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Simple Pendulum-Modeling and Simulation-Lecture Handouts, Lecture notes of Mathematical Modeling and Simulation

Pakistan Institute of Engineering and Applied Sciences, Islamabad (PIEAS)Mathematical Modeling and Simulation

This handout was provided by Dr. Asad Khan for Modeling and Simulation course at Pakistan Institute of Engineering and Applied Sciences, Islamabad (PIEAS). It includes: Simple, Pendulum, Mathematical Model, Angular, Velocity, Particular, Solution, Characteristic, Equation

Typology: Lecture notes

2011/2012

Uploaded on 07/11/2012

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1CIS308‐ModelingandSimulation
Handout#27
SimplePendulum
Anyobjectthatswingsbackandforthiscalledaphysicalpendulum.Thesimplependulumisaspecial
caseofthephysicalpendulumandconsistsofarodoflengthltowhichamassmisattachedatoneend.
Indescribingthemotionofasimplependuluminaverticalplane,wemakethesimplifyingassumptions
thatthemassoftherodisnegligibleandthatnoexternaldampingordrivingforcesactonthesystem.
MathematicalModel

Toderivethedifferentialequationofmotionforthependulum,
webeginwithNewton’ssecondlawinrotationalform
)1(
2
12
2
2
−−−−−= dt
d
I
θ
τ
whereτisthetorque,Iisthemomentofinertia,αistheangular
acceleration,andθistheanglefromthevertical.Inthecaseof
thependulum,thetorqueisgivenby
)2(sin −−−−−−=
θ
τ
mgL
andthemomentofinertiais
)3(
2−−−−−= mLI
Substitutingthevaluesofτfromeq.(2)andIfromeq.(3)ineq.
(1),weget
2
2
2
2
1
sin dt
d
mLmgL
θ
θ
=−
)4(0sin
2
2
−−−−−=+
θ
θ
L
g
dt
d
Forthependulumproblem,thedifferentialequationcanbewrittenintheform
θ’’+ω2sinθ=0,whereω2=g/L‐‐‐‐‐(5)
Becauseofthesinetermthisisanonlineardifferentialequation.Thelinearapproximationforsmall
anglesisobtainedbyreplacingsinθbyθ,thefirsttermintheTaylorseriesexpansionofsinθ
θ’’+ω2θ=0,whereω2=g/L‐‐‐‐‐(6)
Thecharacteristicequationisgivenby
x2+ω2x=0‐‐‐‐‐(7)
Letustakethetrialsolutionx=ert.Substitutinginto(7)gives
r2ert+ω2ert=0
whichimpliesthatr2+ω2=0sor=±iω.Sincethisisalineardifferentialequationthegeneralsolutionof
(7)incomplexformisgivenby
x=c1eiωt+c2e‐iωt‐‐‐‐‐(8)
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CIS308 ‐ Modeling and Simulation 1

Handout#

Simple Pendulum

Any object that swings back and forth is called a physical pendulum. The simple pendulum is a special case of the physical pendulum and consists of a rod of length l to which a mass m is attached at one end. In describing the motion of a simple pendulum in a vertical plane, we make the simplifying assumptions that the mass of the rod is negligible and that no external damping or driving forces act on the system.

Mathematical Model

To derive the differential equation of motion for the pendulum, we begin with Newton’s second law in rotational form

2

2

dt

d

I

where τ is the torque, I is the moment of inertia, α is the angular acceleration, and θ is the angle from the vertical. In the case of the pendulum, the torque is given by

τ =− mgL sin θ −−−−−( 2 )

and the moment of inertia is

I = mL^2 −−−−−( 3 )

Substituting the values of τ from eq. (2) and I from eq. (3) in eq. (1), we get

2

2 2

sin

dt

d

mgL mL

θ

2 sin^0 (^4 )

2

L

g dt

d

For the pendulum problem, the differential equation can be written in the form θ’’+ ω^2 sinθ = 0, where ω^2 = g/L ‐‐‐‐‐ (5)

Because of the sine term this is a nonlinear differential equation. The linear approximation for small angles is obtained by replacing sinθ by θ, the first term in the Taylor series expansion of sinθ θ’’+ ω^2 θ = 0, where ω^2 = g/L ‐‐‐‐‐ (6)

The characteristic equation is given by x^2 + ω^2 x = 0 ‐‐‐‐‐(7)

Let us take the trial solution x = ert. Substituting into (7) gives r 2 ert^ + ω^2 ert^ = 0 which implies that r 2 + ω^2 = 0 so r = ±iω. Since this is a linear differential equation the general solution of (7) in complex form is given by

x = c 1 e iωt^ + c 2 e‐iωt^ ‐‐‐‐‐ (8)

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CIS308 ‐ Modeling and Simulation 2

We can use the identity e iωt^ = cosωt + isinωt to obtain the general solution in real form x = c 1 eiωt^ + c 2 e‐iωt = c 1 (cosωt + i sinωt) + c 2 (cosωt − i sinωt) = (c 1 + c 2 ) cosωt + i(c 1 − c 2 ) sinωt Therefore, the general solution in real form (C and D real) is x = C cosωt + Dsinωt ‐‐‐‐‐ (9)

The particular solution satisfying the initial conditions x(0) = x 0 and x’(0) = v 0 at time t = 0 is obtained by substitution into this equation and its derivative x’ = −ωC sinωt + ω Dcosωt ‐‐‐‐‐ (10)

This gives the values C = x 0 and D = v 0 /ω for the constants C and D, so the particular solution satisfying these initial conditions is given by x = xo cosωt + (νo /ω)sinωt ‐‐‐‐‐ (11)

In the special case that v 0 = 0, corresponding to zero initial angular velocity for the pendulum the particular solution has the simple form x = x 0 cosωt ‐‐‐‐‐ (12)

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