Chapter 2 Kinetics of Particles and of Mass Centers of Bodies

1. 如前所述，Kinetics 是討論到物體受力及運動之關係，而 Newton’s Laws 及

Euler’s Laws 可描述其物理行為
2. 物體（質點、質量中心）受力後之運動情形
3. Newton’s Laws 及 Euler’s Laws 的運動方程式及能量法之應用。

§ 2.1 Newton’s Laws and Euler’s Laws (Text Ch. 12 and Ch. 14)

＊Newton’s Laws：(Text Ch. 12)

relating the external forces on a body to its motion.
1.If the resultant force F on a particle is zero, then the particle has constant
velocity.
2.If F  0, then F is proportional to the time derivative of the particle’s

momentum” mv ”（product of mass and velocity）

i.e. F＝
d
dt
 
mv ＝ m
dv
dt
＝ ma

3.The interaction of two particles is through a pair of self-equilibrating forces.（That

is, they have the same magnitude, opposite directions, and a common line of
action.）

f12 ＝ f 21 f12 ＝  f 21

2-1
System of one particle

2-2
Example of one particle

課本的 Ch. 12 Examples 3~5 自己作作看

2-3
2-4
＊Euler’s Laws (Text Ch. 14)

1.The first Law：

for a body composed of a set of N particles.
For the i -th particle：
N
Fi   f ij = mi ai ------（A）
j 1

Fi ：net external force（by things external to the system）

f ij ：internal force（by other particles of the system）

Note that f ii ＝0

Sum there N equations：

N N N N

 F +  f
i 1
i
i 1 j 1
ij ＝  mi ai
i 1

From Newton’s third law： f ij ＝  f ji

N N
  f ij ＝0
i 1 j 1

2-5
   Fi ＝  mi ai

The sum of the external forces on the system equals the sum of the ma ’s of the
particles making up the system. ------Euler’s first law
For a body whose mass is continuously distributed,

 F ＝  adm ＝  adv ＝  adxdydz

 ：mass density
v ：infinitesimal element of volume

2.The second Law：

take the cross product of ri with both sides of equation（A）

N
ri  Fi   ri  f ij ＝ ri  mi ai
j 1

ri  Fi ：moment about point P of the external force Fi

ri  f ij ：moment w.r.t. P of the force exerted on the i th particle by the j th

particle
Note： Cross product
  
suppose that a ＝ a1i  a 2 j  a3 k

  
b ＝ b1i  b2 j  b3 k

with respect to a right-handed Cartesian coordinate：

    
a  b ＝ a 2 b3  a3b2 i  a3b1  a1b3  j  a1b2  a 2 b1 k

  a a3  a3 a1  a1 a2 
or a b ＝ 2 i j k
b2 b3 b3 b1 b1 b2

sum these N equations

N N N N

 r  F   r  f
i i i ij ＝  ri  mi ai
i 1 i 1 j 1 i 1

since 1. f ij ＝  f ji

2. f ij and f ji share a common line of action join the i th and j th

2-6
 particles.

  ri  Fi ＝  ri  mi ai

or M p ＝  ri  mi ai

The sum of the moments of the external forces about a point equals the sum of
the moments of the ma ’s about that point. ------second law
For a body whose mass is continuously distributed,
 
M p ＝  r  adm

3.Relationship between the external forces acting on a body and the motion of its
mass center.
The definition of the location of the mass center ”C”, is

roc ＝
m R
i i
, where m is the mass（  mi ）of the system.
m
From Euler’s first law：
d 2 Ri
 Fi ＝  mi ai ＝  mi dt 2
assume mass is constant

 m R ＝ dtd mr ＝ m ddtr

2
d2 2
oc
＝ i i oc
dt 2 2 2

d 2 roc
i.e.  Fi ＝ m dt 2
or F ＝ m ac

ac ：acceleration of mass center

For a continuous body,


 d 2R d2R d2
a ＝ 2 ,  F ＝  2 dm ＝ 2  Rdm
dt dt dt

and mroc ＝  Rdm

2-7
§ 2.2 Motions of Particles and of Mass Center of Bodies (Text Ch. 14)

＊Objective：Use the Euler’s first law to analyze the motions of the mass centers of
bodies. Why we study mass centers？because the motion of that point（mass center）
is directly related to the external forces acting on the body（Rigid Body）.

＊Free body diagram： A sketch of a body in which all the external forces and
couples acting upon it are drawn with respect to location, direction and magnitude.

