Chap. 12
Kinematics of a
Particle
Chapter Outline
1.
2.
3.
4.
5.
6.
7.
8.
Rectilinear Kinematics: Continuous Motion
Rectilinear Kinematics: Erratic Motion
General Curvilinear Motion
Curvilinear Motion: Rectangular
Components
Curvilinear Motion: Normal and Tangential
Components
Curvilinear Motion: Cylindrical Components
Absolute Dependent Motion Analysis of Two
Particles
Relative Motion Analysis of Two Particles
Using Translating Axes
12-1-2
An Overview of Mechanics
Mechanics: The study of how
bodies react to forces acting on
them.
Statics: The study of
bodies in equilibrium.
Dynamics:
1. Kinematics – concerned
with the geometric aspects of
motion
2. Kinetics - concerned with
the forces causing the motion
12-1-3
一 分類及定義：
動力學(Dynamics)
運動學(kinematics)：研究運動的幾何關係，例
如：位移(displacement)、速度(velocity)、加速度
(acceleration)、時間。
動力學(kinetics)：討論作用於物體的力、物體的
質量以及物體運動三者之間的關係。用來預測一
已知力造成的運動，或是決定產生一已知運動所
需的力。
12-1-4
質點運動
直線運動 (rectilinear motion)
曲線運動 (curvilinear motion)
12-1-5
二、直線運動：一質點沿著一直線運動
1. 位置 (position)：在
任一時間t，與一固
定原點O的距離
(distance)
＊當一點在任一時間t
的位置座標都知道的
話，則此點的運動即
可決定。即 x = f (t )
為已知。
位移
12-1-6
2. 速度 (velocity)
定義：由時間t到t +Δt的位移為Δx，則
Δx
平均速度 (average velocity) = Δt
瞬間速度 (instantaneous velocity)：
當Δt 極短時，
dx
Δx
v
=
v = lim
或
dt
Δt
Δt →0
v的大小 ( 絕對值 ) 稱為速率 (speed)
12-1-7
3. 加速度 (acceleration)
定義：
平均加速度 (average acceleration)
=
Δv
Δt
12-1-8
瞬間加速度
(instantaneous acceleration)
Δv dv
=
a = lim
Δt →0 Δt
dt
或
dx
d( )
2
d
x
dt
a=
= 2
dt
dt
dv
dv
dv
dx
=
=v
或以 dt =
代入， a =
dt dx v
dx
v
則
⇒ adx = vdv
12-1-9
三、決定質點的運動 ( 求x = f(t) )
Case 1. 已知 a = f(t)
dv
a=
⇒ dv = adt ⇒ dv = f (t ) dt
dt
積分得
∫dv = ∫ f (t )dt
12-1-10
Case 1
令 t = 0 (I.C.) 時，x=x0，v=v0
∴
v
∫v
0
t
dv = ∫ f (t )dt
0
t
v − v0 = ∫ f (t )dt
0
t
⎡
又dx = vdt = ⎢v0 + ∫ f (t )dt ⎤⎥ dt
0
⎣
⎦
再積分一次即可求出 x 和 t 的關係。
12-1-11
Case 2.
已知
a = f (x)
dv
a = v ⇒ vdv = adx = f ( x)dx
dx
v
x
v0
x0
⇒ ∫ vdv = ∫ f ( x)dx
x
1 2 1 2
⇒ v − v0 = ∫ f ( x)dx
x0
2
2
dx
又v = ，將v 代入上式，即可求出 。
dt
12-1-12
Case 3.
