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Chapter 8: Momentum, Impulse, & Collisions linear momentum: Newton’s second law in terms of momentum: impulse: Under what SPECIFIC condition is linear momentum conserved? (The answer does not involve collisions.) List the types of collisions and what defines them in the table below. collisions properties What is the only collision during which you can analyze the energy and why? What is always true about the relative velocities between objects before and after a collision? center of mass:

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Page 1: linear momentum: Newton’s second law in terms of …

Chapter8:Momentum,Impulse,&Collisionslinearmomentum:Newton’ssecondlawintermsofmomentum:impulse:UnderwhatSPECIFICconditionislinearmomentumconserved?(Theanswerdoesnotinvolvecollisions.)Listthetypesofcollisionsandwhatdefinestheminthetablebelow.

collisions properties

Whatistheonlycollisionduringwhichyoucananalyzetheenergyandwhy?Whatisalwaystrueabouttherelativevelocitiesbetweenobjectsbeforeandafteracollision?centerofmass:

Page 2: linear momentum: Newton’s second law in terms of …

Whyisthecenterofmassofacollectionofparticlesoranextended,rigidbodysignificant?Examples:8*IalsorecommendyoudoatleastoneproblemfromeachsectionattheendofthechapterinyourUniversityPhysicsbook.Ex:8.1Acannonballisfiredwithaninitialvelocityvatanangle𝜃withthehorizontal.Atthetopofthetrajectory,theshellexplodesintotwofragmentsofequalmass.Onefragment,whosespeedimmediatelyaftertheexplosioniszero,fallsvertically.Howfarfromthecannondoestheotherfragmentland?Ignoreairresistance.

(Ifv=20m/s,𝜃=60°,thenx=53m.)

𝜃v

explosion

Page 3: linear momentum: Newton’s second law in terms of …

Ex:8.2Abodyofmassmismovingthroughspaceinonedimensionwithaspeedof2vwhen,duetoaninternalexplosion,itbreaksintothreeparts.Onepart,withamassof!

!𝑚movesawayfromthepointofexplosionwithaspeedofvperpendicular

totheoriginaldirectionofvelocity.Asecondpartwithamassof!!𝑚movesinthe

negativedirectionoftheoriginalvelocitywithaspeedof5v.a.Whatisthevelocityofthethirdpart?b.Howmuchenergyisreleasedintheexplosion?Ignoreeffectsduetogravity.(Ifm=20.0kg,v=100m/sinthepositivex-direction,thena.v3x=1.00x103m/s,v3y=-167m/s,orv3=1.01x103m/s,𝜃=9.48°belowthex-axis;b.∆K=3.23x106J.)

Page 4: linear momentum: Newton’s second law in terms of …

Ex:8.3Afteracompletelyinelasticcollision,twoobjectsofthesamemassandsameinitialspeedmoveawaytogetherathalftheirinitialspeed.Findtheanglebetweentheinitialvelocitiesoftheobjects.Ignoreexternalforcessuchasgravity.(𝜃=120°)

Page 5: linear momentum: Newton’s second law in terms of …

Chapter9:RotationofRigidBodiesrigidbody:angularvelocity:Howdoyoudeterminethedirectionofangularvelocity?angularacceleration:Howdoyoudeterminethedirectionofangularacceleration?(Wedidnotcoverthisinclass.)Listtheconstantangularaccelerationequations.Givetheequationoflinearspeedintermsofangularspeed.Givetheequationoflinearaccelerationintermsofangularacceleration.Givetheequationofcentripetalaccelerationintermsofangularvelocity.momentofinertia:rotationalkineticenergy:

Page 6: linear momentum: Newton’s second law in terms of …

Whatisspecialaboutthegravitationalpotentialenergyforanextendedbodythatisdifferentthanthegravitationalpotentialenergyforapointparticle?parallelaxistheorem:Examples:9*IalsorecommendyoudoatleastoneproblemfromeachsectionattheendofthechapterinyourUniversityPhysicsbook.Ex:9.1Awheelhasaconstantangularaccelerationof3.0rad/s2.Duringacertain4.0sinterval,itturnsthroughanangleof120rad.Assumingthatthewheelstartedfromrest,howlonghasitbeeninmotionatthestartofthis4.0-sinterval?(t=-8s;Thewheelstartedfromrest8secondsbeforethestartofthe4.0-sinterval.)

