Lesson 18
System of Particles
I.
A.
System of Particles
Background
If we have a collection of particles, each of the particles will have
different position vectors and may have different masses, be traveling
with different velocities and different accelerations. This leads to the
following question:
If we consider all the particles as a single system, which particle or
location is the one described by Newton’s 2nd Law?

F
Ext

MA
V4
M2
V2

r2
y
V1
M4
V3
M1

r1

r5

r4

r3
M3
M5
V5
x
The motion of a system of particles or extended object is equivalent to
that of a single particle whose mass is the total mass of the system and
is located at a point in space called the center of mass.
B.
Definition of the Center of Mass
The center-of-mass is by definition the average location of the system:
N

rcm 
M
i 1
i

ri
N
M
i 1
i





M 1 r1  M 2 r2  M 3 r3  M 4 r4  ....  M N rN

M 1  M 2  M 3  M 4  ....  M N
C.
Important Facts
1.
The center-of-mass of a system will often reside at a location in which
no particle exists!!
Example: Where is the center-of-mass (average location) of the
following donut?
2.
Analysis of compound objects or systems is often easier using the
concept of the center of mass.
Example: The center-of-mass of a player trying to dunk a basketball
must always stay on the parabolic path for projectile motion that we
worked out in chapter 3.
X
X
The center-of-mass of the player-ball system must stay on the
parabolic path determined by the player’s initial velocity. When the
player kicks their feet downward, the player’s body and the ball must
move upward to keep the center-of-mass on the same path.
PROBLEM 1: Determine the center-of-mass for the following set of point
particles.
y
2 kg
1 kg
2 kg
x
1 kg
3 kg
D.
Continuous Bodies
Although all bodies are actually composed of small discrete particles
(atoms and molecules), it is often easier to make the calculations by
treating the body as a continuous mass distribution. In this case, the
summations can be replaced by integrals.


r dM  r dM


rcm 

M
dM

In order to evaluate these integrals, you must first be able to write the
mass differential dm. This is done using the concept of density.
a)
Linear Mass Density - 
Linear mass density for a differential length dx and mass dm of
an object is defined by the equation
λ
x
dM
dx
dx
L
Rearranging this equation gives us the mass differential in
terms of a linear object’s density:
dM  λ dx
You will either have to be given the functional dependence of
the mass density or the object will have to be an object of
uniform density where you can find the density by taking the
entire length of the object:
λ
M
L
PROBLEM 2: Prove that the center of mass of a uniform density steel thin
rod is at the center of the rod.
y
x
x
dx
L
PROBLEM 3: Find the center of mass of the thin concrete strip shown
below given that its linear density is λ  3 x 2 .
y
x
x
dx
5
b)
Surface (Areal) Mass Density - 
Surface mass density for a differential area dA and mass dm of
an object is defined by the equation
σ
y
dx
x
dM
dA
W
dy
L
Rearranging this equation gives us the mass differential in
terms of an object’s areal mass density:
dM  σ dA
This converts our mass integral into a surface integral.
c)
Volume Mass Density - 
Volume mass density for a differential volume dV and mass dm
of an object is defined by the equation
ρ
dz
dx
dM
dV
dy
Rearranging this equation gives us the mass differential in
terms of an object’s volume density:
dM  λ dV
This converts our mass integral into a volume integral.
Since skill in evaluating multiple integrals is not required for
this course, only the simplest types of area or volume density
problems are done in your textbook. However, it is important
for most students to practice the formal steps in setting up even
these simpler problems since they will see more complicated
problems in other classes which require math skills involving
either vector integration or the evaluation of multiple scalar
integrals. These harder problems include working with nonuniform charge distributions in PHYS2424 and non-uniform
mass distributions in the Sophomore Engineering Principles I
class. It is common for engineers to mix materials (i.e. add steel
rebar to concrete) to produce non-uniform materials with
improved properties or to change the physical dimensions of a
system like thickness in order to handle non-uniform loading of
forces (i.e. water pushes hardest at the bottom of the dam).
E.
Tricks of the Trade
1.
You can often break a problem a complicated non-uniform object into
several uniform shapes whose individual center-of-masses are either
known or can be easily calculated.
2.
You can treat any holes drilled into an object’s by first calculating a
solid object and then accounting for the holes by treating them as
extra particles with “negative” mass.
PROBLEM 4: Find the center-of-mass of the following plate of mass M
and uniform density.
10 cm
10 cm
10 cm
20 cm
30 cm
10 cm
40 cm
3.
The center of mass of many common homogenous objects has already
been calculated. The results of these calculations can be found in
tables in engineering and physics text books.
4.
You can also find the center of mass of irregular objects
experimentally. For everyday objects on the Earth, the pull of gravity
is approximately constant so the center of gravity (the effective point
where the Earth pulls on an extended object) is the same point in
space as the center of mass. By holding an object at different points
and allowing it to rotate under the influence of gravity, it is possible to
find the center of gravity.
This is also the point at which you could place a support to balance
the object!! You may have seen acrobats who were balancing in
strange formations or small toy birds balancing on their beaks. Now,
you should be able to understand how toy works.
II.
Newton II and Center-of-Mass
A.
Velocity of the Center-of-Mass
Using the definition of velocity, we have that the velocity of the
center-of-mass is


d rcm
Vcm 
dt
If we now insert the definition of the center-of-mass into our equation
above, we have

Vcm

 N
  M i ri
d  i 1
  N
dt
  M i
 i 1

 d  N
M
ri 


i
 i 1
dt



N

Mi


i

1


d ri
Mi

dt
i 1
N

Vcm 
N
 Mi
i 1
N
M

Vcm 
i 1
i

vi
N
M
i 1
N

M
i 1
i

vi
N
M
i 1
i





M 1 v1  M 2 v 2  M 3 v 3  M 4 v 4  ....  M N v N

M 1  M 2  M 3  M 4  ....  M N
i
Thus, we see that the velocity of the center-of-mass is just the average
velocity of the system of particles.
B.
Deriving Newton’s 2nd Law for A System of Particles
We will start by finding the total linear momentum of the system of
particles.