a = f (v )
dv
⇒ f (v ) =
dt
1)
dv or 2 )
⇒ dt =
f (v )
先積分求出 1) t~v
dv
f (v ) = v
dx
vdv
⇒ dx =
f (v )
2) x~v
dx
又 v = dt，將v 代入上式，即可求出。
12-1-13
.5
Ex. 已知 a = −60 x −1，且x=4
m時，V = 0
求(a) x=2 m 的速度V
(b) x=1 m 的速度V
(c) x=100 mm 的速度V
12-1-14
解：
dv
v = a = −60 x
dx
X = 4m，V = 0
−1.5
⇒ ∫ vdv = − ∫ 60 x dx
v
x
0
4
−1.5
1
60
⎡ 1 ⎤
= 120 ⎢ ⎥
v =−
x
− 0.5
2
⎣ x⎦
1 1
⇒ v = 240(
− )
x 2
x
x
4
4
− 0.5
2
2
12-1-15
(a) x=2 m 的速度V
x=2m
1 1
v = 240(
− ) = 120( 2 − 1) = 49.706
2 2
⇒ v = 7.05 m / s
2
12-1-16
(b) x=1 m 的速度V
x =1m
1 1
v = 240( − ) = 120 ⇒ v = 10.95 m / s
1 2
2
12-1-17
(c) x=100 mm 的速度V
x = 100 mm = 0.1 m
1
1
2
v = 240(
− ) = 638.9
0.1 2
⇒ v = 25.3 m / s
12-1-18
四、等速(直線)運動：
dv
dx
= a = 0　，　= v = const.
dt
dt
⇒ x = x + vt
0
12-1-19
五、等加速度(直線)運動：
dv
= a = const
dt
1.　v = v + at
0
1
2.　x = x + v t + at
2
dv
or　v = a = const.
dx
vdv = adx ⇒ ∫ vdv = a ∫ dx
2
0
0
v
x
v0
x0
1
⇒ (v − v ) = a ( x − x )
2
2
2
0
0
12-1-20
EXAMPLE
Given:A motorcyclist travels along a straight road at a
speed of 27 m/s. When the brakes are applied,
the motorcycle decelerates at a rate of -6t m/s2.
Find: The distance the motorcycle travels before it
stops.
Plan: Establish the positive coordinates in the direction
the motorcycle is traveling. Since the
acceleration is given as a function of time,
integrate it once to calculate the velocity and again
to calculate the position.
12-1-21
Solution:
EXAMPLE
(continued)
1) Integrate acceleration to determine the vvelocity.
t
a = dv / dt => dv = a dt => ∫ dv = ∫ (−6t )dt
vo
o
=> v – vo = -3t2 => v = -3t2 + vo
2) We can now determine the amount of time required for
the motorcycle to stop (v = 0). Use vo = 27 m/s.
0 = -3t2 + 27 => t = 3 s
3) Now calculate the distance traveled in 3s by integrating the
velocity using so = 0:
s
t
v = ds / dt => ds = v dt => ∫ ds = ∫ (−3t 2 + vo)dt
so
o
=> s – so = -t3 + vot
=> s – 0 = -(3)3 + (27)(3) => s = 54 m
12-1-22
六、圖解法
優點：當圖形很複雜、運動方程式是簡單
的函數，或是由很多段組成時，局
部的變化以圖解法較方便。
原理：s-t 圖的斜率為 v
v-t 圖的斜率為 a
12-1-23
2. 由a-t 到 s-t
12-1-24
3. 已知v-x
12-1-25
七、曲線運動
12-1-26
APPLICATIONS
The path of motion of each plane in
this formation can be tracked with
radar and their x, y, and z coordinates
(relative to a point on earth) recorded
as a function of time.
How can we determine the velocity
or acceleration of each plane at any
instant?
Should they be the same for each
aircraft?
12-1-27
APPLICATIONS
(continued)
A roller coaster car travels down
a fixed, helical path at a constant
speed.
How can we determine its
position or acceleration at any
instant?
If you are designing the track, why is it important to be
able to predict the acceleration of the car?
12-1-28
GENERAL CURVILINEAR MOTION
(Section 12.4)
A particle moving along a curved path undergoes curvilinear
motion. Since the motion is often three-dimensional, vectors
are used to describe the motion.
r = r(x,y,z)
A particle moves along a curve
defined by the path function, s.