2-8
 System of particles

↑ ∝

↓
瀟 smrupuea ngxsin

4
x 0 .

mgx 巡

Note: If the friction coefficients are now swapped, nothing changes until we begin to
analyze the 6 equations. We have

←②
O
f 2 max  2 N 2  0.3(1870)  561  mg sin 25o  871N (as before)
O _

Again, then, xc cannot be zero. Assuming again that the blocks both move, f2 is its
maximum and equation (5) gives

xc  (871  561) / 210  1.48m / s 2 ma

σ
Substituting this acceleration into equation (1), we get
f1  415  100(1.48)  267  1 N1  0.4(889)  356 N
哮
This time we have more friction than we need in order to prevent B1 from slipping on

B2. Thus both xc1 and xc 2 are 1.48m / s 2 #

2-9
十
§ 2.3 Work and kinetic Energy for Particles and Mass Center Motions (Text Ch.
14)

1.Work and Kinetic Energy：

F ＝ m ac

 F  v c ＝ m ac  vc

＝m
d vc
dt
 vc
∞
m 器 √
= wi

＝
m d
2 dt

vc  v c 
d m  d m 2
＝  vc  v c  ＝  vc 
dt  2  dt  2 

Integrating （assume t1 as initial cond. and t 2 as final cond.）

  F  v dt ＝ 2  v t   v t  
t2 m  2 2
c c 2 c 1
t1

  F  v dt ：mass center work

t2
c
t1

Work done by the external forces were each to have a line of

action through the mass center.
m 2 2

 v c t 2   v c t1   ：change of mass center kinetic energy
2
If every point in the body has the same velocity of the
mass center.
mass center work＋change of mass center kinetic energy
＝mass center work and kinetic energy principle
This principle is for mass centers of bodies of bodies.
The work and kinetic energy expressed in this equation do not cover the whole work
and kineticenergy that would be done in this body.

若是一個 rigid body 可以以一 mass center 來表示，則可用此法，事實上

mass center  particle
For a particle：

2-10
 
t2

t1
 F  v dt ＝ m2  v t   v t  
c 2
2
c 1
2


t2

t1
 F  v dt ：Work done on the particle
m 2 2

 v c t 2   v c t1   ：change in the particle’s kinetic energy
2
Work done on the particle＋change in the particle’s kinetic energy
＝Work and kinetic energy principle
or w ＝ T ＝ T2  T1 （This is a scalar equation.）
Note that for a constant force F then
  
t2  t 2 dr 
 r t 2   r t1 
t2
t1 F  vdt ＝ F  t1 v dt ＝ F  t1 dt dt ＝ F

where r is the position vector of particle.
Work done is the dot product of the force with the displacement of the particle.

2.Central force：
The line of action always passes through the same fixed point in the frame of
reference, and a magnitude that depends only upon the distance “r” of the
particle from that fixed point.

suppose v is the velocity of the particle,

 

 d 
v＝ r e r ＝ r er  r er
dt

 
 

 
F v ＝  f r e 
r  r e r  r e r
 
 dr
＝  r f r  ＝  f r 
dt
dr r t 2 
W ＝   f r  f r dr
t2
dt ＝   ------（A）
t1 dt r t1 

d
If  is a function of “r” and f ＝ ------（B）
dr

Work done by F ＝  f r er

Note：
t2   r t 2  
(1)(a) w ＝  F  v dt ＝ F   dr ＝ F  r2  F  r1
t1 r t1 

2-11
 (b) w ＝ KE 2  KE1 ＝
m
2

vt 2   vt1 
2 2

(2)central forces do the work  potential energy

usually F ＝  f r er

dr r2 d
dr ＝  r1    r2 
t2
(a) w ＝   f dt ＝  
t1 dt r1 dr

d
f＝ or  ＝  fdr
dr
(b) 1   2 ＝ KE 2  KE1 ＝ T2  T1
conservative of mechanical（total）energy for conservative forces only.

r t 2  d
w＝ dr ＝   r t 2    r t1  ------（C）
r  t 1 dr

(a)Special case 1：linear spring

f ＝ k r  L0  ＝ k
where L0 ：unstrethed length
k ：spring modulus
k k
 ＝    r  L0 2 ＝   2
 2  2

w ＝   2 t 2    2 t1 
k
2
(b)Special case 2：gravitational force
From Newton’s liw of universal gravitation
2
mgr
f ＝ 2e
r
from （A）

