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Transcript
Spring2016Physics7ALec001(Yildiz)MidtermII
1. (15points)AnobjectisreleasedfromrestatanaltitudehabovethesurfaceoftheEarth.his
comparabletotheradiusofEarth(RE),sogravitationalacceleration(g)isnotconstant.
a) WhatisthevelocityoftheobjectwhenithitsthesurfaceofEarth?
b) WhatisthegravitationalaccelerationoftheobjectatdistancerfromtheEarth’scenter,where
𝑅! < π‘Ÿ < 𝑅! + β„Ž?
c) Whatistherateofchangeofthegravitationalacceleration𝑔 π‘Ÿ asafunctionofthedistancerfrom
theEarth’scenter,where𝑅! < π‘Ÿ < 𝑅! + β„Ž?
Solution:
a) βˆ’βˆ†π‘ˆ = βˆ†πΎ
πΊπ‘€π‘š
π‘ˆ! (π‘Ÿ) = βˆ’
π‘Ÿ
1
1
1
π‘šπ‘£ ! = βˆ’πΊπ‘€π‘š
βˆ’
2
𝑅! + β„Ž 𝑅!
!
𝑣 = 2𝐺𝑀
!!
βˆ’
!
!! !!
b)𝐹! π‘Ÿ = βˆ’
𝑔 π‘Ÿ =βˆ’
c)
!"
!"
=
!"#
!!
= π‘š. 𝑔(π‘Ÿ)
𝐺𝑀
π‘Ÿ!
!!"
!!
2.(20 points) Assume a cyclist of weight mg can exert a
force on the pedals equal to 0.80 mg on the average. The
pedals rotate in a circle of radius 18 cm, the wheels have a
radius of 34 cm, and the front and back sprockets on which
the chain runs have 42 and 19 teeth respectively. The mass
of the bike is 14 kg and that of the rider is 63 kg. Assume
there is no slipping between the ground and the wheel. Use
2
g = 10 m/s for calculations.
a)Howistheangularvelocityoftherearwheelofabicycle
relatedtotheangularvelocityofthefrontsprocketandpedals?Theteetharespacedthesameonboth
sprocketsandtherearsprocketisfirmlyattachedtotherearwheel.
b)Determinethemaximumsteepnessofhillthecyclistcanclimbatconstantspeed.Assumethecyclist's
averageforceisalwaystangentialtopedalmotion.
c)Determinethemaximumsteepnessofhillthecyclistcanclimbatconstantspeed.Assumethecyclist's
averageforceisalwaysdownward.
Iftheriderisridingataconstantspeed,thenthepositiveworkinputbytheridertothe(bicycle+
rider)combinationmustbeequaltothenegativeworkdonebygravityashemovesuptheincline.
Thenetworkmustbe0ifthereisnochangeinkineticenergy.
(a) Iftherider’sforceisdirecteddownwards,thentheriderwilldoanamountofworkequalto
theforcetimesthedistanceparalleltotheforce.Thedistanceparalleltothedownwardforce
wouldbethediameterofthecircleinwhichthepedalsmove.Thenconsiderthatbyusing2
feet,theriderdoestwicethatamountofworkwhenthepedalsmakeonecomplete
revolution.Soinonerevolutionofthepedals,theriderdoestheworkcalculatedbelow.
Wrider = 2 ( 0.90mrider g ) d pedal motion
Inonerevolutionofthefrontsprocket,therearsprocketwillmake 42 19 revolutions,andso
thebackwheel(andtheentirebicycleandrideraswell)willmoveadistanceof
( 42 19) (2Ο€ r ) .Thatisadistancealongtheplane,andsotheheightthatthebicycleand
wheel
riderwillmoveis h = ( 42 19 )( 2Ο€ rwheel ) sin ΞΈ . Finally,theworkdonebygravityinmovingthat
heightiscalculated.
WG = ( mrider + mbike ) gh cos180° = βˆ’ ( mrider + mbike ) gh = βˆ’ ( mrider + mbike ) g ( 42 19 )( 2Ο€ rwheel ) sin ΞΈ
Setthetotalworkequalto0,andsolvefortheangleoftheincline.
Wrider + WG = 0 β†’ 2 [0.90mrider g ] d pedal βˆ’ ( mrider + mbike ) g ( 42 19 )( 2Ο€ rwheel ) sin ΞΈ = 0 β†’
motion
( 0.90mrider ) d pedal
ΞΈ = sin
motion
βˆ’1
( mrider + mbike )( 42 19 )(Ο€ rwheel )
= sin
βˆ’1
0.90 ( 65 kg )( 0.36 m )
( 77 kg )( 42 19 ) Ο€ ( 0.34 m )
= 6.7°
(b)Iftheforceistangentialtothepedalmotion,thenthedistancethatonefootmoveswhile
exertingaforceisnowhalfofthecircumferenceofthecircleinwhichthepedalsmove.The
restoftheanalysisisthesame.
