Download AP Physics Chapter 6 Review Emily Dickinson

APPhysicsChapter6Review
EmilyDickinson
CircularMotionandForces
UniformCircularMotionModel:
Objectsmovingwithconstantspeedinacircularpathareacceleratingcentripetally,or
towardsthecenter.Centripetalmeans“centerseeking.”Aslongastheparticlefollowsthe
circularpath,theaccelerationandforceitexperienceswillbetowardsthecenterofthecircle.
Centripetalaccelerationisdefinedbytheformula:𝑎! =
!!
!
wherevisvelocityandristhe
radiusofthecircularpath.Withacentripetalaccelerationthereisaalsoacentripetalforce.
Newton’sSecondLaw,𝐹!"# = 𝑚𝑎canbereinterpretedas𝐹!"#$%&'"$() = 𝑚𝑎!"#$%&'"$() or
𝐹!"#$%&'"$() =
!! !
!
v
v
𝐹! v
v
Mostproblemsfromthisunitdemonstrate
uniformcircularmotion,andthecentripetalforce
varies.Insituatiationswithobjectsswingingincircles,
attachedwithstringsorcordsetc.,theForceoftension
inthestringprovidesthecentripetalforce.
Inothersituationslikevehiclesgoingaroundacurve,
theforceoffrictionthatoccursbetweenthewheels
andthegroundcreatesthecentripetalforce.Thesame
Newton’sSecondLawapproachshouldbetakenwith
circularmotion.Itisimportanttostartwiththe
summationofallofthecentripetalforces,∑𝐹! = 𝑚𝑎! andthen,workthroughthevarioussteps.
Certainproblemsfromthischapterinvolvethecoefficientoffriction,whichisavalue
eithergivenorcalculatedthatdescribeshowsmoothasurfaceis.Thecoefficientoffriction
canbestaticordynamic,anditisindicatedbytheGreeksymbol,𝜇.
Non-UniformCircularMotion
Whenanobjectmovesinnon-uniform
circularmotion,theaccelerationismadeupof
centripetalaccelerationandtangential
𝐹!"#$"% acceleration.
𝜃
𝑎!"# = 𝑎!"#$"% + 𝑎!"#$%#!&"' 𝐹!"#$"% 𝐹!"#$%#!&"' Theseaccelerationsresultfromradialand
tangentialforces.
𝐹!"# = 𝐹!"#$"% + 𝐹!"#$%#!&"' 𝐹!"#$%#!&"' Problem1:VerticalCircles
Askateboardertriestocompleteafull360-degreepipeturn.The
pipehasaradiusof2meters,andtheskateboarderweighs75kg
withhisskateboard.a)Drawafreebodydiagramwhenthe
skateboarderisperpendicularwiththegroundonhiswayupthe
pipe.b)Theskateboarderisnotwearingahelmetbecausehetrusts
hisphysicsknowledge.Hehascalculatedthathisvelocityneedsto
beatleast3m/satthetopoftheramp.Willhemakeit?Ifnot
determinehisminimumspeedatthetopofthepipe,assumingthe
pipedoesnotapplyanyforceontheskateboarder.
Answers:
Information:mass,m=75kg,radius,r=2m
𝐹!"#$%& 𝐹!"#$%&' Formula:𝐹!"#$%&'"$() =
!! !
!
a)Theskateboarderexperiencesthedownwardforceofgravity,anda
normalforcetowardsthecenterfromthepipe.Weknowthatinthis
situationallofthecentripetalforceisduetothenormalforce,butwe
won’tbeansweringanyspecificquestionsaboutthissituation.
b)
Anotherfreebodydiagram!!!
Atthetopofthepipe,alloftheskateboarder’scentripetalforcecomesfromgravitybecause
thepipedoesnotapplyanynormalforceontheskateboarder.
∑𝐹!"#$%&'"$() = 𝐹!"#$%&' + 𝐹!"#$%& 𝑚𝑣 !
