projectile and circular motion

8/2/2019 Projectile and Circular Motion

1/37

CHAPTER

3Projectile and circular motion

REmEmbER

Before beginning this chapter, youshould be able to:

analyseuniformmotionalongastraightlinealgebraically

performvectoraddition

resolvevectorsintocomponents

applytheenergyconservationmodeltoenergytransfersandtransformations

usetheareaunderaforce-versus-distance(ordisplacement)graphtodetermineworkdonebyaforcewithchangingmagnitude.

KEy idEAs

After completing this chapter, youshould be able to:

analysethemotionofprojectilesnearEarthssurface

analyseuniformcircularmotioninahorizontalplane

applyNewtonssecondlawtonon-uniformcircularmotioninaverticalplane

analysethemotionofplanetsandsatellitesbymodellingtheirorbitsasuniformcircularorbitalmotion

applyNewtonsLawofUniversalGravitationtothemotionofplanetsandsatellites

analyseenergytransformationsasobjectschangepositioninachanginggravitationaleld

distinguishbetweenweightlessnessandapparentweightlessness.

Predicting the path of a projectile can be a matter of life and death.


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47CHAPTER 3 Projectile and circular motion

Projectile motionAny object that is launchedinto the air is a projectile. A basketball throwntowardsagoal,atrapezeartistsoaringthroughtheair,andapackagedroppedfromahelicopterareallexamplesofprojectiles.

Exceptfor thoseprojectileswhosemotion isinitiallystraight upor down,orthosethathavetheirownpowersource(likeaguidedmissile),projectiles

generally follow a parabolic path. Deviations from this path can be causedeitherbyairresistance,byspinningoftheobjectorbywind.Theseeffectsareoftensmallandcanbeignoredinmanycases.Amajorexception,however,istheuseofspininmanyballsports,butthiseffectwillnotbedealtwithinthisbook.

Falling downImagine a ball that has been released somedistance above the ground. Once the ballisset inmotion, the only forces acting onitaregravity(straightdown)and air resistance(straightup).

After the ball is released, the projectiondevice (hand, gun, slingshot or whatever)stopsexertingadownwardsforce.

The net force on the ball in the gure atrightisdownwards.Asaresult,theballaccel-eratesdownwards.Ifthesizeoftheforcesandthemassoftheballareknown,theaccelera-tioncanbecalculatedusingNewtonsSecondLawofMotion.

Often the force exerted on the ball by airresistanceisverysmallincomparisontotheforce ofgravity, and socan beignored.Thismakesitpossibletomodelprojectilemotion

byassumingthattheaccelerationoftheballisdueonlytogravityandisaconstant9.8ms -2downwards.

sample problem 3.1

Ahelicopterdeliveringsuppliestoaood-strickenfarmhovers100m abovetheground.Apackageofsuppliesisdroppedfromrest,justoutsidethedoorofthehelicopter.Airresistancecanbeignored.(a) Calculatehowlongittakesthepackagetoreachtheground.(b) Calculate how farfromits originalposition thepackagehas fallen after

0.5s,1.0s,1.5s,2.0setc.untilthepackagehashittheground.(Youmay

liketouseaspreadsheethere.)Drawascalediagramofthepackagespos-itionathalf-secondintervals.

(a) u=0ms-1,x=100m,a=10ms-2,t=?

x=ut+1

2at

2

100m=0ms-1t+1

2(10ms-2)t2

100

5 0.=t2

t=4.5s

weight

air resistance

velocity

The forces acting on a ball

falling downwards

weight

air resistance

velocity

The forces acting on a ball

falling downwards

Air resistance istheforce

appliedtoanobjectoppositeto

itsdirectionofmotion,bythe

airthroughwhichitismoving.

Air resistance istheforce

appliedtoanobjectoppositeto

itsdirectionofmotion,bythe

airthroughwhichitismoving.

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Solution:Solution:


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UNIT 3 AREA OF STUDY 1 Motion in one and two dimensions48

(Note:thenegativesquarerootcanbeignoredhereasweareinterestedonlyin motionthathasoccurredafterthepackagewasreleasedatt=0,i.e.positivetimes.)

(b) t=0.5s,u=0ms-1,a=10ms-2,x=?

x=ut+1

2at2

=00.5s+

1

2 (10ms-2

)(0.5s)2

=1.23m

Repeatthisfort=1s,1.5s,2setc.togaintheresultslistedinthefol-lowingtableandillustratedatleft.

Table 3.1 Vertical distance travelled over time

TIME (s) 0.5 1 1.5 2 2.5 3 3.5 4 4.5

VERTICAL

DISTANCE (m)1.3 5 11 20 31 45 61 80 100

Revision question 3.1

A camera is dropped by a tourist from a lookout and falls vertically to the ground.The thud of the camera hitting the hard ground below is heard by the tourist3.0 seconds later. Air resistance and the time taken for the sound to reach thetourist can be ignored.

(a) How far did the camera fall?

(b) What was the velocity of the camera when it hit the ground below?

