Chapter 2. MAE002-Kinematics of particle.pdf

The motion of the paraglider can be described in terms of its position, velocity, and
acceleration. When landing, the pilot of the paraglider needs to consider the wind
velocity and the relative motion of the glider with respect to the wind. The study of
motion is known as kinematics and is the subject of this chapte

1. MAE002-Dynamics
Chapter 2 -Kinematics of particles
Themotionoftheparaglidercanbedescribedintermsofitsposition,velocity,and
acceleration.Whenlanding,thepilotoftheparagliderneedstoconsiderthewind
velocityandtherelativemotionofthegliderwithrespecttothewind.Thestudyof
motionisknownaskinematicsandisthesubjectofthischapter.

2. Objectives
•Describetherelationshipsbetweenposition,velocity,
acceleration,andtime.
•Solvekinematicsproblems.
•Analyzetherelativemotionofmultipleparticles.
•Determinethemotionofaparticlethatdependsonthe
motionofanotherparticle.
•Determinewhichcoordinatesystemismostappropriatefor
solvingacurvilinearkinematicsproblem.
•Calculatetheposition,velocity,andaccelerationofaparticle
undergoingcurvilinearmotionusingCartesian,tangentialand
normal,andradialandtransversecoordinates.

3. Contents
1.Introduction
2.Rectilinear Motion
3.Plane Curvilinear Motion
4.Rectangular Coordinates (x-y )
5.Normal and Tangential Coordinates (n-t)
6.Polar Coordinates (r-)
7.Space Curvilinear Motion
8.Relative Motion (Translating Axes)
9.Constrained Motion of Connected Particles
10.Chapter Review

4. Kinematics:studyofthegeometryofmotion.Relatesdisplacement,
velocity,acceleration,andtimewithoutreferencetothecauseof
motion.
Introduction

5. Kinematicrelationshipsareusedto
helpusdeterminethetrajectoryofa
golfball,theorbitalspeedofa
satellite,andtheaccelerationsduring
acrobaticflying.
Introduction

6. Introduction
Particle kinetics includes:
•Rectilinearmotion:position,velocity,andaccelerationofa
particleasitmovesalongastraightline.
•Curvilinearmotion:position,velocity,andaccelerationofa
particleasitmovesalongacurvedlineintwoorthree
dimensions.

7. Rectilinear Motion
•Rectilinearmotion:particlemoving
alongastraightline
•Positioncoordinate:definedby
positiveornegativedistancefroma
fixedoriginontheline.
•Themotionofaparticleisknownif
thepositioncoordinateforparticleis
knownforeveryvalueoftimet.
•Maybeexpressedintheformofa
function,e.g.,32
6ttx 
or in the form of a graph xvs. t.

8. •Instantaneousvelocitymaybepositive
ornegative.Magnitudeofvelocityis
referredtoasparticlespeed.
•Consider particle which occupies position
Pat time tand P’at t+Dt,t
x
v
t
x
tD
D

D
D

D0
lim
Average velocity
Instantaneous velocity
•From the definition of a derivative,dt
dx
t
x
v
t

D
D

D0
lim
e.g.,2
32
312
6
tt
dt
dx
v
ttx


Rectilinear Motion

9. •Consider particle with velocity vat time
tandv’at t+Dt,
Instantaneousaccelerationt
v
a
tD
D

D0
lim t
dt
dv
a
ttv
dt
xd
dt
dv
t
v
a
t
612
312e.g.
lim
2
2
2
0



D
D

D
•From the definition of a derivative,
•Instantaneous acceleration may be:
-positive: increasing positive velocity
or decreasing negative velocity
-negative: decreasing positive velocity
or increasing negative velocity.
Rectilinear Motion

10. •From our example,32
6ttx  2
312tt
dt
dx
v  t
dt
xd
dt
dv
a 612
2
2

-at t= 2 s, x= 16 m, v= v
max= 12 m/s, a= 0
-at t= 4 s, x= x
max= 32 m, v= 0, a= -12 m/s
2
•What are x, v, and aat t= 2 s ?
•Note that v
maxoccurs when a=0, and that the
slope of the velocity curve is zero at this point.
•What are x, v, and aat t= 4 s ?
Rectilinear Motion

