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CBSEClass11physics

ImportantQuestions

Chapter7

SystemofParticlesandRotationalMotion

3MarksQuestions

1.Themomentofinertiaofasolidsphereaboutatangentis .Findthemoment

ofinertiaaboutadiameter?

Ans:AtangentKClisdrawnatpt.CofasolidsphereofmassMandradiusR.Drawa

diameterAOB||toKCl.

ThenaccordingtoTheoremofparallelaxis,I,=I+M(OC)2

I1(M.Iaboutthetangent)=

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2.Fourparticlesofmass1kg,2kg,3kgand4kgareplacedatthefourverticesA,B,Cand

Dofsquareofside1m.Findthepositionofcentreofmassoftheparticle.

Ans:Hence

Thuscentreofmass(0.5m,0.7m)

3.Acircularringofdiameter40cmandmass1kgisrotatingaboutanaxisnormaltoits

planeandpassingthroughthecentrewithafrequencyof10rotationspersecond.

Calculatetheangularmomentumaboutitsaxisofrotation?

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Ans:

4.(a)Whichphysicalquantitiesarerepresentedbythe

(i)Rateofchangeofangularmomentum

(ii)ProductofIand

(b)ShowthatangularmomentumofasatelliteofmassMSrevolvingaroundtheearth

havingmassMeinanorbitofradiusrisequalto

Ans:(a)(1)Torquei.e.

(2)Angularmomentumi.e.L=Iw

(b)Massofsatellite=Ms

Massofearth=Me

Radiusofsatellite=r

Requiredcentripetalforce

Where istheorbitalvelocitywithwhichthesatelliterevolvesroundtheearth.

Gravitationalforcebetweenthesatelliteandtheearth

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Equating(1)and(2)

Nowangularmomentumofthesatellite

HenceProved

5.IntheHClmolecule,theseparationbetweenthenucleiofthetwoatomsisabout1.27

.Findtheapproximatelocationofthe

CMofthemolecule,giventhatachlorineatomisabout35.5timesasmassiveasa

hydrogenatomandnearlyallthemassofanatomisconcentratedinitsnucleus.

Ans.Thegivensituationcanbeshownas:

DistancebetweenHandClatoms=1.27

MassofHatom=m

MassofClatom=35.5m

LetthecentreofmassofthesystemlieatadistancexfromtheClatom.

DistanceofthecentreofmassfromtheHatom=(1.27–x)

Letusassumethatthecentreofmassofthegivenmoleculeliesattheorigin.Therefore,we

canhave:

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Here,thenegativesignindicatesthatthecentreofmassliesattheleftofthemolecule.

Hence,thecentreofmassoftheHClmoleculelies0.037 fromtheClatom.

6.Showthattheareaofthetrianglecontainedbetweenthevectorsaandbisonehalf

ofthemagnitudeofaxb.

Ans.Considertwovectors and ,inclinedatanangleθ,asshowninthe

followingfigure.

InΔOMN,wecanwritetherelation:

=2×AreaofΔOMK

∴AreaofΔOMK

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7.Ametrestickisbalancedonaknifeedgeatitscentre.Whentwocoins,eachofmass5

gareputoneontopoftheotheratthe12.0cmmark,thestickisfoundtobebalanced

at45.0cm.Whatisthemassofthemetrestick?

Ans.LetWand betherespectiveweightsofthemetrestickandthecoin.

Themassofthemetrestickisconcentratedatitsmid-point,i.e.,atthe50cmmark.

Massofthemeterstick=

Massofeachcoin,m=5g

Whenthecoinsareplaced12cmawayfromtheendP,thecentreofmassgetsshiftedby5

cmfrompointRtowardtheendP.Thecentreofmassislocatedatadistanceof45cmfrom

pointP.

ThenettorquewillbeconservedforrotationalequilibriumaboutpointR.

Hence,themassofthemetrestickis66g.

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CBSEClass11physics

ImportantQuestions

Chapter7

SystemofParticlesandRotationalMotion

4MarksQuestions

1.Showthata.(b c)isequalinmagnitudetothevolumeoftheparallelepiped

formedonthethreevectors,a,bandc.

Ans.AparallelepipedwithoriginOandsidesa,b,andcisshowninthefollowingfigure.

Volumeofthegivenparallelepiped=abc

Let beaunitvectorperpendiculartobothbandc.Hence, andahavethesamedirection.

∴

=abccosθ

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=abccos0°

=abc

=Volumeoftheparallelepiped.

2.Ahoopofradius2mweighs100kg.Itrollsalongahorizontalfloorsothatitscentre

ofmasshasaspeedof20cm/s.Howmuchworkhastobedonetostopit?

Ans.Radiusofthehoop,r=2m

Massofthehoop,m=100kg

Velocityofthehoop,v=20cm/s=0.2m/s

Totalenergyofthehoop=TranslationalKE+RotationalKE

Momentofinertiaofthehoopaboutitscentre,I=

Butwehavetherelation,

Theworkrequiredtobedoneforstoppingthehoopisequaltothetotalenergyofthehoop.

∴Requiredworktobedone,

3.Theoxygenmoleculehasamassof kgandamomentofinertiaof

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aboutanaxisthroughitscentreperpendiculartothelinesjoining

thetwoatoms.Supposethemeanspeedofsuchamoleculeinagasis500m/sandthat

itskineticenergyofrotationistwothirdsofitskineticenergyoftranslation.Findthe

averageangularvelocityofthemolecule.

Ans.Massofanoxygenmolecule,m= kg

Momentofinertia,I=

Velocityoftheoxygenmolecule,v=500m/s

Theseparationbetweenthetwoatomsoftheoxygenmolecule=2r

Massofeachoxygenatom=

Hence,momentofinertiaI,iscalculatedas:

Itisgiventhat:

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=

4.Amanstandsonarotatingplatform,withhisarmsstretchedhorizontallyholdinga

5kgweightineachhand.Theangularspeedoftheplatformis30revolutionsper

minute.Themanthenbringshisarmsclosetohisbodywiththedistanceofeachweight

fromtheaxischangingfrom90cmto20cm.Themomentofinertiaofthemantogether

withtheplatformmaybetakentobeconstantandequalto7.6kg .

(a)Whatishisnewangularspeed?(Neglectfriction.)

(b)Iskineticenergyconservedintheprocess?Ifnot,fromwheredoesthechangecome

about?