Terminal velocity

The air resistance on a falling object increases as its velocity increases. Anobject falling fromrest initially experiencesno air resistance. As theobjectaccelerates due to gravity (see the diagram on page 47), the air resistanceincreases. Eventually, if the object doesnt hit a surface rst, the air resist-ancewillbecomeaslargeastheobjectsweight.Thenetforceonitbecomeszeroandtheobjectcontinuestofallwithaconstantvelocity,referredtoasitsterminal velocity.

Moving and fallingIfaballisthrownhorizontally,theonlyforceactingontheballonceithasbeenreleasedisgravity(ignoringairresistance).Astheforceofgravityisthesame regardlessofthemotionof theball, theballwillstillaccelerate down-

wardsatthesamerateasif itweredropped.Therewillnotbeanyhorizontalaccelerationasthereisnonetforceactinghorizontally.Thismeansthatwhiletheballsverticalvelocitywillchange,itshorizontalvelocityremainsthesamethroughoutitsmotion.

Itistheconstanthorizontalvelocityandchangingverticalvelocitythatgiveprojectilestheircharacteristicparabolicmotion.

Noticethat thevertical distance travelled bythe ballin each time periodincreases,butthatthehorizontaldistanceisconstant.

1.3

5.0

11

20

31

45

61

80

0.0

100

Verticaldistanc

e(m)

ground

1.3

5.0

11

20

31

45

61

80

0.0

100

Verticaldistanc

e(m)

ground

Afallingobjectreachesits

terminal velocitywhenthe

upwardsairresistancebecomes

equaltothedownwardforceofgravity.

Afallingobjectreachesits

terminal velocitywhenthe

upwardsairresistancebecomes

equaltothedownwardforceofgravity.


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The vertical

velocity

increases

(i.e. object

accelerates).

The horizontal velocity remains the

same (i.e. there is no acceleration).

Position of a ball at constant time intervals

Keep them separated

Inmodellingprojectilemotion,theverticalandhorizontalcomponentsofthemotionaretreatedseparately.1. The total time taken for the projectile motion is determined by the

verticalpartofthemotionastheprojectilecannotcontinuetomovehori-zontally once it has hit the ground, the target or whatever else it mightcollidewith.

2. Thistotal time can then be usedto calculate the horizontaldistance, orrange,overwhichtheprojectiletravels.

sample problem 3.2Imaginethehelicopterdescribedinsampleproblem3.1isnotstationary,butisyingataslowandsteadyspeedof20ms-1andis100mabovetheground

whenthepackageisdropped.(a) Calculatehowlongittakesthepackagetohittheground.(b) Whatistherangeofthepackage?(c) Calculatetheverticaldistancethepackagehasfallenafter0.5s,1.0s,1.5s,

2.0s,etc.untilthepackagehasreachedtheground.(Youmayliketouseaspreadsheethere.)Thencalculatethecorrespondinghorizontaldistance,andhencedrawascalediagramofthepackagespositionathalf-secondintervals.

Remember,thehorizontalandverticalcomponentsofthepackagesmotionmustbeconsideredseparately.

(a) Inthispartofthequestiontheverticalcomponentisimportant.Verticalcomponent:u=0ms-1,x=100m,a=10ms-2,t=?

x=ut+1

2at

2

100m=0ms-1t+1

2(10ms-2)t2

100

5 0.=t2

t=4.5s

Solution:Solution:


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(Note:again,thepositivesquarerootistakenasweareconcernedonly

withwhathappensaftert=0.)

(b) Therangeofthepackageisthehorizontaldistanceoverwhichittravels.It

isthehorizontalcomponentofvelocitythatmustbeusedhere.

Horizontalcomponent:u=20ms-1(Theinitialvelocityofthepackage

is the same as the velocity of the helicopter in which it has been

travelling.) a=0ms-2 (No forces act horizontally so there is no horizontal

acceleration.)

t=4.5s(frompart(a)ofthisexample)

x=?

x=ut+1

2at2

=20ms-14.5s+0

=90m

(c)

Table 3.2 Vertical and horizontal components of thepackages motion

VERTICAL COMPONENT HORIZONTAL COMPONENT

u=0ms-1,t=0.5s,a=10ms-2,x=?

x=ut+1

2at

2

=0ms-10.5s+1

2(10ms-2)(0.5s)2

=1.3m

u=20ms-1,t=0.5s,a=0ms-2,x=?

x=ut+1

2at

2

=20ms-10.5s+0

=10m

Repeatthecalculationsshownintable3.2fort=1s,1.5s,2s,etc.togaintheresultsshownintable3.3.Thescalediagramofthepackagespositionis

shownonpage51.

Table 3.3 Vertical and horizontal distance travelled over time

TIME (s)

VERTICAL DISTANCE

(m)

HORIZONTAL

DISTANCE (m)

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

4.5

1.3

5.0

11

20

31

45

61

80

100

10

20

30

40

50

60

70

80

90

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Spreadsheets model a package

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1.3

5.0

11

20

31

45

61

80

Verticaldistance(m)

Horizontal distance (m)

0.0

0 10 20 30 40 50 60 70 80 90

100


A ball is thrown horizontally at a speed of 40 m s1 from the top of a cliff into theocean below and takes 4.0 seconds to land in the water. Air resistance can beignored.