11. Determination of the Motion of a Particle
•We often determine accelerations from the forces applied
(kinetics will be covered later)
•Generally have three classes of motion:
-acceleration given as a function of time, a= f(t)
-acceleration given as a function of position, a= f(x)
-acceleration given as a function of velocity, a= f(v)
•Canyouthinkofaphysical
exampleofwhenforceisa
functionofposition?
When force is a function of
velocity?
a spring drag

12. Acceleration as a function of time, position, or velocitya a t 
0
0
vt
v
dv a t dt ()
dv
at
dt
 vdv a x dx 
00
vx
vx
vdv a x dx a a x and
dx dv
dt a
v dt
 
dv
v a v
dx
 
0
0
vt
v
dv
dt
av
 
00
xv
xv
vdv
dx
av
 a a v ()
dv
av
dt

If…. Kinematic relationship Integrate

13. Determine:
•velocityandelevationabovegroundattimet,
•highestelevationreachedbyballand
correspondingtime,and
•timewhenballwillhitthegroundand
correspondingvelocity.
Balltossedwith10m/sverticalvelocity
fromwindow20maboveground.
SOLUTION:
•Integrate twice to find v(t) and y(t).
•Solvefortwhenvelocityequalszero(timeformaximumelevation)and
evaluatecorrespondingaltitude.
•Solve for twhen altitude equals zero (time for ground impact) and evaluate
corresponding velocity.
Sample problem 2.1

14. Sample problem 2.2
 tvtvdtdv
a
dt
dv
ttv
v
81.981.9
sm81.9
0
0
2
0


  ttv 






2
s
m
81.9
s
m
10 
  
2
2
1
0
0
81.91081.910
81.910
0
ttytydttdy
tv
dt
dy
tty
y


 
2
2
s
m
905.4
s
m
10m20 ttty 












SOLUTION:
•Integrate twice to find v(t) and y(t).

15. •Solve for twhen velocity equals zero and
evaluate corresponding altitude.  0
s
m
81.9
s
m
10
2






 ttv s019.1t
•Solve for twhen altitude equals zero and evaluate
corresponding velocity.
   
2
2
2
2
s019.1
s
m
905.4s019.1
s
m
10m20
s
m
905.4
s
m
10m20


























y
ttty m1.25y
Sample problem 2.2

16. •Solve for twhen altitude equals zero and evaluate
corresponding velocity. 0
s
m
905.4
s
m
10m20
2
2












 ttty  
s28.3
smeaningles s243.1


t
t 
   s28.3
s
m
81.9
s
m
10s28.3
s
m
81.9
s
m
10
2
2














v
ttv s
m
2.22v
Sample problem 2.2

17. Brakemechanismusedtoreducegunrecoil
consistsofpistonattachedtobarrelmovingin
fixedcylinderfilledwithoil.Asbarrel
recoilswithinitialvelocityv
0,pistonmoves
andoilisforcedthroughorificesinpiston,
causingpistonandcylindertodecelerateat
rateproportionaltotheirvelocity.
Determinev(t),x(t),andv(x).kva
SOLUTION:
•Integrate a= dv/dt= -kvto find v(t).
•Integrate v(t) = dx/dtto find x(t).
•Integrate a= v dv/dx= -kvto find v(x).
Sample problem 2.2

18. SOLUTION:
•Integrate a= dv/dt= -kvto find v(t).
0
00
ln
vt
v
vtdv dv
a kv k dt kt
dt v v
       
kt
evtv


0
•Integrate v(t) = dx/dtto find x(t).

0
00
000
1
kt
txt
kt kt
dx
v t v e
dt
dx v e dt x t v e
k




  


  
kt
e
k
v
tx

1
0
Sample problem 2.2

19. •Integrate a= v dv/dx= -kvto find v(x).kxvv
dxkdvdxkdvkv
dx
dv
va
xv
v

 
0
0
0 kxvv 
0
•Alternatively, 










0
0
1
v
tv
k
v
tx kxvv 
0 

0
0 or
v
tv
eevtv
ktkt

  
kt
e
k
v
tx

1
0
with
and
then
Sample problem 2.2

20. Problem

21. Problem

22. Uniform rectilinear motion
Foraparticleinuniform
rectilinearmotion,the
accelerationiszeroandthe
velocityisconstant.vtxx
vtxx
dtvdx
v
dt
dx
tx
x





0
0
0
0
constant
Duringfree-fall,aparachutist
reachesterminalvelocitywhenher
weightequalsthedragforce.If
motionisinastraightline,thisis
uniformrectilinearmotion.
Careful –these only apply to
uniform rectilinear motion!

