Ch. 2: Kinematics of Particles -...

Ch. 2: Kinematics of Particles

2.0 Outline

IntroductionRectilinear MotionPlane Curvilinear MotionRectangular Coordinates (x-y)Normal and Tangential Coordinates (n-t)Polar Coordinates (r-θ)Relative Motion (Translating Axes)

2.0 Outline


2.1 Introduction

2.1 Introduction

Kinematics is the study of the motion of bodies with noconsideration to the forces that accompany the motion.It is an absolute prerequisite to kinetics, which is thestudy of the relationships between the motion and thecorresponding forces that cause the motion or aregenerated as a result of the motion.

A particle is a body whose physical dimensions are sosmall compared with the radius of curvature of its path.This makes the body rotation effect insignificant andthe motion of the body can be treated as that of the particle.


2.1 Introduction

Position of Prectangular coordinates x, y, zcylindrical coordinates r, θ, zspherical coordinates R, θ, Φ

Motion of Pabsolute motion analysisrelative motion analysis

Absolute motion analysis: coordinates measured fromfixed reference axes, e.g. motion of the piston describedby the frame fixed to the groundRelative motion analysis: coordinates measured frommoving reference axes, e.g. motion of the piston describedby the frame attached to the car


2.2 Rectilinear Motion: motion along a straight line

2.2 Rectilinear Motion

( )

( )

av

t 0

If change in the position coordinate during t isthe displacement s , v s/ t

ds1 __ instantaneous velocity, v lim s/ t sdt

velocity time rate of change of the position coord., sIf change in th

Δ →

Δ

Δ ± = Δ Δ

= Δ Δ = =

=

( )

( )

av2

2t 0

e velocity during t is v, a v/ t

dv d s2 __ instantaneous acceleration, a lim v/ t v sdt dt

3 __ vdv adsΔ →

Δ Δ = Δ Δ

= Δ Δ = = = =

=



Displacement vs. Distancedisplacement: vector quantity involving

initial and ending positiondistance: positive scalar quantity

Both velocity and acceleration are vector quantities generally their changes include 1) change in magnitudeand 2) change in direction

For rectilinear motion, direction is the constant straightline path algebraic problem

Integration of basic differential relations



( )

2 2

1 1

2 2

1 1

2 2

1 1

s t

2 1s t

v t

2 1v t

v s2 22 1

v s

ds vdt, s s area under v-t curve

dv adt, v v area under a-t curve

vdv ads, v v / 2 area under a-s curve

= − =

= − =

= − =

∫ ∫

∫ ∫

∫ ∫

Relationships among severalmotion quantities

graphic/numericalvs. algebraic approach



( ) ( )d d

kinetic relationdt dt

dt dts v a v, s, t F v, s, t←

∫ ∫

common problems: know a, find s by integration



( )

( )

o

o o

o

o ov t

ov 0

v s2 2

o ov s

s t t2

o o os 0 0

at the beginning, t 0, s s , v v

at time t, dv a dt v v at

vdv a ds v v 2a s s

ds vdt v at dt s s v t at / 2

= = =

= → = +

= → = + −

= = + → = + +

∫ ∫

∫ ∫

∫ ∫ ∫

a) a = constant, e.g. G-force, dry friction force



( ) ( )

( )

( )

o

o

v t t

ov 0 0

s t t t t

o o os 0 0 0 0

o o o

dv f t dt v v f t dt

ds vdt s s vdt s v t f t dtdt

or s f t with i.c. t 0, s , v

= → = +

= → = + = + +

= =

∫ ∫ ∫

∫ ∫ ∫ ∫ ∫

b) a = f(t), e.g. synthetic force, piston force



[ ] ( ) ( ) ( )o

dtt v inv

0 v

dva dv/dt t dt v g t s h tf v

∫= = = → = → =∫ ∫

c) a = f(v), e.g. viscous drag force, damping force

[ ] ( ) ( ) ( )o o o

v s v

ov s v

v vvdv ads dv ds s s dv g vf v f v

= = → = + =∫ ∫ ∫

or



( ) ( ) ( )

[ ] ( ) ( )

o o o

o

v s s2 2

ov s s

s inv

s

vdv f s ds v v 2 f s ds v g s

dsv ds/dt t s h tg s

= → = + → =

= = → =

∫ ∫ ∫

∫

d) a = f(s), e.g. spring force, attraction force



P. 2/19 Small steel balls fall from rest through theopening at A at the steady rate of 2 per second.Find the vertical separation h of two consecutiveballs when the lower one has dropped 3 meters.Neglect air resistance.



P. 2/19

[ ]2

o o o

o o

2

2l l

u l

2u u

a s a g from gravitational force, downward

v v gt and s s v t gt / 2v 0 and define s 0

s gt / 2lower ball: 3 gt / 2, t 0.782 supper ball: t t 0.5 0.282 s

s 3 h gt /

= =

= + = + +

= =

∴ =

= == − =

= − = 2 h 2.61 m→ =

3 m

(3-h) m

@ tl

@ tu



P. 2/20 In traveling a distance of 3 km between pointsA and D, a car is driven at 100 km/h from A to Bfor t seconds and at 60 km/h from C to D alsofor t seconds. If the brakes are applied for 4 sbetween B and C to give the car a uniformdeceleration, calculate t and the distance sbetween A and B.



