CircularMo+on
Chapter5
Circularmo+on
–Rota+ons–
Describethemo-on:
1)  Linearspeed:thedistance
(meters,km,feet,..)travelled
persecond(orminuteor…)
2)  Rota-onal(orangular)speed:
thenumberofrota+onsper
second(orminuteor…)
Todefinecircularmo-onwewillconcentrateonanglesinsteadofdistances.
Physics140,Prof.M.Nikolic
2
Radians
Measuringθindegrees(deginyourcalculator)turnsouttobeapoor
choice.
Radiansareamorenaturalchoiceofangularunit.
0
2π radians = 360 = 1 revolution = 1 rotation
3600
1 radian =
= 57.30
2 ⋅ 3.14
s
θ (rad) =
r
s–arclength
r-radius
Physics140,Prof.M.Nikolic
3
Conceptualques+on–Radians
Q1
Awheelturnsthroughfivecompleterevolu+onsandthenonequarterofa
revolu+on.Throughwhattotalangleinradianshasthewheelturned?
A.
B.
C.
D.
33radians.
50radians.
5radians.
66radians.
1revolu+on=2πradians
→5+¼revolu+ons=(5+¼)x2πradians
→10π+0.5π=10.5π=10.5x3.14=33radians
Physics140,Prof.M.Nikolic
4
Angularposi+onanddisplacement
y
JustlikewedidinChapter3forlinearmo+on,
wewilldefine:
!
rf
θf
Angularposi+on–θ(radians)
→measuredcounterclockwise(CCW)from
!
ri
thex-axis–posi+veangle
θi
x
→θi–ini+alangularposi+on
→θf–finalangularposi+on
Angulardisplacement–Δθ(radians)
Δθ = θ f − θ i
Whenobjectmovesalongacircularpath,posi+onvectorsriandrfhave
magnitudesequaltotheradiusofthecircle.
Physics140,Prof.M.Nikolic
5
Angularvelocity
JustlikewedidinChapter3forlinearmo+on,wewilldefine:
Averageangularvelocity
→angulardisplacementper+me
Δθ θ f − θ i
ω av =
=
Δt t f − ti
Instantaneousangularvelocity
Δθ
ω = lim
Δt →0 Δt
Justlikelinearmo+onbutwithr→θ
Unitsofangularvelocity:
radianspersecond[rad/s]
Counterclockwise(CCW)rota+onrepresentsposi+verota+on.
Clockwiserota+on(CW)representsnega+verota+on.
Physics140,Prof.M.Nikolic
6
Conceptualques+on–Ladybug
Q2
Aladybugsitsattheouteredgeofamerry-go-round,andagentlemanbugsits
halfwaybetweenherandtheaxisofrota+on.Themerry-go-roundmakesacomplete
revolu+ononeachsecond.Thegentlemanbug’sangularvelocityis
A.
B.
C.
D.
Halftheladybug’s.
Thesameasladybug’s.
Twicetheladybug’s.
Itcan’tbedetermined.
Bothbugscoverthesameangle(Δθ)overthe
sameamountof+me→angularvelocityis
thesameforbothbugs
Physics140,Prof.M.Nikolic
7
Rela+onbetweenlinearandangularspeed
y
s
r
θf
Anobjectmovingalongacircularpathduring
angulardisplacement(Δθ)willtravelthedistance
(s)equaltothearclength:
s = r Δθ
θi
Averagelinearspeed(Chapter2):
total distance s
vav =
=
total time
Δt
vav =
⎛ Δθ ⎞
rΔθ
= r⎜ ⎟
⎝ Δt ⎠
Δt
vav = rω av
Physics140,Prof.M.Nikolic
r–radiusofacircle
8
Conceptualques+on–Ladybug
Q3
Aladybugsitsattheouteredgeofamerry-go-round,andagentlemanbugsits
halfwaybetweenherandtheaxisofrota+on.Themerry-go-roundmakesacomplete
revolu+ononeachsecond.Thegentlemanbug’slinearspeedis
A.
B.
C.
D.
Halfastheladybug’s.
Thesameastheladybug’s.
Twicetheladybug’s.
Itcan’tbedetermined.
Bothbugshavethesameangularspeed
(ω)butthelinearspeedsaredifferent
→ladybug:vL=ωR
→gentlemanbug:vG=ωR/2=½vL
Physics140,Prof.M.Nikolic
9
Uniformcircularmo+on
Uniformcircularmo+on
è
whenthespeedofapointmovinginacircleisconstant.
