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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 𝐺 𝑆 = 𝑌(𝑠) 𝐹(𝑠) = 1 𝑚 𝑠 2 +𝑏𝑠+𝑘 f 𝑚 𝑦 =−𝑏 𝑦 −𝑘𝑦+𝑓 𝑚 𝑠 2 +𝑏𝑠+𝑘 𝑌 𝑠 =𝐹(𝑠)
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MATLAB Simulation Mass Spring Dashpot System
Transfer function 𝐺 𝑆 = 𝑌(𝑠) 𝐹(𝑠) = 1 𝑚 𝑠 2 +𝑏𝑠+𝑘 m=1, k=1 Case study b=1 (underdamped <1) b=2 (critically damped =1) b=3 (over damped >1) f num = 1 den = [1 b 1] sys = tf(num, den) step(sys)
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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 diagram
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Cruise Control Model Example 2.1 Applying Newton’s 2nd law
Since v= 𝑥 , 𝑣 = 𝑥 𝑣 + 𝑏 𝑚 𝑣= 𝑢 𝑚 - Transfer function 𝑚 𝑥 =−𝑏 𝑥 +𝑢 𝑥 + 𝑏 𝑚 𝑥 = 𝑢 𝑚 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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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 𝑌(𝑠) 𝑅(𝑠)
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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 𝑀=𝑟×𝐹 𝑀= 𝑥 𝑦 𝑧 𝑟 𝑥 𝑟 𝑦 𝑟 𝑧 𝐹 𝑥 𝐹 𝑦 𝐹 𝑧
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Pendulum Applying Newton’s 2nd law for rotational motion, M=I
Equation of motion Linear equation of motion Is it reasonable assumption?
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SIMULINK of Pendulum Linear model Nonlinear model
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SIMULINK Mass Spring Dashpot System
Dynamic equation of motion Draw the block diagram 𝑚 𝑦 =−𝑏 𝑦 −𝑘𝑦+𝑓 f
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SIMULINK Mass 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) 𝑚 𝑦 =−𝑏 𝑦 −𝑘𝑦+𝑓 f 𝑦 𝑦
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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
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Combined Motion: Rotational and Translational Motion
Position of the center of gravity of the pendulum rod Rotational motion of pendulum Free body diagram
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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
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Combined Motion: Rotational and Translational Motion
For a small angle Free body diagram
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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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