P. 2/20

[ ]

( )

B

A

D

C

C

B

s t/3600

Bs 0

s t/3600

Cs 0

s 4/3600

s 0

C B

B

v ds/dt ds vdt, s t/36

ds vdt, 3 s t/60

ds vdt area under v-t curve,

1s s 4 / 3600 100 60 4 / 452

t 65.5 sec, s s 1.819 km

= = =

= − =

= =

− = × × + =

= = =

∫ ∫

∫ ∫

∫ ∫

t

t

v

a

A B C D

t sec t sec4 sec



P. 2/24 The 350-mm spring is compressed to a 200-mm length, whereit is released from rest and accelerates the sliding block A.The acceleration has an initial value of 130 m/s2 and thendecreases linearly with the x-movement of the block, reachingzero when the spring regains its original 350-mm length.Calculate the time t for the block to go a) 75 mm and b) 150mm.



P. 2/24

130a x 866.7x0.15

= − = −

[ ]

( )

[ ]

( ){ }

v x

0 0.15

2 2 2

t x

0 0.15

1

vdv ads vdv ads

v 866.7x 19.5, v 29.44 0.0225 x assume block move

dsv ds/dt dtv

t 0.034 sin x/0.15 / 2

@ x 0.075 m, t 0.0356 s@ x 0 m, t 0.0534 s

π

−

−

−

= =

= − + = − →

= =

= +

= − == =

∫ ∫

∫ ∫

[ ]a s x 866.7x 0solution of the unforced harmonic equation

x Asin t Bcos t, 866.7 29.44 rad/si.c.: t 0, x 0.15 m, x 0 m/s B 0.15, A 0

x 0.15cos 29.44t

ω ω ω

= + =

= + = == = − = → = − =

∴ = −

a

x

130

-0.15 0



P. 2/26 A train that is traveling at 130 km/h applies its brakes as itreaches point A and slows down with a constant deceleration.Its decreased velocity is observed to be 96 km/h as it passesa point 0.8 km beyond A. A car moving at 80 km/h passespoint B at the same instant that the train reaches point A.In an unwise effort to beat the train to the crossing, the driver‘steps on the gas’. Calculate the constant acceleration a thatthe car must have in order to beat the train to the crossing by4 s and find the velocity v of the car as it reaches the crossing.



P. 2/26

( )

( )

2 2 2o 0 o o o

2 2 2

2o

const acceleration: v v at, v v 2a s s , s s v t at / 2

Train: 96 130 2a 0.8, a 4802.5 km/h1.6 130t 4802.5t / 2, t 0.0189 h or 68.11 s check with v v atCar: to beat the train by 4 sec t

= + = + − = + +

= + × = −

= − = = +

→ =2 2 2

64.11 s or 0.0178 h2 80 0.0178 a 0.0178 / 2, a 3628.3 km/h 0.28 m/sv 80 3628.3 0.0178 144.6 km/h 40.2 m/s= × + × = == + × = =



P. 2/40 The horizontal motion of the plunger and shaft is arrested bythe resistance of the attached disk that moves through theoil bath. If the velocity of the plunger is vo in the position Awhere x = 0 and t = 0, and if the deceleration is proportionalto v so that a = -kv, derive expressions for the velocity vand position coordinate x in terms of the time t. Also expressv in terms of x.



P. 2/40

[ ]

[ ] ( )

[ ]

o

o

v tkt

ov 0

x tkt kto

o0 0

v x

ov 0

dva dv/dt dt, v vkv

vv dx/dt dx v dt, x 1k

vdvvdv ads ds, v v kxkv

e

e e

−

− −

= = =−

= = = −

= = = −−

∫ ∫

∫ ∫

∫ ∫



P. 2/44 The electronic throttle control of a model trainis programmed so that the train speed varieswith position as shown in the plot. Determinethe time t required for the train to completeone lap.



P. 2/44

( ) ( )[ ] ( )

( )t 0.125

0 0.25

0 2 km: constant velocity, t s/v 2 / 0.25 8 sec2 2 / 2 km: dv/ds 0.125 / / 2 0.25 /

vdv ads, a dv/dt v dv/ds dv/dt

0.25 / dt dv/v, t 8.71 sec

lap time 8 2 8.71 4 50.84 sec

π π π

πΔ

− = Δ = =

− + = − = −

= = =

− = Δ =

= × + × =

∫ ∫



P. 2/50 A bumper, consisting of a nest of three springs, is used toarrest the horizontal motion of a large mass that is traveling at40 m/s as it contacts the bumper. The two outer springs causea deceleration proportional to the spring deformation. Thecenter spring increases the deceleration rate when thecompression exceeds 0.5 m as shown on the graph. Determinethe maximum compression x of the outer spring.



P. 2/50

[ ]

( ) ( )( )

0

402

vdv ads vdv area under a-s curve

0 40 1 10.5 1000 x 0.5 1000 1000 4000 x 0.52 2 2

x 0.831 m

= =

−= − × × − × − × + + −

=

∫


2.3 Plane Curvilinear Motion: motion along a curvedpath that lies in a single plane

Vector quantity is independent of any particularcoordinate system

2.3 Plane Curvilinear Motion


Time derivative of a vector (described in fixed coord.)change in both magnitude and direction


at time t, the particle is at A located by at time t t, the particle moves to B located by 'displacement (vector) during time t is (independent of coordinate system)distance traveled (scal

+ Δ = ΔΔ Δ

rr r + r

rar) during time t is s (measured along the path)Δ Δ



av

t 0

average velocity, / taverage speed s / t

dinstantaneous velocity, limt dt

includes the effect of change both in magnitude and direction of as t 0, direction of approaches that