(a) What is the height of cliff above sea level of the throwers hand releases theball from a height of 2.0 metres above the ground?

(b) What horizontal distance did the ball cover?

(c) Calculate the vertical component of the velocity at which the ball hits thewater.

(d) At what angle to the horizontal does the ball strike the water?

What goes up must come downMostprojectilesaresetinmotionwithvelocity.Thesimplestcaseisthatofaballthrowndirectlyupwards.Theonlyforceactingontheballisthatofgravity(ignoringairresistance).Theballacceleratesdownwards.Initially,thisresultsintheballslowingdown.Eventually,itcomestoahalt,thenbeginstomovedownwards,speedingupasitgoes.Noticethat,whenairresistanceisignored,themotionoftheballisiden-

ticalwhetheritisgoinguporcomingdown.Theballwillreturnwiththesamespeedwithwhichitwasprojected.Throughoutthemotionillustratedinthe

Digital doc:

Investigation 3.1:

Predicting the range of

a projectile

The aim of this investigation is

to predict the range (that is, thehorizontal distance travelled) of

a projectile with a known initial

horizontal velocity, and then to

test the prediction.

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Digital doc:

Investigation 3.1:

Predicting the range of

a projectile

The aim of this investigation is

to predict the range (that is, thehorizontal distance travelled) of

a projectile with a known initial

horizontal velocity, and then to

test the prediction.

eBookpluseBookplus


7/37


gurebelow(andforwhichgraphsareshown),theaccelerationoftheballis

aconstant10ms-2downwards.Acommonerrormadebyphysicsstudentsis

tosuggestthattheaccelerationoftheballiszeroatthetopofitsight.Ifthis

weretrue,wouldtheballevercomedown?

The motion of a ball projected

vertically upwards

Graphs of motion for a ball thrown

straight upwards

(a) going up (b) going down

v v

x (m)

t(s)

(a)

v (m s1)

t(s)

(b)

a (m s2)

t(s)

(c)

10

Theaxiomwhatgoesupmustcomedownappliesequallysotobulletsasitdoestoballs.Unfortunately,thismeansthatpeoplesometimesgetkilledwhentheyshootgunsstraightupintotheair.Ifthebulletleftthegunataspeedof60ms-1, itwill return toEarth atroughlythesamespeed.Thisspeediswellandtrulyfastenoughtokillapersonwhoishitbythereturningbullet.

sample problem 3.3

Adancerjumpsverticallyupwardswithaninitialvelocityof4.0ms-1.Assume

thedancerscentreofmasswasinitially1.0mabovetheground,andignore

airresistance.

(a) Howlongdidthedancertaketoreachhermaximumheight?

(b) Whatwasthemaximumdisplacementofthedancerscentreofmass?

(c) Whatistheaccelerationofthedanceratthetopofherjump?

(d) Calculatethevelocityof thedancerscentreofmasswhenitreturnstoits

originalheightabovetheground.

aS a maTTer of facTaS a maTTer of facT


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Thereareseveralwaysofarrivingatthesameanswer.Ashasbeendoneinthisexample,itisalwaysgoodpracticetominimisetheuseofanswersfrompre-viouspartsofaquestion.Thismakesyouranswersmorereliable,preventingamistakemadeearlieronfromdistortingtheaccuracyofyourlatercalcula-tions.Forthisproblem,assignupaspositiveanddownasnegative.

(a) u=4.0ms-1,a= -10ms-2,v=0ms-1(asthedancercomestoahaltatthehighestpointofthejump),t=?

v=u+at

0ms-1=4.0ms-1+ (-10ms-2)t

t=4 0

10

1

2

. m s

m s

-

-

=0.40s

Thedancertakes0.40storeachherhighestpoint.

(b) u=4.0ms-1,a= -10ms-2,v=0ms-1(asthedancercomestoahaltatthehighestpointofthejump),x=?

v2=u2+2ax

(0ms-1)2=(4.0ms-1)2+2(-10ms-2)x

16m=20x

x=0.80m

Themaximumdisplacementofthedancerscentreofmassis0.80m.

(c) Atthetopofthejump,theonlyforceactingonthedanceristheforceofgravity(thesameasatallotherpointsofthejump).Thereforetheaccel-erationofthedancerisaccelerationduetogravity:10ms-2downwards.

(d) For this calculation, only the downwards motion needs to beinvestigated.

u=0ms-1(asthedancercomestoahaltatthehighestpointofthejump),

a= -10ms-2,x= -0.80m(asthemotionisdownwards),v=?

v2=u2+2ax

v2=(0ms1)2+2(10ms2)(0.80m)

v= -4.0ms-1

(Note: here, the negative square root is used, as the dancer is movingdownwards. Remember, the positive and negative signs show directiononly.)

Thevelocityofthedancerscentreofmasswhenitreturnstoitsoriginalheightis4.0ms-1downwards.


