View
1
Download
0
Category
Preview:
Citation preview
Final Exam Review
Prof.BinChenDepartmentofPhysics
Physics111
Exam Information
• Time:Dec18,2018(Tuesday),11:30AM-2:00PM• Location:MechanicalEngineeringCenterBuilding(ME;nexttotheGITC)Room221• Sameformatasbefore,butmorequestions(27intotal)• Youhaveabout5.6minutes,onaverage,toworkoneachquestion
Exam Information: Topics
AllmaterialsinWeeks1-14• RoughlyhalfofthequestionscomefromthoseafterCommonExam#3,i.e.,topicsinWeeks11-14:RotationalDynamics,StaticTorque,FluidMechanics,Gravitation.
• AnotherhalfofthequestionscomefromthosealreadycoveredinCommonExams#1-3,i.e.,topicsinWeeks1-10.
Final Exam Information: Review Sessions
• Thursday(Dec13,ReadingDay)11:30am–1:30pmTiernanLectureHall1withProf.GordonThomas
• Saturday(Dec15)11:00am–1:00pmTiernanLectureHall1withSocietyofPhysicsStudents
FORMULAS – Final Exam Conversion Factors: 1 inch = 2.54 cm; 1 mi =1609.3 m; 1 cm=10-2m; 1 mm= 10-3 m; 1 gram=10-3 kg; Physical constants: 9.8g = m/s2 ; 116.674 10G −= × N m2/kg2 ; 245.97 10EarthM = × kg ; 66.37 10EarthR = × m Math: 360° = 2π radians = 1 revolution. Arc length s rθ= ; 34 / 3sphereV Rπ= ; 24sphereA Rπ= ; 2
circleA Rπ=
quadratic formula to solve 2 0ax bx c+ + = : 2 4
2b b acx
a− ± −=
Vectors: ˆ ˆx yA A i A j= +
r ; cos( )xA A θ=
r ; sin( )yA A θ=
r ; 2 2
x yA A A= +r
; tan y
x
AA
θ =
C A B= +r r r
implies x x xC A B= + ; y y yC A B= +
𝐴 ∙ 𝐵%⃗ = '𝐴''𝐵%⃗ ' cos 𝜃 = 𝐴𝑥 𝐵𝑥 + 𝐴𝑦 𝐵𝑦 + 𝐴𝑧 𝐵𝑧 ; 𝑖̂ ∙ 𝑖̂ = 𝑗̂ ∙ 𝑗̂ = 𝑘5 ∙ 𝑘5 = 1 ; 𝑖̂ ∙ 𝑗̂ = 𝑖̂ ∙ 𝑘5 = 𝑗̂ ∙ 𝑘5 = 0 sinA B A B θ× =
r rr r ; ˆˆ ˆ( ) ( ) ( )y z z y z x x z x y y xA B i A B A B j A B A B k A B A B× = − + − + −r r
ˆ ˆˆ ˆ ˆ ˆ 0i i j j k k× = × = × = ; ˆˆ ˆi j k× = ; ˆˆ ˆj k i× = ; ˆ ˆ ˆk i j× =
1D and 2D motion:
avgxvt
Δ=Δ
; avgvat
Δ=Δ
; dxvdt
= ; 2
2
dv d xadt dt
= =
avgrvt
Δ=Δ
rr ; avgvat
Δ=Δ
rr ; dxvdt
=rr ;
2
2
dv d radt dt
= =r rr
212i ix x v t at= + + ; iv v at= + ; 2 2 2 ( )i iv v a x x= + − ; 21
2i ir r v t at= + +r r r r ; iv v at= +r r r
Circular motion: 2 /T R vπ= ; 2 /T π ω= ; 2 /ca v R= Newtons Laws: F ma=∑
r r ; 12 21F F= −r r
Friction: 𝑓𝑠 ≤ 𝜇𝑠𝑁 ; 𝑓𝑘 = 𝜇𝑘 𝑁
Energies: 𝐾 = 12
𝑚𝑣2 ; 𝑈𝑔 = 𝑚𝑔𝑦 ; 𝑈𝑠 = 12
𝑘𝑥2 ; 𝑊 = ∫ �⃗� ∙ 𝑑𝑟 = �⃗� ∙ ∆𝑟
total g SE K U U= + + ; mech g s sE K U U f dΔ = Δ +Δ +Δ = ; /P dW dt F v= =r rg ; K WΔ =
Momentum and Impulse: p mv=r r ; I Fdt p= = Δ∫r r r
Center of mass: /cm i i ii i
r m r m=∑ ∑r r ; /cm i i ii i
v m v m=∑ ∑r r
Collisions: pr = const and E≠ const (inelastic) or pr = const and E= const (elastic) Rotational motion: 2 / Tω π= ; /d dtω θ= ; /d dtα ω= ; tv rω= ; ta rα= 2 2/c r ta a v r rω= = =
2 2 2tot r ta a a= + ; cmv rω= (rolling, no slipping) ; cma rα=
o tω ω α= + ; 2 / 2f i ot tθ θ ω α= + + ; 2 2 2 ( )f i f iω ω α θ θ= + − 2
pointI MR= ; 2hoopI MR= ; 2 / 2diskI MR= ; 22 / 5sphereI MR= ; 22 / 3shellI MR= ; 2