Ans.(a)58.88rev/min(b)No

(a)Momentofinertiaoftheman-platformsystem=7.6kg

Momentofinertiawhenthemanstretcheshishandstoadistanceof90cm:

=

=8.1kg

Initialmomentofinertiaofthesystem,

Angularspeed,

Angularmomentum, …………(i)

Momentofinertiawhenthemanfoldshishandstoadistanceof20cm:

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= =0.4kg

Finalmomentofinertia,

Finalangularspeed=

Finalangularmomentum, …(ii)

Fromtheconservationofangularmomentum,wehave:

(b)Kineticenergyisnotconservedinthegivenprocess.Infact,withthedecreaseinthe

momentofinertia,kineticenergyincreases.Theadditionalkineticenergycomesfromthe

workdonebythemantofoldhishandstowardhimself.

5.Readeachstatementbelowcarefully,andstate,withreasons,ifitistrueorfalse;

(a)Duringrolling,theforceoffrictionactsinthesamedirectionasthedirectionof

motionoftheCMofthebody.

(b)Theinstantaneousspeedofthepointofcontactduringrollingiszero.

(c)Theinstantaneousaccelerationofthepointofcontactduringrollingiszero.

(d)Forperfectrollingmotion,workdoneagainstfrictioniszero.

(e)Awheelmovingdownaperfectlyfrictionlessinclinedplanewillundergoslipping

(notrolling)motion.

Ans.(a)False

Frictionalforceactsoppositetothedirectionofmotionofthecentreofmassofabody.Inthe

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caseofrolling,thedirectionofmotionofthecentreofmassisbackward.Hence,frictional

forceactsintheforwarddirection.

(b)True

Rollingcanbeconsideredastherotationofabodyaboutanaxispassingthroughthepointof

contactofthebodywiththeground.Hence,itsinstantaneousspeediszero.

(c)False

Whenabodyisrolling,itsinstantaneousaccelerationisnotequaltozero.Ithassomevalue.

(d)True

Whenperfectrollingbegins,thefrictionalforceactingatthelowermostpointbecomeszero.

Hence,theworkdoneagainstfrictionisalsozero.

(e)True

Therollingofabodyoccurswhenafrictionalforceactsbetweenthebodyandthesurface.

Thisfrictionalforceprovidesthetorquenecessaryforrolling.Intheabsenceofafrictional

force,thebodyslipsfromtheinclinedplaneundertheeffectofitsownweight.

6.Twoparticles,eachofmassmandspeedv,travelinoppositedirectionsalong

parallellinesseparatedbyadistanced.Showthatthevectorangularmomentumofthe

twoparticlesystemisthesamewhateverbethepointaboutwhichtheangular

momentumistaken.

Ans.LetatacertaininstanttwoparticlesbeatpointsPandQ,asshowninthefollowing

figure.

AngularmomentumofthesystemaboutpointP:

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……….(i)

Angularmomentumofthesystemaboutpoint :

…………….(ii)

ConsiderapointR,whichisatadistanceyfrompointQ,i.e.,

QR=y

∴PR=d–y

AngularmomentumofthesystemaboutpointR:

………(iii)

Comparingequations(i),(ii),and(iii),weget:

……………(iv)

Weinferfromequation(iv)thattheangularmomentumofasystemdoesnotdependonthe

pointaboutwhichitistaken.

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CBSEClass11physics

ImportantQuestions

Chapter7

SystemofParticlesandRotationalMotion

5MarksQuestions

1.(a)Whyismomentofinertiacalledrotationalinertia?

(b)CalculateM.Iofauniformcirculardiscofmass500gmandradius10cmabout

(i)Diameter(ii)axistangenttothediscandparalleltodiameter

(c)Axispassingtroughcentreandperpendiculartoitsplane?

Ans:(a)Momentofinertiaiscalledrotationalinertiabecauseitmeasuresmomentofinertia

duringitsrotationalmotion.

(b)(i)

(ii)

(iii)

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2.(a)Acatisabletolandonitsfeetafterafall.Why?

(b)Ifangularmomentummomentofinertiaisdecreased,willitsrotational be

alsoconserved?Explain.

Ans:(a)Whencatlandstotheground,ifstretchesitstailasresultM.Iincreases

AsIW=constant

Angularspeedwillbesmallduetoincreaseinmomentofinertiaandthecatisableto

landonitsfeetwithoutanyharm.

(b)LetmomentofinertiaofasystemdecreasefromItoI’

Thenangularspeedincreasefromwtow’

K.E.ofrotationofthesystem

K.Eofthesystemwillincrease.Henceitwillnotbeconserved.

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3.Findthecomponentsalongthex,y,zaxesoftheangularmomentumlofaparticle,

whosepositionvectorisrwithcomponentsx,y,zandmomentumispwithcomponents

.Showthatiftheparticlemovesonlyinthex-yplanetheangular

momentumhasonlyaz-component.

Ans.

Linearmomentumoftheparticle,

Positionvectoroftheparticle,

Angularmomentum,

=

=

Comparingthecoefficientsof weget:

…………..(i)

Theparticlemovesinthex-yplane.Hence,thez-componentofthepositionvectorandlinear

momentumvectorbecomeszero,i.e.,

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z= =0

Thus,equation(i)reducesto:

Therefore,whentheparticleisconfinedtomoveinthex-yplane,thedirectionofangular

momentumisalongthez-direction.

4.Anon-uniformbarofweightWissuspendedatrestbytwostringsofnegligible

weightasshowninFig.7.39.Theanglesmadebythestringswiththeverticalare36.9°

and53.1°respectively.Thebaris2mlong.Calculatethedistancedofthecentreof

gravityofthebarfromitsleftend.

Ans.Thefreebodydiagramofthebarisshowninthefollowingfigure.

Lengthofthebar,l=2m

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and arethetensionsproducedintheleftandrightstringsrespectively.

Attranslationalequilibrium,wehave:

Forrotationalequilibrium,ontakingthetorqueaboutthecentreofgravity,wehave:

Hence,theC.G.(centreofgravity)ofthegivenbarlies0.72mfromitsleftend.

5.Acarweighs1800kg.Thedistancebetweenitsfrontandbackaxlesis1.8m.Itscentre

ofgravityis1.05mbehindthefrontaxle.Determinetheforceexertedbythelevel

groundoneachfrontwheelandeachbackwheel.

