1. Equations of Motion: Kinetic energy: Potential energy: Sin≈
SYSTEM MODELING AND ANALYSIS
Example: Obtain the equations of motion of the simple pendulum shown in the figure. Find the undamped natural frequency of the pendulum.
Kinetic energy:
Potential energy:
There is no external excitation and damping.
Sin≈

2. Kinetic energy: Potential energy: O SYSTEM MODELING AND ANALYSIS
Example: Obtain the equations of motion of the compound pendulum shown in the figure. Find the undamped natural frequency.
Kinetic energy: G O θ
m, L, IO L L1 g
Potential energy:
Where IO is the mass moment of inertia about point O. For small angular displacements
sin θθ

3. Material: Plain carbon steel L1=0.170299 m
SYSTEM MODELING AND ANALYSIS
L1= m
Material:
Plain carbon steel

4. SYSTEM MODELING AND ANALYSIS

5. Yücel Ercan, İleri Dinamik 2014.
SYSTEM MODELING AND ANALYSIS
Example: Obtain the equations of motion of the mechanical system shown in the figure.
Yücel Ercan, İleri Dinamik 2014. θ x M1
Solution: System has two degree of freedom, namely θ and x.
Pendulum rod
Bob
Tranlational mass
Kinetic energy:
Potential energy:
Virtual work:

6. Lagrange’s equation for θ:
SYSTEM MODELING AND ANALYSIS
Potential energy:
Kinetic energy:
Virtual work:
Lagrange’s equation for θ:

7. Lagrange’s equation for x:
(For constant M)
In matrix form;

8. Lagrange’s equation for θ:
SYSTEM MODELING AND ANALYSIS
Example: Obtain the equations of motion of the mechanical system shown in the figure. x
Solution:
(Kinetic energy of the beam is zero due to its negligible mass)
Lagrange’s equation for θ:

9. Lagrange’s equation for x:
SYSTEM MODELING AND ANALYSIS
Lagrange’s equation for x:
In matrix form;

10. Potential energy of pendulum is neglected
SYSTEM MODELING AND ANALYSIS
Example: Obtain the equations of motion of the mechanical system shown in the figure. G
Inputs: f, T, x4
Outputs
Potential energy of pendulum is neglected

11. SYSTEM MODELING AND ANALYSIS

12. SYSTEM MODELING AND ANALYSIS

13. SYSTEM MODELING AND ANALYSIS
In matrix form:

14. EIGENVALUE EQUATION
(Free vibration)
Eigenvalue equation
Eigenvalue equation:

15. Eigenvalues: -34.163+39.572i, -34.163-39.572i, -44.572, -393.02
Eigenvalue equation:
m=0.85 kg, L =0.24 m, k=1200 N/m, c=20 Ns/m
Matlab code:
This value is different in the original presentation
clc;clear
m=0.85;l=0.24;k=1200;c=20;
M=[m*l^2/8,0;0,3*m/4]; %Mass matrix
C=[27*c*l^2/16,-9*c*l/4;-9*c*l/4,6*c]; %Damping matrix
K=[9*k*l^2/8,-3*k*l/2;-3*k*l/2,4*k]; %Stiffness matrix
syms s; %Symbolic variable definition
mtr=M*s^2+C*s+K;
p=solve(det(mtr)) %Eigenvalues of the system
vpa(p,5)
Eigenvalues: i, i, ,

16. Eigenvalues: -34.163+39.572i, -34.163-39.572i, -44.572, -393.02
θh(t) = C1e( i)t + C2e( i)t + A2e t + A3e t
(Complex coefficients)

17. Form of the free vibration response:
Real eigenvalues correspond to an exponential time response
Complex eigenvalues correspond to an exponential-harmonic time response
A1, φ1, A2 and A3 can be calculated from initial conditions
The output reaches to zero as time t goes infinity.
If the real parts of all eigenvalues are negative than the system is stable.
rad/s ω0 iω -σ φ
p=-σ+iω
For the root Δt=0.0071, t∞=0.141
p=-σ
For the root Δt= , t∞=0.016
For the system Δt= , t∞=0.1839
Use smallest t and largest t

18. 1
0.2
0.5
ξ=0.1 t
x(t)
x(t) t 5 3