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Article 5/2

Article 5/2. Center of Mass 237. their respective position vectors r = x i + y j + z k and r = x i + y j + z k. Thus, Eqs. 5/1 b are the components of the single vector equation. J r dm r = m. (5/2). The density p of a body is its mass per unit volume. Thus, the mass

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Article 5/2

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  1. Article 5/2 CenterofMass 237 theirrespectivepositionvectorsr=xi+yj+zkandr=xi+yj+zk. Thus,Eqs.5/1barethecomponentsofthesinglevectorequation Jrdm r= m (5/2) Thedensitypofabodyisitsmassperunitvolume.Thus,themass ofadifferentialelementofvolumedVbecomesdm=pdV.Ifpisnot constantthroughoutthebodybutcanbeexpressedasafunctionofthe coordinatesofthebody,wemustaccountforthisvariationwhencalcu- latingthenumeratorsanddenominators ofEqs.5/1b. Wemaythen writetheseexpressionsas JxpdV JypdV JzpdV (5/3) y= z= x= JpdV JpdV JpdV CenterofMassversusCenterofGravity Equations5/1b,5/2,and5/3areindependentofgravitationaleffects sincegnolongerappears.Theythereforedefineauniquepointinthe bodywhichisafunctionsolelyofthedistributionofmass.Thispointis calledthecenterofmass,andclearlyitcoincideswiththecenterofgrav- ityaslongasthegravityfieldistreatedasuniformandparallel. Itismeaninglesstospeakofthecenterofgravityofabodywhichis removedfromthegravitationalfieldoftheearth,sincenogravitational forceswouldactonit.Thebodywould,however, stillhaveitsunique centerofmass.Wewillusuallyreferhenceforthtothecenterofmass ratherthantothecenterofgravity.Also,thecenterofmasshasaspe- cialsignificanceincalculatingthedynamicresponseofabodytounbal- ancedforces.ThisclassofproblemsisdiscussedatlengthinVol.2 Dynamics. Inmostproblemsthecalculation ofthepositionofthecenterof massmaybesimplifiedbyanintelligentchoiceofreferenceaxes.Ingen- eraltheaxesshouldbeplacedsoastosimplify theequationsofthe boundariesasmuchaspossible.Thus,polarcoordinateswillbeuseful forbodieswithcircularboundaries. Anotherimportantcluemaybetakenfromconsiderationsofsym- metry.Wheneverthereexistsalineorplaneofsymmetryinahomoge- neousbody,acoordinateaxisorplaneshouldbechosentocoincidewith thislineorplane.Thecenterofmasswillalwayslieonsuchalineor plane,sincethemomentsduetosymmetricallylocatedelementswillal- wayscancel,andthebody maybeconsideredcomposedofpairsofthese elements.Thus,thecenterofmassGofthehomogeneousright-circular coneofFig.5/5awillliesomewhereonitscentralaxis,whichisalineof symmetry.Thecenterofmassofthehalfright-circularconeliesonits planeofsymmetry,Fig.5/5b.ThecenterofmassofthehalfringinFig. 5/5cliesinbothofitsplanesofsymmetryandthereforeissituatedon G G (a) (b) A G B (c) Figure 5/5