Ans.Massofthecar,m=1800kg

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Distancebetweenthefrontandbackaxles,d=1.8m

DistancebetweentheC.G.(centreofgravity)andthebackaxle=1.05m

Thevariousforcesactingonthecarareshowninthefollowingfigure.

and aretheforcesexertedbythelevelgroundonthefrontandbackwheels

respectively.

Attranslationalequilibrium:

=mg

=1800 9.8

=17640N…(i)

Forrotationalequilibrium,ontakingthetorqueabouttheC.G.,wehave:

…………..(ii)

Solvingequations(i)and(ii),weget:

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∴ =17640–7350=10290N

Therefore,theforceexertedoneachfrontwheel ,and

Theforceexertedoneachbackwheel

6.(a)Findthemomentofinertiaofasphereaboutatangenttothesphere,giventhe

momentofinertiaofthesphereaboutanyofitsdiameterstobe ,whereMis

themassofthesphereandRistheradiusofthesphere.

(b)GiventhemomentofinertiaofadiscofmassMandradiusRaboutanyofits

diameterstobe ,finditsmomentofinertiaaboutanaxisnormaltothedisc

andpassingthroughapointonitsedge.

Ans.(a)

Themomentofinertia(M.I.)ofasphereaboutitsdiameter=

Accordingtothetheoremofparallelaxes,themomentofinertiaofabodyaboutanyaxisis

equaltothesumofthemomentofinertiaofthebodyaboutaparallelaxispassingthrough

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itscentreofmassandtheproductofitsmassandthesquareofthedistancebetweenthetwo

parallelaxes.

TheM.I.aboutatangentofthesphere=

(b)

Themomentofinertiaofadiscaboutitsdiameter=

Accordingtothetheoremofperpendicularaxis,themomentofinertiaofaplanarbody

(lamina)aboutanaxisperpendiculartoitsplaneisequaltothesumofitsmomentsofinertia

abouttwoperpendicularaxesconcurrentwithperpendicularaxisandlyingintheplaneof

thebody.

TheM.I.ofthediscaboutitscentre=

Thesituationisshowninthegivenfigure.

Applyingthetheoremofparallelaxes:

Themomentofinertiaaboutanaxisnormaltothediscandpassingthroughapointonits

edge=

7.Torquesofequalmagnitudeisappliedtoahollowcylinderandasolidsphere,both

havingthesamemassandradius.Thecylinderisfreetorotateaboutitsstandardaxis

ofsymmetry,andthesphereisfreetorotateaboutanaxispassingthroughitscentre.

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Whichofthetwowillacquireagreaterangularspeedafteragiventime?

Ans.Letmandrbetherespectivemassesofthehollowcylinderandthesolidsphere.

Themomentofinertiaofthehollowcylinderaboutitsstandardaxis,

Themomentofinertiaofthesolidsphereaboutanaxispassingthroughitscentre,

Wehavetherelation:

Where,

α=Angularacceleration

T=Torque

I=Momentofinertia

Forthehollowcylinder,

Forthesolidsphere,

Asanequaltorqueisappliedtoboththebodies,

……….(i)

Now,usingtherelation:

Where,

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=Initialangularvelocity

t=Timeofrotation

=Finalangularvelocity

Forequal andt,wehave:

…(ii)

Fromequations(i)and(ii),wecanwrite:

Hence,theangularvelocityofthesolidspherewillbegreaterthanthatofthehollow

cylinder.

8.(a)Achildstandsatthecentreofaturntablewithhistwoarmsoutstretched.The

turntableissetrotatingwithanangularspeedof40rev/min.Howmuchistheangular

speedofthechildifhefoldshishandsbackandtherebyreduceshismomentofinertia

to2/5timestheinitialvalue?Assumethattheturntablerotateswithoutfriction.

(b)Showthatthechild'snewkineticenergyofrotationismorethantheinitialkinetic

energyofrotation.Howdoyouaccountforthisincreaseinkineticenergy?

Ans.(a)100rev/min

Initialangularvelocity, =40rev/min

Finalangularvelocity=

Themomentofinertiaoftheboywithstretchedhands=

Themomentofinertiaoftheboywithfoldedhands=

Thetwomomentsofinertiaarerelatedas:

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Sincenoexternalforceactsontheboy,theangularmomentumLisaconstant.

Hence,forthetwosituations,wecanwrite:

(b)FinalK.E.=2.5InitialK.E.

Finalkineticrotation,

Initialkineticrotation,

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Theincreaseintherotationalkineticenergyisattributedtotheinternalenergyoftheboy.

9.FromauniformdiskofradiusR,acircularholeofradiusR/2iscutout.Thecentreof

theholeisatR/2fromthecentreoftheoriginaldisc.Locatethecentreofgravityofthe

resultingflatbody.

Ans.R/6;fromtheoriginalcentreofthebodyandoppositetothecentreofthecutportion.

Massperunitareaoftheoriginaldisc=

Radiusoftheoriginaldisc=R

Massoftheoriginaldisc,M=

Thediscwiththecutportionisshowninthefollowingfigure:

Radiusofthesmallerdisc=

Massofthesmallerdisc,M'=

LetOand betherespectivecentersoftheoriginaldiscandthedisccutofffromthe

original.Asperthedefinitionofthecentreofmass,thecentreofmassoftheoriginaldiscis

supposedtobeconcentratedatO,whilethatofthesmallerdiscissupposedtobe

concentratedat .

Itisgiventhat:

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=

Afterthesmallerdischasbeencutfromtheoriginal,theremainingportionisconsideredto

beasystemoftwomasses.Thetwomassesare:

M(concentratedatO),and

concentratedat

(Thenegativesignindicatesthatthisportionhasbeenremovedfromtheoriginaldisc.)

Letxbethedistancethroughwhichthecentreofmassoftheremainingportionshiftsfrom

pointO.

Therelationbetweenthecentersofmassesoftwomassesisgivenas:

Forthegivensystem,wecanwrite:

(ThenegativesignindicatesthatthecentreofmassgetsshiftedtowardtheleftofpointO.)

10.Asolidsphererollsdowntwodifferentinclinedplanesofthesameheightsbut

differentanglesofinclination.(a)Willitreachthebottomwiththesamespeedineach

case?(b)Willittakelongertorolldownoneplanethantheother?(c)Ifso,whichone

andwhy?

