CHAPTER 14
Systems of Particles
14.1. INTRODUCTION
• In the current chapter, you will study the motion
of systems of particles.
• The effective force of a particle is defined as the
product of it mass and acceleration. It will be
shown that the system of external forces acting on
a system of particles is equipollent with the system
of effective forces of the system.
• The mass center of a system of particles will be
defined and its motion described.
• Application of the work-energy principle and the
impulse-momentum principle to a system of
particles will be described. Results obtained are
also applicable to a system of rigidly connected
particles, i.e., a rigid body.
• Analysis methods will be presented for variable
systems of particles, i.e., systems in which the
particles included in the system change.
14.2. APPLICATION OF NEWTON’S LAWS TO THE
MOTION OF A SYSTEM OF PARTICLES.
EFFECTIVE FORCES
• Newton’s second law for each
particle Pi in a system of n
particles,
 n 

Fi   f ij  mi ai
j 1


  n  


ri  Fi   ri  f ij  ri  mi ai
j 1


Fi  external force
f ij  internal forces

mi ai  effective force
• The system of external and internal
forces on a particle is equivalent to
the effective force of the particle.
• The system of external and internal
forces acting on the entire system of
particles is equivalent to the system
of effective forces.
• Summing over all the elements,
n
 n n 

F

f

m
a
 i  ij  i i
n
i 1
i 1 j 1

n
i 1

i 1


n
n
n
 
 


ri  Fi   ri  f ij   ri  mi ai 
i 1 j 1
i 1
• Since the internal forces occur in
equal and opposite collinear pairs,
the resultant force and couple due
to the internal forces are zero,


 Fi   mi ai


 


ri  Fi   ri  mi ai 
• The system of external forces
and the system of effective
forces are equipollent by not
equivalent.
14.3. LINEAR AND ANGULAR MOMENTUM OF A
SYSTEM OF PARTICLES
• Linear momentum of the system
of particles,
• Angular momentum about fixed
point O of system of particles,
 n

L   mi vi
n



H O   ri  mi vi 
n
 n


L   mi vi   mi ai
n
n





H O   ri  mi vi   ri  mi vi
i 1
i 1
i 1
i 1
• Resultant of the external
forces is equal to rate of
change of linear momentum
of the system of particles,
 
F  L


i 1


i 1
n



H O   ri  mi ai 
i 1
• Moment resultant about fixed
point O of the external forces is
equal to the rate of change of
angular momentum of the system
of particles,


 M O  HO
14.4. MOTION OF THE MASS CENTER OF A SYSTEM
OF PARTICLES
• Mass center G of system of particles is

defined by position vector rG which
satisfies
n


mrG   mi ri
i 1
• Differentiating twice,
n


mrG   mi ri
i 1
n

 
mvG   mi vi  L
i 1



m aG  L   F
• The mass center moves as if the entire
mass and all of the external forces were
concentrated at that point.
14.5. ANGULAR MOMENTUM OF A SYSTEM OF
PARTICLES ABOUT ITS MASS CENTER
• The angular momentum of the system of
particles about the mass center,
n



H G   ri mi vi
i 1
n
n




 
H G   ri mi ai    ri mi ai  aG 
i 1
  
ai  aG  ai
• Consider the centroidal
frame of reference
Gx’y’z’, which translates
with respect to the
Newtonian frame Oxyz.
• The centroidal frame is
not, in general, a
Newtonian frame.
i 1
n


  n  
H G   ri mi ai     mi ri   aG
i 1
 i 1


n
n



 
H G   ri mi ai    ri Fi
i 1
i 1


H G   M G

• The moment resultant about G of the
external forces is equal to the rate of
change of angular momentum about G of
the system of particles.
• Angular momentum about G of particles
in their absolute motion relative to the
Newtonian Oxyz frame of reference.
n



H G   ri  mi vi 
i 1
n


 
H G   ri  mi vG  vi
  
vi  vG  vi
• Angular momentum about G
of the particles in their
motion relative to the
centroidal Gx’y’z’, frame of
reference,
n



H G   ri  mi vi
i 1
i 1
n

 


 n
H G    mi ri   vG   ri  mi vi
i 1
 i 1




H G  H G   M G
• Angular momentum about G of the
particle momenta can be calculated
with respect to either the Newtonian
or centroidal frames of reference.
14.6. CONSERVATION OF MOMENTUM FOR A SYSTEM
OF PARTICLES
• If no external forces act on
the particles of a system,
then the linear momentum and
angular momentum about the
fixed point O are conserved.