The position of the particle at any instant is designated by the
vector
r = r(t). Both the magnitude and direction of r may vary with time.
If the particle moves a distance Ds
along the curve during time interval Dt,
the displacement is determined by
vector subtraction: D r = r’ - r
12-1-29
VELOCITY
Velocity represents the rate of change in the position of a
particle.
The average velocity of the particle
during the time increment Δt is
vavg = Δr/Δt .
The instantaneous velocity is the
time-derivative of position
v = dr/dt .
The velocity vector, v, is always
tangent to the path of motion.
The magnitude of v is called the speed. Since the arc length Δs
approaches the magnitude of Δr as t→0, the speed can be
obtained by differentiating the path function (v = ds/dt). Note
that this is not a vector!
12-1-30
ACCELERATION
Acceleration represents the rate of change in the
velocity of a particle.
If a particle’s velocity changes from v to v’ over a
time increment Δt, the average acceleration during
that increment is:
aavg = Δv/Δt = (v - v’)/Δt
The instantaneous acceleration is the timederivative of velocity:
a = dv/dt = d2r/dt2
A plot of the locus of points defined by the arrowhead
of the velocity vector is called a hodograph. The
acceleration vector is tangent to the hodograph, but
not, in general, tangent to the path function.
12-1-31
CURVILINEAR MOTION: RECTANGULAR COMPONENTS
(Section 12.5)
It is often convenient to describe the motion of a particle in
terms of its x, y, z or rectangular components, relative to a fixed
frame of reference.
The position of the particle can be
defined at any instant by the
position vector
r=xi+yj+zk .
The x, y, z components may all be
functions of time, i.e.,
x = x(t), y = y(t), and z = z(t) .
The magnitude of the position vector is: r = (x2 + y2 + z2)0.5
The direction of r is defined by the unit vector: ur = (1/r)r
12-1-32
RECTANGULAR COMPONENTS: VELOCITY
The velocity vector is the time derivative of the position vector:
v = dr/dt = d(xi)/dt + d(yj)/dt + d(zk)/dt
Since the unit vectors i, j, k are constant in magnitude and
direction, this equation reduces to v = vxi + vyj + vzk
•
•
•
x
y
z
where vx =
= dx/dt, vy =
= dy/dt, vz =
= dz/dt
The magnitude of the velocity
vector is
v = [(vx)2 + (vy)2 + (vz)2]0.5
The direction of v is tangent
to the path of motion.
12-1-33
RECTANGULAR COMPONENTS: ACCELERATION
The acceleration vector is the time derivative of the
velocity vector (second derivative of the position vector):
a = dv/dt = d2r/dt2 = axi + ayj + azk
•
•
••
where ax = vx = x = dvx /dt, ay = vy = y
= dvy /dt,
•
••
••
az = vz = z = dvz /dt
The magnitude of the acceleration vector is
a = [(ax)2 + (ay)2 + (az)2 ]0.5
The direction of a is usually
not tangent to the path of the
particle.
12-1-34
EXAMPLE
Given:The motion of two particles (A and B) is described by
the position vectors
rA = [3t i + 9t(2 – t) j] m
rB = [3(t2 –2t +2) i + 3(t – 2) j] m
Find: The point at which the particles collide and their
speeds just before the collision.
Plan: 1) The particles will collide when their position
vectors are equal, or rA = rB .
2) Their speeds can be determined by differentiating
the position vectors.