PeriodandFrequency
Thefrequency(f)isthenumberofcompleterevolu+onspersecond.
Units:Hertz[Hz(rev/s)]
Theperiod(T)isthe+meittakesapointtomakeonerevolu+on.
T=
Linearspeed:
v=
1
f
distance traveled 2π r
=
= 2π rf
time
T
Physics140,Prof.M.Nikolic
v
2π
ω = = 2π f =
r
T
10
Exercise:Speedincentrifuge
Acentrifugeisspinningat5400rpm.
a)Findtheperiodandfrequencyofmo+on.
1rpm=1revolu+onsperminute→thisisfrequency
!
rev $ ! 1min $
rev
= 90 Hz
First,convertrpmtorev/s=Hz: f = # 5400
&×#
& = 90
"
min % " 60s %
s
Period:
T=
1
f
T=
1
= 0.011 s
90 Hz
b)Findangularspeed.
ω=
2π
= 2πf
T
Physics140,Prof.M.Nikolic
ω = 2π rad ⋅ 90Hz = 180π
rad
rad
= 565.2
s
s
11
Radialaccelera+on
Consideranobjectmovinginacircularpathof
radiusratconstantspeed.
r
Linearvelocityisalwaysatangent
tothemo+onpath.
!
vi
r
!
vf
!
Δv
Direc+onofvectorvelocityischanging
→velocityischanging→objecthasnon
zeroaccelera+on
! ! !
Δv = v f − vi = aΔt
→aisinthesamedirec+onasΔv
WhenΔt→0,instantaneousaccelera+onpointstowardsthecenterofthecircle
(radiallyinward)→andiscalledradialaccelera-onar.
Physics140,Prof.M.Nikolic
12
Magnitudeofradialaccelera+on
Similarly,to:s = rΔθ
r
r
Δθ
Wecanalsosay:
s
!
vf
!
vi
Δv = vΔθ
Δθ
!
Δv
Δv = vωΔt
!
!
vi = v f = v
Magnitudeofradialaccelera+on:
Δv
ar =
Δt
Physics140,Prof.M.Nikolic
v2
ar = vω = = ω 2 r
r
Units:[m/s2]
13
Q4
Conceptualques+on–Radialaccelera+on
AnLPrecord,10cminradius,spinswithangularvelocityof20rad/s.Iftheangular
speedoftherecordisdoubled,whathappenstoitsradialaccelera+onattherims?
A.
B.
C.
D.
Radialaccelera+ondoubles.
Radialaccelera+onquadruples.
Radialaccelera+onquarters.
Radialaccelera+onhalves.
v2
ar = vω = = ω 2 r
r
ωdoubles→arquadruples
Ormathema+cally:
ar1 = ω12 r = (20rad / s)2 ⋅ 0.1m = 40 m/s2
ar 2 = ω 22 r = (40rad / s)2 ⋅ 0.1m = 160 m/s2
ar 2 = 4ar1
Physics140,Prof.M.Nikolic
14
ApplyingNewton’ssecondlaw
Radialaccelera+onhasconstantmagnitudeandisdirectedtowardthecircle’scenter.
!
!
Newton’ssecondlaw: F = ma
net
→Somethingmustprovidetheforce–Centripetalforce
CentripetalforceisNOTaforceofnature.Anyoftheforceswediscussedin
Chapter4canplayaroleofcentripetalforcewhenobject’smovingalongacircle.
Tosolveproblemsinuniformcircularmo+on→applythesamesetofrules
providedinChapter4
→setx-axisalongradialaccelera+on
→therewillbenoaccelera+onydirec+onforuniformcircularmo+on
2
F
=
ma
=
m
ω
r
∑ x r
Physics140,Prof.M.Nikolic
and
∑F
y
=0
15
Exercise:Amusementparkride
Therotorisanamusementparkridewherepeoplestandagainsttheinsideofa
cylinder.Oncethecylinderisspinningfastenough,thefloordropsout.
(a)Whatforcekeepsthepeoplefromfallingoutthebotomofthecylinder?
First,drawafreebodydiagram(FBD)ofonepersonstandingagainstthewall.
y
ar
fs
N
Makesurethat:
→Weightpointsstraightdown
→Normalforceisperpendiculartothesurface
→Fric+onforceisparalleltothesurfaceandopposesthemo+on
x
w
Xdirec+on:fsx=0
Nx=N
Wx=0
Ydirec+on:fsy=fs
Ny=0
Wy=-W=-mg Theforceofsta+cfric+onkeepspeople
fromfallingoutthebotom.