23. Uniformly accelerated rectilinear motion
Ifforcesappliedtoabodyare
constant(andinaconstant
direction),thenyouhave
uniformlyacceleratedrectilinear
motion.
Anotherexampleisfree-fall
whendragisnegligible

24. Uniformly accelerated rectilinear motion
Foraparticleinuniformlyacceleratedrectilinearmotion,the
accelerationoftheparticleisconstant.Youmayrecognizethese
constantaccelerationequationsfromyourphysicscourses.0
0
0
constant
vt
v
dv
a dv a dt v v at
dt
      
0
21
0 0 0 0 2
0
xt
x
dx
v at dx v at dt x x v t at
dt
        
00
22
00
constant 2
vx
vx
dv
v a vdv a dx v v a x x
dx
     
Careful –these only apply to uniformly accelerated
rectilinear motion!

25. Motion of several particles
Wemaybeinterestedinthemotionofseveraldifferentparticles,
whosemotionmaybeindependentorlinkedtogether.

26. Motion of Several Particles: Relative Motion
•Forparticlesmovingalongthesameline,timeshouldberecorded
fromthesamestartinginstantanddisplacementsshouldbe
measuredfromthesameorigininthesamedirection.
ABAB xxx
relativepositionofBwith
respecttoAABAB xxx  
ABAB vvv
relative velocity of Bwith
respect to AABAB vvv  
ABAB aaa
relative acceleration of B
with respect to AABAB aaa 

27. Sample Problem 11.4
Ballthrownverticallyfrom12mlevel
inelevatorshaftwithinitialvelocityof
18m/s.Atsameinstant,open-platform
elevatorpasses5mlevelmoving
upwardat2m/s.
Determine(a)whenandwhereballhits
elevatorand(b)relativevelocityofball
andelevatoratcontact.
SOLUTION:
•Ball:uniformlyacceleratedrectilinear
motion.
•Elevator:uniformrectilinearmotion.
•Writeequationforrelativepositionof
ballwithrespecttoelevatorandsolve
forzerorelativeposition,i.e.,impact.
•Substitute impact time into equation for
position of elevator and relative
velocity of ball with respect to elevator.

28. SOLUTION:
•Ball:uniformlyacceleratedrectilinearmotion.2
2
2
2
1
00
2
0
s
m
905.4
s
m
18m12
s
m
81.9
s
m
18
ttattvyy
tatvv
B
B




















•Elevator:uniformrectilinearmotion.ttvyy
v
EE
E








s
m
2m5
s
m
2
0
Sample Problem 11.4

29. •Writeequationforrelativepositionofball
withrespecttoelevatorandsolveforzero
relativeposition,i.e.,impact.
•Substituteimpacttimeintoequationsforpositionofelevatorand
relativevelocityofballwithrespecttoelevator.65.325
Ey m3.12
Ey  
65.381.916
281.918

 tv
EB s
m
81.19
EBv
Sample Problem 11.42
/
(12184.905)(52)0
BE
y t t t  0.39 (meaningless)
3.65
ts
ts



30. Sample Problem 11.5
CarAistravellingataconstant90mi/hwhenshepassesaparked
policeofficerB,whogiveschasewhenthecarpassesher.Theofficer
acceleratesataconstantrateuntilshereachesthespeedof105
mi/h.Thereafter,herspeedremainsconstant.Thepoliceofficer
catchesthecar3mifromherstartingpoint.Determinetheinitial
accelerationofthepoliceofficer.
?