βŽ›
⎞
Wrider = 2 ( 0.90mrider g ) ⎜ Ο€ rpedal ⎟ ; Wrider + WG = 0 β†’
⎝
motion
⎠
( 0.90mrider ) Ο€ rpedal
ΞΈ = sin βˆ’1
motion
( mrider + mbike )( 42 19 )(Ο€ rwheel )
= sin βˆ’1
0.90 ( 65 kg )( 0.18 m )
( 77 kg )( 42 19 )( 0.34 m )
3.(20points)TwoblockofmassesMand2Mareconnectedtoaspringof
springconstantkthathasoneendfixed,asshown.Thehorizontalsurface
= 10.5° β‰ˆ 10°
andthepulleyarefrictionlessandthepulleyhasnegligiblemass.Theblocksarereleasedfromrestwith
thespringrelaxed.
a) Whatisthevelocityoftheblockswhenthehangingblockhasfallenadistanceh?
b) Whatmaximumdistancehmaxdoesthehangingblockfallbeforemomentarilystopping?
c) Intheabsenceoffriction,thesystemisexpectedtooscillatebackandforthinfinitely.However,if
thereiskineticfrictionbetweenblockandthehorizontalsurface(coefficientofkineticfrictionisµk),the
systemwilleventuallycometoacompletestop.Whatistheequilibriumpositionofthespring’s
extensionwhenthesystemcomestoacompletestop?Whatisthetotaldistancetraveledbythe
hangingblockbeforethesystemcomestoacompletestop?(Assumethatthefrictionalforceonthe
massMonthehorizontalsurfaceisnegligiblewhenthesystemcomestoacompletestop.)
Solution
a) π‘ˆ! + 𝐾! = π‘ˆ! + 𝐾! 0=
1 !
1
1
π‘˜β„Ž βˆ’ 2π‘€π‘”β„Ž + 𝑀𝑣 ! + 2𝑀𝑣 ! 2
2
2
𝑣=
b) π‘ˆ! + 𝐾! = π‘ˆ! + 𝐾! and
4π‘€π‘”β„Ž βˆ’ π‘˜β„Ž!
3𝑀
𝑣! = 0 π‘€β„Žπ‘’π‘› β„Ž = β„Ž!"# 0=
1
!
π‘˜β„Ž
βˆ’ 2π‘€π‘”β„Ž!"# + 0
2 !"#
β„Ž!"# =
4𝑀𝑔
π‘˜
c) Inthepresenceofanon-conservativeforce,π‘ˆ! + 𝐾! + π‘Š!" = π‘ˆ! + 𝐾! π‘Š!" = βˆ’πΉ!" 𝑑!"! 𝐹!" = πœ‡! . 𝑁 = πœ‡! π‘šπ‘”
𝐾! = 0 π‘ π‘¦π‘ π‘‘π‘’π‘š 𝑖𝑠 π‘ π‘‘π‘Žπ‘‘π‘–π‘œπ‘›π‘Žπ‘Ÿπ‘¦ , π‘‘π‘œ 𝑓𝑖𝑛𝑑 π‘ˆ! , 𝑀𝑒 𝑛𝑒𝑒𝑑 π‘‘π‘œ 𝑓𝑖𝑛𝑑 π‘‘β„Žπ‘’ π‘’π‘žπ‘’π‘–π‘™π‘–π‘π‘Ÿπ‘–π‘’π‘š π‘π‘œπ‘ π‘–π‘‘π‘–π‘œπ‘›.
𝐴𝑑 π‘’π‘žπ‘’π‘–π‘™π‘–π‘π‘Ÿπ‘–π‘’π‘š, 𝐹! = 0,
π‘‘β„Žπ‘’π‘Ÿπ‘’π‘“π‘œπ‘Ÿπ‘’ 𝑇 βˆ’ π‘˜π‘₯!" = 0
π΅π‘’π‘π‘Žπ‘’π‘ π‘’ π‘‘β„Žπ‘’ β„Žπ‘Žπ‘›π‘”π‘–π‘›π‘” π‘šπ‘Žπ‘ π‘  𝑖𝑠 π‘Žπ‘™π‘ π‘œ π‘ π‘‘π‘Žπ‘‘π‘–π‘œπ‘›π‘Žπ‘Ÿπ‘¦,
𝑇 𝑖𝑠 π‘‘β„Žπ‘’ π‘‘π‘’π‘›π‘ π‘–π‘œπ‘› π‘œπ‘› π‘‘β„Žπ‘’ π‘Ÿπ‘œπ‘π‘’ 𝐹! = 0,
π‘‘β„Žπ‘’π‘Ÿπ‘’π‘“π‘œπ‘Ÿπ‘’ 𝑇 βˆ’ 2𝑀𝑔 = 0
2𝑀𝑔 = π‘˜π‘₯!" π‘ˆ! + 𝐾! + π‘Š!" = π‘ˆ! + 𝐾! 0 βˆ’ πœ‡! π‘šπ‘”π‘‘!"! =
1
π‘˜π‘₯ ! βˆ’ 2𝑀𝑔π‘₯!" + 0
2 !"