A basketball player jumps directly upwards so that his centre of mass reaches amaximum displacement of 50 cm.

(a) What is the velocity of the basketballers centre of mass when it returns to itsoriginal height above the ground?

(b) For how long was the basketballers centre of mass above its originalheight?

Solution:Solution:


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Hagg md ar

Sometimesdancers,basketballersandhighjumpersseemtohanginmidair.Itisasthoughtheforceofgravityhadtemporarilystoppedactingonthem.Ofcoursethisisnotso!Itisonlythepersonscentreofmassthat

moves inaparabolic path.Thearrangement of thepersonsbodycanchangethepositionofthecentreofmass,causingthebodytoappeartobehanginginmidaireventhoughthecentreofmassisstillfollowingitsoriginalpath. Highjumperscanusethiseffecttoincreasetheheightoftheirjumps.Bybendingherbodyasshepassesoverthebar,PetrinaPricecancausehercentreofmasstobeoutsideherbody!Thisallowsherbodytopassoverthebar,whilehercentreofmasspassesunderit.TheamountofenergyavailabletoraisePetrinascentreofmassislimited,soshecanraisehercentre ofmass only by a certain amount.This technique allowsher toclearahigherbarthanothertechniquesforthesameamountofenergy.

Australian high jumper Petrina Prices centre of mass passes

under the bar, while her body passes over the bar!

Shooting at an angleGenerally, projectilesareshot, thrownordrivenat some angle to thehori-

zontal.Inthesecasestheinitialvelocitymayberesolvedintoitshorizontal

andverticalcomponentstohelpsimplifytheanalysisofthemotion.

Ifthevelocity and the angleto the horizontal are known, the sizeof the

componentscanbecalculatedusingtrigonometry.

Themotion ofprojectiles withaninitial velocityat an angleto the hori-zontalcanbedealtwithinexactlythesamemannerasthosewitha velocity

straightuporstraightacross.However,theinitialvelocitymustbeseparated

intoitsverticalandhorizontalcomponents.

sample problem 3.4

Astuntdriveristryingtodriveacaroverasmallriver.Thecarwilltravelupa

ramp(atanangleof40)andleavetheramptravellingat22ms-1.Theriveris

50mwide.Willthecarmakeit?

pHySicS in focuS pHySicS in focuS

The velocity can be resolved

into a vertical and a horizontal

component.

Q

vhorizontal

= v cos Q

vvertical

= vsin Qv

The velocity can be resolved

into a vertical and a horizontal

component.

Q

vhorizontal

= v cos Q

vvertical

= vsin Qv


10/37


40n 40n

50 m

river

velo

city

=22

ms1

Assignupaspositiveanddownasnegative.

Beforeeitherpartofthemotioncanbeexamined,itisimportanttocalculatetheverticalandhorizontalcomponentsoftheinitialvelocity.

40n

vhorizontal

= 22 cos 40n

= 17 m s1

= 14 m s1

vvertical

= 22 sin 40n

v= 22 m s1

Therefore the initial vertical velocity is 14ms-1 and the initial horizontalvelocityis17ms-1.

Inordertocalculatetherangeofthecar(howfaritwilltravelhorizontally),itisclearthatthehorizontalpartofitsmotionmustbeconsidered.However,theverticalpartisalsoimportant.Theverticalmotionisusedtocalculatethetimeintheair.Then,thehorizontalmotionisusedtocalculatetherange.

Table 3.4 Calculating the horizontal and vertical components

VERTICAL COMPONENT HORIZONTAL COMPONENT

(Usethersthalfofthemotionfromtake-offuntilthecarhasreachedits

highestpoint.)

u=14ms-1,a=-10ms-2,

v=0ms-1(asthecarcomestoa

verticalhaltatitshighestpoint),

t=?

v=u+at

0=14ms-1+(-10ms-2)t

t=14

10

1

2

m s

m s

-

-

=1.4s

Asthisisonlyhalfthemotion,thetotal

timeintheairis2.8s.(Itispossible

todoublethetimeinthissituation

becausewehaveignoredairresistance.

Thetwopartsofthemotionare

symmetrical.)

u=17ms-

1,t=2.8s(beingtwicethetimetakentoreachmaximumheightas

calculatedfortheverticalcomponent),

a=0ms-2,x=?

x=ut

=17ms-12.8s

=48m

Therefore,theunluckystuntdriverwillfallshortofthesecondrampandwillland intheriver.Maybethestudyofphysicsshouldbeaprerequisiteforallstuntdrivers!

Soution:Soution:

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Modelling a stunt driver

Worksheets build on the Falling

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A hockey ball is hit towards the goal at an angle of 25 to the ground with aninitial speed of 32 km h1.

(a) What are the horizontal and vertical components of the initial velocity of theball?

(b) How long does the ball spend in ight?(c) What is the range of the hockey ball?

Projectile motion calculationsHerearesometipsforprojectilemotioncalculations.Ithelpstodrawadiagram.Alwaysseparatethemotionintoverticalandhorizontalcomponents.Remembertoresolvetheinitialvelocityintoitscomponentsifnecessary.Thetimeinightisthelinkbetweentheseparateverticalandhorizontalcomponentsofthemotion.Attheendofanycalculation,checktoseeifthequantitiesyouhavecalcu-

latedarereasonable.

