T1-S_Solution_marking_scheme

　　IPhO 2016 Q1 Solution

　　Problem1:Solution/markingscheme??TwoProblemsinMechanics(10points)

　　PartA.TheHiddenDisk(3.5points)A1(0.8pt)Findanexpressionforbasafunctionofthequantities(1),theangleφandthetiltingangleΘofthebase.SolutionA1:

　　Geometricsolution:usethattorquewithrespecttopointofcontactis0?cen-terofgravityhastobeverticallyabovepointofcontact.

　　sinφ=

　　sinΘ=DbD

　　r1[0.8]0.30.3

　　HereDmaybecalledanothername.Solvethis:

　　sinφ=r1r1sinΘsinΘ?b=bsinφ0.2

　　Alternative:Torqueandforceswithrespecttoanotherpoint:CorrectequationfortorqueCorrectequationforforceCorrectsolution[0.8]0.30.30.2A2(0.5pt)Findtheequationofmotionfor?.ExpressthemomentofinertiaISofthecylinderarounditssymmetryaxisSintermsofT,bandtheknownquantities(1).Youmayassumethatweareonlydisturbingtheequilibriumpositionbyasmallamountsothat?isalwaysverysmall.SolutionA2:

　　Writesomeequationoftheform?=?ω2?

　　Writinganequationoftheform?=Acosωtisalsocorrect.Twosolutions:

　　1.Kineticenergy:1˙2andpotentialenergy:?bMgcos?.Totalenergyiscon-IS?

　　served,anddi??erentiationw.r.t.timegivestheequationofmotion.

　　2.Angularequationofmotionfromtorque,τ=IS?=?Mgbsin?.

　　Correctequation(eitherenergyconservationortorqueequationofmotion)Finalanswer??ISMgbT2

　　T=2π?IS=Mgb4π[0.5]0.10.30.1

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　　MasswrongFactor1/2wronginformulaformomentofinertiaofadiskCorrectanswerformomentofinertiaof`excess'disk:

　　14I2=πh2(ρ2?ρ1)r22

　　UsingSteiner'stheorem:

　　2IS=I1+I2+d2πr2h2(ρ2?ρ1)[0.7]0.2-0.1-0.10.20.1

　　correctsolution:

　　11b2M244IS=πh1ρ1r1+πh2(ρ2?ρ1)r2+22πr2h2(ρ2?ρ1)0.2

　　2

　　itcorrectly:2ρhM?πr111h2=πr2(ρ2?ρ1)0.2PartB.RotatingSpaceStation(6.5points)

　　B1(0.5pt)AtwhatangularfrequencyωssdoesthespacestationrotatesothattheastronautsexperiencethesamegravitygEasontheEarth'ssurface?

　　SolutionB1:

　　Anequationforthecentrifugalforcealongthelinesof

　　Fce=mω2r[0.5]0.1

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　　IPhO 2016 Q1 Solution

　　Balancingtheforces,correctequation

　　2gE=ωssR0.2

　　0.2Correctsolutionωss??=gE/R

　　B2(0.2pt)AssumingthatonEarthgravityisconstantwithaccelerationgE,whatwouldbetheangularoscillationfrequencyωEthatapersononEarthwouldmeasure?SolutionB2:

　　RealizethatresultisindependentofgE

　　Correctresult:??ωE=k/m[0.2]0.10.1B3(0.6pt)WhatangularoscillationfrequencyωdoesAlicemeasureonthespacestation?SolutionB3:

　　someversionofthecorrectequationforforce

　　2F=?kxmωssx[0.6]0.2

　　0.2gettingthesignright2xF=?kx+mωss

　　Findcorrectdi??erentialequation

　　2mx+(k?mωss)x=00.1

　　0.1Derivecorrectresult??ω=k/m?ωss

　　http://www.wendangwang.cominggE/Rinsteadofωss

　　B4(0.8pt)DeriveanexpressionofthegravitygE(h)forsmallheightshabovethesurfaceoftheEarthandcomputetheoscillationfrequencyω?E(linearapproximationisenough).TheradiusoftheEarthisgivenbyRE.

　　4

　　B6(1.1pt)Calculatethehorizontalvelocityvxandthehorizontaldisplacementdx(relativetothebaseofthetower,inthedirectionperpendiculartothetower)ofthemassatthemomentithitsthe??oor.YoumayassumethattheheightHofthetowerissmall,sothattheaccelerationasmeasuredbytheastronautsisconstantduringthefall.Also,youmayassumethatdx??H.

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　　IPhO 2016 Q1 Solution

　　SolutionB6:

　　Thereareseveralpossiblesolutions.Solutionone??UsingCoriolisforce

　　?Velocityvx

　　EquationforCoriolisforcewithcorrectvelocity:

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　　FC(t)=2mωssRtωss=2mωssRt

　　[1.1]

　　0.10.20.20.1

　　Integratethis,orrealizethatitislikeuniformaccelerationforthevelocity:

　　3

　　vx(t)=ωssRt2

　　plugincorrectvaluefor

　　t=

　　overallcorrectresult

　　??

　　R2H/ωss

　　vx=2Hωss

　　?Thedisplacementdx:

　　133dx=Rωsst

　　3

　　Insteadofintegrating,studentsmaysimply`average'bytaking1ofthe??nal

　　1

　　velocity.Thisgivesafactorof1insteadof.Deductatotalof0.1ptsforthis.Pluginvaluefort

　　11132

　　dx=Rωss(2H/ωssR)3/2=23/2H3/2R?1/2=

　　333

　　??