12-1-35
EXAMPLE
(continued)
Solution:
1) The point of collision requires that rA = rB, so xA =
xB and yA = yB .
x-components: 3t = 3(t2 – 2t + 2)
Simplifying: t2 – 3t + 2 = 0
Solving:
t = {3 ± [32 – 4(1)(2)]0.5}/2(1)
=> t = 2 or 1 s
y-components: 9t(2 – t) = 3(t – 2)
Simplifying:
3t2 – 5t – 2 = 0
Solving: t = {5 ± [52 – 4(3)(–2)]0.5}/2(3)
=> t = 2 or – 1/3 s
So, the particles collide when t = 2 s. Substituting this
value into rA or rB yields
xA = xB = 6 m
and yA = yB = 0
12-1-36
EXAMPLE
(continued)
2) Differentiate rA and rB to get the velocity vectors.
.
.
vA = drA/dt = .xA i + yA j = [3i + (18 – 18t)j] m/s
At t = 2 s: vA = [3i – 18j] m/s
•
•
vB = drB/dt = xBi + yBj = [(6t – 6)i + 3j] m/s
At t = 2 s: vB = [6i + 3j] m/s
Speed is the magnitude of the velocity vector.
vA = (32 + 182) 0.5 = 18.2 m/s
vB = (62 + 32) 0.5 = 6.71 m/s
12-1-37
例題
例：拋體運動中 (motion of projectile)，假設空
氣阻力不算，此物體只承受重力，則
a = &x& = 0　, a = &y& = − g　, a = &z& = 0
x
y
z
12-1-38
若 (x0, y0, z0) 為砲口的座標
(vx)0, (vy)0, (vz)0 為初速v0的分量座標
則 經兩次積分可得：
,　v = y =(v ) − gt　,　v = z =(v )
⎧v = x =(v ) ⎪
⎨
1
,　y = y +(v ) t − gt ,　z = z +(v ) t
⎪⎩x = x +(v ) t　2
x
x 0
y
y 0
z
z 0
2
0
x 0
0
y 0
0
z 0
12-1-39
若拋體僅在xy平面上運動，
且x0 = y0 = z0 =0，則：
, v = (v ) − gt　, v = (v )
⎧v = (v ) ⎪
⎨
1
, y = (v ) t − gt , z = 0
⎪⎩ x = (v ) t　2
x
x
0
y
y
z
0
z
0
2
x
0
y
0
12-1-40
結論
1) 拋體在xy平面上運動。
2) 在水平方向上為等速運動。
3) 垂直方向上為等加速度運動。
4) 彈道為拋物線。
12-1-41
p.47, 12-83
The roller coaster car travels down the helical path at constant speed such
that the parametric equations that define its position are x = csinkt, y = ccoskt,
z = h - bt,wherec,h,and b are constants. Determine the magnitudes of its
velocity and acceleration.
12-1-42
p.48, 12-92
Water is discharged from the hose with a speed of 12 m/s. Determine the
two possible angles θ the fireman can hold the hose so that the water
strikes the building at B. Take s = 6 m.
12-1-43
12-1-44
12-1-45
p.51, 12-104
The football is kicked over the goalpost with an initial velocity of vA = 24 m/s
as shown.Determine the point B(x, y) where it strikes the bleachers.