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Ans.(a)Yes(b)Yes(c)Onthesmallerinclination

(a)Massofthesphere=m

Heightoftheplane=h

Velocityofthesphereatthebottomoftheplane=v

Atthetopoftheplane,thetotalenergyofthesphere=Potentialenergy=mgh

Atthebottomoftheplane,thespherehasbothtranslationalandrotationalkineticenergies.

Hence,totalenergy=

Usingthelawofconservationofenergy,wecanwrite:

………(i)

Forasolidsphere,themomentofinertiaaboutitscentre,

Hence,equation(i)becomes:

Butwehavetherelation,

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Hence,thevelocityofthesphereatthebottomdependsonlyonheight(h)andacceleration

duetogravity(g).Boththesevaluesareconstants.Therefore,thevelocityatthebottom

remainsthesamefromwhicheverinclinedplanethesphereisrolled.

(b),(c)Considertwoinclinedplaneswithinclinations and ,relatedas:

<

Theaccelerationproducedinthespherewhenitrollsdowntheplaneinclinedat is:

gsin

Thevariousforcesactingonthesphereareshowninthefollowingfigure.

isthenormalreactiontothesphere.

Similarly,theaccelerationproducedinthespherewhenitrollsdowntheplaneinclinedat

is:

gsin

Thevariousforcesactingonthesphereareshowninthefollowingfigure.

isthenormalreactiontothesphere.

> ;sin >sin ...(i)

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∴ > …(ii)

Initialvelocity,u=0

Finalvelocity,v=Constant

Usingthefirstequationofmotion,wecanobtainthetimeofrollas:

v=u+at

11.Asolidcylinderrollsupaninclinedplaneofangleofinclination30°.Atthebottom

oftheinclinedplanethecentreofmassofthecylinderhasaspeedof5m/s.

(a)Howfarwillthecylindergouptheplane?

(b)Howlongwillittaketoreturntothebottom?

Ans.Asolidcylinderrollingupaninclinationisshowninthefollowingfigure.

Initialvelocityofthesolidcylinder,v=5m/s

Angleofinclination,θ=30°

Heightreachedbythecylinder=h

(a)EnergyofthecylinderatpointA:

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EnergyofthecylinderatpointB=mgh

Usingthelawofconservationofenergy,wecanwrite:

Momentofinertiaofthesolidcylinder,

Butwehavetherelation,

InΔABC:

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Hence,thecylinderwilltravel3.82muptheinclinedplane.

(b)ForradiusofgyrationK,thevelocityofthecylinderattheinstancewhenitrollsbackto

thebottomisgivenbytherelation:

Forthesolidcylinder,

Thetimetakentoreturntothebottomis:

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Therefore,thetotaltimetakenbythecylindertoreturntothebottomis(2 0.764)1.53s.

12.AsshowninFig.7.40,thetwosidesofastepladderBAandCAare1.6mlongand

hingedatA.AropeDE,0.5mistiedhalfwayup.Aweight40kgissuspendedfroma

pointF,1.2mfromBalongtheladderBA.Assumingthefloortobefrictionlessand

neglectingtheweightoftheladder,findthetensionintheropeandforcesexertedby

thefloorontheladder.(Takeg=9.8 )

(Hint:Considertheequilibriumofeachsideoftheladderseparately.)

Ans.Thegivensituationcanbeshownas:

NB=ForceexertedontheladderbythefloorpointB

NC=ForceexertedontheladderbythefloorpointC

T=Tensionintherope

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BA=CA=1.6m

DE=0.5m

BF=1.2m

Massoftheweight,m=40kg

DrawaperpendicularfromAonthefloorBC.ThisintersectsDEatmid-pointH.

ΔABIandΔAICaresimilar

∴BI=IC

Hence,Iisthemid-pointofBC.

DE||BC

BC=2 DE=1m

AF=BA–BF=0.4m…(i)

Disthemid-pointofAB.

Hence,wecanwrite:

………….(ii)

Usingequations(i)and(ii),weget:

FE=0.4m

Hence,Fisthemid-pointofAD.

FG DHandFisthemid-pointofAD.Hence,Gwillalsobethemid-pointofAH.

ΔAFGandΔADHaresimilar

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InΔADH:

Fortranslationalequilibriumoftheladder,theupwardforceshouldbeequaltothe

downwardforce.

=mg=392…(iii)

Forrotationalequilibriumoftheladder,thenetmomentaboutAis:

Addingequations(iii)and(iv),weget:

ForrotationalequilibriumofthesideAB,considerthemomentaboutA.

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13.Twodiscsofmomentsofinertia and abouttheirrespectiveaxes(normaltothe

discandpassingthroughthecentre),androtatingwithangularspeeds and are

broughtintocontactfacetofacewiththeiraxesofrotationcoincident.(a)Whatisthe

angularspeedofthetwo-discsystem?(b)Showthatthekineticenergyofthecombined

systemislessthanthesumoftheinitialkineticenergiesofthetwodiscs.Howdoyou

accountforthislossinenergy?Take ≠ .

Ans.(a)

Momentofinertiaofdisc

Angularspeedofdisc

Angularspeedofdisc

Angularmomentumofdisc

Angularmomentumofdisc

Angularmomentumofdisc

Totalinitialangularmomentum,

Whenthetwodiscsarejoinedtogether,theirmomentsofinertiagetaddedup.

Momentofinertiaofthesystemoftwodiscs,

Let betheangularspeedofthesystem.

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Totalfinalangularmomentum,

Usingthelawofconservationofangularmomentum,wehave:

(b)KineticenergyofdiscI,

KineticenergyofdiscII,

Totalinitialkineticenergy,

Whenthediscsarejoined,theirmomentsofinertiagetaddedup.

Momentofinertiaofthesystem,

Angularspeedofthesystem=

Finalkineticenergy :

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AllthequantitiesonRHSarepositive.

ThelossofKEcanbeattributedtothefrictionalforcethatcomesintoplaywhenthetwo

discscomeincontactwitheachother.

14.(a)Provethetheoremofperpendicularaxes.

(Hint:Squareofthedistanceofapoint(x,y)inthex–yplanefromanaxisthroughthe

originperpendiculartotheplaneis ).

(b)Provethetheoremofparallelaxes.

(Hint:Ifthecentreofmassischosentobetheorigin ).