L F 0

L  constant


HO   M O  0

H O  constant
• In some applications, such as
problems involving central
forces,


L F 0

L  constant


HO   M O  0

H O  constant
• Concept of conservation of
momentum also applies to the
analysis of the mass center motion,


L  F 0


L  mvG  constant

vG  constant


HG   M G  0

H G  constant
14.7. KINETIC ENERGY OF A SYSTEM OF PARTICLES
• Kinetic energy of a system of
particles,
 
T   mi vi  vi  
n
1
2
i 1
n
1
2
m v
i 1
2
i i
• Expressing the velocity in terms of
the centroidal reference frame,
T
   


m
v
 i G  vi  vG  vi
n
1
2
i 1
 1 n
 n
 2  n
T    mi vG  vG   mi vi  2  mi vi 2
i 1
i 1
 i 1 
1
2
  
vi  vG  vi
T  mv 
1
2
2
G
n
1
2
 m v
i 1
2
i i
• Kinetic energy is equal to kinetic
energy of mass center plus kinetic
energy relative to the centroidal
frame.
14.8. WORK-ENERGY PRINCIPLE. CONSERVATION OF
ENERGY FOR A SYSTEM OF PARTICLES
• Principle of work and energy can be applied to each particle Pi ,
T1  U 1 2  T2

represents the work done by the internal forces f ij
where U 1 2
and the resultant external force Fi acting on Pi .
• Principle of work and energy can be applied to the entire system
by adding the kinetic energies of all particles and considering the
work done by all external and internal forces.


• Although f ij and f ji are equal and opposite, the work of these
forces will not, in general, cancel out.
• If the forces acting on the particles are conservative, the work
is equal to the change in potential energy and
T1  V1  T2  V2
which expresses the principle of conservation of energy for the
system of particles.
14.9. PRINCIPLE OF IMPULSE AND MOMENTUM FOR A
SYSTEM OF PARTICLES
 
F  L


 M O  HO

 
  Fdt  L2  L1



  M O dt  H 2  H 1
t2
t1
t2



L1    Fdt  L2
t1
t2
t1
t2



H 1    M O dt  H 2
t1
• The momenta of the particles at time t1 and the impulse of
the forces from t1 to t2 form a system of vectors equipollent
to the system of momenta of the particles at time t2 .
14.10. VARIABLE SYSTEMS OF PARTICLES
• Kinetics principles established so far were derived for constant
systems of particles, i.e., systems which neither gain nor lose
particles.
• A large number of physics and engineering applications require
the consideration of variable systems of particles, e.g., hydraulic
turbine, rocket engine, etc.
• For analyses, consider auxiliary systems which consist of the
particles instantaneously within the system plus the particles
that enter or leave the system during a short time interval.
The auxiliary systems, thus defined, are constant systems of
particles.
14.11. STEADY STREAM OF PARTICLES
• System consists of a steady stream of
particles against a vane or through a
duct.
• Define auxiliary system which includes
particles which flow in and out over  t.
• The auxiliary system is a constant
system of particles over  t.
t2



L1    Fdt  L 2



 mi vi  m v A   F t 


t1
 dm  
 F  dt vB  v A 



 mi vi  m vB

Steady Stream of Particles. Applications
• Fluid Stream Diverted by
Vane or Duct
• Fan
• Fluid Flowing Through a
Pipe
• Jet Engine
• Helicopter
14.12. STREAMS GAINING OR LOSING MASS
• Define auxiliary system to include
particles of mass m within system at
time t plus the particles of mass m
which enter the system over time
interval  t.
• The auxiliary system is a constant
system of particles.
t2



L1    Fdt  L 2
t1





mv  m va    F t  m  m v  v 


 