Physics140,Prof.M.Nikolic
16
Exercise:Amusementparkride
Therotorisanamusementparkridewherepeoplestandagainsttheinsideofa
cylinder.Oncethecylinderisspinningfastenough,thefloordropsout.
(b)Ifμs=0.40andthecylinderhasr=2.5m,whatistheminimumangularspeedof
thecylindersothatthepeopledon’tfallout?
y
ApplyNewton’s2ndLaw:
fs
N
∑ F = ma
∑F = 0
x
y
x
r
N = mω 2 r
f s −W = 0
fs = W
w
Aslongasforceofsta+cfric+onis
largerorequaltoforceofgravity
(weight)èpeoplewillnotfallout
µ s N = µ s mω 2 r = mg
9.8 m/s2
ω=
=
= 3.13 rad/s
µs r
(0.40)(2.5 m)
g
Physics140,Prof.M.Nikolic
17
Takingexams
Fromnowon,Iwillprovideeveryonewith
equa+onsnecessaryfortheexam.
Iwillposttheequa+onsheetonCanvasaweek
beforetheexamsoeveryonecanseeitango
overit.
YouareNOTallowedtobringanythingtothe
exam(exceptforpen/pencilandcalculator).The
equa+onsheetwillbeatachedtoyouexam.
Physics140,Prof.M.Nikolic
18
Exercise:Slippingcoin
A20-gcoinisplacedonarecordthatisrota+ngat33.3rpm.Ifμs=0.1,howfar
fromthecenteroftherecordcanthecoinbeplacedwithouthavingitslipoff?
DrawanFBDforthecoin.
y
ar
N
Xdirec+on:fsx=fs
Nx=0
Wx=0
Ydirec+on:fsy=0
Ny=N
Wy=-W=-mg Forceofsta+cfric+onkeepsthecoinfromslippingoff.
fs
ApplyNewton’s2ndLaw:
x
∑F
x
w
= mar
fs = mω 2 r
Weneedtofindfsfirstanduseittofindtheradius.
∑F
y
=0
N −W = 0
fs = µ s N
ay=0(Thereisnoaccelera+oninver+caldirec+on).
N = W = mg
N = 20 ×10 −3 kg ⋅ 9.8 m/s2 = 0.196 N
19
Exercise:Slippingcoin
A20-gcoinisplacedonarecordthatisrota+ngat33.3rpm.Ifμs=0.1,howfar
fromthecenteroftherecordcanthecoinbeplacedwithouthavingitslipoff?
ar
y
fs = µ s N = 0.1⋅ 0.196 N = 0.0196 N
N
fs = mω 2 r
fs
fs
r=
mω 2
Youaregivenfrequencyf=33.3rpm→convertittoHz→andthen
calculateangularspeedω=2πf
x
ω = 2πf = 33.3
w
r=
rev ⎛ 2π rad ⎞⎛ 1 min ⎞
⎜
⎟⎜
⎟ = 3.5 rad/s
min ⎝ 1 rev ⎠⎝ 60 sec ⎠
0.0196 kg m/s2
−3
20 ×10 kg ⋅ (3.50 rad/s)
2
= 0.08 m
Notethatthesameprincipleapplieswhenyoudriveacarinacircularpathalonganunbankedroad.
20
Bankedcurves
Topreventcarsfromgoingintoskid
èroadsarebanked(+ltedataslightangle)
y
N
ApplyNewton’s2ndLaw:
∑F
ar
y
x
∑F
x
=0
N cosθ = mg
= max
v2
N sin θ = mar = m
r
v2
mg tan θ = m
r
v2
tan θ =
gr
v 2 = gr tan θ
Fric+onforcenegligiblecomparedhorizontalcomponentofthenormalforce.
Physics140,Prof.M.Nikolic
21
Circularorbits
ConsiderasatelliteinacircularorbitabouttheEarth.
Butthereisnostringjoiningthesatellitetothe
Earthnoristhereanythingtohavefric+onagainst
Gravity
Whatforceisholdingthesatelliteinacircularorbit?