Ans.(a)Thetheoremofperpendicularaxesstatesthatthemomentofinertiaofaplanarbody

(lamina)aboutanaxisperpendiculartoitsplaneisequaltothesumofitsmomentsofinertia

abouttwoperpendicularaxesconcurrentwithperpendicularaxisandlyingintheplaneof

thebody.

AphysicalbodywithcentreOandapointmassm,inthex–yplaneat(x,y)isshowninthe

followingfigure.

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Momentofinertiaaboutx-axis,Ix=

Momentofinertiaabouty-axis,Iy=

Momentofinertiaaboutz-axis,Iz=

=m

=

Hence,thetheoremisproved.

(b)Thetheoremofparallelaxesstatesthatthemomentofinertiaofabodyaboutanyaxisis

equaltothesumofthemomentofinertiaofthebodyaboutaparallelaxispassingthrough

itscentreofmassandtheproductofitsmassandthesquareofthedistancebetweenthetwo

parallelaxes.

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Supposearigidbodyismadeupofnparticles,havingmasses ,at

perpendiculardistances respectivelyfromthecentreofmassOoftherigid

body.

ThemomentofinertiaaboutaxisRSpassingthroughthepointO:

IRS=

Theperpendiculardistanceofmassmi,fromtheaxisQP=a+ri

Hence,themomentofinertiaaboutaxisQP:

Now,atthecentreofmass,themomentofinertiaofalltheparticlesabouttheaxispassing

throughthecentreofmassiszero,thatis,

Also,

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M=Totalmassoftherigidbody

Hence,thetheoremisproved.

15.Provetheresultthatthevelocityvoftranslationofarollingbody(likearing,disc,

cylinderorsphere)atthebottomofaninclinedplaneofaheighthisgivenby

.

Usingdynamicalconsideration(i.e.byconsiderationofforcesandtorques).Notekis

theradiusofgyrationofthebodyaboutitssymmetryaxis,andRistheradiusofthe

body.Thebodystartsfromrestatthetopoftheplane.

Ans.Abodyrollingonaninclinedplaneofheighth,isshowninthefollowingfigure:

m=Massofthebody

R=Radiusofthebody

K=Radiusofgyrationofthebody

v=Translationalvelocityofthebody

h=Heightoftheinclinedplane

g=Accelerationduetogravity

Totalenergyatthetopoftheplane, =mgh

Totalenergyatthebottomoftheplane,

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But

Fromthelawofconservationofenergy,wehave:

Hence,thegivenresultisproved.

16.Adiscrotatingaboutitsaxiswithangularspeed isplacedlightly(withoutany

translationalpush)onaperfectlyfrictionlesstable.TheradiusofthediscisR.What

arethelinearvelocitiesofthepointsA,BandConthediscshowninFig.7.41?Willthe

discrollinthedirectionindicated?

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Ans. =R ; =R ; ;Thediscwillnotroll

Angularspeedofthedisc=

Radiusofthedisc=R

Usingtherelationforlinearvelocity,v= R

ForpointA:

=R ;inthedirectiontangentialtotheright

ForpointB:

=R ;inthedirectiontangentialtotheleft

ForpointC:

;inthedirectionsameasthatof

ThedirectionsofmotionofpointsA,B,andConthediscareshowninthefollowingfigure

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Sincethediscisplacedonafrictionlesstable,itwillnotroll.Thisisbecausethepresenceof

frictionisessentialfortherollingofabody.

17.Asoliddiscandaring,bothofradius10cmareplacedonahorizontaltable

simultaneously,withinitialangularspeedequalto10πrad .Whichofthetwowill

starttorollearlier?Theco-efficientofkineticfrictionis =0.2.

Ans.Disc

Radiioftheringandthedisc,r=10cm=0.1m

Initialangularspeed, =10πrad

Coefficientofkineticfriction, =0.2

Initialvelocityofboththeobjects,u=0

Motionofthetwoobjectsiscausedbyfrictionalforce.AsperNewton'ssecondlawofmotion,

wehavefrictionalforce,f=ma

mg=ma

Where,

a=Accelerationproducedintheobjects

m=Mass

∴a= g…(i)

Asperthefirstequationofmotion,thefinalvelocityoftheobjectscanbeobtainedas:

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v=u+at

=0+ gt

= gt…(ii)

Thetorqueappliedbythefrictionalforcewillactinperpendicularlyoutwarddirectionand

causereductionintheinitialangularspeed.

Torque,T=–Iα

α=Angularacceleration

=–Iα

…………….(iii)

Usingthefirstequationofrotationalmotiontoobtainthefinalangularspeed:

………(iv)

Rollingstartswhenlinearvelocity,v=r

………(v)

Equatingequations(ii)and(v),weget:

…….(vi)

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………….(vii)

……………(viii)

Since ,thediscwillstartrollingbeforethering.

18.Acylinderofmass10kgandradius15cmisrollingperfectlyonaplaneof

inclination30°.Thecoefficientofstaticfriction =0.25.

(a)Howmuchistheforceoffrictionactingonthecylinder?

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(b)Whatistheworkdoneagainstfrictionduringrolling?

(c)Iftheinclination oftheplaneisincreased,atwhatvalueof doesthecylinder

begintoskid,andnotrollperfectly?

Ans.Massofthecylinder,m=10kg

Radiusofthecylinder,r=15cm=0.15m

Co-efficientofkineticfriction, =0.25

Angleofinclination,θ=30°

Momentofinertiaofasolidcylinderaboutitsgeometricaxis,

Thevariousforcesactingonthecylinderareshowninthefollowingfigure:

Theaccelerationofthecylinderisgivenas:

(a)UsingNewton'ssecondlawofmotion,wecanwritenetforceas:

=ma

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(b)Duringrolling,theinstantaneouspointofcontactwiththeplanecomestorest.Hence,the

workdoneagainstfrictionalforceiszero.

(c)Forrollingwithoutskid,wehavetherelation:

19.SeparationofMotionofasystemofparticlesintomotionofthecentreofmassand

motionaboutthecentreofmass:

(a)Show

WherepiisthemomentumoftheIthparticle(ofmassmi)and is

thevelocityoftheIthparticlerelativetothecentreofmass.

