Chapter(5v2 r2 =G M r3 τ=orbital period= 2πr v ∴τ=2π r3 GM τ Earth =2π (1.5×1011) 3...

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Chapter 5 Circular Motion and Gravitation

Transcript of Chapter(5v2 r2 =G M r3 τ=orbital period= 2πr v ∴τ=2π r3 GM τ Earth =2π (1.5×1011) 3...

Chapter(5

Circular(Motion(andGravitation

Uniform(Circular(Motion

DEFINITION(OF(UNIFORM(CIRCULAR(MOTION

Uniform(circular(motion(is(the(motion(of(an(object(traveling(at(a(constant(speed(on(a(circular(path.

Uniform(Circular(Motion

Let$T be$the$time$it$takes$for$the$object$totravel$once$around$the$circle.

T ! period of$the$circular$motionv

rT

=2π

r

Uniform(Circular(Motion

Example:))A)Tire.Balancing)Machine

The$wheel$of$a$car$has$a$radius$of$0.29$m$and$is$being$rotatedat$830$revolutions$per$minute$on$a$tire=balancing$machine.$$Determine$the$speed$at$which$the$outer$edge$of$the$wheel$ismoving.

revolutionmin102.1minsrevolution830

1 3−×=

s 072.0min 102.1 3 =×= −T

v = 2π rT

=2π 0.29 m( )

0.072 s= 25m s = 56 mi/hr

Centripetal*Acceleration

In#uniform#circular#motion,#the#speed#is#constant,#but#the#direction#of#the#velocity#vector#is#not$constant.

90=+ βα90=+θα

θβ =

P

Centripetal*Acceleration

Δvv≈vΔtr

ΔvΔt

≈v2

r

rvac2

=(instantaneous)+Centripetal+acceleration

Δt→ 0

Centripetal*Acceleration

The$direction$of$the$centripetal$acceleration$is$towards$the$center$of$the$circle2$in$the$same$direction$as$the$change$in$velocity.

rvac2

=

Centripetal*Acceleration

Conceptual+Example:++Which+Way+Will+the+Object+Go?

An#object#is#in#uniform#circular#motion.##At#point#O it#is#released##from#its#circular#path.##Does#the#object#move#along#the#straightpath#between#O and#A or#along#the#circular#arc#between#pointsO and#P$?

Centripetal*Acceleration

Example:)The)Effect)of)Radius)on)Centripetal)Acceleration

The$bobsled$track$contains$turns$with$radii$of$33$m$and$24$m.$$Find$the$centripetal$acceleration$at$each$turn$for$a$speed$of$34$m/s.$$Express$answers$as$multiples$of$ .sm8.9 2=g

Centripetal*Accelerationrvac

2=

ac =34m s( )2

33 m= 35m s2 = 3.6g

ac =34m s( )2

24 m= 48m s2 = 4.9g

Centripetal*Force

∑ = aF m!

a =

!F∑m

Recall&Newton�s&Second&Law

When&a&net&external&force&acts&on&an&objectof&mass&m,&the&acceleration&that&results&is&directly&proportional&to&the&net&force&and&hasa&magnitude&that&is&inversely&proportional&tothe&mass.&&The&direction&of&the&acceleration&isthe&same&as&the&direction&of&the&net&force.

Centripetal*Force

Thus,&in&uniform&circular&motion&there&must&be&a&netforce&to&produce&the&centripetal&acceleration.

The&centripetal&force&is&the&name&given&to&the&net&force&required&to&keep&an&object&moving&on&a&circular&path.&&

The&direction&of&the&centripetal&force&always&points&towardthe&center&of&the&circle&and&continually&changes&direction&as&the&object&moves.

rvmmaF cc

2

==

Centripetal*Force

Example:)The)Effect)of)Speed)on)Centripetal)Force

The$model$airplane$has$a$mass$of$0.90$kg$and$moves$at$constant$speed$on$a$circle$that$is$parallel$to$the$ground.$The$path$of$the$airplane$and$the$guideline$lie$in$the$samehorizontal$plane$because$the$weight$of$the$plane$is$balancedby$the$lift$generated$by$its$wings.$$Find$the$tension$in$the$17$mguideline$for$a$speed$of$19$m/s.

rvmTFc2

==

T = 0.90 kg( )19m s( )2

17 m=19 N

Centripetal*Force

Conceptual+Example:+A+Trapeze+Act

In#a#circus,#a#man#hangs#upside#down#from#a#trapeze,#legsbent#over#and#arms#downward,#holding#his#partner.##Is#it#harderfor#the#man#to#hold#his#partner#when#the#partner#hangs#straight#down#and#is#stationary#or#when#the#partner#is#swingingthrough#the#straight;down#position?