The real world including air resistanceSofarinthischapter,theeffectsofairresistancehavebeenignoredsothatwecaneasilymodelprojectilemotion.Thereasontheforceofairresistancecom-plicatesmatters somuchis that itis not constantthroughoutthemotion. Itdependsonthevelocityoftheprojectile,thesurfaceareathatisbeinghitbytheair,thetypeofsurfaceandeventhespinoftheprojectile.Forobjectswiththesamesurfaceandspin,airresistanceincreasesasthespeedoftheobjectincreases.Nomatterwhataffectstheamountofairresistance,onethingisalwaystrue

airresistanceopposesthemotionoftheprojectile.Logintowww.jacplus.

com.autolocatetheProjectilemotionappletweblinkforthischapter.

path ofprojectilewithout airresistance

path of a projectilewith air resistanceF

a.r.

Fa.r.

Fa.r.

w

w

w

While the magnitude of air resistance changes throughout the motion, it always

opposes the direction of the motion.

Weblink:

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Uniform circular motionHumansseemtospendalotoftimegoingaroundincircles.Trafcatround-abouts,childrenonmerry-go-rounds,cyclists invelodromes.Ifyoustop tothinkabout it, you are always goingaround incirclesas a result ofEarthsrotation.ThesatellitesorbitingEarth,includingtheMoon,travelinellipses.However,

theirorbitscanbemodelledascircularmotion.

The motion of satellites around Earth can be modelled as circular motion with a

constant speed.

Getting nowhere fastRalphhasbeenabaddogandhasbeenchainedup.Toamusehimself,herunsincircles.Ralphsowner,Julie,isa physics teacher. She knows thatnomatter howgreatRalphs averagespeed is,healwaysends up inthesameplace,sohisaveragevelocityisalwayszero.

Instantaneous velocity

Although Ralphs average velocity for a single lapis zero, his instantaneous velocity is continuallychanging.Velocityisavectorandhasamagnitudeanddirection.WhilethemagnitudeofRalphsvelocitymaybeconstant,thedirectioniscontinuallychanging.Atonepoint,Ralphistravellingeast,sohisinstantaneousvelocity is in aneasterlydirection. A short time later, he will be travellingsouth,sohisinstantaneousvelocityisinasoutherlydirection.IfRalph couldmaintain a constant speed, themagnitude ofhis velocity

wouldnotchange,butthedirectionwouldbecontinuallychanging.Thespeed is therefore constant andcanbecalculatedusing theformula

v=x

t,wherevistheaveragespeed,xisthedistancetravelledandtisthetime

interval.Itismostconvenienttousetheperiodoftheobjecttravellinginacircle.Thus:

v=x

t

=circumference

period

=2pr

wherer=radiusofthecircleT=period.

Instantaneous velocityisthe

velocityataparticularinstant

oftime.

Instantaneous velocityisthe

velocityataparticularinstant

oftime.

Theperiodofarepeatedcircularmotionisthetime

takenforacompleterevolution.

Theperiodofarepeatedcircularmotionisthetime

takenforacompleterevolution.


13/37


sample problem 3.5

Ralphschainis7.0mlongandattachedtoasmallpostinthemiddleofthegarden.Ittakesanaverageof9stocompleteonelap.(a) WhatisRalphsaveragespeed?(b) WhatisRalphsaveragevelocityafterthreelaps?

(c) WhatisRalphsinstantaneousvelocityatpointA?(Assumehetravelsataconstantratearoundthecircle.)

(a) To calculate Ralphs speed, we need to know how far hehas travelled.Usingtheformulaforthecircumferenceofacircle(distance=2pr):

distance=2p7.0m

=44m.

Now,theaveragespeedcanbecalculated.

v=x

t

=44

9

m

s

=5ms-1

Ralphtravelswithanaveragespeedof5ms-1.

(b) Afterthreelaps,Ralph isinexactlythesameplaceashe started,sohisdisplacementiszero.Nomatterhowlonghetooktoruntheselaps,hisaveragevelocitywouldstillbezero,asvav=

D

D

x

t.

(c) Ralphsvelocityisaconstant5ms-1ashetravelsaroundthecircle.Attheinstant inquestion, themagnitudeofhis instantaneous velocity isalso5ms-1.ThismeansRalphsvelocityis5ms-1north.

Reviion quetion 3.5

A battery operated toy car completes a single lap of a circular track in 15 s withan average speed of 1.3 m s1. Assume that the speed of the toy car is constant.

(a) What is the radius of the track?

(b) What is the magnitude of the toy cars instantaneous velocity halfwaythrough the lap?

(c) What is the average velocity of the toy car after half of the lap has beencompleted?

(d) What is the average velocity of the toy car over the entire lap?

Changing velocities and accelerationsAnyobjectmovinginacirclehasacontinuallychangingvelocity.Rememberthat although the magnitude of the velocity is constant, the direction ischanging.Asallobjectswithchanging velocitiesare experiencing anaccel-eration,thismeansallobjectsthataremovinginacircleareaccelerating.