 F t  mv  mv  va   m v



d
v
dm
 F  m dt  dt u
 dm 

ma   F 
u
dt
Work Some Example Problems
14.92 A chain of length l and mass m falls through a small hole in a plate.
Initially, when y is very small, the chain is at rest. In each case
shown on this slide and a later slide, determine (a) the acceleration
of the first link as a function of y and (b) the velocity of the chain
as the last link passes through the hole. In this first case assume
that the individual links are at rest until they fall through the hole
and in the later case assume that at any instant all links have the
same speed. Ignore the effect of friction in both cases.
Define a mass/length
 m/l
y
The mass of length y is
y
m'   y
Apply the principle of momentum-impulse.
Direction is positive downward.
y
 y v   y g t   ( y  y )( v  v )
y
  y v   y  v   v y    y v
dy
dv
y g  y v
dt
dt
d ( yv )
yg
dt
Can you see what we
can do to simplify this?
d( y v )
yg
dt
y
y g dt  d ( y v )
y
dy
v
dt
or
dy
dt 
v
Therefore
Remember that
dy
yg
 d( y v )
v
g y 2 dy  ( y v ) d ( y v )
In general this integrates to
g
y3
3

( y v )2
2
g
y
y
a  v dv
dy
y3
3

( y v )2
2
v  23 g y
When y = l
v  23 g l
g
dv 
dy
3
2g y
3
g
a
3
WOW
y
y
y
y
m v   y g  t  m( v   v )
m
y g t  m  v
l
dv
dv
y
v
g 
dy
dt
l
y
y
dv
y
g v
dy
l
dv
y
g v
dy
l
v2
y2
g 
2
2l
g
a y
l
For y = l
v gl
v
g
y
l
A 1200 lb spacecraft is designed to include a two stage rocket
with stages A and B, each weighing 21,300 lb, including 20,000 lb
of fuel. The fuel is consumed at a rate of 500 lb/s and ejected
with a relative velocity of 12,000 ft/s. Knowing that when stage A
expels its last particle of fuel, its casing is released and
jettisoned, determine the altitude at which (a) stage A of the
rocket is released and (b) the fuel of both stages has been
consumed.
Given:
Total Weight
W  1200 lb  2( 21,300 lb )
W  43 ,800 lb  mg
dW  500 lb / s  qg
dt
u  12 ,000 ft / s
Velocity of the exhaust
ve ˆj  ( v  u ) ˆj
Time to burn one tank of fuel
t
20 ,000 lb
 40 s
500 lb / s
Apply the principle of momentum-impulse.
Direction is positive upward.
( m  qt )v  ( m  qt ) g t  ( m  q( t  t ))( v  v )  mve
( m  qt )v  ( m  qt ) g t  ( m  qt )v  ( m  qt )v
 qvt  qtv  qt ( v  u )
 ( m  qt ) g  ( m  qt ) dv  qv  qv  qu
dt
 ( m  qt ) g  ( m  qt ) dv  qu
dt
qu
dv  {
 g }dt
( m  qt )
y
There is a corresponding
v
y
dv  {
v
t
0
0
qu
 g }dt
( m  qt )
 dv    {
qu
 g }dt
( m  qt )
v  u{ln[( m  qt )]  ln[ m ]}  gt
v  u ln[
( m  qt )
]  gt
m
At t = 40 s
v A  6031 ft / s
This is needed later.
v  u ln[
{ u ln[
( m  qt )
dy
]  gt 
m
dt
( m  qt )
]  gt }dt  dy
m
y
t
0
0
 dy   { u ln[
( m  qt )
]  gt }dt
m


( m  qt ) ( m  qt ) ( m  qt )
y A    mu {
ln[
]
}  1 gt 2 
m
m
m
2 0
 q
4 0s
y A  20.0 mi
To get the height at the second stage burnout use
v  u ln[
( m  qt )
dy
]  gt  v A 
m
dt
You would again integrate from t = 0 to t = 40 s
and use
W  1200 lb  ( 21,300 lb )
W  22 ,500 lb  mg


( m  qt ) ( m  qt ) ( m  qt )
y    mu {
ln[
]
}  1 gt 2  v At   y A
m
m
m
2
 q
0
4 0s
y  126.8 mi