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
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. a=[0.0039,3.75,847,141834,2799360];p=roots(a);vpa(p,4)
Eigenvalue equation:
m=0.85 kg, L =0.24 m, k=1200 N/m, c=38 Ns/m
Matlab code:
a=[0.0039,3.75,847,141834, ];p=roots(a);vpa(p,4)
clc;clear
m0=0.85;l0=0.24;k0=1200;c0=38;
m=[m0*l0^2/8,0;0,3*m0/4];
c=[27*c0*l0^2/16,-9*c0*l0/4;-9*c0*l0/4,6*c0];
k=[9*k0*l0^2,-3*k0*l0/2;-3*k0*l0/2,4*k0];
syms s;p=solve(det(m*s^2+c*s+k));vpa(p,4)
Eigenvalues: i, i, -22.4,

16. Eigenvalues: -104.3+181.4i, -104.3-181.4i, -22.4, -730.2
Form of the free vibration response:
A1, φ1, A2 and A3 can be calculated from initial conditions
The output reaches zero as time t goes infinity. The value of the steady-state reponse will be zero.
If the real parts of all eigenvalues are negative than the system is stable.
rad/s ω0 iω -σ φ
p=-σ+iω
f0=1/T0
For the root Δt= , t∞=0.28
p=-σ
For the root için Δt= , t∞=0.0086
For the system Δt= , t∞=0.28

17. x(t) t 5 3 1
0.2
0.5
ξ=0.1 t
x(t)