Banked'Curves

On#an#unbanked#curve,#the#static#frictional#forceprovides#the#centripetal#force.

Example: Find%the%maximum%speed%thata%car%can%drive%around%a%circular%curveof%radius%100%m%if%the%static%coefficientof%friction%between%the%tires%and%road%is%1.0%and%the%mass%of%the%car%is%1000%kg.

Fr = fSMAX =mac =

mv2

r∑

fSMAX = µSFN = µSmg =

mv2

r⇒ µSg =

v2

r∴v = µSgr = 1.0( ) 9.80( ) 100( ) = 31 m/s

= 69 mi/hr

Independentof%the%massof%the%car!

Banked'Curves

On#a#frictionless#banked#curve,#the#centripetal#force#is#the#horizontal#component#of#the#normal#force.##The#vertical#component#of#the#normal#force#balances#the#car�s#weight.

Banked'Curves

rvmFF Nc

2

sin == θ mgFN =θcos

Banked'Curves

rvmFN2

sin =θ

mgFN =θcosrgv2tan =θ

Banked'CurvesExample:)The)Daytona)500

The$turns$at$the$Daytona$International$Speedway$have$a$maximum$radius$of$316$m$and$are$steeply$banked$at$31degrees.$$Suppose$these$turns$were$frictionless.$$At$what$speed$would$the$cars$have$to$travel$around$them?

rgv2tan =θ θtanrgv =

v = 316 m( ) 9.8m s2( ) tan31! = 43m s 96 mph( )

Vertical)Circular)Motion

In#vertical#circular#motion#the#gravitationalforce#must#also#be#considered.

An#example#of#vertical#circular#motion#is#thevertical#�loop9the9loop� motorcycle#stunt.Normally,#the#motorcycle#speed#will#varyaround#the#loop.

The#normal#force,#FN,#and#the#weight#ofthe#cycle#and#rider,#mg,#are#shown#at#fourlocations#around#the#loop.

Vertical)Circular)Motion

rvmmgFN21

1 =−

rvmmgFN23

3 =+

rvmFN22

2 =

rvmFN24

4 =

1)

2)

3)

4)

There%is%a%minimum%speed%the%ridermust%have%at%point%3%in%order%to%stay%onthe%loop.

This%speed%may%be%found%by%settingFN3 = 0 in%the%centripetal%force%equationfor%point%3,%i.e.%in%FN3 + mg = mv3

2/r

! v3min = (rg)1/2

e.g.%for%a%track%with%r = 10 m,

v3min = ((10)(9.8))1/2 =%9.9%m/s%=%22%mph

Vertical)Circular)Motion

Example: Find%the%orbital%periods%of%Mercury%(~0.4%AU%from%the%Sun)%and%Neptune%(~30%AU%from%the%Sun)%and%compare%with%that%of%the%Earth.M = Sun mass, m = planet mass, r = distance to Sun, v = planet speed in orbit

1%AU%=%distance%from%Earth%to%Sun%=1.5%x%1011 m

FC =mv2

r=G mM

r2 ⇒v2

r2 =GMr3

τ = orbital period = 2πrv

∴τ = 2π r3

GM

τ Earth = 2π1.5×1011( )

3

6.67×10−11( ) 1.99×1030( )= 3.2×107 s

τMercury = 2π0.4×1.5×1011( )

3

6.67×10−11( ) 1.99×1030( )= 8.0×106 s = 0.25τ Earth

τNeptune = 2π30×1.5×1011( )

3

6.67×10−11( ) 1.99×1030( )= 5.2×109 s =160τ Earth