∑𝐹!"#$%&'"$() =
= 𝑚𝑔 + 0
𝑟
𝐹!"#$%& 𝑣 = 𝑔𝑟
𝐹!"#$%&' 𝑣=
9.8 (2)
v=4.42m/s
Basedonthatminimumvelocityvalue,whichisgreaterthanhispredicted3m/sweknow
thattheskateboarderwillfallonhisheadandprobablyhaveanawfulconcussion.:(
Problem2:GoCart
Inagocartracewithacirculartrackofradius50m,onecartmightslipoffoftheroad.The
road’scoefficientoffrictionis.3.Themassofthecartis90kg.a)Assumingthattheonly
centripetalforceistheforceoffriction,whatisthetotalcentripetalforceonthecart?b)How
fastdoesthecartneedtogoinordertomaketheturn?
m=
90kg
𝑣
Information: Mass,m=90kg
r=50
m
Formuals:
𝐹!"#$%#&' = 𝜇𝐹!"#$%& !! !
Radius,r=50m
∑𝐹!"#$%&'"$() = ! Coefficientoffriction,𝜇=.3
ForceofFriction
𝐹!"#$%& 𝐹!"#$%&' 𝐹!"#$%#&'!!"#$%&'"$() 𝐹!"#$%#&' = 𝜇𝐹!"#$%& 𝐹!"#$%#&' = 𝜇(𝑚𝑔)
𝐹!"#$%#&' = (.3)(90)(9.8)
𝐹!"#$%#&' = 264.4 𝑁
b)velocity
∑𝐹!"#$%&'"$() =
∑𝐹!"#$%&'"$() = 𝐹!"#$%#&'
𝑣=
𝑣=
𝑚𝑣 !
𝑟
𝑚𝑣 !
= 264.4 𝑁 =
𝑟
𝐹!"#$%#&' 𝑟
𝑚
264.4 (50)
90
𝑣 = 12.12 𝑚/𝑠
Problem3:Non-UniformCircularMotion
Aballisattachedtotheceilingwithathread,and
movesinnon-uniformcircularmotion.Theballhas
amassof3kg,avelocityof10m/s,aradiusof1
𝐹!"#$"% 𝜃 meter,andathetavalueof30degrees.
𝐹!"#$"% 𝐹!"#$%#!&"' Whatisthetensioninthestring,andwhatarethe
tangentialandradialforcesactingontheball?
𝐹!"#$%#!&"' 𝐹!"#$%&# Answers:
Freebodydiagram:
𝐹!"#$"%
𝐹!"#$%&' = 𝑚𝑔𝑐𝑜𝑠(𝜃)
𝐹!"#
= 𝑚𝑔𝑠𝑖𝑛(𝜃)
Theradialforceisthecentripetalforce,sowestartwithasummationoftheforcesthatare
centripetal.Itisimportantthatwebreakuptheforceofgravityintotheradialandtangential
componentsofgravity.𝐹!"#$"% = 𝑚𝑔𝑐𝑜𝑠(𝜃)and𝐹!"#$%#!&"' = 𝑚𝑔𝑠𝑖𝑛(𝜃)
𝑚𝑣 !
∑𝐹!"#$"% =
= 𝐹!"#$%&# − 𝐹!"#$%&' !"#$"% = 𝐹!"#$%&# − 𝑚𝑔𝑐𝑜𝑠(𝜃)
𝑟
𝑚𝑣 !
= 𝐹!"#$%&# − 𝑚𝑔𝑐𝑜𝑠(𝜃)
𝑟
!
3 (10)
𝐹!"#$%&# =
+ 3 9.8 cos 30 = 325.5 𝑁
1
𝐹!"#$"% =
!! !
!
= 300 𝑁
∑𝐹!"#$%#!&"' = 𝐹!"#$%&'!!"#$%!&"' = 𝑚𝑔𝑠𝑖𝑛 𝜃 𝐹!"#$%#!&"' = 3 9.8 sin 30 = 14.7 𝑁