Also,proveusingthedefinitionofthecentreofmass

(b)ShowK=

WhereKisthetotalkineticenergyofthesystemofparticles, isthetotalkinetic

energyofthesystemwhentheparticlevelocitiesaretakenwithrespecttothecentreof

massand isthekineticenergyofthetranslationofthesystemasawhole(i.e.

ofthecentreofmassmotionofthesystem).TheresulthasbeenusedinSec.7.14.

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(c)Show

Where istheangularmomentumofthesystemaboutthecentreof

masswithvelocitiestakenrelativetothecentreofmass.Remember rest

ofthenotationisthestandardnotationusedinthechapter.Note and can

besaidtobeangularmomenta,respectively,aboutandofthecentreifmassofthe

systemofparticles.

(d)Show

Furthershowthat

Where isthesumofallexternaltorquesactingonthesystemaboutthecentreof

mass.

(Hint:UsethedefinitionofcentreofmassandNewton’sThirdLaw.Assumethe

internalforcesbetweenanytwoparticlesactalongthelinejoiningtheparticles.)

Ans.(a)Takeasystemofimovingparticles.

Massoftheithparticle=

Velocityoftheithparticle=

Hence,momentumoftheithparticle,

Velocityofthecentreofmass=V

Thevelocityoftheithparticlewithrespecttothecentreofmassofthesystemisgivenas:

…(1)

Multiplying throughoutequation(1),weget:

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Where,

=Momentumoftheithparticlewithrespecttothecentreofmassofthesystem

Wehavetherelation:

Takingthesummationofmomentumofalltheparticleswithrespecttothecentreofmassof

thesystem,weget:

Where,

=Positionvectorofitsparticlewithrespecttothecentreofmass

Asperthedefinitionofthecentreofmass,wehave:

(b)Wehavetherelationforvelocityoftheparticleas:

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…(2)

Takingthedotproductofequation(2)withitself,weget:

Here,forthecentreofmassofthesystemofparticles,

Where,K= =Totalkineticenergyofthesystemofparticles

K'= =Totalkineticenergyofthesystemofparticleswithrespecttothecentreof

mass

=Kineticenergyofthetranslationofthesystemasawhole

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CBSEClass11Physics

Chapter-7(SystemofParticlesandRotationalMotion)

NUMERICALS

1. Threemasses3kg,4kgand5kgarelocatedatthecornersofanequilateraltriangle

ofside1m.Locatethecentreofmassofthesystem.

Ans.(x,y)=(0.54m,0.36m)

2. Twoparticlesmass100gand300gatagiventimehavevelocities and

ms-1respectively.DeterminevelocityofCOM.

Ans.VelocityofCOM=

3. FromauniformdiscofradiusR,acirculardiscofradiusR/2iscutout.Thecentreof

theholeisatR/2fromthecentreoforiginaldisc.Locatethecentreofgravityofthe

resultantflatbody.

Ans.COMofresultingportionliesatR/6fromthecentreoftheoriginaldiscinadirection

oppositetothecentreofthecutoutportion.

4. Theangularspeedofamotorwheelisincreasedfrom1200rpmto3120rpmin16

seconds,(i)Whatisitsangularacceleration(assumetheaccelerationtobeuniform)

(ii)Howmanyrevolutionsdoesthewheelmakeduringthistime?

Ans.

n=576

5. Ametrestickisbalancedonaknifeedgeatitscentre.Whentwocoins,eachofmass

5gareputoneontopoftheotheratthe12.0cmmark,thestickisfoundtobe

balancedat45.0cm,whatisthemassofthemetrestick?

Ans.m=66.0g

6. A3mlongladderweighting20kgleansonafrictionlesswall.Itsfeetrestonthe

floor1mfromthewallasshowninfigure.FindthereactionforcesF1andF2ofthe

wallandthefloor.

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Ans.

IfF2makesanangle withthehorizontalthen

tan =

=80°

7. Calculate the ratio of radii of gyration of a circular ring and a disc of the same

radiuswithrespecttotheaxispassingthroughtheircentresandperpendicularto

theirplanes.

Ans.

8. Anautomobilemovesonaroadwithaspeedof54kmh-1.Theradiusofitswheelsis

0.35m.Whatistheaveragenegativetorquetransmittedbyitsbrakestoawheelif

thevehicleisbroughttorestin15s?Themomentofinertiaofthewheelaboutthe

axisofrotationis3kgm2.

Ans.

9. ArodoflengthLandmassMishingedatpoint0.Asmallbulletofmassmhitsthe

rod, as shown in figure. The bullet get embedded in the rod. Find the angular

velocityofthesystemjustaftertheimpact.

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Ans.Usingconservationofangularmomentum

Linitial=Linitial

MVL=l

OrMVL=

Or

10. A solid disc and a ring, both of radius 10 cm are placed on a horizontal table

simultaneously,withinitialangularspeedequalto10.Whichofthetwowillstartto

rollearlier?Thecoefficientofkineticfrictionispk=0.2-1πrads

Ans.Thediscbeginstorollearlierthanthering.

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CBSEClass11thPhysics

Chapter-7(SystemofParticlesandRotationalMotion)

2MARKSQUESTIONS

1. Showthatintheabsenceofanyexternalforce,thevelocityofthecentreofmassremains

constant.

2. Statethefactorsonwhichthepositionofcentreofmassofarigidbodydepends.

Ans.

(i)Shapeofbody

(ii)massdistribution

3. Whatistheturningeffectofforcecalledfor?Onwhatfactorsdoesitdepend?

Ans.Torque

Factors:

(i)Magnitudeofforce

(ii)Perpendiculardistanceofforcevectorfromaxisofrotation.

4. Statethefactorsonwhichthemomentofinertiaofabodydepends.

Ans.

(I)Massofbody

(ii)Sizeandshapeofbody

(iii)Massdistributionw.r.t.axisofrotation

(iv)positionandorientationofrotationalaxis

5. Onwhatfactorsdoesradiusofgyrationofbodydepend?

Ans.Massdistribution.

6. Whydoweprefertouseawrenchoflongerarm?

Ans.toincreasetorque.

7. CanabodybeInequilibriumwhileinmotion?Ifyes,giveanexample

Ans.Yes,ifbodyhasnolinearandangularacceleration.Henceabodyinuniform

straightlinemotionwillbeinequilibrium.