  2. Chapter5 DistributedForces 238 lineAB.ItiseasiesttofindthelocationofG byusingsymmetrywhenit exists. 5/3 CentroidsofLines,Areas,andVolumes Whenthedensitypofabodyisuniformthroughout,itwillbea constantfactorinboththenumeratorsanddenominators ofEqs.5/3 andwillthereforecancel.Theremainingexpressions defineapurely geometricalpropertyofthebody,sinceanyreferencetoitsmassprop- ertieshasdisappeared.Thetermcentroidisusedwhenthecalculation concernsageometricalshapeonly.Whenspeakingofanactualphysical body,weusethetermcenterofmass.If thedensityisuniformthrough- outthebody,thepositionsofthecentroid andcenterofmassareiden- tical,whereasifthedensityvaries,thesetwopointswill,ingeneral,not coincide. Thecalculation ofcentroidsfallswithinthreedistinctcategories, dependingonwhetherwecanmodeltheshapeofthebodyinvolvedasa line,anarea,oravolume. (1)Lines. ForaslenderrodorwireoflengthL,cross-sectionalarea A,anddensityp,Fig.5/6,thebodyapproximatesalinesegment,and dm=pAdL.IfpandAareconstantoverthelengthoftherod,thecoor- dinatesofthecenterofmassalsobecomethecoordinatesofthecentroid Cofthelinesegment,which,fromEqs.5/1b,maybewritten z dL L y C z JxdL JydL JzdL x= y= z= L L L – z x– (5/4) x y x Notethat,ingeneral,thecentroidCwillnotlieontheline.Iftherod lieson asingleplane,suchasthex-yplane,onlytwocoordinatesneedto becalculated. (2)Areas.Whenabodyofdensityphasasmallbut constantthick- nesst,wecanmodelitasasurfaceareaA,Fig.5/7.Themassofanele- mentbecomesdm=ptdA.Again,ifpandtareconstantovertheentire area,thecoordinatesofthecenterofmassofthebodyalsobecomethe coordinatesofthecentroidCofthesurfacearea,andfromEqs.5/1bthe coordinatesmaybewritten Figure 5/6 z dA C z– A JxdA JydA JzdA x= y= z= A A A z y (5/5) x– x y– t y ThenumeratorsinEqs.5/5arecalledthefirstmomentsofarea.*If the surfaceiscurved,asillustratedinFig.5/7withtheshellsegment,all threecoordinateswillbeinvolved.ThecentroidCforthecurvedsur- facewillingeneralnotlieonthesurface.If theareaisaflatsurfacein, x Figure 5/7 *Secondmomentsofareas(momentsoffirstmoments)appearlaterinourdiscussionof areamomentsofinertiainAppendixA.

  3. Article 5/3 CentroidsofLines,Areas,andVolumes 239 say,thex-yplane,onlythecoordinatesofCinthatplaneneedtobe calculated. (3)Volumes.ForageneralbodyofvolumeVanddensityp,theele- menthasamassdm=pdV.Thedensitypcancelsifitisconstantover theentirevolume,andthecoordinatesofthecenterofmassalsobecome thecoordinatesofthecentroidCofthebody.From Eqs.5/3or5/1bthey become JxdV JydV JzdV x= y= z= (5/6) V V V KEYCONCEPTS ChoiceofElementforIntegration Theprincipaldifficultywithatheoryoftenliesnotinitsconcepts butintheproceduresforapplyingit.Withmasscentersandcentroids theconceptofthemomentprincipleissimpleenough;thedifficultsteps arethechoiceofthedifferential elementandsettinguptheintegrals. Thefollowingfiveguidelineswillbeuseful. (1)Order ofElement.Wheneverpossible,afirst-orderdifferential elementshouldbeselectedinpreferencetoahigher-order elementso thatonlyoneintegration willberequiredtocovertheentirefigure. Thus, inFig.5/8aafirst-orderhorizontalstripofareadA=ldywillre- quireonlyoneintegration withrespecttoytocovertheentirefigure. Thesecond-orderelementdxdywillrequiretwointegrations,firstwith respecttoxandsecondwithrespecttoy,tocoverthefigure.Asafurther example,forthesolidcone inFig.5/8bwechooseafirst-orderelementin theformofacircularsliceofvolumedV=r2dy.Thischoicerequires onlyone integration,andthusispreferabletochoosingathird-orderele- mentdV=dxdydz,whichwouldrequirethreeawkwardintegrations. (2)Continuity.Wheneverpossible,wechooseanelementwhichcan beintegratedinonecontinuousoperationtocoverthefigure.Thus,the horizontalstripinFig.5/8awouldbepreferabletothevertical stripin Fig.5/9,which,ifused,wouldrequiretwoseparateintegralsbecauseof thediscontinuityintheexpressionfortheheightofthestripatx=x1. (3)DiscardingHigher-OrderTerms. Higher-ordertermsmayal- waysbedroppedcomparedwithlower-orderterms(seeArt.1/7). Thus, theverticalstripofareaunderthecurveinFig.5/10 isgivenbythe y y dy y x x1 x dx Figure 5/9 Figure 5/10 y y l dy dx dy x x (a) y y dy r x z (b) Figure 5/8