31. Motion of Several Particles: Dependent Motion
•Position of a particle may dependon position of one or
more other particles.
•PositionofblockBdependsonpositionofblockA.
Sinceropeisofconstantlength,itfollowsthatsumof
lengthsofsegmentsmustbeconstant.
BAxx2
constant (one degree of freedom)
•Positions of three blocks are dependent.
CBA xxx22
constant (two degrees of freedom)
•For linearly related positions, similar relations hold
between velocities and accelerations.022or022
022or022


CBA
CBA
CBA
CBA
aaa
dt
dv
dt
dv
dt
dv
vvv
dt
dx
dt
dx
dt
dx

32. PulleyDisattachedtoacollar
whichispulleddownat3in./s.At
t=0,collarAstartsmovingdown
fromKwithconstantacceleration
andzeroinitialvelocity.Knowing
thatvelocityofcollarAis12in./s
asitpassesL,determinethe
changeinelevation,velocity,and
accelerationofblockBwhen
collarAisatL.
SOLUTION:
•Defineoriginatupperhorizontalsurface
withpositivedisplacementdownward.
•CollarAhasuniformlyaccelerated
rectilinearmotion.Solveforacceleration
andtimettoreachL.
•PulleyDhasuniformrectilinearmotion.
Calculatechangeofpositionattimet.
•BlockBmotionisdependentonmotions
ofcollarAandpulleyD.Writemotion
relationshipandsolveforchangeofblock
Bpositionattimet.
•Differentiatemotionrelationtwiceto
developequationsforvelocityand
accelerationofblockB.
Sample problem 11.5

33. Sample problem 11.5
SOLUTION:
•Defineoriginatupperhorizontalsurfacewith
positivedisplacementdownward.
•CollarAhasuniformlyacceleratedrectilinear
motion.Solveforaccelerationandtimetto
reachL.  

2
2
0
2
0
2
s
in.
9in.82
s
in.
12
2







AA
AAAAA
aa
xxavv 
s 333.1
s
in.
9
s
in.
12
2
0


tt
tavv
AAA

34. •Pulley Dhas uniform rectilinear motion.
Calculate change of position at time t.
  in. 4s333.1
s
in.
3
0
0








DD
DDD
xx
tvxx
•BlockBmotionisdependentonmotionsof
collarAandpulleyD.Writemotionrelationship
andsolveforchangeofblockBpositionattimet.
Total length of cable remains constant,
     
  0in.42in.8
02
22
0
000
000



BB
BBDDAA
BDABDA
xx
xxxxxx
xxxxxx  in.16
0

BBxx
Sample problem 11.5

35. •Differentiatemotionrelationtwiceto
developequationsforvelocityand
accelerationofblockB.0
s
in.
32
s
in.
12
02
constant2














B
BDA
BDA
v
vvv
xxx s
in.
18
Bv 0
s
in.
9
02
2







B
BDA
v
aaa 2
s
in.
9
Ba
Sample problem 11.5

36. Group problem solving
SliderblockAmovestotheleftwitha
constantvelocityof6m/s.Determine
thevelocityofblockB.
SOLUTION STEPS:
•Sketchyoursystemandchoose
coordinatesystem
•Writeoutconstraintequation
•Differentiatetheconstraintequationto
getvelocity

37. Given: v
A= 6 m/s left Find: v
B
x
A
y
B
This length is constant no matter how the
blocks move
•Sketchyoursystemandchoose
coordinates
•Differentiatetheconstraintequation
togetvelocityconst nts3 a
AB
x y L  
•Defineyourconstraintequation(s)6 m/s + 3 0
B
v 2 m/s
B
v
Note that as x
Agets bigger, y
Bgets smaller.
Group problem solving

38. Graphical Solution of Rectilinear-Motion Problems
Engineers often collect position, velocity, and acceleration data.
Graphical solutions are often useful in analyzing these data.Data Fideltity / Highest Recorded Punch
0
20
40
60
80
100
120
140
160
180
47.76 47.77 47.78 47.79 47.8 47.81
Time (s)
Acceleration (g)
Accelerationdatafroma
headimpactduringaround
ofboxing.

39. •Given the x-tcurve, the v-tcurve is equal to the x-tcurve slope.
•Given the v-tcurve, the a-tcurve is equal to the v-tcurve slope.
Graphical Solution of Rectilinear-Motion Problems

40. •Giventhea-tcurve,thechangeinvelocitybetweent
1andt
2is
equaltotheareaunderthea-tcurvebetweent
1andt
2.
•Given the v-tcurve, the change in position between t
1and t
2is equal
to the area under the v-tcurve between t
1and t
2.
Graphical Solution of Rectilinear-Motion Problems

41. Curvilinear Motion: Position, Velocity & Acceleration
The softball and the car both undergo curvilinear motion.
•A particle moving along a curve other than a straight line is in
curvilinear motion.

