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1Q20-001 Boulderrollingdownahill-rotationalenergy PROMPTSAFTER Aboulderontopofahillbreaksfreeandbeginstorolldownthehillwithoutslipping. Approximatingtheboulderasasolidspherewithradius3m,whatisthespeedofthe boulderatthebottomofthehillafterithasundergoneaverticaldisplacementof100m? STRATEGY Wewillusetheprincipleofconservationofmechanicalenergytosolveforthespeedofthe boulderatthebottomofthehill. 𝑈! + 𝐾! = 𝑈! + 𝐾! Conservationofmechanicalenergyappliestothisproblem, because,althoughfrictionalforceisactingontheboulder r causingittoroll,noworkisdonebyfrictionontheboulder. y r=3m h IMPLEMENTATION x h = 100 m Energy Energy First,weneedtodeterminethetypesofmechanicalenergyin Initial Final boththeinitial(topofthehill)andthefinal(bottomofthe hill)states.Sincetheboulderisinitiallyatrest,ithasonly Kr gravitationalpotentialenergy.Thetotalmechanicalenergy Kt 𝐸! atthetopofthehillisgivenby 𝐸! = 𝑈! = 𝑚𝑔ℎ! K U K U Aftertheboulderhasundergoneaverticaldisplacementof100 m,thegravitationalpotentialenergyhasbeenconvertedtotranslationalandrotational kineticenergy.Inthefinalstate,theboulderwillhaveacombinationofgravitational potentialenergy,translationalkineticenergyandrotationalkineticenergy.Ifwetakethe bottomofthehilltobewherethegravitationalpotentialenergyiszero,thetotalmechanic energy𝐸! inthefinalstatebecomes 1 1 𝐸! = 𝐾!"#$% + 𝐾!"# = 𝑚𝑣 ! + 𝐼𝜔 ! 2 2 Thetotalmechanicalenergyatthetopandatthebottomofthehillisthesame(conserved), soourconservationofmechanicalenergyequationbecomes 1 1 𝑚𝑣 ! + 𝐼𝜔 ! 2 2 Nowwecansolveforthetranslationalspeedoftheboulderatthebottomofthehill.The rotationalkineticenergyisdependentontherotationalinertiaandangularvelocityofthe boulder.Wearetoldthattheboulderis(a)asolidsphereand(b)thatisnotslippingasit ! rolls–thismeansthatwecanusetherotationalinertiaofasolidsphere𝐼 = 𝑚𝑟 ! andthe 𝑚𝑔ℎ! = ! ! relationshipbetweenangularspeedandtranslationalspeed𝜔 = toputrotationalkinetic ! energyintermsofthemass,radius,andtranslationalspeedoftheboulder. 𝑚𝑔ℎ! = 1 1 2 𝑚𝑣 ! + 𝑚𝑟 ! 2 2 5 𝑣 𝑟 ! = 7 𝑚𝑣 ! 10 Thisequationcanbesimplifiedfurtherasmassoftheboulderappearsonbothsidesofthe equationandcanbecancelled. 1Q20-001 Boulderrollingdownahill-rotationalenergy PROMPTSAFTER CALCULATION Solvingforthetranslationalspeedoftheboulderatthebottomofthehillbecomes: 10 ∙ 9.8 m ! ∙ 100 m 10𝑔ℎ! s = = 37 m s 7 7 Note:Thespeedoftheboulderisindependentofboththemassandtheradiusofthe boulder. SELF-EXPLANATIONPROMPTS 𝑣= 1. Frictionalforceisneededforthebouldertoroll,andnotslide,downthehill.Explain why“noworkisdonebyfrictionontheboulder”. 2. Explainwhytheboulderhasonlyrotationalandtranslationalkineticenergyatthe bottomofthehill. 3. Howwouldthefinalspeedchangeiftheboulderwasslidingdownthehillinsteadof rolling?Deriveanexpressionforthespeedoftheboulderatthebottomofthehillifit wasslidinginsteadofrolling. 1Q20-001 Boulderrollingdownahill-rotationalenergy PROMPTSAFTER