βˆ’πœ‡! π‘šπ‘”π‘‘!"!
1 2𝑀𝑔
= π‘˜
2
π‘˜
𝑑!"! =
!
!!"
!!!
βˆ’ 2𝑀𝑔
2𝑀𝑔
π‘˜
4.(25points)Supposea
spacecraftwithinitialmassof
mi.Withoutitspropellant,the
spacecrafthasamassofmf=
mi/3.Therocketthatpowers
thespacecraftisdesignedto
ejectthepropellantwitha
speedofurelativetothe
rocketataconstantrateofR.Thespacecraftisinitiallyatrestinspaceandtravelsinastraightline.
a) Howlongwouldittaketherockettoreleaseallofitspropellant?
b) Whatism(t),themassoftherocketasafunctionoftime?
c) Whatisv(t),thespeedoftherocketasafunctionoftime?
d) Howfarwillthespacecrafttravelbeforeitsrocketusesallthepropellantandshutsdown?
Hint: ln 1 βˆ’ π‘Žπ‘₯ 𝑑π‘₯ =
!"!!
!
ln 1 βˆ’ π‘Žπ‘₯ βˆ’ π‘₯
Solution:
a) 𝑑! =
!! !!!
!
=
!
!! ! !
!
!
! !!
=!
!
b) π‘š 𝑑 = π‘š! βˆ’ 𝑅𝑑
c) Because 𝑭!"# = 0
!"
!"
π‘š
!"
!"
= 𝑣!"#
!!
!"
= βˆ’π‘…π‘£!"# = βˆ’π‘’ (π‘Ÿπ‘’π‘”π‘Žπ‘Ÿπ‘‘π‘™π‘’π‘ π‘  π‘œπ‘“ π‘‘β„Žπ‘’ 𝑠𝑝𝑒𝑒𝑑 π‘œπ‘“ π‘‘β„Žπ‘’ π‘Ÿπ‘œπ‘π‘˜π‘’π‘‘)
π‘š! βˆ’ 𝑅𝑑
!
𝑣=
!
𝑒𝑅
𝑑𝑑 π‘š! βˆ’ 𝑅𝑑
𝑣 = 𝑒𝑙𝑛
d) π‘₯! =
𝑑𝑓
!
𝑣𝑑𝑑 = βˆ’π‘’
𝑑𝑓
!
𝑙𝑛 1 βˆ’
!"
π‘šπ‘–
𝑑𝑣
= 𝑒𝑅
𝑑𝑑
π‘šπ‘–
π‘š! βˆ’ 𝑅𝑑
𝑑𝑑 UsingtheHintgiveninthequestion,π‘₯! = βˆ’π‘’
π‘‘βˆ’
π‘šπ‘–
!
𝑙𝑛 1 βˆ’
!"
π‘šπ‘–
βˆ’π‘‘
𝑑𝑓
!
and𝑑! =
! !!
! !
π‘₯! = βˆ’π‘’
2 π‘šπ‘– π‘šπ‘–
𝑅 2 π‘šπ‘–
2 π‘šπ‘–
π‘šπ‘–
𝑅. 0
βˆ’
𝑙𝑛 1 βˆ’
βˆ’
βˆ’ βˆ’
𝑙𝑛 1 βˆ’
βˆ’0 3𝑅
𝑅
π‘šπ‘– 3 𝑅
3𝑅
𝑅
π‘šπ‘–
π‘₯! = βˆ’π‘’
1 π‘šπ‘–
2
2 π‘šπ‘–
𝑙𝑛 1 βˆ’
βˆ’
3𝑅
3
3𝑅
π‘₯! =
𝑒 π‘šπ‘– 2
𝑅 3
βˆ’
1
𝑙𝑛3 3
5.(20points)Acylindricallysymmetricspoolofmassm
andradiusRsitsatrestonahorizontaltablewith
friction.Withyourhandonamasslessstringwrapped
aroundtheaxleofradiusr,youpullonthespoolwitha
constanthorizontalforceofmagnitudeTtotheright.As
aresult,thespoolrollswithoutslippingadistanceL
alongthetable.Assumethatthespoolisasoliduniform
cylinder.
a)Whatisthedistancethatyourhandtravelsasthe
spoolmovesadistanceL?
b)Findthefinaltranslationalspeedofthecenterofmassofthespoolusingthework-energyprinciple?
c)Findthevalueofthefrictionforcefandaccelerationofthespoolusingtheequationsfordynamicsof
themotion(e.g.forceandtorque)?
c.Inlectures,weusedF=maandΟ„=IΞ±tofindfrictionalforceandacceleration.Hereisanalternative
solutionusingImpulse-momentumtheorem.