kinematics of particles
DESCRIPTION
LectureTRANSCRIPT
![Page 1: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/1.jpg)
Ch. 7: Kinematics of Particles
7.0 Outline 316
Introduction 317 Rectilinear Motion 319 Plane Curvilinear Motion 341 Rectangular Coordinates (x-y) 349 Normal and Tangential Coordinates (n-t) 365 Polar Coordinates (r-θ) 381 Relative Motion (Translating Axes) 404
7.0 Outline
316
![Page 2: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/2.jpg)
Ch. 7: Kinematics of Particles
7.1 Introduction
7.1 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.
317
![Page 3: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/3.jpg)
Ch. 7: Kinematics of Particles
7.1 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
318
![Page 4: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/4.jpg)
Ch. 7: Kinematics of Particles
7.2 Rectilinear Motion: motion along a straight line
7.2 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
∆ →
∆
∆ ± = ∆ ∆
= ∆ ∆ = =
=
( )
( )
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 ads∆ →
∆ ∆ = ∆ ∆
= ∆ ∆ = = = =
=
319
![Page 5: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/5.jpg)
Ch. 7: Kinematics of Particles
7.2 Rectilinear Motion
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
320
![Page 6: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/6.jpg)
Ch. 7: Kinematics of Particles
7.2 Rectilinear Motion
( )
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
321
![Page 7: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/7.jpg)
Ch. 7: Kinematics of Particles
7.2 Rectilinear Motion
( ) ( )d d
kinetic relationdt dt
dt dts v a v, s, t F v, s, t←
∫ ∫
common problems: know a, find s by integration
322
![Page 8: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/8.jpg)
Ch. 7: Kinematics of Particles
7.2 Rectilinear Motion
( )
( )
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
323
![Page 9: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/9.jpg)
Ch. 7: Kinematics of Particles
7.2 Rectilinear Motion
( ) ( )
( )
( )
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
324
![Page 10: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/10.jpg)
Ch. 7: Kinematics of Particles
7.2 Rectilinear Motion
[ ] ( ) ( ) ( )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
325
![Page 11: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/11.jpg)
Ch. 7: Kinematics of Particles
7.2 Rectilinear Motion
( ) ( ) ( )
[ ] ( ) ( )
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
326
![Page 12: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/12.jpg)
Ch. 7: Kinematics of Particles
7.2 Rectilinear Motion
P. 7/1 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.
327
![Page 13: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/13.jpg)
Ch. 7: Kinematics of Particles
7.2 Rectilinear Motion
P. 7/1
[ ]2
o o o
o o2
2l l
u l2
u 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
328
![Page 14: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/14.jpg)
Ch. 7: Kinematics of Particles
7.2 Rectilinear Motion
P. 7/2 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.
329
![Page 15: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/15.jpg)
Ch. 7: Kinematics of Particles
7.2 Rectilinear Motion
P. 7/2
[ ]
( )
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
330
![Page 16: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/16.jpg)
Ch. 7: Kinematics of Particles
7.2 Rectilinear Motion
P. 7/3 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.
331
![Page 17: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/17.jpg)
Ch. 7: Kinematics of Particles
7.2 Rectilinear Motion
P. 7/3
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
332
![Page 18: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/18.jpg)
Ch. 7: Kinematics of Particles
7.2 Rectilinear Motion
P. 7/4 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.
333
![Page 19: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/19.jpg)
Ch. 7: Kinematics of Particles
7.2 Rectilinear Motion
P. 7/4
( )
( )
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= × + × = == + × = =
334
![Page 20: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/20.jpg)
Ch. 7: Kinematics of Particles
7.2 Rectilinear Motion
P. 7/5 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.
335
![Page 21: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/21.jpg)
Ch. 7: Kinematics of Particles
7.2 Rectilinear Motion
P. 7/5
[ ]
[ ] ( )
[ ]
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
−
− −
= = =−
= = = −
= = = −−
∫ ∫
∫ ∫
∫ ∫
336
![Page 22: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/22.jpg)
Ch. 7: Kinematics of Particles
7.2 Rectilinear Motion
P. 7/6 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.
337
![Page 23: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/23.jpg)
Ch. 7: Kinematics of Particles
7.2 Rectilinear Motion
P. 7/6
( ) ( )[ ] ( )
( )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
π∆
− = ∆ = =
− + = − = −
= = =
− = ∆ =
= × + × =
∫ ∫
338
![Page 24: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/24.jpg)