Gms M e
It’sgravity: Fg =
r2
ApplyNewton’s2ndlaw
Fg = ms ar
Physics140,Prof.M.Nikolic
Gms M e
v2
= ms
2
r
r
GM e
v=
r
Thespeedofasatelliteina
circularorbitdoesnotdepend
onmassofthesatellite
22
Kepler’s1stLawofPlanetaryMo+on
Thelawoforbits:Allplanetsmoveinellip+calorbits,withtheSunatonefocus
Luckilyforus,wewillonlyworkwith
perfectcirclesinthisclass.
Physics140,Prof.M.Nikolic
23
Kepler’s2ndLawofPlanetaryMo+on
Thelawofareas:AlinethatconnectsaplanettotheSunsweepsout
equalareasintheplaneoftheplanet’sorbitinequal+mes.
EartharoundSum
-closestdistanceè
v=30.3km/s
-furthestdistanceè
v=29.3km/s
PlanetsaremovingfasterwhentheyareclosertotheSun
GSCI101,Prof.M.Nikolic
24
Kepler’s3rdLawofPlanetaryMo+on
Thelawofperiods:Thesquareoftheperiodofanyplanetis
propor+onaltothecubeofthesemimajoraxisofitsorbit.
OrinEnglish:Outerplanetshavefurthertogoand
movemoreslowlyintheirorbitsaroundSun
2π 3 / 2
T=
r
GM
GSCI101,Prof.M.Nikolic
or
2
4
π
T2 =
r3
GM
25
Exercise:Kepler’slaws
TheHubbleSpaceTelescopeorbitsEarth613kmaboveEarth’ssurface.Whatisthe
periodofthetelescope’sorbit?
Whatisgiven:
MassoftheEarth:ME=5.98x1024kg
RadiusoftheEarth:RE=6371km
h=613km
G=6.67x10-11Nm2/kg
2
4
π
T2 =
r3
GM E
DistancerismeasuredfromthecenteroftheEarthtothetelescope:r=RE+handhastobe
convertedtometers:
→r=6371km+613km=6984km=6984x103m
4π 2
T=
( Re + h)3
GM E
4 ⋅ 3.14 2 ⋅ (6984 ×10 3 m)3
7
T=
=
4
.
2
×
10
s
−11
2
2
24
6.67 ×10 Nm /kg ⋅ 5.98 ×10 kg
Physics140,Prof.M.Nikolic
26
Nonuniformcircularmo+on
Whathappenswhenthespeedincircularmo+onisnotconstant?
Then,whenΔt→0,Δvdoesnotpointtowardsthecenterofthecircle
→Thereisnowanothercomponenttoaccelera+ontangenttothe
pathofthecircle→tangen-alaccelera-onat
a
Thenetaccelera+onis:
at
2
a = ar + at
2
atchangesthemagnitudeofv
archangesthedirec-onofv
ar
v
AndNewton’s2ndlaws+llapplies:
∑ F = ma ∑ F = ma
r
Physics140,Prof.M.Nikolic
r
t
t
27
Exercise:Childonaswing
A35-kgchildswingsonaropewithalengthof6.5mthatishangingfromatree.At
thebotomoftheswing,thechildismovingataspeedof4.2m/s.Whatisthe
tensionintherope?
DrawanFBDforthechildatthebotomoftheswing.
y
T
ar
at
→radialaccelera+onpointstowardsthecenterofthecircle
(alongyaxis)
→tangen+alaccelera+onistangenttothemo+on
ApplyNewton’s2ndLaw:
∑F
x
r
w
Xdirec+on: Ydirec+on:
Tx=0
Ty=T
Wx=0
Wy=-W=-mg
Physics140,Prof.M.Nikolic
= mar
v2
T − W = mω r = m
r
v2
T = mg + m
r
2
(4.2m / s) 2
T = 35kg ⋅ 9.8m / s + 35kg
= 438 N
6.5m
2
28
Tangen+alandangularaccelera+on
Duringnonuniformcircularmo+onangularvelocity(ω)ischanging
→thereshouldbeanangularaccelera-on(α)
Averageangularaccelera-on
Δω ω f − ω i
α av =
=
Δt
t f − ti
Instantaneousangularaccelera-on
Δω
α = lim
Δt →0 Δt
Justlikelinearmo+onbutwithv→ω
Unitsofangularaccelera+on:
radianspersecond[rad/s2]
Physics140,Prof.M.Nikolic
29
Angularmo+onvs.linearmo+on
Tangen+alaccelera+on:
Δv
Δω
at =
=r
= rα
Δt
Δt
Linearmo-on
Δvx = v fx − vix = a x Δt
1
Δx = x f − xi = vix Δt + ax Δt 2
2
v 2fx − vix2 = 2ax Δx
Δx =
1
(vix + v fx )Δt
2
Physics140,Prof.M.Nikolic
v = rω
Angularmo-on
Δω = ω f − ω i = αΔt
1
Δθ = θ f − θ i = ω i Δt + αΔt 2
2
ω 2f − ωi2 = 2αΔθ
Δθ =
1
(ωi + ω f )Δt
2
30
Exercise:StoppingtheEarth
SupposetheEarthstartedtoundergoanangularaccelera+onof-1.2×10-10rad/s2in
theoppositedirec+ontoitscurrentrota+on.