  4. 240 Chapter5 DistributedForces y y r x θ x Figure 5/11 first-ordertermdA=ydx,andthesecond-ordertriangulararea1dxdy isdiscarded.Inthelimit,ofcourse,thereisnoerror. (4) ChoiceofCoordinates.Asageneralrule,wechoosethecoordi- natesystemwhichbestmatchestheboundaries ofthefigure.Thus,the boundariesoftheareainFig.5/11aaremosteasilydescribedinrectan-gularcoordinates,whereastheboundariesofthecircularsector ofFig. 5/11barebestsuitedtopolarcoordinates. (5)CentroidalCoordinateofElement.Whenafirst-orsecond- orderdifferentialelementischosen,itisessentialtousethecoordinate ofthecentroidoftheelementforthemomentarminexpressingthemo- mentofthedifferentialelement.Thus,forthehorizontalstripofareain Fig.5/12a, themomentofdAaboutthey-axisisxcdA,wherexcisthe x-coordinateofthecentroidCoftheelement. Notethatxcisnotthex whichdescribeseitherboundaryofthearea.Inthey-directionforthis elementthemomentarmycofthecentroidoftheelementisthesame, inthelimit,asthey-coordinatesofthetwoboundaries. Asasecondexample,considerthesolidhalf-coneofFig.5/12bwith thesemicircularsliceofdifferentialthicknessastheelementofvolume. Themomentarmfortheelementinthex-directionisthedistancexcto thecentroidofthefaceoftheelementandnotthex-distance tothe boundaryoftheelement.Ontheotherhand, inthez-directionthemo- mentarmzcofthecentroidoftheelementisthesameasthez-coordinate oftheelement. Withtheseexamplesinmind,werewriteEqs.5/5and5/6intheform JxcdA JycdA JzcdA A A A y x xc zc C x y z (a) (b) Figure 5/12 2 x=ky (a) (b) 2 x= y= z= (5/5a) yc xc C

  5. Article 5/3 CentroidsofLines,Areas,andVolumes 241 and JxcdV JycdV JzcdV V V V Itisessentialtorecognizethatthesubscriptcservesasareminderthat themomentarmsappearinginthenumeratorsoftheintegralexpres- sionsformomentsarealwaysthecoordinates ofthecentroidsofthe particularelementschosen. Atthispointyou shouldbecertaintounderstandclearlytheprinci- pleofmoments,whichwasintroducedinArt.2/4.Youshouldrecognize thephysicalmeaningofthisprincipleasitisappliedtothesystemof parallel weightforcesdepictedinFig.5/4a.Keepinmindtheequiva- lencebetweenthemomentoftheresultantweightWandthesum(inte- gral)ofthemomentsoftheelementalweightsdW, toavoidmistakesin settingupthenecessarymathematics.Recognitionoftheprincipleof momentswillhelpinobtainingthecorrectexpression forthemoment armxc,yc,orzcofthecentroidofthechosendifferentialelement. Keepinginmindthephysicalpictureoftheprincipleofmoments, wewill recognizethatEqs.5/4,5/5,and5/6,whicharegeometricrela- tionships,aredescriptivealsoofhomogeneousphysicalbodies,because thedensitypcancels.Ifthedensityofthebodyinquestionisnotcon- stantbutvaries throughoutthebodyassomefunctionofthecoordi- nates, thenitwillnotcancelfromthenumeratoranddenominatorof themass-centerexpressions.Inthisevent,wemustuseEqs.5/3asex- plainedearlier. x= y= z= (5/6a) SampleProblems5/1 through5/5whichfollowhavebeencarefully chosentoillustratetheapplicationofEqs.5/4,5/5,and5/6forcalculat- ingthelocationofthecentroidforlinesegments(slenderrods),areas (thinflatplates), andvolumes(homogeneoussolids).Thefiveintegra- tionconsiderationslistedaboveareillustratedindetailinthesesample problems. SectionC/10 ofAppendixCcontainsatableofintegralswhichin- cludesthoseneededfortheproblemsinthisandsubsequentchapters.A summaryofthecentroidalcoordinatesforsomeofthecommonlyused shapesisgiveninTablesD/3andD/4,AppendixD.

