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APPhysicsChapter6Review EmilyDickinson CircularMotionandForces UniformCircularMotionModel: Objectsmovingwithconstantspeedinacircularpathareacceleratingcentripetally,or towardsthecenter.Centripetalmeansโcenterseeking.โAslongastheparticlefollowsthe circularpath,theaccelerationandforceitexperienceswillbetowardsthecenterofthecircle. Centripetalaccelerationisdefinedbytheformula:๐! = !! ! wherevisvelocityandristhe radiusofthecircularpath.Withacentripetalaccelerationthereisaalsoacentripetalforce. NewtonโsSecondLaw,๐น!"# = ๐๐canbereinterpretedas๐น!"#$%&'"$() = ๐๐!"#$%&'"$() or ๐น!"#$%&'"$() = !! ! ! v v ๐น! v v Mostproblemsfromthisunitdemonstrate uniformcircularmotion,andthecentripetalforce varies.Insituatiationswithobjectsswingingincircles, attachedwithstringsorcordsetc.,theForceoftension inthestringprovidesthecentripetalforce. Inothersituationslikevehiclesgoingaroundacurve, theforceoffrictionthatoccursbetweenthewheels andthegroundcreatesthecentripetalforce.Thesame NewtonโsSecondLawapproachshouldbetakenwith circularmotion.Itisimportanttostartwiththe summationofallofthecentripetalforces,โ๐น! = ๐๐! andthen,workthroughthevarioussteps. Certainproblemsfromthischapterinvolvethecoefficientoffriction,whichisavalue eithergivenorcalculatedthatdescribeshowsmoothasurfaceis.Thecoefficientoffriction canbestaticordynamic,anditisindicatedbytheGreeksymbol,๐. Non-UniformCircularMotion Whenanobjectmovesinnon-uniform circularmotion,theaccelerationismadeupof centripetalaccelerationandtangential ๐น!"#$"% acceleration. ๐ ๐!"# = ๐!"#$"% + ๐!"#$%#!&"' ๐น!"#$"% ๐น!"#$%#!&"' Theseaccelerationsresultfromradialand tangentialforces. ๐น!"# = ๐น!"#$"% + ๐น!"#$%#!&"' ๐น!"#$%#!&"' Problem1:VerticalCircles Askateboardertriestocompleteafull360-degreepipeturn.The pipehasaradiusof2meters,andtheskateboarderweighs75kg withhisskateboard.a)Drawafreebodydiagramwhenthe skateboarderisperpendicularwiththegroundonhiswayupthe pipe.b)Theskateboarderisnotwearingahelmetbecausehetrusts hisphysicsknowledge.Hehascalculatedthathisvelocityneedsto beatleast3m/satthetopoftheramp.Willhemakeit?Ifnot determinehisminimumspeedatthetopofthepipe,assumingthe pipedoesnotapplyanyforceontheskateboarder. Answers: Information:mass,m=75kg,radius,r=2m ๐น!"#$%& ๐น!"#$%&' Formula:๐น!"#$%&'"$() = !! ! ! a)Theskateboarderexperiencesthedownwardforceofgravity,anda normalforcetowardsthecenterfromthepipe.Weknowthatinthis situationallofthecentripetalforceisduetothenormalforce,butwe wonโtbeansweringanyspecificquestionsaboutthissituation. b) Anotherfreebodydiagram!!! Atthetopofthepipe,alloftheskateboarderโscentripetalforcecomesfromgravitybecause thepipedoesnotapplyanynormalforceontheskateboarder. โ๐น!"#$%&'"$() = ๐น!"#$%&' + ๐น!"#$%& ๐๐ฃ ! โ๐น!"#$%&'"$() = = ๐๐ + 0 ๐ ๐น!"#$%& ๐ฃ = ๐๐ ๐น!"#$%&' ๐ฃ= 9.8 (2) v=4.42m/s Basedonthatminimumvelocityvalue,whichisgreaterthanhispredicted3m/sweknow thattheskateboarderwillfallonhisheadandprobablyhaveanawfulconcussion.:( Problem2:GoCart Inagocartracewithacirculartrackofradius50m,onecartmightslipoffoftheroad.The roadโscoefficientoffrictionis.3.Themassofthecartis90kg.a)Assumingthattheonly centripetalforceistheforceoffriction,whatisthetotalcentripetalforceonthecart?b)How fastdoesthecartneedtogoinordertomaketheturn? m= 90kg ๐ฃ Information: Mass,m=90kg r=50 m Formuals: ๐น!"#$%#&' = ๐๐น!"#$%& !! ! Radius,r=50m โ๐น!"#$%&'"$() = ! Coefficientoffriction,๐=.3 ForceofFriction ๐น!"#$%& ๐น!"#$%&' ๐น!"#$%#&'!!"#$%&'"$() ๐น!"#$%#&' = ๐๐น!"#$%& ๐น!"#$%#&' = ๐(๐๐) ๐น!"#$%#&' = (.3)(90)(9.8) ๐น!"#$%#&' = 264.4 ๐ b)velocity โ๐น!"#$%&'"$() = โ๐น!"#$%&'"$() = ๐น!"#$%#&' ๐ฃ= ๐ฃ= ๐๐ฃ ! ๐ ๐๐ฃ ! = 264.4 ๐ = ๐ ๐น!"#$%#&' ๐ ๐ 264.4 (50) 90 ๐ฃ = 12.12 ๐/๐ Problem3:Non-UniformCircularMotion Aballisattachedtotheceilingwithathread,and movesinnon-uniformcircularmotion.Theballhas amassof3kg,avelocityof10m/s,aradiusof1 ๐น!"#$"% ๐ meter,andathetavalueof30degrees. ๐น!"#$"% ๐น!"#$%#!&"' Whatisthetensioninthestring,andwhatarethe tangentialandradialforcesactingontheball? ๐น!"#$%#!&"' ๐น!"#$%&# Answers: Freebodydiagram: ๐น!"#$"% ๐น!"#$%&' = ๐๐๐๐๐ (๐) ๐น!"# = ๐๐๐ ๐๐(๐) Theradialforceisthecentripetalforce,sowestartwithasummationoftheforcesthatare centripetal.Itisimportantthatwebreakuptheforceofgravityintotheradialandtangential componentsofgravity.๐น!"#$"% = ๐๐๐๐๐ (๐)and๐น!"#$%#!&"' = ๐๐๐ ๐๐(๐) ๐๐ฃ ! โ๐น!"#$"% = = ๐น!"#$%&# โ ๐น!"#$%&' !"#$"% = ๐น!"#$%&# โ ๐๐๐๐๐ (๐) ๐ ๐๐ฃ ! = ๐น!"#$%&# โ ๐๐๐๐๐ (๐) ๐ ! 3 (10) ๐น!"#$%&# = + 3 9.8 cos 30 = 325.5 ๐ 1 ๐น!"#$"% = !! ! ! = 300 ๐ โ๐น!"#$%#!&"' = ๐น!"#$%&'!!"#$%!&"' = ๐๐๐ ๐๐ ๐ ๐น!"#$%#!&"' = 3 9.8 sin 30 = 14.7 ๐