a)Azerhowlongwoulditcometoastop(inrota+on)?
Whatisgiven:
α=-1.2×10-10rad/s2–TheEarthisslowingdown→accelera+onisnega+ve
ωf=0–theEarthstopped
Δω = ω f − ω i = αΔt
Wedonotknowini+al(current)angularvelocitybut,hopefully,
everyoneknowstheperiodofEarth’srota+on:
→T=1day=24h=86400s
ωi =
ω f − ωi
Δt =
α
Physics140,Prof.M.Nikolic
2π
T
ωi =
2 ⋅ 3.14
rad
= 7.28 ×10−5
86400 s
s
0 − ωi 0 − 7.28 ×10 −5 rad / s
Δt =
=
= 6.1×10 5 s ≈ 7 days
−10
2
α
−1.2 ×10 rad / s
31
Exercise:StoppingtheEarth
SupposetheEarthstartedtoundergoanangularaccelera+onof1.2×10-10rad/s2in
theoppositedirec+ontoitscurrentrota+on.
b)Howmanyrevolu+onswouldtheEarthmakebeforeitstopped?
Whatisgivenoralreadyfound:
α=-1.2×10-10rad/s2–TheEarthisslowingdown→accelera+onisnega+ve
ωf=0–theEarthstopped
ωi=7.28×10-5rad/s
Δt=6.1x105s
Revolu+onsareexpressedinangleθ→1revolu+on=2πradians
Δθ =
Δθ =
1
(ωi + ω f )Δt
2
1
7.28 ×10 −5 rad/s + 0 ⋅ 6.1 ×10 5 s = 22.2 rad
2
(
)
⎛1 revolution ⎞
⎟ = 3.54 revolutions
Δθ = 22.2 rad × ⎜
⎝ 2π rad ⎠
Physics140,Prof.M.Nikolic
32
Apparentweightandar+ficialgravity
Trytorecallproblemswiththeapparentandtrueweight(personinanelevator)
→personfeelsweightlesswhena=g
Similarsitua+onforastronautsinspace
→Wecansimulategravitybyrota+ngthespacesta+on
→astronauthitsthewalls
→normalforcemakeshimrotatewiththespacesta+onwith
radialaccelera+onar=mω2r
Foraspacesta+onwewant: ar = g
N
Physics140,Prof.M.Nikolic
2
v
Fr = mar = mω 2 r = m = mg
r
33
Exercise:Ar+ficialgravity
Ifawashingmachine’sdrumwithradiusof25cmproducesar+ficialgravityonthe
clothesof16g,whatisthefrequencyinrevolu+onsperminuteofthedrum?
Whatisgiven:
r=25cm=0.25m
ar=16g=16x9.8m/s2
ar = ω 2 r
16 g
ω=
r
16 ⋅ 9.8m / s 2
ω=
= 25 rad/s
0.25m
Tofindthefrequency:
f =
ω
2π
Physics140,Prof.M.Nikolic
f =
⎛ 60s ⎞
25 rad/s
⎟ = 240 rpm
= 4 Hz = 4 rev/s × ⎜
⎝
6.28
1min ⎠
34
Hollywoodmoviesbusted
StarTrek:USSEnterprise–notpossible
2001:AspaceOdyssey–possible
StarWars:Millenniumfalcon–notpossible
Physics140,Prof.M.Nikolic
35
WeightlessintheISS
TheInterna+onalSpaceSta+onorbitstheEarthevery91minutesata
distanceof353kmabovethesurfaceoftheEarth
Interna+onalspacesta+ondoesnotrotateènoapparentgravity
Physics140,Prof.M.Nikolic
36