42. •Theposition vectorof a particle at time tis defined by a vector
between origin Oof a fixed reference frame and the position
occupied by particle.
•Consider a particle which occupies positionP defined by at time t
and P’defined by at t + Dt, r
 r

Curvilinear Motion: Position, Velocity & Acceleration

43. 0
lim
t
s ds
v
t dt
D
D

D Instantaneous velocity (vector)Instantaneous speed (scalar)0
lim
t
r dr
v
t dt
D
D

D
Curvilinear Motion: Position, Velocity & Acceleration

44. 0
lim
t
v dv
a
t dt
D
D
  
D instantaneous acceleration (vector)
•Consider velocity of a particle at time tand velocity at t + Dt,v
 v


•In general, the acceleration vector is not tangent to the particle path and velocity
vector.
Curvilinear Motion: Position, Velocity & Acceleration

45. Rectangular Components of Velocity & Acceleration
•When position vector of particle Pis given by its
rectangular components,kzjyixr


•Velocity vector,kvjviv
kzjyixk
dt
dz
j
dt
dy
i
dt
dx
v
zyx










•Acceleration vector,kajaia
kzjyixk
dt
zd
j
dt
yd
i
dt
xd
a
zyx










2
2
2
2
2
2

46. •Rectangularcomponentsparticularlyeffective
whencomponentaccelerationscanbeintegrated
independently,e.g.,motionofaprojectile,00  zagyaxa
zyx 
with initial conditions,0,,0
000000 
zyx vvvzyx
Integrating twice yields 
  0
0
2
2
1
00
00


zgtyvytvx
vgtvvvv
yx
zyyxx
•Motion in horizontal direction is uniform.
•Motion in vertical direction is uniformly
accelerated.
•Motion of projectile could be replaced by two
independent rectilinear motions.
Rectangular Components of Velocity & Acceleration

47. Sample Problem 11.7
Aprojectileisfiredfromthe
edgeofa150-mcliffwithan
initialvelocityof180m/satan
angleof30°withthe
horizontal.Neglectingair
resistance,find(a)thehorizontal
distancefromtheguntothe
pointwheretheprojectilestrikes
theground,(b)thegreatest
elevationabovetheground
reachedbytheprojectile.
SOLUTION:
•Considertheverticalandhorizontal
motionseparately(theyare
independent)
•Applyequationsofmotioniny-
direction
•Applyequationsofmotioninx-
direction
•Determinetimetforprojectiletohit
theground,usethistofindthe
horizontaldistance
•Maximum elevation occurs when v
y=0

48. SOLUTION:
Given: (v)
o=180 m/s(y)
o=150 m
(a)
y= -9.81 m/s
2
(a)
x= 0 m/s
2
Vertical motion –uniformly accelerated:
Horizontal motion –uniform rectilinear:
Choose positive x to the right as shown
Sample Problem 11.7

49. SOLUTION:
Horizontal distance
Projectile strikes the ground at:
Solving for t, we take the positive root
Maximum elevation occurs when v
y=0
Substitute into equation (1) above
Substitute t into equation (4)
Maximum elevation above the ground =
Sample Problem 11.7

50. Group Problem Solving
Abaseballpitchingmachine“throws”baseballswithahorizontal
velocityv
0.Ifyouwanttheheighthtobe42in.,determinethevalueof
v
0.
SOLUTION:
•Considertheverticalandhorizontalmotionseparately(theyare
independent)
•Apply equations of motion in y-direction
•Apply equations of motion in x-direction
•Determine time tfor projectile to fall to 42 inches
•Calculate v
0=0

51. Group Problem Solving
•Analyze the motion in the
y-direction
Given: x= 40 ft, y
o= 5 ft, y
f= 42 in. Find: v
o2
0
1
(0)
2
f
y y t gt   221
1.5 ft (32.2 ft/s )
2
t   21
3.5 5
2
gt 0.305234 st
•Analyze the motion in the
x-direction00
0 ( )
x
x v t v t   0
40 ft ( )(0.305234 s)v 0
131.047 ft/s 89.4 mi/hv

52. Motion Relative to a Frame in Translation
Asoccerplayermustconsider
therelativemotionoftheball
andhisteammateswhen
makingapass.
Itiscriticalforapilotto
knowtherelativemotionof
hisaircraftwithrespectto
theaircraftcarriertomake
asafelanding.

