Simple Pendulum-Stochastic Process-Lecture Slides, Slides of Stochastic Processes

Main topics for this course are Stochastic process, random variables, linear congruent generators, pdfs and cdfs, rejection method, metropolis methods, sampling techniques, random walks and genetic algorithm. This lecture includes: Simple, Pendulum, Rigid, Rod, Torque, Model, Equation, Moment, Inertia, Pivot, System, Motion, Derivative, Undamped, Conditions, Lorentz, Strange, Attractors

Typology: Slides

2011/2012

Uploaded on 08/12/2012

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A simple Pendulum
A a mass m is attached to a rigid rod and the mass is at distance L from the
frictionless pivot. The system moves in a plane. The motion of is governed by
the equation for torque:
a(t)
I-
=
τ
dt
d
mL sin mgL- 2
2
2
θ
θ
=
where, τ(t), is the torque around the pivot point;
I is the moment of inertia about the pivot point, and
a(t) is the angular acceleration.
We use θ
as a angular displacement; v(t)
as its time rate of
change, and a(t)
as the second derivative of θ(t).
Then torque is mgLsinθ
with I = mL2.
The model equation is then
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Download Simple Pendulum-Stochastic Process-Lecture Slides and more Slides Stochastic Processes in PDF only on Docsity!

A simple Pendulum A a mass

m

is attached to a rigid rod and the mass is at distance

L

from the

frictionless pivot. The system moves in a plane. The motion of is governed bythe equation for torque:

a(t)

I

d^ dt

mL

sin

mgL

2 2

2

θ

θ =

where,

(t),

is the torque around the pivot point;

I

is the moment of inertia about the pivot point, and

a(t)

is the angular acceleration.

We use

as a angular displacement;

v(t)

as its time rate of

change, and

a(t)

as the second derivative of

(t).

Then torque is

mgLsin

with

I

= mL

2

The model equation is then

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Physical Pendulum

L

g

s

m

g

g L

dt d

/

/

8 .

9

sin

=

Ω

=

=

θ

θ

This equation can be simplified by dividing by

mL

Also, choose

L = 9.8 m

θ

θ

sin

2 2

=

d^ dt

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mass = 1 and L = 9.8, θ

(0) = 10 and d

θ

/dt(0) = 0.

Example 1: Undamped System of simple pendulum

0

5 0

1 0 0

1 5 0

2 0 0

2 5 0

3 0 0

(^43210) - 1 - 2 - 3 - 4

θ angular displacement,

tim e , t

u n d a m p e d m o tio n o f p e n d u lu m

in itia l c o n d itio n s : θ = − 3.

a n d

θ ' = 0.

XY Graph

sin

Trigonometric

Function

simout

To Workspace

simout

To Workspace

1 s

Integrator

1 s

Integrator

-1 Gain

0

5 0

1 0 0

1 5 0

2 0 0

2 5 0

3 0 0

-1 -2 -3 - 4 3 2 1 0

dtθ/ angular velocity, d

tim e , t

u n d a m p e d m o tio n o f p e n d u lu m

in itia l c o n d itio n s : θ = − 3.

a n d

θ ' = 0.

sin

2

2 2

L

g

as

d dt

θ

θ

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mass = 1 and L = 9.8,

Example 1: Undamped System of simple pendulum

0 . 1

/

; 0

sin

2

2 2

=

=

Ω

=

L

g

as

d dt

θ

θ

XY Graph

sin

Trigonome tric

Function

simout

To Workspace 1

simout

To Workspace

1 s

Integrator

1 s

Inte grator

-1 Gain

  • 6
  • 5
  • 4
  • 3
  • 2
  • 1

0

1

2

3

4

5

6

  • 1 - 2 - 3

3 2 1 0

/dtθd

t im e , t

Phase portrait for given initialconditions: θ

and

d

θ

/dt(0) = 0.

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mass = 1 and L = 9.8,

Example 1: Undamped System of simple pendulum

0 . 1

/

; 0

sin

2

2 2

=

=

Ω

=

L

g

as

d dt

θ

θ

XY Graph

sin

Trigonome tric

Function

simout

To Workspace 1

simout

To Workspace

1 s

Integrator

1 s

Inte grator

-1 Gain

  • 6
  • 5
  • 4
  • 3
  • 2
  • 1

0

1

2

3

4

5

6

  • 1 - 2

2 1 0

/dtθd

t im e , t

Phase portrait for given initialconditions: θ

and

d

θ

/dt(0) = 0.