Δ →

= Δ Δ

= Δ Δ

Δ= = =

Δ

Δ → Δ

v r

r rv r

v rr

av

of the tangent to the pathaverage velocity velocity,

is always a vector tangent to the path

Consider only the magnitude of the velocitydsspeed, v sdt

as t 0, A A'average speed speed,

∴ → →

∴

= = =

Δ → →

∴ → Δ →Δ

v vv

v

r av avs, , and v→ →v v v



Magnitude of the derivativedmagnitude of the velocity s v speeddt

Derivative of the magnituded dr r rate at which the length of position vector is changingdt dt

Derivative of the direction

Der

= = = = = =

= = =

r v r

rr

ivative of the magnitude and Derivative of the direction of the vectorcontribute to Derivative of that vector



av

t 0

average acceleration, / tdinstantaneous acceleration, lim

t dt

includes the effect of change both in magnitude and direction of

Because the magnitude at any point can be arbitrar

Δ →

= Δ Δ

Δ= = =

Δ

a vv va v

a v

v y, generallythe direction of the acceleration is niether tangent nor normal to the pathbut its normal component always points towardthe center of curvature of the path



The acceleration bears the same relation to the velocityas the velocity bears to the position vector.



Derivatives and Integrations of vectors

( )

( )

( )

( )( ) ( ) ( )

x y z

x y z

d P P Pdtd u

u udt

ddt

ddt

if x, y, z and an element of volume, d dxdydz

d V x, y, z d V x, y, z d V x, y, z d

τ

τ τ τ τ

=

= +

=

×= × ×

=

= + +∫ ∫ ∫ ∫

P P = i + j+ k

PP P

P QP Q + P Q

P QP Q + P Q

V = V

V i j k

ii i



Three common coordinate systems for plane curvilinearmotion – rectangular, normal & tangential, and polar

Choose the appropriate coordinate system accordingto the manner in which the motion is generatedor by the form in which the data are specified


2.4 Rectangular Coordinates (x-y)

Suit for motions where the x- and y- componentsof the acceleration are independently generatedor determined, i.e. x- and y- coordinates at a specificpoint are related by the same instant of time only.It looks like the superposition of two perpendicularrectilinear motions, in x- and y- directions,simultaneously. Their combination generate thecurvilinear motion.




( ) ( )x y x y

x y x y

y x

2 2x y

2 2x y

x y x t y tx y v v (v and v are independent)

x y a a (a and a are independent)

direction of velocity, tan v / v dy / dx

speed, v v v

a a a

θ = =

= +

= +

r = i + j = i + jv = r = i + j = i + j

a = v = r = i + j = i + j

( ) ( )( )

1 2If x t and y t , elimination of time t between these two

parametric equations gives the equation of the curved path y x

f f

f

= =

=


Projectile motion: motion of the thrown object

Neglect aerodynamic drag and the earth’s curvature androtation, and assume the altitude range is small enough


x ya 0 and a g= = −



( ) ( )( ) ( )

( ) ( )

( ) ( )

x x x o xo o

2y y y o yo o

22y y oo

a 0 v v x x v t

a g v v gt y y v t gt / 2

v v 2g y y

x- and y- motions are independentelimination of time t of x t and y t gives the parabolic path

= = = +

= − = − = + −

= − −



P. 2/73 A particle is ejected from the tube at A with a velocity v at anangle θwith the vertical y-axis. A strong horizontal wind givesthe particle a constant horizontal acceleration a in the x-direction.If the particle strikes the ground at a point directly under itsreleased position, determine the height h of point A. Thedownward y-acceleration may be taken as the constant g.



P. 2/73

( )

2x x

2y y

since a is constant, the instant formulas can be used1a a, v vsin at, x vtsin at2

1a g, v vcos gt, y vtcos gt2

when x 0 & y h,0 t at/2 vsin , t 2vsin /a

substitute t into y-coord

θ θ

θ θ

θ θ

= = − + = − +

= = + = +

= =

= − =22v gequation, h sin cos sin

a aθ θ θ⎛ ⎞= +⎜ ⎟⎝ ⎠

ay = g

ax = a



P. 2/76 Electrons are emitted at A with a velocity v at the angle θintothe space between two charged plates. The electric fieldbetween the plates is in the direction E and repels the electronsapproaching the upper plate. The field produces an accelerationof the electrons in the E-direction of eE/m, where e is theelectron charge and m is its mass. Determine the field strengthE that will permit the electrons to cross one-half of the gapbetween the plates. Also find the distance s.



P. 2/76

( )

( )

( )

x y

2 2y 0 y o

22 2

2

Ea 0 am

v v 2a y y

at the peak, y b/2E mu0 usin 2 b/2, E sin

m b

1 Ex utcos y utsin t2 m

sat s, 0 : s utcos , t and substitue into y-equationucos

s0 stan 1 tan , s 02b

e

ee

e

θ θ

θ θ

θθ

θ θ

= = −

= + −

=

= − × =

= = − ×

= =

⎛ ⎞= − =⎜ ⎟⎝ ⎠

2b, tanθ

ay = -eE/m

u

x

y



P. 2/78 Water is ejected from the nozzle with a speed vo = 14 m/s.For what value of the angle θwill the water land closest tothe wall after clearing the top? Neglect the effects of wallthickness and air resistance. Where does the water land?