53. •Designateoneframeasthefixedframeof
reference.Allotherframesnotrigidlyattached
tothefixedreferenceframearemovingframesof
reference.
•Vector joining Aand Bdefines the position of Bwith respect to the
moving frame Ax’y’z’andABr
 ABAB rrr


•Differentiating twice,
ABv

velocity of Brelative to A.ABAB vvv

 
ABa

acceleration of Brelative to A.ABAB aaa


•Absolute motion of Bcan be obtained by combining motion of Awith
relative motion of Bwith respect to moving reference frame attached to A.
Motion Relative to a Frame in Translation
•PositionvectorsforparticlesAandBwith
respecttothefixedframeofreferenceOxyzare and
AB
rr

54. Motion Relative to a Frame in Translation

55. Sample Problem 11.9
AutomobileAistravelingeastatthe
constantspeedof36km/h.As
automobileAcrossestheintersection
shown,automobileBstartsfromrest
35mnorthoftheintersectionand
movessouthwithaconstant
accelerationof1.2m/s
2
.Determine
theposition,velocity,andacceleration
ofBrelativetoA5safterAcrosses
theintersection.
SOLUTION:
•Defineinertialaxesforthesystem
•Determinetheposition,speed,and
accelerationofcarAatt=5s
•Usingvectors(Eqs11.31,11.33,and
11.34)oragraphicalapproach,
determinetherelativeposition,
velocity,andacceleration
•Determinetheposition,speed,and
accelerationofcarBatt=5s

56. SOLUTION:
•Define axes along the road
Given:v
A=36 km/h, a
A= 0, (x
A)
0= 0
(v
B)
0= 0, a
B= -1.2 m/s
2
, (y
A)
0= 35 m
Determine motion of Automobile A:
We have uniform motion for A so:
At t= 5 s
Sample Problem 11.9

57. SOLUTION:
Determine motion of Automobile B:
We have uniform acceleration for B so:
At t= 5 s
Sample Problem 11.9

58. SOLUTION:
We can solve the problems geometrically, and apply the arctangent relationship:
Or we can solve the problems using vectors to obtain equivalent results:
B A B/ A
r r r 
B A B/ A
v v v 
B A B/ A
a a a 20 50
20 50 (m)


B/ A
B/ A
j i r
r j i 6 10
6 10 (m/s)
  
  
B/ A
B/ A
j i v
v j i 2
1.2 0
1.2 (m/s )
  

B/ A
B/ A
j i a
a j
Physically, a driver in car A would “see” car B travelling south and west.
Sample Problem 11.9

59. Tangential and Normal Components
Ifwehaveanideaofthepathofavehicle,itisoftenconvenient
toanalyzethemotionusingtangentialandnormalcomponents
(sometimescalledpathcoordinates).
Non-rectangularcomponents:radialdistanceandangular
displacement

60. •Thetangentialdirection(e
t)istangenttothepathoftheparticle.
Thisvelocityvectorofaparticleisinthisdirection
x
y
e
t
e
n
•The normal direction (e
n) is perpendicular to e
tand points towards
the inside of the curve.
v= v
te
t
= the instantaneous radius of curvature2
tn
dvv
a e e
dt
 t
vve
•The acceleration can have components in both the e
nand e
t directions
Tangential and Normal Components
22
v
a
dt
dv
ae
v
e
dt
dv
a
ntnt 


61. Amotorististravelingonacurved
sectionofhighwayofradius2500ft
atthespeedof60mi/h.Themotorist
suddenlyappliesthebrakes,causing
theautomobiletoslowdownata
constantrate.Knowingthatafter8s
thespeedhasbeenreducedto45
mi/h,determinetheaccelerationofthe
automobileimmediatelyafterthe
brakeshavebeenapplied.
SOLUTION:
•Defineyourcoordinatesystem
•Calculatethetangentialvelocity
andtangentialacceleration
•Determineoverallacceleration
magnitudeafterthebrakeshave
beenapplied
•Calculate the normal acceleration
Tangential and Normal Components