( ) /12rod centerI ML= 2
( ) / 3rod endI ML= ; 2i i
iI m r=∑ ; 2
cmI I Mh= + ; r Fτ = ×rr r ; Iτ α=∑ ; L r p= ×
r r r ; L Iω=r r
Energy: 2 / 2rotK Iω= ; rot cmK K K= + ; 0K UΔ + Δ = ; W τ θ= Δ ; instP τω=
Fluid: 𝜌 = 𝑀𝑉
; 𝑃 = 𝑃𝑜 + 𝜌𝑔ℎ ; 𝐴1𝑣1 = 𝐴2𝑣2; 𝑃1 + 𝜌𝑔𝑦1 + 12
𝜌𝑣12 = 𝑃2 + 𝜌𝑔𝑦2 + 12
𝜌𝑣22 ; 𝐵 = 𝜌𝑓𝑙𝑢𝑖𝑑 𝑉𝑜𝑏𝑗𝑒𝑐𝑡 𝑔
Gravitation: 1 2122ˆg
Gm mF rr
= −r
; 2( ) /g r GM r= ; 1 2 /U Gm m r= − ; 2
2 34T aGMπ=
Fluid
Cubic water balloon
26. Aliquidhasadensityof150kg/m3.Itfillsacubictankthatmeasures0.1mineachside.Howmuchdoesitweigh,inN?
a.0.15b.1.5c.15d.150e. 1500
Density=m/VW=mgW=density*V*g=1.5
Drowning a Balloon
27.Ascientisthasaballoonthatweighs0.001kg.Sheimmersesitinwaterbypushingdownwith15N.Thedensityofwateris1000kgm-3.Howbigistheballoon,in(1/1000)m3?a.1.5b.0.65c.6.5d.0.15e.Can’ttell.
Water Pressure in the Home • Waterentersahousethroughapipewithaninside
diameterof2.0cmatanpressureof4×105Pa.A2.0-cm-diameterpipeleadstothe2ndfloorbathroom5mabove.Whentheflowspeedattheinletpipeis1.5m/s,Findthepressureonthesecondfloorin105pa.
A)3B)1C)4.5D)4E)3.5
p2 = p1 −
12ρ(v2
2 − v12)− ρg( y2 − y1)=3.5×105Pa
v2 =
A1A2v1 =1.5m/s
2222
2111 2
121 vgypvgyp ρρρρ ++=++
Gravity
Gravitation field
Aplanethasagravitationalfieldof15.7N/kgonitssurface.While
descendingontotheplanet,anastronautmeasuresagravitational
fieldof0.63N/kg.Whatisthedistanceoftheastronautabovethe
planet’ssurface,intermsoftheplanet’sradius?a. 2b. 25c. 16d. 4e. 5 g=GM/R2
Elevator on another planet
25.Anemptyelevator,M=100kg,hasacounterweightm=90kgconnectedbyacableoverapulleywithcoefficientoffriction0.05.Theelevatorfalls20mandlandswithavelocityof1m/s.Theaccidenthappensnotonearthbutonadifferentplanet.WhatisginN/kg?• a.48• b.24• c.9.5• d.4.3• e.2.1
F.d=(M+m)vv/2=95F=(M-m)g–mu(M+m)gg=F/[M-m-mu(M+m)]=F/[0.5]g=(95/20)/.5=9.5
Astronaut in Orbit
• Iweigh600Nonthesurfaceoftheearth.IfItravelonaspaceshuttleorbitingatadistanceof3timestheEarth’sradiusaboveground.Whatismymassinkgatthespaceshuttle?
a.3.83b.6.80c.15.3d.3.83e.61.2
Massisanintrinsicmeasureoftheobject,anddoesnotchangewithdistance!
MorePracticeProblems
Testing a weapon
14.A0.007kgbulletisfiredintoastationaryblockwithmass3.5kg,onafrictionless,horizontalsurface.Afterthecollision,thebulletgetsstuckintheblockandtheymovetogetherat1m/s.Findtheinitialspeedofthebulletinm/s.a.500b.200c.150d.125e.10
mV=(M+m)vV=v(M/m)approx.V=500
Rock climber hangs on
18.Arockclimberonaledgepushesablockofmassm=2.1kgupagainstaverticalwallatanangleof40degrees.Thecoefficientofstaticfrictionbetweentheblockandthewallis0.5.Whatistheminimumforce,inN,neededtokeeptheblockfromslidingdown?a. 15b. 41c. 25d. 19e. 0
Fup=FdownFup=Fcos40+m*Fsin40Fdown=mgF=mg/(cos40+m*sin40)=19
Walking in a Wind storm
22.AstrongwindisblowingSouthwithaforceof15N.Awomanmoves4meastandthen3mNorth.Howmuchworkisdonebythewindduringherwalk?a. +45Jb. -45Jc. +60Jd. -75Je. +15J
W=F.dF=-15d=3oppositeW=-45
Rotating disk
• Astudentdropsaringonarotatingdiskwithanangularspeedof10radians/s.Thediskhasmomentofinertiaof4.0kgm2.Afterthedrop,thediskandringarerotatingtogetherwithanangularspeedof5kgm2radians/s.Whatisthemomentumofinertiaofthering?• A.4• B.2• C.1• D.5• E.16
Inthefigure,pointPisatrestwhenitisonthex-axis.ThelinearspeedofpointPwhenitreachesthey-axisisclosestto
• A)0.18m/s.• B)0.24m/s.• C)0.35m/s.• D)0.49m/s.• E)0.71m/s.
Recommended