  6. 242 Chapter5 DistributedForces SAMPLEPROBLEM5/1 Centroidofacirculararc. Locatethecentroidofacirculararcasshownin the figure. r a C a Solution. Choosingtheaxisofsymmetryasthex-axismakesy=0.Adiffer- entialelementofarchasthelengthdL=rdOexpressedinpolarcoordinates, andthex-coordinateoftheelementisrcosO. ApplyingthefirstofEqs.5/4andsubstituting L=2argive O [Lx=JxdL] (2ar)x=J a y rcos0 d0 a 0 x a r C C 2r/π r r (rcosO)rdO rd0 -a 2arx=2r2sina x=rsina Ans. a Forasemicirculararc2a=,whichgivesx= 2r/.Bysymmetrywesee immediatelythatthisresultalsoappliestothequarter-circulararcwhenthe measurement ismadeasshown. HelpfulHint OItshouldbe perfectlyevidentthatpolarcoordinatesarepreferabletorectan- gularcoordinates toexpressthelengthofacirculararc. SAMPLEPROBLEM5/2 y dy h x y x Centroidofatriangulararea. Determinethedistancehfromthebaseofa triangleofaltitudehtothecentroidofitsarea. Solution. Thex-axisistakentocoincidewiththebase.Adifferentialstripof areadA=xdyischosen. Bysimilartrianglesx/(h-y)=b/h.Applyingthesec- ondofEqs.5/5agives O b [Ay=JycdA] h dy=bh2 y 2 6 h 0 HelpfulHint OWe save one integration here by usingthefirst-orderelementofarea.RecognizethatdAmustbeexpressedintermsofthe integration variabley;hence,x=ƒ(y)isrequired. y=h Ans. and 3 Thissameresultholdswithrespecttoeitheroftheothertwosidesofthe triangleconsideredanewbasewithcorrespondingnewaltitude.Thus,thecen- troidliesattheintersectionofthemedians,sincethedistanceofthispointfrom anysideisone-thirdthealtitude ofthetriangle withthatsideconsideredthe base.

  7. Article 5/3 CentroidsofLines,Areas,andVolumes 243 SAMPLEPROBLEM5/3 Centroidofthe areaofacircularsector. Locatethecentroidofthearea ofacircularsectorwithrespecttoitsvertex. r a C a SolutionI. Thex-axisischosenastheaxisofsymmetry,andyistherefore automaticallyzero.Wemaycovertheareabymovinganelementin theformof apartialcircular ring,asshowninthefigure,fromthecentertotheouterpe- riphery.TheradiusoftheringisrOanditsthicknessisdrO,sothatitsarea is dA=2rOadrO. Thex-coordinatetothecentroidoftheelementfromSampleProblem5/1 is xc=rOsina/a, whererOreplacesrintheformula.Thus,thefirstofEqs.5/5a gives y r dr0 x a r0sina xc=——––– SolutionI O a r0 8 J 2a(r2)x=J r rOsina a ( O ) [Ax= xcdA] (2rOadrO) 2 a r2ax=2r3sina 3 x=2rsina Ans. 3 a HelpfulHints ONotecarefullythatwemust distin- guishbetweenthevariable rO andtheconstantr. 8BecarefulnottouserOasthecen- troidalcoordinatefortheelement. SolutionII. Theareamayalsobecoveredbyswingingatriangleofdifferen- tialareaaboutthevertexandthroughthetotalangleofthesector.Thistriangle, shownintheillustration, hasanareadA=(r/2)(rdO), wherehigher-order terms areneglected.FromSampleProblem5/2thecentroidofthetriangularelement ofareaistwo-thirdsofitsaltitudefromitsvertex,sothatthex-coordinatetothe centroidoftheelementisxc=2rcosO.ApplyingthefirstofEqs.5/5agives 3 [Ax=JxcdA] (r2a)x=J 3rcosO)(2r2dO) a (2 1 y xc=2 3 d0 a 0 x a r SolutionII 4r/3π C C r r -a r2ax=2r3sina 3 x=2rsina Ans. andasbefore 3 a Forasemicirculararea2a=,whichgivesx=4r/3.Bysymmetrywesee immediatelythatthisresultalsoappliestothequarter-circularareawherethe measurement ismadeasshown. Itshouldbenotedthat,ifwehadchosenasecond-order elementrOdrO dO, oneintegrationwithrespecttoOwouldyieldtheringwithwhich SolutionI began.Ontheotherhand,integrationwith respecttorOinitially wouldgivethetriangularelementwithwhichSolutionIIbegan.