Anacceleration canbecausedonlybyanunbalancedforce, sonon-zeronet force isneeded tomove anobject ina circle. For example, a hammerthrowermustapplyaforcetothehammertokeepitmovinginacircle.Whenthehammerisreleased,itmovesoffwiththevelocityithadattheinstantofrelease.Thenetforceonthehammeristhegravitationalforceonit(neglectingthesmallamountofairresistance),andthehammerwillexperienceprojec-tilemotion.

7.0 m

N

A

S

EW

7.0 m

N

A

S

EW

Solution:Solution:


14/37


The hammer is always accelerating while it moves in a circle.

In which direction is the force?

Thegure at left showsdiagrammatically the head of the hammer movinginacircleattwodifferenttimes.Ittakes tsecondstomovefromAtoB.To

determinetheacceleration,thechangeinvelocitybetweenthesetwopointsmustbedetermined.Vectoradditionmustbeusedtodothis.

Dv=v2-v1

Dv=v2+(-v1).

Notice that when the Dv vector is transferredback to the original circlehalfwaybetweenthetwopointsintime,itispointingtowardsthecentreofthecircle.(See thegurebelow.) (Such calculationsbecomemoreaccuratewhen very small time intervals areused;however, a large time interval hasbeenusedheretomakethediagramclear.)

Asa=Dv

t,theaccelerationvectorisinthesamedirectionasDv,buthasa

differentmagnitudeanddifferentunits.

AB

v2

v2

v1

v1

$v

$v

(a) (b)

(a) Vector addition (b) The change in velocity is towards the centre of the circle.

No matter which time interval is chosen, the acceleration vector alwayspoints towards the centre of the circle. So, in order for an object to haveuniformcircularmotion,theaccelerationoftheobject must be towards thecentre ofthe circle.Such anacceleration iscalled centripetal acceleration.Theword centripetal literallymeans centre-seeking. As stated in NewtonsSecondLawofMotion,thenetforceonanobjectisinthesamedirectionastheacceleration(Fnet=ma).Therefore,thenetforceonanobjectmovingwithuniformcircularmotionistowardsthecentreofthecircle.Duetothecentre-seekingnatureofthenetforce,itiscalledthe centripetal force.

Centripetal accelerationisthe

centre-directedaccelerationof

anobjectmovinginacircle.

Centripetal accelerationisthe

centre-directedaccelerationof

anobjectmovinginacircle.

Centripetal forceisthecentre-

directedforceactingonan

objectmovinginacircle.Inthe

caseofacircularmotionwith

constantspeed,thecentripetal

forceisequaltothenetforce.

Centripetal forceisthecentre-

directedforceactingonan

objectmovinginacircle.Inthe

caseofacircularmotionwith

constantspeed,thecentripetal

forceisequaltothenetforce.

As long as the thrower keeps

turning, the hammer moves in

a circle. When the hammer is

released, it moves in a straight line.

The direction in which thehammer moves while beingspun around

The direction in which thehammer moves if let go

Velocity vectors for a hammer

moving in an anticlockwise circle

AB

v2

v1


15/37


Rememberthatwhilethehammerthrowerisexertingaforceonthehammerheadtowardsthecentreofthecircle,thehammerheadmustbeexertinganequalandoppositeforceonthe throwerawayfromthecentreofthecircle(accordingtoNewtonsThirdLawofMotion).

Whenreadingaboutcircularmotion,youmayencounterthetermcentri-fugal force. Centrifugal means centre-eeing. As it has the oppositemeaningtocentripetal,thetwowordsmust notbeusedinterchangeably.Whichwordyouuse todescribethe forcesof circularmotiondependson your frameof reference. If you are outside the circularmotion ina stationary frameof reference, you can easily identify that the forceonthemovingobjectistowardsthecentreofthecircleacentripetalforce.However,ifyouaretheobjectthatismoving,thingsappearalittledifferently. Imaginethatyouaresittingonthebackseatofabusmovingquicklyaroundacorner.Youfeelasthoughyouarebeingpushedoutfromthecorner; that is, you thinka forceis pushingyou away fromthe centreofthecirclebecauseyourbodymovesoutwards.Inrealityyourbodyis

continuingtomoveinastraightlineyouareinamovingframeofreference.Infact,yourbottomisbeingpushedtowardsthecentreofthecircleduetotheforceof friction.Thisresultsinyourleaningoutwards.Sowhatyouareexperiencingistheeffectofinertia;however,fromyourframeofreference(insidethebus),theonlyexplanationyoucanlogi-callyofferistheexistenceofanunbalancedforcepushingyououtwards.Thisforceis calledthecentrifugal force.Anobserverin thestationaryframeof referenceoutside the bus can see you aremoving ina circle(otherwiseyouwouldyoutofthebus!).Sothenetforceonyourbodymustbeacentripetalforce.