Ch. 7: Kinematics of Particles
7.2 Rectilinear Motion
P. 7/7 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.
339
![Page 25: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/25.jpg)
Ch. 7: Kinematics of Particles
7.2 Rectilinear Motion
P. 7/7
[ ]
( ) ( )( )
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
= =
−= − × × − × − × + + −
=
∫
340
![Page 26: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/26.jpg)
Ch. 7: Kinematics of Particles
2.3 Plane Curvilinear Motion: motion along a curvedpath that lies in a single plane
Vector quantity is independent of any particularcoordinate system
7.3 Plane Curvilinear Motion
341
![Page 27: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/27.jpg)
Ch. 7: Kinematics of Particles
Time derivative of a vector (described in fixed coord.)change in both magnitude and direction
7.3 Plane Curvilinear Motion
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
+ ∆ = ∆∆ ∆
rr r + r
rar) during time t is s (measured along the path)∆ ∆
342
![Page 28: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/28.jpg)
Ch. 7: Kinematics of Particles
7.3 Plane Curvilinear Motion
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
∆ →
= ∆ ∆= ∆ ∆
∆= = =
∆
∆ → ∆
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,
∴ → →∴
= = =
∆ → →
∴ → ∆ →∆
v vv
v
r
av avs, , and v→ →v v v
343
![Page 29: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/29.jpg)
Ch. 7: Kinematics of Particles
7.3 Plane Curvilinear Motion
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
344
![Page 30: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/30.jpg)
Ch. 7: Kinematics of Particles
7.3 Plane Curvilinear Motion
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
∆ →
= ∆ ∆∆
= = =∆
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
345
![Page 31: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/31.jpg)
Ch. 7: Kinematics of Particles
7.3 Plane Curvilinear Motion
The acceleration bears the same relation to the velocityas the velocity bears to the position vector.
346
![Page 32: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/32.jpg)
Ch. 7: Kinematics of Particles
7.3 Plane Curvilinear Motion
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
347
![Page 33: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/33.jpg)
Ch. 7: Kinematics of Particles
7.3 Plane Curvilinear Motion
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
348
![Page 34: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/34.jpg)
Ch. 7: Kinematics of Particles
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.
7.4 Rectangular Coordinates (x-y)
349
![Page 35: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/35.jpg)
Ch. 7: Kinematics of Particles
7.4 Rectangular Coordinates (x-y)
( ) ( )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 + ja = 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
= =
=
350
![Page 36: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/36.jpg)
Ch. 7: Kinematics of Particles
Projectile motion: motion of the thrown object
Neglect aerodynamic drag and the earth’s curvature androtation, and assume the altitude range is small enough
7.4 Rectangular Coordinates (x-y)
x ya 0 and a g= = −
351
![Page 37: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/37.jpg)
Ch. 7: Kinematics of Particles
7.4 Rectangular Coordinates (x-y)
( ) ( )( ) ( )
( ) ( )
( ) ( )
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
= = = +
= − = − = + −
= − −
352
![Page 38: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/38.jpg)
Ch. 7: Kinematics of Particles
7.4 Rectangular Coordinates (x-y)
P. 7/8 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.
353
![Page 39: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/39.jpg)
Ch. 7: Kinematics of Particles
7.4 Rectangular Coordinates (x-y)
P. 7/8
( )
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
354
![Page 40: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/40.jpg)
Ch. 7: Kinematics of Particles
7.4 Rectangular Coordinates (x-y)
P. 7/9 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.
355
![Page 41: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/41.jpg)
Ch. 7: Kinematics of Particles
7.4 Rectangular Coordinates (x-y)
P. 7/9
( )
( )
( )
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
356
![Page 42: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/42.jpg)
Ch. 7: Kinematics of Particles
7.4 Rectangular Coordinates (x-y)
P. 7/10 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)
357
![Page 43: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/43.jpg)
Ch. 7: Kinematics of Particles
7.4 Rectangular Coordinates (x-y)
P. 7/10
( )( )
( )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
358
![Page 44: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/44.jpg)
Ch. 7: Kinematics of Particles
7.4 Rectangular Coordinates (x-y)
P. 7/11 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
359
![Page 45: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/45.jpg)
Ch. 7: Kinematics of Particles
7.4 Rectangular Coordinates (x-y)
P. 7/11 ( )
( )
( ) ( )
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
∞ ∞
− − − = = + − −
= = −
∫ ∫
360
![Page 46: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/46.jpg)
Ch. 7: Kinematics of Particles
7.4 Rectangular Coordinates (x-y)
P. 7/12 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°.
361
![Page 47: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/47.jpg)
Ch. 7: Kinematics of Particles
7.4 Rectangular Coordinates (x-y)
P. 7/12 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
= +
+=
362
![Page 48: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/48.jpg)
Ch. 7: Kinematics of Particles
7.4 Rectangular Coordinates (x-y)
P. 7/13 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.
363
![Page 49: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/49.jpg)
Ch. 7: Kinematics of Particles
7.4 Rectangular Coordinates (x-y)
P. 7/13
( )
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
− + =
= −
∴ →
364
![Page 50: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/50.jpg)
Ch. 7: Kinematics of Particles
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.