  8. 244 Chapter5 DistributedForces SAMPLEPROBLEM5/4 Locatethecentroidoftheareaunderthecurvex=ky3fromx=Otox=a. y x=ky3 – C b – a x x SolutionI. AverticalelementofareadA=ydxischosenasshowninthefig- ure.Thex-coordinateofthecentroidisfoundfromthefirstofEqs.5/5a.Thus, y [Ax=JxcdA] xJydx=J a a O xydx O O y b x=ky3 y y =y 2 x y b 2 x=ky3 dy x a–x y a x Substituting y=(x/k)1/3 andk=a/b3andintegratinggive 3abx=3a2b x=4a c – x dx Ans. 7 4 7 InthesolutionforyfromthesecondofEqs.5/5a,thecoordinatetothe centroidoftherectangularelementisyc=y/2,whereyistheheightofthestrip governedbytheequationofthecurvex=ky3.Thus,themomentprinciplebe- comes [Ay=Jy dA] 3aby=J a y (2) yd x c 4 a O Substituting y=b(x/a)1/3 andintegratinggive 3aby=3ab2 y=2b Ans. 5 4 1O xc=–a–+––x SolutionII. Thehorizontalelementofareashowninthelowerfiguremaybe employedinplaceoftheverticalelement.Thex-coordinatetothecentroidofthe rectangularelementisseentobexc=x+1(a-x)=(a+x)/2,whichissimply 2 theaverageofthecoordinates aandxoftheendsofthestrip.Hence, [Ax=JxcdA] xJ(a-x)dy=J b b (a+x )(a-x)dy 2 O O Thevalueofyisfoundfrom [Ay=JycdA] yJ(a-x)dy=Jy(a-x)dy b b O O HelpfulHint ONote thatxc =x for the vertical element. whereyc=yforthehorizontalstrip.Theevaluationoftheseintegralswillcheck thepreviousresultsforxandy.

  9. Article 5/3 CentroidsofLines,Areas,andVolumes 245 SAMPLEPROBLEM5/5 Hemisphericalvolume. Locatethe centroidofthe volumeofahemisphere ofradiusrwithrespecttoitsbase. SolutionI. Withtheaxeschosenasshownin thefigure,x=z=Obysymme- z y2+z2 =r2 z y r dy x yc=y SolutionI z yc=y/2 dz z y r x SolutionII z dθ r rdθ θ y SolutionIII try.Themostconvenientelementisacircularsliceofthicknessdyparallelto thex-zplane.Sincethehemisphereintersectsthey-zplaneinthecircley2+z2= r2,the radiusofthe circularsliceisz=+Jr2-y2.Thevolumeofthe elemental slicebecomes dV=(r2-y2)dy 0 ThesecondofEqs.5/6arequires I yI I r r (r2-y2)dy= y(r2-y2)dy [Vy= yedV] O O whereye=y.Integrating gives 2r3y=lr4 y=3r Ans. 3 4 8 y SolutionII. Alternativelywemayuseforourdifferentialelementacylindrical shelloflengthy, radiusz,andthicknessdz,asshown inthe lowerfigure.Byex- pandingtheradiusoftheshellfromzerotor,wecovertheentirevolume.By symmetrythecentroidoftheelementalshellliesatitscenter,sothatye=y/2. ThevolumeoftheelementisdV=(2zdz)(y).Expressingyintermsofzfrom theequationofthecirclegivesy=+Jr2-z2.Usingthevalueof2r3computed 3 inSolutionIforthevolumeofthehemisphereandsubstitutinginthesecondof Eqs.5/6agiveus I 3r3)y=I r 2 2 Jr -z (2 (2zJr2-z2)dz [Vy= y dV] e 2 O =I (r2z-z3)dz=r4 r 4 O y=3r Ans. 8 SolutionsIandIIareofcomparableusesinceeachinvolves anelementof simpleshapeandrequiresintegrationwithrespecttoonevariableonly. SolutionIII. Asanalternative,wecouldusetheangle0asourvariablewith limitsofOand/2.Theradiusofeitherelementwouldbecomersin0,whereas thethicknessofthesliceinSolutionIwouldbedy=(rd0)sin0andthatofthe shellinSolutionIIwouldbedz=(rd0)cos0.Thelengthoftheshellwouldbe y=rcos0. HelpfulHint 0Canyouidentifythehigher-orderel- ementof volumewhichisomitted fromtheexpressionfordV?

