
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Combined Motion:Rotational and Translational Motion
• Inverted pendulum mounted car– 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 a small angle
Free body diagram
Flexible Read/Write for a Disk Drive
• Examples– Find the equations of motion– Find transfer function
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Flexible Read/Write for a Disk Drive
• Equations of motion
• Transfer function
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