2 
2
r t 2  mgre 1
w＝ dr ＝ mgre  
r t1 
 r t 2   r t1  
2
r

3.Conservative force：
A force whose work is independent of the path traveled by the point on which it
acts.
i.e. The work depends only on the initial and final positions, and for a constant

force,  ＝  F  r

Now from the work and kinetic energy principle w ＝ T

2-12
  t1    t 2  ＝ T t 2   T t1 
or T t 2    t 2  ＝ T t1    t1 
or T   ＝constant＝total energy
＝mechanical energy
For a conservative system, the total energy is constant.
Note：
Non-conservative force：friction force

2-13
 = 立 lmq ^~ )
m
OT

r = Herotrioea
= lole

皆L
v,

T
Tz
Ψ, φ
,
=
-
e

在 { mlor
T =
U

背浸是
向
00
ψ =mglsin {
io

Ψ z
=
mglsin θ

姚
( )

25 lb 4 V ,
Vz =0

lb
↓
=
25
作城的 mg
11pe 9 8
mls
“

sIg
=
.

鬣 ”
= 9 goculs
U 03= .

含
a GIy 32 t
2
= .
F kS F 向上待 , V “

0
=

器
“

U = 正向 可 勳能變化
W=4T E T 1
{ in /
s ( × 12 )
=

1 . Fr 克 E - 比 N = O
1

最
然 -0z
] Appy 在 = OT
m =
slag
[ 是 0 是 .

frictom 被功 [ 比f ) =
2. 2
N = mg 告 zolb
= [ + I + c .
O

[ ( ri viy ] / )
-
1 ω+ 0
6 ( 10 ) + 18
=
f = MN = 6 ub -
+L

500 =
定攀 ( 30)
-
2
比 = - b (0 + 0 ) -

3
mg 做功 g
:
1 64 in
.
=
.

瓦mp = 25 季
×
Clo + c ) Fmap = k -
ε = [ 64 db
4. 作功比3p
SPrimy
p =
令 ( 3?
-

六
^ = ) 量

2-14
二
。
。
、
一
、
一
牛
一
一
心
民
十
二
力
§ 2.4 Momentum Form of Euler’s First law (Text Ch. 14)

回顧 Text Ch. 12 單一質點:

Text Ch. 12 System of one particle

2-15
2-16
接下來, 我們要考慮多質點力系(系統)…

2-17
 Text Ch. 14 System of particles
lolzr
Momentum（ L ）of a particle is defined to be the product of its mass and its velocity.
N
L ＝  mi vi
i 1
連續体

Note：For a body of continuously distributed mass. L   v dm ⑤→
d Ri
If Ri is a position vector for the i th particle, vi 
dt
L ftmzv
 m R 
= m
d Ri d
.
N
L   mi  i i Ψ limear momentum
i 1 dt dt :

=
( ansular momentam

→
質 樂登中
m R
0r

 mroc
,

recall i i

L
d
dt
dr
 
m roc  m oc  mv c
dt

dL d

dt dt

mv c  ma c   F 
 歐拉第 上式
 dL
 F  ------ （A） momentum form of Euler’s first Law.
dt
momentumy
of Lipsqr
COnsfraatim

Jq d你tn
E ol tot

超 的
= o

. [ bxitVoy }
omeme 8
[ = m. 特 MoA . Dparticles Mz的 + Mo
8
— — 3M

Integrate equation （A） “ 3

Mz ( Ani Vzcyj }
動量改变量
+

量
   
  F dt  Lt 2   Lt1  ＝change in momentum ＝ L lhorpieVizyj
t2 Ma ( zz + 的y) 看

t1 “
@
"
=
Loefficieut ofrestitaton
彈性璃撞妥疑
 Impulse-Momentum Principle 衝量 加時間的 √ 第 e =

嚴器鴒想瓷菲 leanmun
E 後上前, 假設AB , 碰撞⼀定前 "isintialstate
J Fat 碰撞⼀定有兩質點 ,
ε
=

後 f final srate
"
"

impulse 衝量
.

mo mentam change
⼩ dint
Poblem
[
)
平⾯碰撞 must be vectr '
s
Ai fixt Viyj +

dirj ie
=

y
-

于 .

let
ry L
O .
I ,

⼀axis
.