7.5 Normal and Tangential Coordinates (n-t)
n
t
At point A,unit vector in the n-directionunit vector in the t-directionradius of curvature of the pathρ
===
ee
365
![Page 51: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/51.jpg)
Ch. 7: Kinematics of Particles
7.5 Normal and Tangential Coordinates (n-t)
( )
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
366
![Page 52: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/52.jpg)
Ch. 7: Kinematics of Particles
7.5 Normal and Tangential Coordinates (n-t)
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 = ea = i + j = e e
367
![Page 53: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/53.jpg)
Ch. 7: Kinematics of Particles
7.5 Normal and Tangential Coordinates (n-t)
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
368
![Page 54: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/54.jpg)
Ch. 7: Kinematics of Particles
7.5 Normal and Tangential Coordinates (n-t)
P. 7/14 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.
369
![Page 55: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/55.jpg)
Ch. 7: Kinematics of Particles
7.5 Normal and Tangential Coordinates (n-t)
P. 7/14
( )
( )
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
∆→ = = = − =
∆=
= + = + =
= =∴ !
370
![Page 56: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/56.jpg)
Ch. 7: Kinematics of Particles
7.5 Normal and Tangential Coordinates (n-t)
P. 7/15 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.
371
![Page 57: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/57.jpg)
Ch. 7: Kinematics of Particles
7.5 Normal and Tangential Coordinates (n-t)
P. 7/15
( )
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
372
![Page 58: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/58.jpg)
Ch. 7: Kinematics of Particles
7.5 Normal and Tangential Coordinates (n-t)
P. 7/16 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
373
![Page 59: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/59.jpg)
Ch. 7: Kinematics of Particles
7.5 Normal and Tangential Coordinates (n-t)
P. 7/16
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
374
![Page 60: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/60.jpg)
Ch. 7: Kinematics of Particles
7.5 Normal and Tangential Coordinates (n-t)
P. 7/17 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.
375
![Page 61: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/61.jpg)
Ch. 7: Kinematics of Particles
7.5 Normal and Tangential Coordinates (n-t)
P. 7/17
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
( )
n2
t 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
= = − = − += − = − −
= = =
= = + = +
∴
aa n a i j
a a a i j
il is knownβ
y
x
tn
van
at
376
![Page 62: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/62.jpg)
Ch. 7: Kinematics of Particles
7.5 Normal and Tangential Coordinates (n-t)
P. 7/18 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?
377
![Page 63: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/63.jpg)
Ch. 7: Kinematics of Particles
7.5 Normal and Tangential Coordinates (n-t)
P. 7/18
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
+ =
= = × =
= = ± ± → = =
= = = → =
378
![Page 64: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/64.jpg)
Ch. 7: Kinematics of Particles
7.5 Normal and Tangential Coordinates (n-t)
P. 7/19 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.
379
![Page 65: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/65.jpg)
Ch. 7: Kinematics of Particles
7.5 Normal and Tangential Coordinates (n-t)
P. 7/19
( )
( ) ( ) ( )
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
= = = = =
+ = =
= = =
∴
= + = +
= → = → = →
= + = =
= =
∫ ∫ ∫
∫
2t
2 2 2n
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= = = =
= = → =
380
![Page 66: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/66.jpg)
Ch. 7: Kinematics of Particles
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.
7.6 Polar Coordinates (r-θ)
381
![Page 67: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/67.jpg)
Ch. 7: Kinematics of Particles
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
7.6 Polar Coordinates (r-θ)
382
![Page 68: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/68.jpg)
Ch. 7: Kinematics of Particles
( ) ( ) ( ) ( )
( )
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
7.6 Polar Coordinates (r-θ)
383
![Page 69: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/69.jpg)
Ch. 7: Kinematics of Particles
P. 7/20 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.
7.6 Polar Coordinates (r-θ)
r θ
384
![Page 70: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/70.jpg)
Ch. 7: Kinematics of Particles
P. 7/20
7.6 Polar Coordinates (r-θ)
r
2 2 2r
and of the rocket is in the vertical directionv 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
385
![Page 71: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/71.jpg)
Ch. 7: Kinematics of Particles
P. 7/21 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
.