  10. 246 Chapter5 DistributedForces PROBLEMS IntroductoryProblems 5/1Withyourpencil,makeadotonthepositionofyour bestvisualestimateofthecentroidofthetriangular area.Checkthepositionofyourestimatebyreferring totheresultsofSample Problem5/2andtoTable D/3. 5/3Specifythex-,y-,andz-coordinatesofthemasscen- terofthehomogeneoussemicylinder. y 360 mm 120 mm x 10 z Problem5/3 8 Specifythex-,y-,andz-coordinatesofthemasscenter ofthequadrantofthehomogeneoussolidcylinder. 5/4 6 4 240mm z y 2 120 mm 0 0 2 4 6 8 10 Problem5/1 x 5/2Withyourpencil,makeadotonthepositionofyour bestvisualestimateofthecentroidoftheareaofthe circularsector.Checkyourestimatebyusingthere- sultsofSampleProblem5/3. Problem5/4 Auniformsemicircular rodofradiusrissupportedin abearingatits upperendandisfreetoswinginthe verticalplane.Calculatetheangle0madebythedi- ameterwiththeverticalfortheequilibrium position. 5/5 y 200mm 30° 30° O θ x r Problem5/2 Problem5/5

  11. Article 5/3 Problems 247 5/6Determinethey-coordinateofthecentroidofthearea bydirectintegration. Determinethex-andy-coordinatesofthecentroidof theshadedarea. 5/9 y y y= x3 1+— R 8 R/2 x 1 Problem5/6 5/7Determinethey-coordinateof thecentroidof the shadedarea.Checkyourresultforthespecialcase a=O. x 0 0 1 2 Problem5/9 y Determinethecoordinatesof thecentroidof the shadedarea. 5/10 h 60° y b a 60° x x=ky2 a Problem5/7 5/8Determinethex-andy-coordinatesofthecentroidof thetrapezoidalarea. x y Problem5/10 Determinethecoordinatesof thecentroidof the shadedarea. 5/11 y a πx y=bsin —– b 2a x h Problem5/8 b x a Problem5/11

  12. 248 Chapter5 DistributedForces RepresentativeProblems 5/12Determinethex-andy-coordinatesofthecentroid oftheshadedarea. Locatethecentroidoftheshadedareashown. 5/15 y y 3 x 0 a x2 ––––4 y= x 4 a 2 –3 Problem5/12 –4 5/13Determinethecoordinatesof thecentroidof the shadedarea. Problem5/15 Determinethex-andy-coordinatesofthecentroid ofthetrapezoidalarea. 5/16 y y x=ky2 a a h x b Problem5/13 x 5/14Findthedistancezfromthevertexoftheright- b circularconetothecentroidofitsvolume. Problem5/16 Locatethecentroidoftheshadedarea. 5/17 – z y h C x=a(1–—– ) 2 y b2 b Problem5/14 x a Problem5/17

  13. Article 5/3 Problems 249 Determinethex-coordinateofthemasscenterof thetaperedsteelrodoflengthLwherethediameter atthelargeendistwicethediameteratthesmall end. 5/18Determinethecoordinatesof thecentroidof the shadedarea. 5/21 y b y=kx2 y L x –b– 2 Dia.=2D Dia.=D Problem5/21 x 0 a 0 Determinethex-andy-coordinatesofthecentroid oftheshadedarea. 5/22 Problem5/18 y 5/19Themassperunitlengthoftheslenderrodvaries withpositionaccordingtop=pO(l-x/2), wherexis infeet.Determinethelocationofthecenterofmass oftherod. a y=k2 x y=k1x3 a 1′ x x Problem5/22 Problem5/19 Determinethex-andy-coordinatesofthecentroid oftheshadedareashown. 5/23 5/20Calculatethecoordinatesofthecentroidoftheseg- mentofthecirculararea. y y a x2 y2 a2 b2 —+—=1 b x x a Problem5/20 Problem5/23

