EG260 Solutions-example_2

EG260 Solutions-example_2 - EG-260 @Avril 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 be angle of rotation Then moment( )( )2kasinaak=(small angle approximation) So that the equation of motion is 2=+kaI&&or 2=+Ika&&( )22=++bamka&&mkbaan+=ii).Rotation is about OMoment of inertia about pivot = I, moment of inertia about cg = IGUsing the parallel axis therefore, 2mrIIG+=Restoring moment (decreasing , negative) = mgrmgrM=sinsfor small Newtons Second Law: mgrsinmgrI=&&mg r OEG-260 @Avril Slone 2008 So ( )2=++mgrmrIG&&2mrImgrGn+=2.First determine the moment of inertia of the disc Consider an annular element ( )hxxxmxI222==where h,are the density and thickness. So ===rrGhrxhdxhxI4432422But the mass of the plate 2hrm=So that 22mrIG=Now the disc rotates by such thatlrThe tension supporting in any one wire is T where...
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This note was uploaded on 05/18/2010 for the course ENGINEERIN EG 260 taught by Professor Stone during the Spring '10 term at Swansea UK.

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EG260 Solutions-example_2 - EG-260 @Avril Slone 2008 EG260...

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