12-1-46
12-1-47
十、曲線運動：切線及法線
1．質點的平面運動
p為一質點在任意時間的
位置
et為通過p與路徑(path)
相切的單位向量
et’為通過p’ 的單位向量
e t , en
(ut , un)
12-2-1
將et ,et’ 表示在同一原點之圖形：
12-2-2
Δθ
)
2
Δe
2 sin( Δθ / 2) sin( Δθ / 2)
∴
=
=
Δθ
Δθ
Δθ / 2
Δe
⎛ de
⎞
當Δθ → 0時， 垂直於e ⊥e ⎟
⎜Q
Δθ
⎝ dθ
⎠
Δe
sin( Δθ / 2)
(為一垂直於e 的單位向量)
lim
= lim
= 1　Δθ / 2
Δθ
Δe = e '−e = 2 sin(
t
t
t
t
t
t
t
t
t
Δt →0
Δt →0
t
12-2-3
Δe de
Def. e = lim
(法線單位向量)
=
θ
Δθ dθ
(皆垂直於路徑)
⇒ 速度　v = ve t
n
t
Δ →0
t
dv dv
de
加速度　a =
= e +v
dt dt
dt
t
t
12-2-4
by chain rule
1
v
de de dθ ds
⋅
= e ⋅ ⋅v = e
又　=
ρ
ρ
dt dθ ds dt
t
t
n
n
dθ 1
= =κ
其中
ds ρ
ρ：曲率半徑　radius　of　curature
κ：曲率　curvature
dv
v
∴a = e + e = a e + a e
dt
ρ
2
t
n
t
t
n
n
12-2-5
at 為a的切線分量
(tangential component)
等於速率變化
⎛ dv ⎞
⎜ ⎟
⎝ dt ⎠
an 為a的法線分量
(normal component)
等於 ⎛ v 2 ⎞
⎜⎜ ⎟⎟
⎝ρ⎠
an 永遠朝著曲率中心C
12-2-6
意義和應用
意義：
1) 反映質點速率的變化。
2) 反映質點運動方向的變化。
3) 和同時為0時，加速度才為0。
4) 一質點以等速率沿著一曲線移動，其加速度不為0，除非
是通過反曲點(ρ=∞)，或此路徑為直線。
應用： 機翼、火車軌道、凸輪設計時，避免曲率的突然變化
(ρ太小，太大)
12-2-7
SPECIAL CASES OF MOTION
The particle moves along a path expressed as y = f(x).
The radius of curvature, ρ, at any point on the path can be
calculated from
2 ]3/2
[
1
(dy/dx)
+
ρ = ________________
d2y/dx 2
12-2-8
2. 質點的空間運動
Def. et 和e n 所在的平面稱為 osculating plane
(O.P.), e n 稱為p點的 ”principal normal”
binormal： e b = e t × e n , e b ⊥ O.P.
12-2-9
p.61, 12-116.
The automobile has a speed of 24 m/s at point A and an acceleration a
having a magnitude of 3 m/s2, acting in the direction shown. Determine the
radius of curvature of the path at point A and the tangential component of
acceleration.
12-2-10
p. 63, 12-132
The car is traveling at a speed of 30 m/s.The driver applies the brakes at A
and thereby reduces the speed at the rate of a, = (-t/8) m/s2, where t is in
seconds. Determine the acceleration of the car just before it reaches point C
on the circular curve. It takes 15 s for the car totravel from A toC.
12-2-11
12-2-12
12-2-13
p.66, 12-148
A spiral transition curve is used on railroads to connect a straight portion of
the track with a curved portion. If the spiral is defined by the equation y =
9(10-6)x3, where x and y are in meters, determine the magnitude of the
acceleration of a train engine moving with a constant speed of 12 m/s when it
is at point x = 180 m.