  14. 250 Chapter5 DistributedForces 5/24Locatethecentroidoftheareashowninthefigure bydirectintegration.(Caution:Observecarefullythe propersignoftheradicalinvolved.) If theshadedareaof Prob. 5/26isrotated360° aboutthey-axis,determinethey-coordinateofthe centroidoftheresultingvolume. 5/27 y Determinethex-andy-coordinatesofthecentroid oftheshadedarea. 5/28 y a a 2 a x — Problem5/24 x 5/25UsetheresultsofSampleProblem5/3tocompute thecoordinatesofthemasscenteroftheportionof thesolidhomogeneouscylindershown. z Problem5/28 Determinethey-coordinateofthecentroidofthe shadedarea. 5/29 y a 45° a 2 45° 6′′ 10′′ x y x Problem5/25 Problem5/29 5/26Determinethey-coordinateofthecentroidofthe shadedareashown. Determinethez-coordinateofthecentroidofthe volume generatedby revolving the shadedarea aroundthez-axisthrough360°. 5/30 y b Parabolic z a a –b– 5 r Problem5/30 x 0 a 0 Problem5/26

  15. Article 5/3 Problems 251 Determinethex-coordinateofthecentroidofthe solidsphericalsegment.Evaluateyourexpression forh=R/4andh=0. The figurerepresentsaflatpieceof sheetmetal symmetricalaboutaxisA-Aandhavingaparabolic upperboundary.Chooseyourowncoordinatesand 5/31 5/34 calculatethedistancehfromthebasetothecenter ofgravityofthepiece. y h A x 3′′ R O 2′′ 3′′ A Problem5/34 Problem5/31 Determinethex-coordinateofthemasscenterof theportionof thesphericalshellof uniformbut smallthickness. 5/35 Locatethecentroidoftheshadedareabetweenthe ellipseandthecircle. 5/32 y y R — 4 b –3—R 4 x b a x Problem5/32 Calculatethedistancehmeasuredfromthebaseto thecentroidofthevolumeofthefrustumofthe right-circular cone. 5/33 y Problem5/35 h —h 2 r z Problem5/33

  16. 252 Chapter5 DistributedForces Determinethey-coordinateofthecentroidofthe volumegenerated byrevolvingtheshadedareaof therighttriangleaboutthez-axisthrough90° as showninthefigure. 5/36Thethicknessofthetriangularplatevarieslinearly withyfromavaluet0alongitsbasey=0to2t0at y=h.Determinethey-coordinateofthecenterof massoftheplate. 5/39 y y 2t 0 h h h x x b t0 b Problem5/36 z 5/37Locatethemasscenterofthehomogeneoussolid bodywhosevolumeisdeterminedbyrevolvingthe shadedareathrough360°aboutthez-axis. Problem5/39 Determinethez-coordinateofthemasscenterofthe homogeneousquarter-sphericalshell,whichhasa radiusr. �5/40 r 300mm z 200mm y r=kz3 z r 0 0 x Problem5/40 Problem5/37 �5/41 Determinethey-coordinateofthecentroidofthe planeareashown.Seth=0 inyourresultandcom- 5/38Thehomogeneousslenderrodhasauniformcross sectionandisbentintotheshapeshown.Calculate the y-coordinateof the masscenter of the rod. y=4a parewiththeresult forafullsemicircular 3 (Reminder: A differential arc length is dL= area(seeSampleProblem5/3andTableD/3).Also evaluateyourresultfor theconditionsh=aand J(dx)2+(dy)2=Jl+(dx/dy)2dy.) y 4 h=a. 2 x=ky2 y 100mm a h x x 100mm Problem5/41 Problem5/38

  17. Article 5/3 Problems 253 �5/42 �5/45 Determinethex-coordinateofthemasscenterofthe cylindricalshellofsmalluniformthickness. The thickness of the semicircular plate varies linearlywithyfromavalue2t0alongitsbasey=0 tot0 aty=a.Determinethey-coordinateof the masscenteroftheplate. 2R y t0 R a x 2t0 x Problem5/42 4R �5/43 Determinethecoordinatesofthecentroidofthevol- umeobtainedbyrevolvingtheshadedareaaboutthe z-axisthroughthe90°angle. Problem5/45 z �5/46 Determinethex-andy-coordinatesofthecentroid of thevolumegeneratedbyrotatingtheshaded areaaboutthez-axisthrough90°. a z a x y y a Problem5/43 �5/44 Determinethex-coordinateofthemasscenterofthe solidhomogeneousbodyshown. 2R x R Problem5/46 x 4R Problem5/44

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