(1)

(2)



P. 2/78

( )( )

( )x o

2 2 2o y yo

2 2 2 2

Treat water as stream of particles. At one particular particle,@ just above the wall 19, 1

x v t 19 14t cos

1 1 1y y v t a t 1 0.3 14t sin gt , 0.7 gt 14t sin2 2 2

114 t 19 0.7 gt2

θ

θ θ

⎡ ⎤= =⎣ ⎦⎡ ⎤= + + = + − + =⎢ ⎥⎣ ⎦

⎛= + +

( )

2

2

, t 2.14, 1.81 sec and 50.64 , 41.43

50.64 makes the water land closest to the wallwater will land at x, 0

10 0.3 14t sin 50.64 gt , t 2.234 sec2

x 14t cos50.64 19.835 mwater lands 0.835

θ

θ

⎞ = = ° °⎜ ⎟⎝ ⎠

= °

= + − =

= =∴ m to the right of B



P. 2/94 A projectile is ejected into an experimental fluid at time t = 0.The initial speed is vo and the angle to the horizontal is θ.The drag on the projectile results in an acceleration term

, where k is a constant and is the velocity of theprojectile. Determine the x- and y-components of both thevelocity and displacement as functions of time. What is theterminal velocity? Include the effects of gravitational acceleration.

D k= −a v v



P. 2/94 ( )

( )

( ) ( )

x

o

y

o

x y

xx x

v tkt

x x oxv cos 0

t xkt kto

o0 0

yy y

v tkt

y y oyv sin 0

o

k k v v

dva kvdt

1 dxdv dt, v v coskv dt

v cosv cos dt dx, x 1k

dva kv g

dt1 g g dydv dt, v v sin

kv g k k dt

gv sink

e

e e

e

e

θ

θ

θ

θθ

θ

θ

−

− −

−

− −

= − =

= = =−

= = −

= − − =

⎛ ⎞= = + − =⎜ ⎟− − ⎝ ⎠

⎛ ⎞+⎜ ⎟⎝ ⎠

∫ ∫

∫ ∫

∫ ∫

a = v = i + j

( )

( ) ( )

ytkt kt

o0 0

x yt t

g 1 g gdt dy, y v sin 1 tk k k k

gterminal velocity: v 0, vk

eθ

∞ ∞

− −⎧ ⎫ ⎛ ⎞− = = + − −⎨ ⎬ ⎜ ⎟⎝ ⎠⎩ ⎭

= = −

∫ ∫



P. 2/95 A projectile is launched with speed vo frompoint A. Determine the launch angle θthatresults in the maximum range R up the incline ofangle α (where 0 ≤ α≤ 90°). Evaluate yourresults for α = 0, 30, and 45°.



P. 2/95 x x o o2

y y o o

oo

2

oo o

2

2 2o

2 2 2o o

a 0, v v cos , x v t cos

a g, v v sin gt, y v t sin gt / 2

Rcosat B, Rcos v t cos tv cos

Rcos g RcosRsin v sinv cos 2 v cos

gRcossin cos tan2v cos

2v cos tan 2v sin

θ θ

θ θ

αα θθ

α αα θθ θ

αα α θθ

θ α

= = =

= − = − = −

= → =

⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= −

=

( ) ( )

( )

( )

2 2o o

2o

1 1

cos gRcosdRchange in causes change in R, max R when 0d

2v tan 2cos sin d v 2cos 2 d gcos dR

2vdR cos 2 sin 2 tan 0d gcostan 2 1/ tan

1 12 tan 180 tantan tan

180 90

θ θ α

θθ

α θ θ θ θ θ α

θ θ αθ α

θ α

θα αα

− −

−

∴ =

− = −

= + =

= −

⎛ ⎞ ⎛ ⎞= − = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= − − 9090

2

ααθ

= +

+=



P. 2/96 Determine the equation for the envelope a of the parabolictrajectories of a projectile fired at any angle but with a fixedmuzzle velocity u. (Hint: Substitute m = tanθ, where θis thefiring angle with the horizontal, into the equation of the trajectory.The two roots m1 and m2 of the equation written as a quadraticin m give the two firing angles for the two trajectories shownsuch that the shells pass through the same point A. Point Awill approach the envelope a as the two roots approach equality.)Neglect air resistance and assume g is constant.



P. 2/96

( )

x x2

y y

2

2 2

2 2 2

2 2 2

a 0, v ucos , x utcos

a g, v usin gt, y utsin gt / 2

eliminate t to get the trajectory equation

g xy xtan2 u cos

given point x, y , find by letting m tan

1 tan sec 1 m

gx m 2xu

θ θ

θ θ

θθ

θ θ

θ θ

= = =

= − = − = −

⎛ ⎞= − ⎜ ⎟

⎝ ⎠=

+ = = +

−

∵

( )( )

( )

2 2m 2yu gx 0

two roots point x, y can be reached from two distinct paths

point x, y will approach the envelope a as those two pathsapproach each other two distinct roots become two repeated roots

disc

+ + =

∴

→∴

∵

( ) ( )

( )

22 2 2 2

2 2

2

riminant must be zero

2xu 4gx 2yu gx 0

u gxy2g 2u

point x, y on the envelope must obey this relation envelope equation

− + =

= −

∴ →


2.5 Normal and Tangential Coordinates (n-t)

It is a moving coordinate system, along the pathwith the particle. The +n direction points toward thecenter of curvature.


n

t

At point A,unit vector in the n-directionunit vector in the t-directionradius of curvature of the pathρ

==

=

ee



( )

t t

t t

position of the particle 0, 0ds dvdt dtv along the t-axis

0 when particle moves along t directiond v vdt

ρ β

ρβ

β

=

= =

=

> +

= +

v = e e

va = e e



t

t t n n t n

tn n t

n t2

2n

t

t

x y t

x y n n t t

Determine

d d dddetour : and d

v v

va v

a v svdv a ds

v v v

a a a a

β β β

ββ

β

β ρβρ

ρβ ρβ

= = → =

= = −

∴ +

= = =

= = = +=

+

e

e e e e e eee e e

a = e e

v = i + j = e

a = i + j = e e



n

n 2n

n

t

tt

t

change in direction of d vda vdt dt

component a always directed toward the center of curvaturechange in magnitude of d dva v sdt dt

component a will be in t direction if spee

β β ρβ

=

= = = =

=

= = = =

+

a vv

a vv

t

d v is increasingcomponent a will be in t direction if speed v is decreasing

has the direction along the motion of the object may not have the direction along the motion of the object

−

va



P. 2/118 The camshaft drive system of a four-cylinderautomobile engine is shown. As the engineis revved up, the belt speed v changesuniformly from 3 m/s to 6 m/s over a 2 secondinterval. Calculate the magnitudes of theaccelerations of points P1 and P2 half waythrough this time interval.