62. SOLUTION:
•Define your coordinate system
e
t
e
n
•Determine velocity and acceleration in the
tangential direction
•The deceleration constant, therefore
•Immediately after the brakes are applied, the
speed is still 88 ft/s2 2 2 2
2.75 3.10
nt
a a a   
Tangential and Normal Components

63. In2001,aracescheduledattheTexasMotorSpeedwaywas
cancelledbecausethenormalaccelerationsweretoohighand
causedsomedriverstoexperienceexcessiveg-loads(similarto
fighterpilots)andpossiblypassout.Whataresomethingsthat
couldbedonetosolvethisproblem?
Somepossibilities:
•Reducetheallowedspeed
•Increasetheturnradius
(difficultandcostly)
•Havetheracerswearg-suits
Tangential and Normal Components

64. Radial and Transverse Components
Byknowingthedistancetotheaircraft
andtheangleoftheradar,airtraffic
controllerscantrackaircraft.
Firetruckladderscanrotateaswellas
extend;themotionoftheendofthe
laddercanbeanalyzedusingradialand
transversecomponents.

65. •ThepositionofaparticlePisexpressedasadistancerfromthe
originOtoP–thisdefinestheradialdirectione
r.Thetransverse
directione
isperpendiculartoe
rr
v r e r e


•The particle velocity vector is
•The particle acceleration vector is  
2
2
r
a r r e r r e

      rerr


Radial and Transverse Components
Polar coordinates (r,)

66. Ifyouaretravellinginaperfectcircle,whatisalwaystrueabout
radial/transversecoordinatesandnormal/tangentialcoordinates?
a)The e
rdirection is identical to the e
ndirection.
b)The e
direction is perpendicular to the e
ndirection.
c)The e
direction is parallel to the e
rdirection.
Concept Quiz

67. RotationofthearmaboutOis
definedby=0.15t
2
whereisin
radiansandtinseconds.CollarB
slidesalongthearmsuchthatr=
0.9-0.12t
2
whererisinmeters.
Afterthearmhasrotatedthrough
30
o
,determine(a)thetotalvelocity
ofthecollar,(b)thetotal
accelerationofthecollar,and(c)
therelativeaccelerationofthe
collarwithrespecttothearm.
SOLUTION:
•Evaluate time tfor = 30
o
.
•Evaluateradialandangular
positions,andfirstandsecond
derivativesattimet.
•Calculate velocity and acceleration
in cylindrical coordinates.
•Evaluate acceleration with respect
to arm.
Sample Problem 11.12

68. SOLUTION:
•Evaluate time tfor = 30
o
.s 869.1rad524.030
0.15
2


t
t
•Evaluate radial and angular positions, and
first and second derivatives at time t.2
2
sm24.0
sm449.024.0
m 481.012.09.0



r
tr
tr

 2
2
srad30.0
srad561.030.0
rad524.015.0







 t
t
Sample Problem 11.12

69. •Calculate velocity and acceleration.  
r
r
r
v
v
vvv
rv
srv





122
tan
sm270.0srad561.0m481.0
m449.0





  0.31sm524.0 v   
    
r
r
r
a
a
aaa
rra
rra






122
2
2
2
22
2
tan
sm359.0
srad561.0sm449.02srad3.0m481.0
2
sm391.0
srad561.0m481.0sm240.0









  6.42sm531.0 a
Sample Problem 11.12

70. Sample Problem 11.12
•Evaluateaccelerationwithrespectto
arm.
Motionofcollarwithrespecttoarm
isrectilinearanddefinedby
coordinater.2
sm240.0ra
OAB 

71. Sample Problem
Duringaparasailingride,theboatistravelingataconstant30km/hr
witha200-mlongtowline.Attheinstantshown,theanglebetween
thelineandthewateris30°andisincreasingataconstantrateof2°/s.
Determinethevelocityandaccelerationoftheparasaileratthisinstant.

72. Chapter review