　　Integratevx(t):

　　0.3-0.1

　　8H3

　　R

　　0.2

　　Solutiontwo??UsinginertialframeThissolutionissimilartothewaytosolveB7,butneedsmorecomplicatedapproximationsthanSolutionone.

　　?vx

　　Hereφdenotestheanglesweptbythemassandαtheangletheastronauts(andtower)hasrotatedwhenthemasslandsonthe??oor,see

　　Initiallythevelocityofthemassinaninertialframeisvx=ωss(R?H).

　　Whenthemasslands,thex-directionhasbeenrotatedbyφsothenewhorizontalvelocitycomponentisthen

　　ωss(R?H)cosφ

　　(Studentmayalsowritecosαinsteadofcosφ,sincedx??H.)

　　cosφ=

　　HR?H

　　=1?RR

　　0.10.10.10.1

　　Transformingtotherotatingreferenceframe,oneneedstosubtractωssR.Finallyinthereferenceframeoftheastronauts

　　?? ??

　　H2H

　　vx=ωssR1??ωssR≈ωssR1?2?ωssR=?2ωssH

　　RR

　　Thesignofthevelocitydependonthechoiceofreferencedirection,soapositive

　　signisalsocorrect.

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　　0.2

　　Themasstravelsthedistancel,andduringthefallthespacestationrotatesbyφ,seeFigure2.Accordingtotheintersectingchordtheorem,

　　l2=H(2R?H)

　　Therotatedangleisφ=ωsstwhere

　　t=

　　isthefalltime.Thus

　　φ=??lR?H0.10.1-0.10.10.10.3R?H0.1

　　0.1????dl=φ?arcsin=?arcsinx(2?x)RRR?H

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　　IPhO 2016 Q1 Solution

　　Figure1:Notationforsolution

　　two

　　Figure2:Notationforsolutionthree.Denotex≡H/Randy≡Since

　　y3

　　arcsiny≈y+6

　　onegetsd2≈y(1+x)?y?y3/6=y(x?y2/6)≈2xy/3≈2x3=R3

　　Finalanswer√??2HR0.1B7(1.3pt)Findalowerboundfortheheightofthetowerforwhichitcanhappenthatdx=0.

　　SolutionB7:

　　Thekeyistouseanon-rotatingframeofreference.Ifthemassisreleasedcloseenough[1.3]

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　　IPhO 2016 Q1 Solution

　　tothecenter,itslinearvelocitywillbesmallenoughforthespacestationtorotatemorethan2πbeforeithitstheground.Thevelocityisgivenby

　　v=ωss(R?H)

　　distancedthatthemass??iesbeforehittingthespacestation

　　d2=R2?(R?H)2

　　usenon-rotatingframeofreferencetoobtaintimetuntilimpact??t=d/v=ωss(R?H)

　　NowthereareseveralpossiblewaystorelateHandtherotatedangleφofthespacestation:Solutionone

　　Rsinφ

　　ωssRcosφ

　　Thistimemustmatcht=φ/ωss.Obtaintheequationt=

　　φ=tanφ

　　Realizingthatthereisanin??nitenumberofsolutions.Thisequationhasonetrivialsolutionφ=0,nextsolutionisslightlylessthan3π/2whichcorrespondstothecaseHR(andisthusnotcorrect).TheonethatgivesalowerboundforHisthethirdsolution

　　φ≈5π/2

　　Theequationφ=tanφcanbesolvedgraphicallyornumericallytoobtainaclosevalue(φ=7.725rad)whichmeans

　　H/R=(1?cosφ)≈0.871

　　Givepointsifthemethodiscorrect,dependingonthevalueofH/Rfound,accordingtotheseintervals:

　　0.85≤H/R≤0.88:0.4pts

　　0.5≤H/R0.85:0.3pts

　　00.5orH0.88:0.2pts

　　H=0ormethodisincorrect:0pts

　　SolutiontworelationbetweenHandrotatedangleφ

　　R?H=cosφR

　　??????H=1?cosR1?H/R

　　√??1?(1?x)Figure3givesaplotoff(x)=1?cos.Thegoalisto??ndanapproximatesolutionforthesecondintersection.The??rstintersectionisdiscarded??itisintroducedbecauseofcosφ=cos(?φ)andcorrespondstoasituationwithHR.Realizingthatthereisanin??nitenumberofsolutions.obtainequationoftheform0.10.10.10.20.20.20.40.20.20.2

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　　Usethesamepointsforthenumericalansweraswasmentionedinsolutionone.Ifthestudentplotsfratherthang,??ndsolutiontof=1:isequivalenttothesolutionabove.Givesamenumberofpoints.????√Itisalsopossibletousecos=sin(1/x).0.4

　　B8(1.7pt)Alicepullsthemassadistanceddownwardsfromtheequilibriumpointx=0,y=0,andthenletsitgo(see??gure4).

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　　IPhO 2016 Q1 Solution

　　[1.7]

　　-0.1-0.1

　　0.1

　　0.20.1

　　0.20.10.10.1

　　0.2

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　　IPhO 2016 Q1 Solution

　　Correctqualitativesketch:periodicmotion

　　overallconstantmovement

　　B):cusps0.10.10.1

　　Andadditionallycorrectquantitativesketch:

　　A)+B):peaksandcuspsareaty=d

　　C):cuspsareatdistance?x=4πωssdfromeachotherω0.10.2

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