^
UBi
UBmxd M

c VAYJ 女知⾸撞 poplem

B
v
= 能量交換必在 LD ⼯
.
,

sVoits ^
VAf =VAfxi⑤
能量在y -
axis 不变

[
+
VA
VAfyJ VAiyj
VA
A

↓
=
”

Vbfxi ]
2-18Ubf
+
mmj Ub Vbfyj UAiy5
= → =

ontact point
<

fangline
、
士
心
心
力
力
一
 ＊Coefficient of Restitution：
A measure of the capacity for colliding bodies to rebound off each other.
Typical case： apply impulse-momentum principle：
t2 
  F dt  m A vc  v Ai
t1
 


t1
t2 
F dt  mB vc  v Bi  
tipy
:

判斷 V碰是否為常數 t3 
  F dt  m A v Af  vc
t2
 
.
if =
0

V = 0,
9
,

) Fdt o

 了
ε
=

a =
0 .
t3 
= 6後 = (
前 ( . 0 . 1 H
. ] 
t2
F dt  mB v Bf  vc
⼀
碰撞係数定新
VInt
VAt
tip
-

"C
coefficient of restitution “e” VAI
=
1
-

0 = e =
,

OAO ( 三 7 單鞋能意中
“
求碰撞後速度
,
J

RA = 排 1B
rBixi
F
<

碰撞系数 )
T const , 3
8 = 0
OE
0.
e=
=
( =
a

V 说像
yj
U. 以
I M P
o
C . .

iyj
.
.

你
.

前 扎後 =

上不做功 ⼉不变
的⽅向能量保留
,

了 VAix
^

能量只在
交换
,

Lo I .
L M
APPly
C .
0 .
.

①
前 ③
8 後 ⼀ ⼀

mttmVBt
☆
:

)
缓 :
VAi + 的 Bi
tos 30g
UAt
媒
4 ( sin 30i
= .

VAf = Vin + j 六 46

= zi + 3 46 j fs
.

N y
-
dir onsenmed
⼀

- Ʃµ - j )
^

VBa =5
f - v
(

V = j

3 V

m (2 ) + m ( 4)
-

mV1 + mvi
号
= .

=
-

φi
-
> j xdir =

d唱
→ v

,
wth ⾶ =
0 . 8 解聯立 ,

2-19
2-20
 ⾓動量守恆 rxF =

P
§ 2.5 Momentum Forms of Euler’s Second Law (Text Ch. 14)
1
%
⼒矩
1.Moment of momentum （angular momentum） w.r.t. P（arbitrary point）
上
-

卵
H

H p   ri  mi vi
week
“ 品
Next -

剛体的⾓動量
H p ：moment of momentum w.r.t. P （angular momentum）
freerotation

r m v
i i i ：the sum of the moments （w.r.t. P）of the

moment of the individual particles making up the body.

H c    i  mi v i ri  rpc   i


 H p    i  rpc  mi vi 
＝  i  mi vi  rpc   mi vi


since  i i  mv c
m v L


 H p  H c  rpc  L ------（A）
相对物理量 ( 消動量 ]
H p ：moment of momentum of a body about any point P 任意點

H c ：moment of momentum about its mass center C 質⼼ ×

↓
則剛体轉是以質⼼系中⼼
繞著轉的
 
rpc  L ：moment of its momentum L about P 物体轉動 皆相對於質⼼ ,

于故⼼相对於剛体为固定點,
2.consider a point O fixed in the inertial frame of ref.

H o   Ri  mi vi
CO . A
. M ( ⾓動量守恒了
⾓⼒量 d Ho    += 0 于⾓動量守恒
变化率    Ri  mi vi  Ri  mi ai  ⼒矩 = ,
0

dt  
 

＝  Ri  mi ai   M o

d Ho
  Mo  momentum form of Euler’s 2nd Law.
dt

2-21
(a)From equation （A）

H o  H c  roc  L

d H o d H c d roc  dL
   L  roc 
dt dt dt dt

d Hc dL
＝  roc 
dt dt
consider the moment about “O”

 

 O  c oc 
M  M  r  F

d Ho  dL
note that  M O  and  F  dt
dt
d Hc
 Mc 
dt
(b)For an arbitrary point P

M p   M c  rpc   F

d Hc dL
  rpc 
dt dt


 M p  H c  rpc  mac

2-22
2-23
讓我們來複習一下…

2-24
 五

瘤
R Yij :
”
= Nomomemt
melght 国
巧

,
Ʃ M = 0 =d
社
=e

器濕喆 iatt
,

7δ
dl
快
=ω
= H

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e1

myij )

i ^ dLj ) (
tl -
+

等在 zu

洲男在 _
= r
γm

=
YR YL =

2-25
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