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mass = 1 and L = 9.8,

Example 1: Undamped System of simple pendulum

0 . 1

/

; 0

sin

2

2 2

=

=

Ω

=

L

g

as

d dt

θ

θ

XY Graph

sin

Trigonome tric

Function

simout

To Workspace 1

simout

To Workspace

1 s

Integrator

1 s

Inte grator

-1 Gain

  • 6
  • 5
  • 4
  • 3
  • 2
  • 1

0

1

2

3

4

5

6

  • 1 - 2 - 3

3 2 1 0

/dtθd

t im e , t

Phase portrait for given initialconditions: θ

and

d

θ

/dt(0) = 2.

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The Lorentz Strange Attractors

Phase space can have

fixed points

(x, v)

such that they satisfy the

model (as

f(x, v) = 0

and

g(x, v) = 0).

This corresponds to steady state.

The set of points in the phase space are identified as

orbit or trajectory

If the set of points in the simulation repeat itself after some time

(T),

then the orbit is said to be

periodic

that is

x(t + T) = x(t)

. The orbit of

mass-spring system in a friction free environment is an ellipse in phasespace. A closed curve is called a

limit cycle

in phase space towards which an

orbit evolves as time goes to large values. It has property that all othercurves move towards it or away from it. When all the neighboring trajectories are going towards the limit cycle it iscalled a stable or

attracting cycle

, otherwise it is an unstable or repelling

one.

Then orbit is called attractor.

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The Lorentz Strange Attractors

Phase space can have

fixed points

(x, v)

such that they satisfy the

model (as

f(x, v) = 0

and

g(x, v) = 0).

This corresponds to steady state.

The set of points in the phase space are identified as

orbit or trajectory

If the set of points in the simulation repeat itself after some time

(T),

then the orbit is said to be

periodic

that is

x(t + T) = x(t)

. The orbit of

mass-spring system in a friction free environment is an ellipse in phasespace. A closed curve is called a

limit cycle

in phase space towards which an

orbit evolves as time goes to large values. It has property that all othercurves move towards it or away from it. When all the neighboring trajectories are going towards the limit cycle it iscalled a stable or

attracting cycle

, otherwise it is an unstable or repelling

one.

Then orbit is called attractor.

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The Lorentz Strange Attractors The model equations are

t
y
t
x
dx dt

σ

t ( y ) t ( x r ) t ( z ) t ( x

dy dt

) ( ) ( ) ( t z b t y t x

dz dt

=

Let us use Simulink

for coupled ODEs

and trace the projection on xz-plane

with parameter values,

b

= 10, and

r

= 28. The initial values are

x(0)

y(0)

= 5., and

z(0)

= 25. The resulting solutions are illustrated as

Figures.

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The Lorentz Strange Attractors

x

dx/dt

-10x

10y

y

dx/dt = -10x + 10y

z

x

xz

28x -xz

y

dy/dt

dy/dt = y + 28x - xz

x

y

xy

z

-8z/

dz/dt = -8z/3 + xy

XY Graph

z

To Workspace

y

To Workspace

x

To Workspace

Product Product

1 s

Integrator

1 s

Integrator

1 s

Integrator

-8/3Gain

28 Gain3-1 Gain

10 Gain

-10 Gain

The patchdiagram forSimulink

using

coupled ODEs

in

Lorentz

model

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Pendulum with damping

dt d

C

dt d

θ

θ

θ

=

sin

2

2

The

damped

pendulum

is

an

example

of

a

dissipative

system

because energy is being lost to the surroundings. A term

Cv(t)

is

used on the right-hand side of the equation of motion to take intoaccount the air resistance. The equation for the angular accelerationbecomes Where,

mass is unity

and

g/L = 1.

Let us make a patch diagram for MATLAB/Simulink.

The results

of simulation in terms of phase diagram are shown in Figure. At time t = zero, the pendulum is given a clockwise swing from aninitial angle of 3

and various initial velocities.

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Pendulum with damping

dt d

q

dt d

θ

θ

θ

Ω

=

2

2

2

The underdamped

case:

The overdamped

case

The critical case:

4 /

4 /

4 /

2

2

2

2

2

2

q q q

=

Ω

<

Ω

Ω

s

l

s

m

g

/

13 .

3

1 ; / 8. 9

2

=

Ω

=

=

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mass = 1 and L = 9.8, θ

(0) = 9 and d

θ

/dt(0) = -2.

Example 1: Undamped System of simple pendulum

0 . 1

/

; 0

sin

1 . 0

2

2 2

= = Ω = + +

L

g

as

d dt

d dt

θ

θ

θ

0

5

10

15

20

25

30

10 8 6 4 2 0 -2 -4 -6 - angular displacement

time, t

damped pendulum

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Solution of Damped Pendulum

t

i

e

Trial Solution:

q

iq

q

q

iq iq

Special case: critical point

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