7.6 Polar Coordinates (r-θ)
β
r, r, , and θ θ
386
![Page 72: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/72.jpg)
Ch. 7: Kinematics of Particles
P. 7/21
7.6 Polar Coordinates (r-θ)
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 60vcos30 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
387
![Page 73: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/73.jpg)
Ch. 7: Kinematics of Particles
P. 7/22 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.
7.6 Polar Coordinates (r-θ)
θ α=
388
![Page 74: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/74.jpg)
Ch. 7: Kinematics of Particles
P. 7/22
7.6 Polar Coordinates (r-θ)
( )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
389
![Page 75: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/75.jpg)
Ch. 7: Kinematics of Particles
P. 7/23 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.
7.6 Polar Coordinates (r-θ)
ω θ=
390
![Page 76: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/76.jpg)
Ch. 7: Kinematics of Particles
P. 7/23
7.6 Polar Coordinates (r-θ)
( )
( )
( )
( )
o o o2
o 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
391
![Page 77: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/77.jpg)
Ch. 7: Kinematics of Particles
P. 7/24 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.
7.6 Polar Coordinates (r-θ)
r and θ
s
392
![Page 78: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/78.jpg)
Ch. 7: Kinematics of Particles
P. 7/24
7.6 Polar Coordinates (r-θ)
( )
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
−= =
= =
−= =
+ +
393
![Page 79: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/79.jpg)
Ch. 7: Kinematics of Particles
P. 7/25 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):
7.6 Polar Coordinates (r-θ)
r, , r, , r, and θ θ θ
394
![Page 80: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/80.jpg)
Ch. 7: Kinematics of Particles
P. 7/25
7.6 Polar Coordinates (r-θ)
( )
( )
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
θv
a
395
![Page 81: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/81.jpg)
Ch. 7: Kinematics of Particles
P. 7/26 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.
7.6 Polar Coordinates (r-θ)
Kθ =
Kθ =
396
![Page 82: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/82.jpg)
Ch. 7: Kinematics of Particles
P. 7/26
7.6 Polar Coordinates (r-θ)
2 2 2n t
K 02 2K 0
motion of pin P is the circular path with radius bv 2bK constanta v / 4bK a v 0, a 4bK constant
θ θ
β θ β β
ρβ
ρ
= =
= = =
= = =
= = = = = =
rb
b
θ
θβ
397
![Page 83: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/83.jpg)
Ch. 7: Kinematics of Particles
P. 7/27 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 .
7.6 Polar Coordinates (r-θ)
rv
398
![Page 84: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/84.jpg)
Ch. 7: Kinematics of Particles
P. 7/27
7.6 Polar Coordinates (r-θ)
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
399
![Page 85: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/85.jpg)
Ch. 7: Kinematics of Particles
P. 7/28 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 .
7.6 Polar Coordinates (r-θ)
β
r, r, , and θ θ
400
![Page 86: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/86.jpg)
Ch. 7: Kinematics of Particles
P. 7/28
7.6 Polar Coordinates (r-θ)
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
tθ30
401
![Page 87: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/87.jpg)
Ch. 7: Kinematics of Particles
P. 7/29 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.)
7.6 Polar Coordinates (r-θ)
402
![Page 88: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/88.jpg)
Ch. 7: Kinematics of Particles
P. 7/29
7.6 Polar Coordinates (r-θ)
( )/ 6 t
0 0
0.125
0 0
pin A must always be in the slot 40 rev/min CCWif 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=
403
![Page 89: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/89.jpg)
Ch. 7: Kinematics of Particles
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
7.7 Relative Motion (Translating Axes)
404
![Page 90: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/90.jpg)
Ch. 7: Kinematics of Particles
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.
7.7 Relative Motion (Translating Axes)
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
405
![Page 91: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/91.jpg)
Ch. 7: Kinematics of Particles
7.7 Relative Motion (Translating Axes)
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
→
406
![Page 92: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/92.jpg)
Ch. 7: Kinematics of Particles
7.7 Relative Motion (Translating Axes)
P. 7/30 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.
407
![Page 93: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/93.jpg)
Ch. 7: Kinematics of Particles
7.7 Relative Motion (Translating Axes)
P. 7/30
[ ]
( )
[ ]
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
408
![Page 94: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/94.jpg)
Ch. 7: Kinematics of Particles
7.7 Relative Motion (Translating Axes)
P. 7/31 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?
409
![Page 95: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/95.jpg)
Ch. 7: Kinematics of Particles
7.7 Relative Motion (Translating Axes)
P. 7/31
( )
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
410
![Page 96: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/96.jpg)
Ch. 7: Kinematics of Particles
7.7 Relative Motion (Translating Axes)
P. 7/32 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
411
![Page 97: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/97.jpg)
Ch. 7: Kinematics of Particles
7.7 Relative Motion (Translating Axes)
P. 7/32
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
412
![Page 98: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/98.jpg)
Ch. 7: Kinematics of Particles
7.7 Relative Motion (Translating Axes)
P. 7/33 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.
413
![Page 99: Kinematics of Particles](https://reader033.vdocuments.site/reader033/viewer/2022042503/55cf9bf4550346d033a7fefe/html5/thumbnails/99.jpg)
Ch. 7: Kinematics of Particles
7.7 Relative Motion (Translating Axes)
P. 7/33
( ) ( ) ( )
( )
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
414