P. 2/118

( )

( )

2t

22 2 2 2 2P1 n t

2P2 t

dv vuniform speed change a 6 3 / 2 1.5 m/sdt t

belt speed v 4.5 m/s at the time half of the interval

a a a 4.5 / 0.06 1.5 337.5 m/s

a a 1.5 m/sthe sprocket and the belt must be very strong

Δ→ = = = − =

Δ=

= + = + =

= =∴ !



P. 2/120 A baseball player releases a ball with theinitial conditions shown in the figure.Determine the radius of curvature of thetrajectory (a) just after release and (b) at theapex. For each case, compute the time rateof change of the speed.



P. 2/120

( )

2 2

n

2t

22

n

2t

Just after release:v 30a gcos30 , 105.9 m

v a gsin30 4.905 m/s

At the apex:

30cos30va g , 68.8 m

v a 0 m/s

ρρ ρ

ρρ ρ

= = = =

= = − = −

= = = =

= =

v = 30 m/s

a = g

n

t

v = 30cos30 m/sa = g

n

t



P. 2/125 Pin P in the crank PO engages the horizontalslot in the guide C and controls its motion onthe fixed vertical rod. Determine the velocity

and the acceleration of guide C for agiven value of the angle if (a) and

and (b) if and .

yθ θ ω=

0θ = 0θ = θ α=

y



P. 2/125

2n t

2

(a) , 0for pin P: v r , a r , a 0guide C must have its velocity and acceleration, y and y,transmitted by those of pin P along y direction

y r sin , y r cos

(b) 0, for pin P:

θ ω θ

ω ω

ω θ ω θ

θ θ α

= =

= = =

∴ = =

= =

n t v 0, a 0, a rguide C must have its velocity and acceleration, y and y,transmitted by those of pin P along y direction

y 0, y r sin

α

α θ

= = =

∴ = =

θ

θ

θ

θ

tn

tn

tn

tn

a = at = rα

v = 0

a = an = rω2

v = rω



P. 2/128 The pin P is constrained to move in the slotted guides thatmove at right angles to one another. At the instant represented,A has a velocity to the right of 0.2 m/s which is decreasingat the rate of 0.75 m/s each second. At the same time, B ismoving down with a velocity of 0.15 m/s which is decreasingat the rate of 0.5 m/s each second. For this instantdetermine the radius of curvature ρ of the path followed by P.Is it possible to determine also the time rate of change of ρ.



P. 2/128

t n t

Guides A and B move perpendicular to each other,motion of guides A and B are independent

and their motions are imparted to pin P.0.2 0.15

0.75 0.5in n-t description

0.8 0.6 , and al

∴

= −−

= − ⊥

v i ja = i + j

n i j n n

∵

( )

n

2t t t

n t2 2

n

t

ong

a 0.9 m/s , 0.72 0.540.03 0.04

v 0.25a , 1.25 m0.05

appears only in eq.: a v v/

which has 2 unknowns: and

cannot be determined unt

ρρ

ρ ρβ ρβ ρβ ρ ρ

ρ β

ρ

= = − = − += − = − −

= = =

= = + = +

∴

a

a n a i ja a a i j

i

il is knownβ

y

x

tn

van

at



P. 2/129 As a handling test, a car is driven through theslalom course shown. It is assumed that thecar path is sinusoidal and that the maximumlateral acceleration is 0.7g. If the testers wishto design a slalom through which the maximumspeed is 80 km/h, what cone spacing Lshould be used?



P. 2/129

n

2

2

n n min

max a 0.7g and max speed 80 km/h

sinusoidal path y 3sin x 3sin xL

y' 3 cos x and y'' 3 sin xvmax a at peak of the wave a and at peak

by sense: fast car needs big L use max speed to

πω

ω ω ω ω

ρρ

= =

⎛ ⎞→ = = ⎜ ⎟⎝ ⎠

= = −

=

→

∵

( ) ( )

3/ 22

2min

2

2

find lower bound of L

1 y'y''

At peak, 80 10 / 36 / 0.7g 71.9 m

x n /2, n 1, 3, y' 0, y'' 3171.9 , 0.0681 L 46.14 m

3 L

ρ

ρ ρ

ω π ωπω

ω

⎡ ⎤⎡ ⎤+⎣ ⎦⎢ ⎥=⎢ ⎥⎣ ⎦

= = × =

= = ± ± → = =

= = = → =

… ∓

∓∓



P. 2/130 A particle starts from rest at the origin andmoves along the positive branch of the curvey = 2x3/2 so that the distance s measured fromthe origin along the curve varies with the time taccording to s = 2t3, where x, y, and s are inmillimeters and t is in seconds. Find themagnitude of the total acceleration of theparticle when t = 1 s.