8. Thereisastickhalfofwhichiswoodenandhalfisofsteel,

i. itispivotedatthewoodenendandaforceisappliedatthesteelendatright

angletoitslength

ii. itispivotedatthesteelendandthesameforceisappliedatthewoodenend.In

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whichcaseistheangularaccelerationmoreandwhy?

Ans.I(firstcase)>l(Secondcase)

(firstcase)< (Secondcase)

9. Ifearthcontractstohalfitsradiuswhatwouldbethelengthofthedayatequator?

Ans.

L=l1w1=l2w2

or

or hours

10. Aninternalforcecannotchangethestateofmotionofcentreofmassofabody.How

doestheinternalforceofthebrakesbringavehicletorest?

Ans.Inthiscasetheforcewhichbringthevehicletorestisfriction,anditisanexternal

force.

11. Whendoesarigidbodysaidtobeinequilibrium?Statethenecessaryconditionfor

abodytobeinequilibrium.

Ans.Fortranslationequilibrium

Forrotationalequilibrium

12. Howwillyoudistinguishbetweenahardboiledeggandaraweggbyspinningitona

tabletop?

Ans.Forsameexternaltorque,angularaccelerationofraweggwillbesmallthanthatof

Hardboiledegg

13. Whatarebinarystars?Discusstheirmotioninrespectoftheircentreofmass.

14. InwhichconditionabodyingravitationalfieldIsinstableequilibrium?

Ans.Whenverticallinethroughcentreofgravitypassesthroughthebaseofthebody.

15. Givethephysicalsignificanceofmomentofinertia.

Ans.Itplaysthesameroleinrotatorymotionasthemassdoesintranslatorymotion.

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CBSEClass11physics

ImportantQuestions

Chapter7

SystemofParticlesandRotationalMotion

1MarksQuestions

1.Awheel0.5minradiusismovingwithaspeedof12m/s.finditsangularspeed?

Ans:

2.Statetheconditionfortranslationalequilibriumofabody?

Ans:Fortranslationsequilibriumofabodythevectorsumofalltheforcesactingonthebody

mustbezero.

3.Howisangularmomentumrelatedtolinearmomentum?

Ans:

Where istheanglebetween

4.Whatisthepositionofthecentreofmassofauniformtriangularlamina?

Ans:Atthecentroidofthetriangularlamina.

5.Whatisthemomentofinertiaofasphereofmass20 andradius aboutits

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diameter?

Ans:

I=0.5kgm2

6.Whatarethefactorsonwhichmomentofinertiaofabodydepends?

Ans:(1)Massofthebody

(2)Shapeandsizeofthebody

(3)Positionoftheaxisofrotation

7.Twoparticlesinanisolatedsystemundergoheadoncollision.Whatisthe

accelerationofthecentreofmassofthesystem?

Ans:Accelerationiszeroasforce,areinternalforces.

8.Whichcomponentofaforcedoesnotcontributetowardstorque?

Ans:Theradialcomponentofaforcedoesnotcontributetowardstorque.

9.Whatisthepositionofcentreofmassofarectangularlamina?

Ans:Thecentreofmassofarectangularlaminaisthepointofintersectionofdiagonals.

10.Givethelocationofthecentreofmassofa(i)sphere,(ii)cylinder,(iii)ring,and(iv)

cube,eachofuniformmassdensity.Doesthecentreofmassofabodynecessarilylie

insidethebody?

Ans.Geometriccentre;No

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Thecentreofmass(C.M.)isapointwherethemassofabodyissupposedtobeconcentrated.

Forthegivengeometricshapeshavingauniformmassdensity,theC.M.liesattheir

respectivegeometriccenters.

Thecentreofmassofabodyneednotnecessarilyliewithinit.Forexample,theC.M.of

bodiessuchasaring,ahollowsphere,etc.,liesoutsidethebody.

11.AchildsitsstationaryatoneendofalongtrolleymovinguniformlywithaspeedV

onasmoothhorizontalfloor.Ifthechildgetsupandrunsaboutonthetrolleyinany

manner,whatisthespeedoftheCMofthe(trolley+child)system?

Ans.Nochange

Thechildisrunningarbitrarilyonatrolleymovingwithvelocityv.However,therunningof

thechildwillproducenoeffectonthevelocityofthecentreofmassofthetrolley.Thisis

becausetheforceduetotheboy'smotionispurelyinternal.Internalforcesproducenoeffect

onthemotionofthebodiesonwhichtheyact.Sincenoexternalforceisinvolvedintheboy-

trolleysystem,theboy'smotionwillproducenochangeinthevelocityofthecentreofmass

ofthetrolley.

12.Tomaintainarotoratauniformangularspeedof200rad ,anengineneedsto

transmitatorqueof180Nm.Whatisthepowerrequiredbytheengine?

(Note:uniformangularvelocityintheabsenceoffrictionimplieszerotorque.In

practice,appliedtorqueisneededtocounterfrictionaltorque).Assumethattheengine

is100%efficient.

Ans.Angularspeedoftherotor, =200rad/s

Torquerequired, =180Nm

Thepoweroftherotor(P)isrelatedtotorqueandangularspeedbytherelation:

P=

=180 200=36 103

=36kW

Hence,thepowerrequiredbytheengineis36kW.

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CBSEClass11Physics

Chapter-7(SystemofParticlesandRotationalMotion)

1MARKQUESTIONS

1. WhatIsarigidbody?

2. Statetheprincipleofmomentsofrotationalequilibrium.

3. Iscentreofmassofabodynecessarilylieinsidethebody?Giveanyexample

Ans.No.examplering

4. Canthecoupleactingonarigidbodyproducetranslatorymotion?

Ans.No.Itcanproduceonlyrototorymotion.

5. Whichcomponentoflinearmomentumdoesnotcontributetoangularmomentum?

Ans.RadialComponent

6. Asystemisinstableequilibrium.WhatcanwesayaboutItspotentialenergy?

Ans.P.Eisminimum.

7. Isradiusofgyrationaconstantquantity?

Ans.No,itchangeswiththepositionofaxisofrotation.

8. Twosolidspheresofthesamemassaremadeofmetalsofdifferentdensities.Which

ofthemhasalargemomentofinertiaaboutthediameter?

Ans.Sphereofsmalldensitywillhavelargemomentofinertia.

9. ThemomentofinertiaoftworotatingbodiesAandBarelAandlB(lA>lB)andtheir

angularmomentaareequal.Whichonehasagreaterkineticenergy?

