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EG260 Solutions-example_2

EG260 Solutions-example_2 - [email protected] Slone 2008 EG260...

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Unformatted text preview: EG-260 @Avril Slone 2008 EG260 Dynamics - Example Sheet 2 Single degree of freedom systems Answers 1.i).Moment of inertia ( )2bamI+=Let θbe angle of rotation Then moment( )( )θθ2kasinaak−=××−(small angle approximation) So that the equation of motion is 2=+θθkaI&&or 2=+θθIka&&( )22=++θθbamka&&mkbaan+=ωii).Rotation is about OMoment of inertia about pivot = I, moment of inertia about cg = IGUsing the parallel axis therefore, 2mrIIG+=Restoring moment (decreasing θ, negative) = θθmgrmgrM−≈−=sinsfor small θ Newton’s Second Law: θθθmgrsinmgrI−≈−=&&mg r θ OEG-260 @Avril Slone 2008 So ( )2=++θθmgrmrIG&&2mrImgrGn+=ω2.First determine the moment of inertia of the disc Consider an annular element ( )hxxxmxIρδπδδ222==where h,ρare the density and thickness. So ∫=⎥⎦⎤⎢⎣⎡==rrGhrxhdxhxI4432422πρπρπρBut the mass of the plate 2hrmπρ=So that 22mrIG=Now the disc rotates by θsuch thatφθlr≈The tension supporting in any one wire is T where...
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EG260 Solutions-example_2 - [email protected] Slone 2008 EG260...

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