12-2-14
12-2-15
十一、徑向 (radial) 和
橫向(transverse) 分量
12-2-16
1. 極座標：r、θ
e , eθ 為通過 p 點的單位向量
r
e = (cosθ , sin θ )
r
eθ = (− sin θ , cosθ )
⎧ de
⎪⎪ dθ = eθ
⎨
⎪ de θ = − e
⎪⎩ dθ
r
r
12-2-17
dθ &
⎧ de r de r dθ
⎪⎪ dt = dθ dt = eθ dt = θeθ = e& r
Dot : ⎨
⎪ deθ = deθ dθ = −e dθ = −θ&e = e&
r
r
θ
⎪⎩ dt
dθ dt
dt
dr d (re r ) dr
de r
速度v =
=
= er + r
= r&e r + re& r = r&e r + rθ&eθ
dt
dt
dt
dt
= vr e r + vθ eθ
r& = vr
rθ& = v
θ
12-2-18
v = r&e r + re& r
dv
= &r&e r + r&e& r + r&θ&eθ + rθ&&eθ + rθ&e& θ
加速度a =
dt
= &r&e r + r&θ&eθ + r&θ&eθ + rθ&&eθ + rθ&(−θ&e r )
= (&r& − rθ& 2 )e + (rθ&& + 2r&θ&)e
r
θ
⇓ ⇓
ar aθ
⎧⎪vr = r&　, vθ = rθ&
⇒⎨
⎪⎩ar = &r& − rθ& 2 , aθ = rθ&& + 2r&θ&
dv
dv
∗ ar ≠ r , aθ ≠ θ
dt
dt
12-2-19
2. 圓周運動：
•
••
r = r, r = r = 0
⎧ v = rθ&eθ
∴⎨
&
&
&
a
r
θ
e
r
θ
eθ
=
−
+
⎩
2
r
12-2-20
3. Cylindrical Coordinates圓柱座標系 (R,θ, z)
r = Re R + z k
dr &
v=
(Q e& R = θ&eθ )
= R e R + Rθ&eθ + z& k dt
d v &&
a=
= Re R + R& e& R + R& θ&eθ + Rθ&&eθ + Rθ&e&θ + &z&k
dt
&& − Rθ& 2 )e R + ( Rθ&& + 2 R& θ&)eθ + &z&k
= (R
12-2-21
p.76, 12-166
The slotted arm OA rotates counterclockwise about O
with a constant angular velocity of θ& . The motion of pin
B is constrained such that it moves on the fixed circular
surface and along the slot in OA. Determine the
magnitudes of the velocity and acceleration of pin B as a
function of θ.
θ&
12-2-22
12-2-23
12-2-24
p. 80, 12-188
The partial surface of the cam is that of a logarithmic spiral r = (40e0.05θ),
where θ is in radians. If the cam rotates at a constant angular velocity
of θ& = 4 rad/s ,determine the magnitudes of the velocity and
acceleration ofthe point on the cam that contacts the follower rod at the
instant θ = 30°.
12-2-25
12-2-26
12-2-27
十二、 相依運動
( Dependent Motion )
12-2-28
S + l + S = l (total　length)
A
CD
B
T
S + S = l − l = constant
A
B
T
CD
dS dS
⇒
+
= 0 ⇒ v + v = 0 ⇒ v = −v
dt
dt
同理　a = − a
A
B
A
A
B
A
B
B
12-2-29
A → B ↑ 往下為正
2S + h + S + (r + r ) = l
B
1
A
2
T
∴ 2S + S = l − h − (r + r ) = constant
B
A
⇒ 2v = −v
B
2a = − a
B
1
T
2
A
A
不同基準線　, 往上為正
2(h − S ) + h + S = l
B
A
T
⇒ 2v = v , 2a = a
B
A
B
A
12-2-30
十三、相對運動
12-2-31
相對位置：
r =r −r
B/ A
B
A
r =r +r
B
A
B/ A
12-2-32
相對速度 /相對加速度
相對速度：
v =v +v
B
A
B/ A
相對加速度：
a =a +a
B
A
B/ A
12-2-33
12-2-34
12-2-35
p. 94, 12-196
Determine the displacement of the log if the truck at C pulls the
cable 1.2 m to the right.
12-2-36
12-2-37
p. 97, 12-212
The man pulls the boy up to the tree limb C by walking backward at a
constant speed of 1.5 m/s. Determine the speed at which the boy is being
lifted at the instant xA = 4 m. Neglect the size of the limb. When xA = 0,
yB = 8 m , so that A and B are coincident, i.e., the rope is 16 m long.
12-2-38
12-2-39
12-2-40
p. 100, 2-228
At the instant shown car A is traveling with a velocity of 30 m/s and has an
acceleration of 2 m/s2 along the highway. At the same instant B is traveling
on the trumpet interchange curve with a speed of 15 m/s, which is
decreasing at 0.8 m/s2. Determine the relative velocity and relative
acceleration of B with respect to A at this instant.
12-2-41
12-2-42
12-2-43