P. 2/130

( )

( ) ( ) ( )

3 2t

3/ 22

2n

3/2

2 2 2

x

0

s 2t , v s 6t , a v 12t

1 y'a v / &

y''3y 2x , y' 3 x , y''

2 xx must be known to determine

ds dx dy 1 y' dx

t 0 1, s 0 2, x 0 x

2 1 9xdx, x 0.913 mm y 1.746 mm

At t 1 s, v 6 m

ρ ρ

ρ

= = = = =

⎡ ⎤+⎣ ⎦= =

= = =

∴

= + = +

= → = → = →

= + = =

= =

∫ ∫ ∫

∫2

t2 2 2

n

m/s, a 12 mm/s , y' 2.8665, y'' 1.57, 17.8 mm

a 6 /17.8 2.02 mm/s a 12.17 mm/s

ρ= = = =

= = → =


2.6 Polar Coordinates (r-θ)

The particle is located by the radial distance r froma fixed pole and by an angle θto fixed radial line.



r

r r

r r

r r r

r

2 2r

rfrom the figure, d d and d d

and

r r r rv r due to the stretch of

v r due to the rotation of

v v v

θ θ

θ θ

θ

θ

θ

θ θ

θ θ

θ

θ

= = −

∴ = = −

+ = +=

=

= +

r = ee e e e

e e e e

v = r = e e e er

r



( ) ( ) ( ) ( )

( )

2r r r

2r

2

2 2r

r r

r r r r r r r r 2r

a r r1 da r 2r r , 2r combines two effectsr dt

a a aNote: a v and a v , i.e. we must also account for the

change i

θ θ θ θ

θ

θ

θ θ

θ θ θ θ θ θ

θ

θ θ θ θ

+ + + + = − + +

= −

= + =

= +

≠ ≠

a = v = e e e e e e e

r

2r

n direction of both v and v , which are r along

and r along , respectivelyθ θθ

θ−

e

e



P. 2/138 The rocket is fired vertically and tracked bythe radar shown. When θreaches 60°,other corresponding measurements give thevalues r = 9 km, = 21 m/s2, and = 0.02 rad/s.Calculate the magnitudes of the velocity andacceleration of the rocket at this position.


r θ


P. 2/138


r

2 2 2r

and of the rocket is in the vertical direction

v r 180 m/s vsin30 v 360 m/sv r vcos30 311.77 m/s

a r r 17.4 m/s acos30 a 20.09 m/s

θ θ

θ

= = = → == = =

= − = = → =

v a

r

θ

30


P. 2/141 Link AB rotates through a limited range of theangle β, and its end A causes the slotted linkAC to rotate also. For the instant representedwhere β= 60°and = 0.6 rad/s constant,determine the corresponding values of

.


β

r, r, , and θ θ


P. 2/141


t

2 2n

Description of position, velocity, and acceleration of Ain r- when they are given in n-t coordinate systemv 0.15 0.6 0.09 m/s along t-direction

0 a 0

a a 0.054 m/s along n-direction

θ

ρβ

β

ρβ

= = × = +

= → =

= = = +

( ) ( )

r r r

2r r r

2 2

From geometry, r 0.15 m and 60

vcos30 vsin30 r r

r 0.078 m/s 0.3 rad/s

asin30 acos30 r r r 2r

r 0.0135 m/s 2.31E-4 rad/s

θ θ θ

θ θ θ

θ

θ

θ

θ θ θ

θ

= = °

+ = − = +

= = −

+ = − − = − + +

= − =

v = v v e e e e

a = a a e e e e

r

t

n

θ

30 30


P. 2/149 The slotted arm OA forces the small pin tomove in the fixed spiral guide defined byr = Kθ. Arm OA starts from rest at θ=π/4and has a constant counterclockwise angularacceleration . Determine the magnitudeof the acceleration of the pin when θ= 3π/4.


θ α=


P. 2/149


( )2 2o o

2

pin is constrained to move in the guidepath of the pin: r K

and differential relations r K and r K hold

/ 4 3 / 4, 0 , constant

2

At 3 / 4, ,

θ

θ θ

θ π π θ θ θ α

θ θ θ θ θ

θ π θ πα θ

∴ =

= =

= → = → = =

⎡ ⎤= + −⎣ ⎦= =

( ) ( )2r

r 3K /4, r K , r K

r r r 2r

a 10.753Kθ

α

π πα α

θ θ θ

α

=

= = =

− + +

=

a = e e


P. 2/151 The circular disk rotates about its center O with a constantangular velocity and carries the two spring-loadedplungers shown. The distance b that each plunger protrudesfrom the rim of the disk varies according to b = bosin2πnt,where bo is the maximum protrusion, n is the constantfrequency of oscillation of the plungers in the radial slots,and t is the time. Determine the maximum magnitudes of ther- and θ-components of the acceleration of the ends Aof the plungers during their motion.


ω θ=


P. 2/151


( )

( )

( )

( )

o o o

2o o

2 2 2 2 2r o o

o

2 2 2 2r o omax max

location of point A described by r, r r b r b sin 2 nt

r 2 nb cos 2 nt, r 2 n b sin 2 nt

, 0

a r r r 4 n b sin 2 nt

a r 2r 4 n b cos 2 nt

a r 4 n b a 4θ

θ

θπ

π π π π

θ ω θ

θ ω π ω π

θ θ π ω π

ω π ω

= + = +

= = −

= =

= − = − − +

= + =

= + + = on bπ ω


P. 2/155 The small block P starts from rest at time t = 0at point A and moves up the incline withconstant acceleration a. Determine as a function of time.


r and θ

s


P. 2/155


( )