Calculating accelerations and forcesUsingvectordiagramsandtheformulaea=

Dv

tandFnet=ma,itispossibleto

calculatethe accelerationsand forcesinvolved incircularmotion.However,doingcalculations thisway is tedious,and results can be inaccurate if thevectordiagramsarenotdrawncarefully.Itismuchsimplertohaveaformulathatwillavoidthese difculties.Thederivationofsucha formula isa littlechallenging,butitisworththeeffort!Byre-examiningthetwopreviousgures(seep.59),itispossibletoseethat

theybothcontainisoscelestriangles.Theseareshownatleft.Figure (a) is a diagram showing distances. It has the radius of the circle

markedintwice.Theseradiiformtwosidesofanisoscelestriangle.Thethirdsideisformedbyaline,orchord,joiningpointAwithpointB.Itisthedis-

tancebetweenthetwopoints.Whentheangle qisverysmall,thelengthofthechordisvirtuallythesameasthelengthofthearcwhichalsojoinsthesetwopoints.Asthisisadistance,itslengthcanbecalculatedusingx=vt.Figure(b)isadiagramshowingvelocities.Astheobjectwasmovingwith

uniformcircularmotion,thelengthofthevectorsv2and-v1areidenticalandformtwosidesofanisoscelestriangle.Asbothpartsofthegureatleftarederivedfromthebottomgureonpage59,bothoftheanglesmarked asqarethe samesize.Therefore, thetrianglesareboth isosceles triangles,con-tainingthesameangle,q.Thismeanstheyaresimilartrianglestheycanbethoughtofasthesametriangledrawnontwodifferentscales.Thegureonpage61showsthesetrianglesredrawntomakethismoreobvious.

aS a maTTer of facTaS a maTTer of facT

The triangles shown in parts (a)

and (b) are both isosceles triangles.

A

rr

Q

B

(a)

v2

v1

v1

v2

$v

(b)

Q

The triangles shown in parts (a)

and (b) are both isosceles triangles.

A

rr

Q

B

(a)

v2

v1

v1

v2

$v

(b)

Q


16/37


Asthetrianglesaresimilar,theratiooftheirsidesmustbeconstant,so:

Dv

vt=

v

r.

Multiplyingbothsidesbyv:

Dv

t=

v

r

2

.

Asa=Dv

t :

a=v

r

2

.

Thisformulaprovidesawayofcalculatingthecentripetalaccelerationofamassmovingwithuniformcircularmotionhavingspeedvandradiusr.

Iftheaccelerationofaknownmassmovinginacirclewithconstantspeedhasbeencalculated,thenetforcecanbedeterminedbyapplyingFnet=ma.

Themagnitudeofthenetforcecanalsobecalculatedusing:

Fnet=ma=mv

r

2

.

sample problem 3.6

Acarisdrivenaroundaroundaboutataconstantspeedof20kmh-1(5.6ms-1).Theroundabouthasaradiusof3.5mandthecarhasamassof1200kg.(a) Whatisthemagnitudeanddirectionoftheaccelerationofthecar?(b) Whatisthemagnitudeanddirectionoftheforceonthecar?

(a) v=5.6ms-1,r=3.5m,a=?

a=v

r

2

=( . )

.

5 6

3 5

1 2m s

m

-

=9.0ms-2

Thecaracceleratesat9.0ms-2towardsthecentreoftheroundabout.

(b) There are two different formulae that can be used to calculate thisanswer.

(i) Usetheanswerto(a)andsubstituteintoFnet=ma.

a=9.0ms-2,m=1200kg,Fnet=?

Fnet=ma

=1200kg9.0ms-2

=1.1104N

(ii) UsetheformulaFnet=mv

r

2

.

v=5.6ms-1,r=3.5m,m=1200kg,Fnet=?

Fnet=mv

r

2

=1200 5 6

3 5

1 2kg m s

m

( . )

.

-

=1.1104N

Bothmethodsgivetheforceonthecaras1.1104Ntowardsthecentreoftheroundabout.

The two triangles aresimilar

triangles.

A

rr

Q

B

(a)

v v

$v(b)

Q

vt

The two triangles aresimilar

triangles.

A

rr

Q

B

(a)

v v

$v(b)

Q

vt

Solution:Solution:


17/37


Sometimesit isnoteasy tomeasurethevelocityoftheobjectunder-goingcircularmotion.However,thiscanbecalculatedfromtheradiusofthecircleandthetimetakentocompleteonecircuitusingtheequationv= 2pr

T.

a =v

r

2

a=2

2

prT

r

a=4 2

2

p rT

SubstitutingthisintoFnet=ma:

Fnet=m r

T

4 2

2

p.


Kwong (mass 60 kg) rides the Gravitron at the amusement park. This ride movesKwong in a circle of radius 3.5 m, at a rate of one rotation every 2.5 s.

(a) What is Kwongs acceleration?

(b) What is the net force acting on Kwong? (Include a magnitude and a directionin your answer.)

(c) Draw a labelled diagram showing all the forces acting on Kwong.

Examples of centripetal forceWheneveranobjectisinuniformcircularmotion,thenetforceonthatobject

must be towards thecentreof thecircle(that is,a centripetal force).Someexamplesofcommonsituationsinvolvingcentripetalforcesfollow.