Ans.

10. Aparticlemovesonacircularpathwithdecreasingspeed.Whathappenstoits

angularmomentum?

Ans.as = i:emagnitude decreasesbutdirectionremainsconstant.

11. Whatisthevalueofinstantaneousspeedofthepointofcontactduringpurerolling?

Ans.zero

12. Whichphysicalquantityisconservedwhenaplanetrevolvesaroundthesun?

Ans.Angularmomentumofplanet.

13. Whatisthevalueoftorqueontheplanetduetothegravitationalforceofsun?

Ans.zero.

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14. Ifnoexternaltorqueactsonabody,willitsangularvelocitybeconstant?

Ans.No.wotY.

15. Whytherearetwopropellersinahelicopter?

Ans.duetoconservationofangularmomentum

16. AchildsitsstationaryatoneendofalongtrolleymovinguniformlywithspeedVon

asmoothhorizontalfloor.Ifthechildgetsupandrunsaboutonthetrolleyinany

manner,thenwhatistheeffectofthespeedofthecentreofmassofthe(trolley-t-

child)system?

Ans.Nochangeinspeedofsystemasnoexternalforceisworking.

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CBSEClass11physics

ImportantQuestions

Chapter7

SystemofParticlesandRotationalMotion

2MarksQuestions

1.Aplanetrevolvesaroundonmassivestarinahighlyellipticalorbitisitsangular

momentumconstantovertheentireorbit.Givereason?

Ans:Aplanetrevolvesaroundthestarundertheeffectofgravitationalforcesincetheforce

isradialanddoesnotcontributetowardstorque.Thusintheabsenceofanexternaltorque

angularmomentumoftheplanetremainsconstant.

2.Obtaintheequation ?

Ans:Since

Integratingwithinthelimits

3.Whatisthetorqueoftheforce actingatthepoint

abouttheorigin?

Ans:

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4.Whatisthevalueoflinearvelocityif

Ans:

5.Establishthethirdequationofrotationalmotion

Ans:

Multiplyanddivideby

Integratingweget

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Henceprove.

6.Findtheexpressionforradiusofgyrationofasolidsphereaboutoneofitsdiameter?

Ans:M.Iofasolidsphere

Aboutitsdiameter=

K=RadiusofGyration

7.Provethatthecentreofmassoftwoparticlesdividesthelinejoiningtheparticlesin

theinverseratiooftheirmasses?

Ans:

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Ifcentreofmassisattheorigin

Intermsofmagnitude

8.Showthatcrossproductoftwoparallelvectorsiszero?

Ans:

If and areparalleltoeachother

9.Provetherelation

Ans:Weknow

Differentiatingwrt.Time

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From(1)and(2)

10.Showthatforanisolatedsystemthecentreofmassmoveswithuniformvelocity

alongastraightlinepath?

Ans:Let bethetotalmassconcentratedatcentreofmasswhosepositionvectoris

Foranisolatedsystem

11.Theangleθcoveredbyabodyinrotationalmotionisgivebytheequationθ=6t+5t2

+2t3.Determinethevalueofinstantaneousangularvelocityandangularacceleration

attimet=2S.

Ans:

Angularvelocity

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Againangularacceleration

12.Asolidcylinderofmass20kgrotatesaboutitsaxiswithangularspeed100rad .

Theradiusofthecylinderis0.25m.Whatisthekineticenergyassociatedwiththe

rotationofthecylinder?Whatisthemagnitudeofangularmomentumofthecylinder

aboutitsaxis?

Ans.Massofthecylinder,m=20kg

Angularspeed, =100rad

Radiusofthecylinder,r=0.25m

Themomentofinertiaofthesolidcylinder:

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∴Kineticenergy

=

∴Angularmomentum,L=I

=6.25 100

=62.5Js

13.Aropeofnegligiblemassiswoundroundahollowcylinderofmass3kgandradius

40cm.Whatistheangularaccelerationofthecylinderiftheropeispulledwithaforce

of30N?Whatisthelinearaccelerationoftherope?Assumethatthereisnoslipping.

Ans.Massofthehollowcylinder,m=3kg

Radiusofthehollowcylinder,r=40cm=0.4m

Appliedforce,F=30N

Themomentofinertiaofthehollowcylinderaboutitsgeometricaxis:

I=

=

Torque,

=30 0.4=12Nm

Forangularacceleration ,torqueisalsogivenbytherelation:

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Linearacceleration=rα=0.4 25=10m

14.Abulletofmass10gandspeed500m/sisfiredintoadoorandgetsembedded

exactlyatthecentreofthedoor.Thedooris1.0mwideandweighs12kg.Itishingedat

oneendandrotatesaboutaverticalaxispracticallywithoutfriction.Findtheangular

speedofthedoorjustafterthebulletembedsintoit.

(Hint:Themomentofinertiaofthedoorabouttheverticalaxisatoneendis .)

Ans.Massofthebullet,m=10g=

Velocityofthebullet,v=500m/s

Thicknessofthedoor,L=1m

Radiusofthedoor,

Massofthedoor,M=12kg

Angularmomentumimpartedbythebulletonthedoor:

α=mvr

…….(i)

Momentofinertiaofthedoor:

=

But

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15.ExplainwhyfrictionisnecessarytomakethediscinFig.7.41rollinthedirection

indicated.

(a)GivethedirectionoffrictionalforceatB,andthesenseoffrictionaltorque,before

perfectrollingbegins.

(b)Whatistheforceoffrictionafterperfectrollingbegins?

Ans.Atorqueisrequiredtorollthegivendisc.Asperthedefinitionoftorque,therotating

forceshouldbetangentialtothedisc.SincethefrictionalforceatpointBisalongthe

tangentialforceatpointA,africtionalforceisrequiredformakingthediscroll.

(a)ForceoffrictionactsoppositetothedirectionofvelocityatpointB.Thedirectionof

linearvelocityatpointBistangentiallyleftward.Hence,frictionalforcewillacttangentially

rightward.Thesenseoffrictionaltorquebeforethestartofperfectrollingisperpendicular

totheplaneofthediscintheoutwarddirection.

(b)SincefrictionalforceactsoppositetothedirectionofvelocityatpointB,perfectrolling

willbeginwhenthevelocityatthatpointbecomesequaltozero.Thiswillmakethefrictional

forceactingonthedisczero.

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