2

2

2

22 2 2 2 2 4

22 3

22 2

1block moves with constant acceleration s at2

atby geometry, x R scos R cos , x atcos2

aty ssin sin , y atsin2

ar x y R Rat cos t4at 2Rcos at

2rr 2Ratcos a t , ra2 R Rat cos t4

α α α

α α α

α

αα

α

→ =

= + = + =

= = =

= + = + +

+= + =

+ + 4

22

22

2 2

222 2 4

y xy yxtan , secx x

1 rseccos x

xy yx Ratsinar R Rat cos t4

θ θ θ

θθ

αθα

−= =

= =

−= =

+ +


P. 2/158 The block P slides on the surface shown withconstant speed v = 0.6 m/s and passes point Oat time t = 0. If R = 1.2 m, determine thefollowing quantities at time t = 2(1+π/3):


r, , r, , r, and θ θ θ


P. 2/158


( )

( )

2 1 /3s

0 0

2 2 1

r

dsv ds vdt, s 1.2 0.4dt

0.4block P moves up the quarter for / 3 601.2

at this position, x R Rcos30 2.239 m y R Rsin30 0.6

r x y 2.318 m tan y/x 15v vcos45 r 0.424

π

π

π π

θ

+

−

⎡ ⎤= = = +⎢ ⎥⎣ ⎦

= = °

= + = = − =

= + = = = °

= = =

∫ ∫

2 2

t n

2 2r

2

m/s

v vsin45 r , 0.183 rad/s

v 0.6a v 0 a 0.3 a1.2

a 0.3cos 45 r r , r 0.134 m/s

a 0.3sin 45 r 2r , 0.0245 rad/s

θ

θ

θ θ

ρθ

θ θ θ

= = =

= = = = = =

= − = − = −

= = + =

15

15

30t

n r

θ

v

a


P. 2/161 The slotted arm OA oscillates about O within the limits shownand drives the crank CP through the pin P. For an interval ofthe motion, , a constant. Determine the magnitude ofthe corresponding total acceleration of P for any value of θwithin the range for which . Use polar coordinates rand θ. Show that the magnitudes of the velocity andacceleration of P in its circular path are constant.


Kθ =

Kθ =


P. 2/161


2 2 2n t

K 0

2 2K 0motion of pin P is the circular path with radius b

v 2bK constanta v / 4bK a v 0, a 4bK constant

θ θ

β θ β β

ρβ

ρ

= =

= = =

= = =

= = = = = =

rb

b

θ

θβ


P. 2/163 The earth satellite has a velocity v = 17,970 km/h as it passesthe end of the semiminor axis at A. Gravitational attractionproduces an acceleration a = ar = -1.556 m/s2 as calculatedfrom the gravitational law. For this position, calculatethe rate at which the speed of the satellite is changing andthe quantity .


rv


P. 2/163


2t

2r

2

a v acos60 0.778 m/s

a a 1.556 r r but is unknown too

v r vcos30, 2.7E-4 rad/s, r 0.388 m/sθ

θ θ

θ θ

= = − = −

= = − = −

= = = = −

r

θ

t

n

60°

a

v


P. 2/164 Pin A moves in a circle of 90 mm radius ascrank AC revolves at the constant rate =60 rad/s. The slotted link rotates about point Oas the rod attached to A moves in and outof the slot. For the position β= 30°,determine .


β

r, r, , and θ θ


P. 2/164


2 2 2

22A

A n

r A

A

2r n

r 300 90 2 300 90cos30, r 226.57 mm90 r , 11.456

sin sin30vv 5.4 m/s and a a 324 m/s

v r v cos 48.54 3.575 m/s

v r v sin 48.54, 17.86 rad/s

a a cos 41.456 r r , r 315 m/sθ

θθ

ρβρ

θ θ

θ

= + − × × =

= = °

= = = = =

= = =

= = =

= = − = 2

2na a sin 41.456 r 2r , 1510 rad/sθ θ θ θ= − = + = −

θr

v

a

r

n

tθ

30


P. 2/243 If the slotted arm is revolving CCW at the constant rate of40 rev/min and the cam is revolving clockwise at the constantrate of 30 rev/min, determine the magnitude of the accelerationof the center of the roller A when the cam and arm are in therelative position for which θ= 30°. The limacon has thedimensions b = 100 mm and c = 75 mm. (Caution: Redefinethe coordinates as necessary after noting that the θin theexpression r = b – ccosθis not the absolute angle appearingin Eq. 2/14.)



P. 2/243


( )/ 6 t

0 0

0.125

0 0

pin A must always be in the slot 40 rev/min CCW

if is the angle the cam is revolving at the rate 30 rev/min CW,then r b ccos

d d dt, t 0.125 sdt

d dt,

π

β

θ

β βθ β

θω θ θ

β β

→ =

=

= − +

⎡ ⎤= = =⎢ ⎥⎣ ⎦

=

∫ ∫

∫ ∫

( )( ) ( )

( ) ( )

( )( )

2

22 2r

2

2 2r

0.3927 rad 22.5

0 and 0r 0.1 0.075cos 30 22.5 54.3 mm

r c sin 0.436

r c cos 2.453

a r r 2.453 0.0543 40 2 / 60 1.5 m/s

a r 2r 2 0.436 40 2 / 60 3.653 m/s

a a a

θ

θ

β

θ β

θ β θ β

θ β θ β

θ π

θ θ π

= = °

= =

= − + =

= + + =

= + + =

= − = − × =

= + = × × × =

= + 23.95 m/s=


2.7 Relative Motion (Translating Axes)

It is in many cases easy and practical to observemotion of an object w.r.t. a moving reference system.And its absolute motion can be determined bycombining the relative motion with the absolutemotion of the moving reference frame.

