Post on 27-Feb-2020
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.$$Determine$the$speed$at$which$the$outer$edge$of$the$wheel$ismoving.
revolutionmin102.1minsrevolution830
1 3−×=
s 072.0min 102.1 3 =×= −T
v = 2π rT
=2π 0.29 m( )
0.072 s= 25m s = 56 mi/hr
Centripetal*Acceleration
In#uniform#circular#motion,#the#speed#is#constant,#but#the#direction#of#the#velocity#vector#is#not$constant.
90=+ βα90=+θα
θβ =
P
Centripetal*Acceleration
The$direction$of$the$centripetal$acceleration$is$towards$the$center$of$the$circle2$in$the$same$direction$as$the$change$in$velocity.
rvac2
=
Centripetal*Acceleration
Conceptual+Example:++Which+Way+Will+the+Object+Go?
An#object#is#in#uniform#circular#motion.##At#point#O it#is#released##from#its#circular#path.##Does#the#object#move#along#the#straightpath#between#O and#A or#along#the#circular#arc#between#pointsO and#P$?
Centripetal*Acceleration
Example:)The)Effect)of)Radius)on)Centripetal)Acceleration
The$bobsled$track$contains$turns$with$radii$of$33$m$and$24$m.$$Find$the$centripetal$acceleration$at$each$turn$for$a$speed$of$34$m/s.$$Express$answers$as$multiples$of$ .sm8.9 2=g
Centripetal*Force
∑ = aF m!
a =
!F∑m
Recall&Newton�s&Second&Law
When&a&net&external&force&acts&on&an&objectof&mass&m,&the&acceleration&that&results&is&directly&proportional&to&the&net&force&and&hasa&magnitude&that&is&inversely&proportional&tothe&mass.&&The&direction&of&the&acceleration&isthe&same&as&the&direction&of&the&net&force.
Centripetal*Force
Thus,&in&uniform&circular&motion&there&must&be&a&netforce&to&produce&the¢ripetal&acceleration.
The¢ripetal&force&is&the&name&given&to&the&net&force&required&to&keep&an&object&moving&on&a&circular&path.&&
The&direction&of&the¢ripetal&force&always&points&towardthe¢er&of&the&circle&and&continually&changes&direction&as&the&object&moves.
rvmmaF cc
2
==
Centripetal*Force
Example:)The)Effect)of)Speed)on)Centripetal)Force
The$model$airplane$has$a$mass$of$0.90$kg$and$moves$at$constant$speed$on$a$circle$that$is$parallel$to$the$ground.$The$path$of$the$airplane$and$the$guideline$lie$in$the$samehorizontal$plane$because$the$weight$of$the$plane$is$balancedby$the$lift$generated$by$its$wings.$$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?
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'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.
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