Page 7: linear momentum: Newton’s second law in terms of …

Ex:9.2Aflywheelwithadiameterof1.20misrotatingatanangularspeedof200rev/min.a.Whatistheangularspeedoftheflywheelinradianspersecond?(𝜔=20.9rad/s)b.Whatisthelinearspeedofapointontherimoftheflywheel?(v=12.5m/s)c.Whatconstantangularacceleration(inrevolutionsperminute-squared)willincreasethewheel’sangularspeedto1000rev/minin60.0s?(𝛼=800rev/min2)d.Howmanyrevolutionsdoesthewheelmakeduringthat60.0s?(𝜃=600rev)

Ex:9.3Calculatetherotationalinertiaofawheelthathasakineticenergyof24,400Jwhenrotatingat602rev/min.(I=12.3kg∙m2)

Page 8: linear momentum: Newton’s second law in terms of …

Ex:9.4Calculatetherotationalinertiaofameterstickwithmass0.56kgaboutanaxisperpendiculartothestickandlocatedatthe20cmmark.(Treatthestickasathinrod.)(I=9.7x10-2kg∙m2)

Page 9: linear momentum: Newton’s second law in terms of …

Chapter10:DynamicsofRotationalMotionWhatisthedifferencebetweentranslationalandrotationalmotion?torque:Howdoyoudeterminethedirectionoftorque?GivetheequationfortherotationalanalogofNewton’ssecondlawforarigidbody.Givetheequationforthekineticenergyduetorollingwithoutfriction.Givetheequationfortheconditionforrollingwithoutslipping.Listtheequationsforthesummationofforces(inequilibrium,withlinearacceleration,andwithcentripetalacceleration)andthesummationoftorques(inequilibriumandwithangularacceleration).

Page 10: linear momentum: Newton’s second law in terms of …

Givetheequationforworkdonebyaconstanttorque.Givetheequationfortherotationalanalogofthework-energytheorem.Givetheequationforpowerduetoatorque.angularmomentum:Howdoyoudeterminethedirectionofangularmomentum?UnderwhatSPECIFICconditionisangularmomentumconserved?Givetheequationfortheangularmomentumofarigidbodyrotatingaroundasymmetryaxis.precession:precessionangularspeed:

Page 11: linear momentum: Newton’s second law in terms of …

Examples:10*IalsorecommendyoudoatleastoneproblemfromeachsectionattheendofthechapterinyourUniversityPhysicsbook.Ex:10.1Athinuniformrodhaslength2.0mandcanpivotaboutahorizontal,frictionlesspinthroughoneend.Itisreleasedfromrestatangle𝜃=40°abovethehorizontal.Usetheprincipleofconservationofenergytodeterminetheangularspeedoftherodasitpassesthroughthehorizontalposition.(𝜔=3.1rad/s)

pin

𝜃

Page 12: linear momentum: Newton’s second law in terms of …

Ex:10.2A140-kghooprollsalongahorizontalfloorsothatthehoop’scenterofmasshasaspeedof0.150m/s.Howmuchworkmustbedoneonthehooptostopit?(W=-3.15J)Ex:10.3Atoneinstant,force𝑭 = 4.0! 𝑁actsona0.025-kgobjectthathaspositionvector𝒓 = (2.0!− 2.0𝒌)m and velocity vector𝒗 = (−5.0!+ 5.0𝒌)m/s. About theoriginandinunitvectornotation,whatarea.theobject’sangularmomentumandb.thetorqueactingontheobject?(a.𝑳=0;b.𝝉=(8.0N∙m)!+(8.0N∙m)𝒌)

Page 13: linear momentum: Newton’s second law in terms of …

Ex:10.4A3.0-kgparticlewithvelocity𝒗=(5.0m/s)!–(6.0m/s)!isat𝑥 = 3.0 𝑚,𝑦 = 8.0 𝑚.Itispulledbya7.0Nforceinthenegativex-direction.Abouttheorigin,whatarea.theparticle’sangularmomentum,b.thetorqueactingontheparticle,andc.therateatwhichtheangularmomentumischanging?(a.𝑳=(-1.7x102kg∙m2/s)𝒌;b.𝝉=(56N∙m)𝒌;c.!𝑳

!"=(56N∙m)𝒌)