Fixed coordinate system ? has no motion in space-- earth-ground-- non-rotating coordinate system with origin on the

earth’s axis of rotation-- non-rotating coordinate system fixed to the sun



At this moment, we will study the relative plane motionwhich employs the pure-translation moving referencesystem. That is the relative motion will be observed onthe moving reference frame that has no rotation.


Observe motion of A from point B. Measurements are madereferenced with the moving coordinate.Find the absolute motion of A described in fixed coordinate.

Two frames involved:1.

pure translating

chosen fixed frame X-Y, e.g.

2. pure translating moving frame x-y, e.g.: Coordinates of X-Y and x-y may not parallel to each other

++Separation of the frame, vector, and its description++Consistency of th

Note

e description for the vector quantities in a particular equation++Transformation between two descriptions



A B

A/B

A B A/B A B A/B A B A/B

A

and are measured referenced with the fixed coordinate frame is measured referenced with the moving coordinate frame.

describe

pure translating

= + = + = +

r rr

r r r v v v a a a

r /B

A/B A/B A/B

A/B B/A A/B B/A

in a specific pure translating moving frame x-yx y x y x y

For the relative motion using the pure translating coordinate system

= = =

= − = −

r i + j v i + j a i + j

r r v v A/B B/A

B

A A/B

if the pure-translating moving frame has constant velocity, determination of the absolute acceleration can be made

on the system, a pure-trainertial

= −

=∴ →

a a

a 0a = a

nd

nslating system that has no acceleration Newton's 2 law holds in the as well as in the reference frame

inertialfixed

→



P. 2/191 The car A has a forward speed of 18 km/h andis accelerating at 3 m/s2. Determine thevelocity and acceleration of the car relative toobserver B, who rides in a nonrotating chairon the ferris wheel. The angular rateω= 3 rev/min of the ferris wheel is constant.



P. 2/191

[ ]

( )

[ ]

A B A/B A

B

A/B A B

A B A/B

observer B rides in a nonrotating chaircar A is observed through the moving coord frame

10 18 5 m/s36

cos 45 sin 45 2 2 m/s3 2 m/s

pure translating

ρβ

→

= × =

= − = −

= − = +

v = v + v v i i

v i j i j v v v i j

a = a + a

( )

2A

2 2B

2A/B A B

3 m/s

cos 45 sin 45 0.628 0.628 m/s

3.628 0.628 m/s

ρβ

=

= − − = − −

= − =

a i

a i j i j

a a a i + j



P. 2/193 Hockey player A carries the puck on his stickand moves in the direction shown with a speedvA = 4 m/s. In passing the puck to his stationaryteammate B, by what shot angle α should thedirection of his shot trail the line of sight if helaunches the puck with a speed of 7 m/srelative to himself?



P. 2/193

( )

P A P/A

player A saw the puck moves in 45 directionbut in fact, the puck moved to stationary player Bwhich stood in 45 direction

7 4 , 23.8sin 45 sin

α

αα

+ °

°

= = °

v = v + v



P. 2/204 The aircraft A with radar detection equipment is flyinghorizontally at 12 km and is increasing its speed at the rate of1.2 m/s each second. Its radar locks onto an aircraft flyingin the same direction and in the same vertical plane at analtitude of 18 km. If A has a speed of 1000 km/h at the instantthat θ= 30°, determine the values of at this sameinstant if B has a constant speed of 1500 km/h

r

θ

r and θ



P. 2/204

B A

B/A B A r

r B/A

B/A

2B A

B/A B A

r 6000 / sin 30 12000 m10 101500 416.67 m/s 1000 277.78 m/s36 36

138.89 m/s r rv r v cos30 r 120.28 m/s

v r v sin 30 5.787E-3 rad/s

1.2 m/s

θ

θ

θ

θ θ

= =

× = × =

− = = +

= = → =

= = − → = −

= =

−

v = = i v i

v = v v i e e

a 0 a i

a = a a ( ) ( )2 2r

2 2r B/A

2B/A

1.2 m/s r r r 2r

a r r a cos30 r 0.637 m/s

a r 2r a cos 60 1.66E-4 rad/s

θ

θ

θ θ θ

θ

θ θ θ

= − = − + +

= − = − → = −

= + = → =

i e e



P. 2/205 A batter hits the baseball A with an initial velocity of vo = 30 m/sdirectly toward fielder B at an angle of 30°to the horizontal;the initial position of the ball is 0.9 m above the ground level.Fielder B requires ¼ s to judge where the ball should be caughtand begins moving to that position with constant speed.Because of great experience, fielder B chooses his runningspeed so that he arrives at the “catch position” simultaneouslywith the baseball. The catch position is the field location atwhich the ball altitude is 2.1 m. Determine the velocity of theball relative to the fielder at the instant the catch is made.



P. 2/205

( ) ( ) ( )

( )

22 2 2o o y

y x A

2 2o o

Ball: g constant

v v 2a s s v 30sin 30 2g 2.1 0.9

14.19 m/s 30cos30 25.98 m/s 25.98 14.19 m/s

1 1s s v t at 2.1 0.9 30sin 30 t gt , t 2.976 s2 2

horizontal

⎡ ⎤= + − = − −⎣ ⎦= − = = ∴ = −

⎡ ⎤= + + = + − =⎢ ⎥⎣ ⎦∴

a = - j =

v j v i i v i j

B

A/B A B

displacement of the ball 2.976 30cos30 77.32 m from battercatcher B must move 77.32 65 12.32 m to the right,

in 2.976 0.25 2.726 s 4.52 m/s21.46 14.19 m/s

= × =∴ − =

− = → == − = −

v iv v v i j

Ch. 2: Kinematics of Particles -...

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Transcript of Ch. 2: Kinematics of Particles -...