Tension

Inphysics, tensionisusedtodescribetheforceappliedbyanobjectthatisbeingpulledorstretched.

(b)

Fnet

T1

T2

W

(a)

Centripetal force is the source of fun in many theme park rides. (a) Tension contributes

to the centripetal force in many amusement park rides. (b) The net force acting on a

compartment


18/37


tension (T)

normalreaction (N)

weight (W)

Fnet

Not to scale

A component of the tension is the net force (or centripetal force) acting on the female

skater when she is performing a death roll.

Friction

Whena carroundsa corner, thesidewaysfrictional forcescontributeto thecentripetal force.The forwards frictional forces of the ground on the tyres

keepthecarmoving,butifthesidewaysfrictionalforcesarenotsufcient,the

netforceonthecarwillnotbetowardsthecentreofthecurve.Inthissitu-

ation,thenetforceislessthanthecentripetalforcerequiredtokeepthecar

movinginacircleanditwillnotmakeitaroundthecorner!

TheformulaFnet=mv

r

2

showsthatasthevelocityincreases,theforceneeded

tomoveinacirclegreatlyincreases(Fnetv2).Thisiswhyitisvitalthatcars

donotattempttocornerwhiletravellingtoofast.

weight (W)

FfrictionFfriction

Ffriction

Ffriction

Fnet

normalreaction (N)

(N)

(N)(N)

The sideways frictional forces of the ground on the tyres enable a car to move around

a corner.

Trackathletes, cyclists andmotorcyclists also rely on sideways frictional

forcestoenablethemtomovearoundcorners.Toincreasethesizeoftheside-

waysfrictionalforce,whichwill thereforeallowthemtocornermorequickly,

they often lean into thecorner.The lean alsomeans that they are pushing

onthesurface atanangle,so the reaction forceisnolongernormalto the

ground.Ithasacomponenttowardsthecentreoftheircircularmotion.


19/37


it

Fnt

ratin

Leaning into a corner increases the size of the centripetal force, allowing a higher

speed while cornering. The sideways friction is greater, and the reaction force of the

ground has a component towards the centre of the circular motion.

In velodromes, the track is banked so that a component of the normalreaction acts towards the centre of the velodrome, thus increasing the netforceinthisdirection.Asthecentripetalforceislarger,thecyclistscanmove

aroundthecornersfasterthaniftheyhadtorelyonfrictionalone.

Going around the bend

Whenavehicletravelsaroundabend,orcurve,atconstantspeed,itsmotioncanbeconsideredtobepartofa circularmotion.Thecurvemakesupthearcofacircle.Inorderforacartotravelaroundacornersafely,thenetforceactingonitmustbetowardsthecentreofthecircle.Part(a)ofthenextgureshowstheforcesactingonavehicleofmass m

travellingaroundacurvewitharadius,r,ataconstantspeed,v.Theforcesactingonthecarareweight,W,frictionandthenormalreaction,N.

N

W

(a)

Sias

ritin (Fr)F

nt

()

Fr

Fnt

N sin

Frs

W

(a) For the vehicle to take the corner safely, the net force must be towards the centre of

the circle. (b) Banking the road allows a component of the normal reaction to contribute

to the centripetal force.


20/37


Onalevelroadtheonlyforcewithacomponenttowardsthecentreofthe

circleis thesidewaysfriction.Thissidewaysfrictionmakesupthe wholeof

the magnitude ofthe net force (and therefore the centripetal force) on the

vehicle.Thatis:

Fnet=sidewaysfriction

=mv

r

2

.

Ifyoudrivethevehiclearoundthecurvewithaspeedsothatmv

r

2

isgreater

thanthe sidewaysfriction,the motion isno longer circularand thevehicle

willskidoffthe road.If theroadiswet,sidewaysfriction islessanda lower

speedisnecessarytodrivesafelyaroundthecurve.

Iftheroadisbankedatanangleqtowardsthecentreofthecircle,acom-

ponentofthenormalreactionNsinqcanalsocontributetothecentripetal

force.Thisisshownindiagram(b)onpage64.

Fnet=Frcosq + Nsinq

Thelargercentripetalforcemeansthat,foragivencurve,bankingtheroad

makesahigherspeedpossible.

Loosegravelonbendsinroadsisdangerousbecauseitreducestheside-waysfrictionforce.Atlowspeedsthisisnotaproblem,butavehicletravelling

athighspeedislikelytolosecontrolandrunofftheroadinastraightline.

sample problem 3.7

Acarof mass1280kgtravelsaroundabendwitharadiusof12.0m.Thetotal

sidewaysfrictiononthewheelsis16400N.Theroadisnotbanked.Calculate

themaximumconstantspeedatwhichthecarcanbedrivenaroundthebend

withoutskiddingofftheroad.

Thecarwillmaintainthecircularmotionaroundthebendif:

Fnet=mv

r

2

where

v=maximumspeed.

Ifv were toexceed this speed,Fnet

projectile and circular motion

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