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modelling of simple pendulum
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ModelingofaSimplePendulum
SimpleLinearModel:
Thesimplestanalyticalmodelforastandardpendulumcanbederivedbyassumingnodrag forceand smallvariations in theangleof thependulum.Assume that theentiremassof thependulum is concentrated in theboband that the length from thepivotpointtothecenterofgravityofthebobisgivenby.
FindthesumofthemomentsaroundthepointOtakingthecounterclockwisedirectionaspositive.Notethatthetension,,inthethinroddoesnotcontributetothemomentaboutthepointO.
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0
Themassmomentofinertiaofthependulum,,isgivenby:
Theangularaccelerationofthependulum,,isthesecondderivativeoftheangle.
Thedifferentialequationgoverningthedynamicsofthependulumbecomes:
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Dividing throughby and rearranging termsyields the following secondorderdifferentialequation:
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Although this equation appears simple enough, it is nonlinear in nature due to thepresence of a transcendental function. This nonlinear equation can, however, belinearizedbyassumingsmallvariationsintheangleofthependulum,.
Forsmallangles(expressedinradians),isapproximatelyequalto.Forexample:
22.5 0.3927 , 0.3827
Therefore for small variations in angle, the lineardifferentialequationdescribing thedynamicsofasimplependulumcanbeapproximatedby:
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ThisequationcanbesolvedanalyticallybyconsideringthebasicpropertiesofLaplacetransforms. Recall that the second derivative of a function can be expressed in theLaplacedomainasfollows:
0 0
Thedifferentialequationdescribingthedynamicsofthependulumbecomes:
0
Assume that thependulum is released from some initial angle, ,no initial velocity( 0).
0
Solvingforyields:
ThiscaneasilybeputintotheformofthefollowingbasicLaplacetransform:
Theangleof thependulum, ,asa functionof timebasedona simple, linearmodelbecomes:
Although it is not necessary to use MatlabSimulink to visualize this function, forcomparisonpurposestheSImulinkmodelbecomes:
SimpleNonLinearModel:
Althoughfindingananalyticalsolutiontothesimplenonlinearmodelisnotevident,itcanbesolvednumericallywiththeaidofMatlabSimulink.Startbyisolatingthehighestorderderivativefromthelinearmodel:
0
TheMatlabSimulinkmodelthatsolvesthisequationisgivenby:
NonLinearModelIncludingaConstantAerodynamicDragForce:
Assumethattheaerodynamicdragactsonlyonthebob;i.e.,thedragonthethinrodisnegligible.Thedragforce,,imposesanadditionalmomentaboutthepointO.
Thedifferentialequationgoverningthedynamicsofthependulumbecomes:
0
Isolatingthehighestorderderivativeofyields:
1
TheMatlabSimulinkmodelthatsolvesthisequationisgivenby:
Note that the drag force must change sign based on the angular velocity of thependulum.Inotherwords,althoughthedragforceistakenasconstant,itmustalwaysactoppositetothedirectionofmotion.
NonLinearModelIncludingAerodynamicDragForcethatisafunctionofVelocity:
Theaerodynamicdragforincompressibleflowoveranybodyisgivenby:
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whereisthevelocityoftheincompressiblefluid,isthedensityofthefluid,isthefrontalarea(note:notthesurfacearea),andisthedragcoefficient.
ThedragcoefficientforacylindricalbodyisafunctionoftheReynoldsnumberasshowninthefigureonthefollowingpage.
RecallthattheReynoldsnumberisgivenby:
whereisthediameterofthecylinderandistheviscosityoftheincompressiblefluid(inthiscaseair).
TbheMatlabSy:
Simulinkmoodel for thependulumwwithvariableeaerodynam
micdrag isggiven
NonLinearModelIncludingVariableAerodynamicDragandFrictionatthePivot:
Therecouldpotentiallybeasignificantamountoffrictionatthepivotpoint. TheMatlabSimulinkmodel including aerodynamic drag and Coulomb friction at the
pivotpointisgivenby:
Theresponseofthesystembasedonaninitialpendulumangleof22.5degreesisillustratedbelow.
Thefollowingtablesummarizestheparametersusedinthesimulation:
T0 = 22.5*pi/180; %% Initial pendulum angle (radians) L = 0.6; %% Length of rod (distance to CG) (m) g = 9.81; %% Acceleration due to gravity (m/s^2) M = 0.06; %% Mass of bob (kg) FD = 0.0007; %% Drag Force (N) DB = 0.0333; %% Diameter of bob (m) HB = 0.0508; %% Height of bob (m) A = DB*HB; %% Frontal area of bob (m^3) RHO = 1.23; %% Density of air (kg/m^2) MEW = 1.79E-5; %% Viscosity of air (kg/(m s)) CD = 1; %% Drag Coefficient KD = 0.5*CD*A*RHO; %% Drag constant (kg/m) TFF = 0.0005; %% Friction torque at pivot (N-m)
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