P
N

N
p M
i 1
i
i 1
i






v i  M1 v1  M 2 v 2  M 3 v 3  M 4 v 4  ....  M N v N
The time-rate-of-change of the total linear momentum of the system is
then given by




d P d M Vcm


dt
dt

d pi

i 1 dt
N
Newton’s 2nd Law for a particle says that the time-rate-of-change of
momentum for a particular particle is equal to the net external force
applied to that particle so we can apply this to each particle in the
system

dP

dt
N

F
i
i 1
We can now break the forces applied to each particle into two parts:
(those due to agents external to the system and those due to
interactions with other particles in the system).

dP

dt

F
 i (external) 
N
i 1

F
 ij
N
i 1
j 1 , i  j







dP 
 F1(external)  F2 (external)  ...  FN (external)  F12  F21  F13  F31  .........
dt


Newton’s 3rd Law tells us that Fij   Fji so the second summation of
internal forces is zero. Thus, we have our final result that

dP

dt




F

F

F

...

F
 i (external) 1(external) 2 (external)
N (external)
N
i 1




d P d M Vcm 

FExternal
dt
dt
This is Newton’s 2nd Law for a system of particles with the total linear
momentum of the system only changed by the forces outside the
system even though the particles are allowed to move with respect to
each other like in spilled water or an exploding bomb!!
We now consider some examples where the center-of-mass concept
helps us to analyze complicated problems.
EXAMPLE 1: A uniform chain of length L and mass M is hanging from a
table of height L as shown below. If a student pulls the chain up onto the
table, how much work does the student do in increasing the chain’s
gravitational potential energy?
L
A.
Work Method Using Calculus
B
Using Center of Mass to Simplify Calculations
EXAMPLE 2 : At t = 0 a ball of mass M is traveling at speed V in the +x
direction inside a box of mass M and length 2L which is at rest as shown
below. Graph as a function of time a) the position of the ball; b) the position
of the box; and c) the position of the center of mass.
V
x=0
x = 2L
B.
Energy for Systems of Particles
It is important in dealing with systems of particles to be careful and
work from first principles and not just memorize. For instance, many
students incorrectly assume that the kinetic energy of a system of
particles is just the kinetic energy of the center of mass.
K Total  K cm
An object whose center-of-mass is at rest may still have particles who
have speed, but whose net velocity is zero. For instance, consider the
two 1 kg balls moving below:
y
2 m/s
2 m/s
x
The center-of-mass velocity is zero so the problem’s center-of-mass
motion is equivalent to the following:
y
0 m/s
2 kg
x
Thus, the kinetic energy of the center of mass is zero while the total
kinetic energy of the system is 4 J!!
The actual kinetic energy of a system can be related to the kinetic
energy of the center of mass by using our knowledge of reference

frames. The velocity of the ith particle, v i , can be written in terms of

the velocity of the center of mass, Vcm , and the velocity of the ith

particle with respect to the center-of-mass reference frame, Vi , as

 
v i  Vcm  Vi

Ri

ri

rcm
We can then find the total kinetic energy as
K Total 






 
1 N
1 N
m i v i  v i    m i Vcm  Vi  Vcm  Vi

2 i 1
2 i 1

Applying the dot product and remembering that the order of the
vectors doesn’t matter for dot products, we have after some algebra



  m V
K Total 


 
1 N
1 N
m i Vcm  Vcm   m i Vi  Vi 

2 i 1
2 i 1
K Total 

1 
Vcm  Vcm
2
K Total 

 m
N
i 1

i


i
i 1
cm
  m V
 
1 N
m i Vi  Vi 

2 i 1


N


N
i
i 1
cm

 Vi


 Vi


N


 


1
1 N
M Vcm  Vcm   m i Vi  Vi  Vcm   m i Vi
2
2 i 1
i 1
The last sum is a sum of linear momentum as measured in the centerof-mass reference frame and is always zero by the definition of the
center-of-mass!!
K Total 





 
1
1 N
M Vcm  Vcm   m i Vi  Vi
2
2 i 1

Thus, the total energy is reduced to sum of two kinetic energy terms
which can be interpreted as the kinetic energy of the center of mass
and the kinetic energy with respect to the center of mass. This second
term is usually called internal kinetic energy.
K cm 



1
M Vcm  Vcm
2
K Internal 


 
1 N
m i Vi  Vi

2 i 1
K Total  K cm  K Internal

In our previous, example the system had 0 J of center-of-mass kinetic
energy and 4 J of Internal Kinetic Energy.
A similar discussion can be performed for potential energy leading to
U Total  U cm  U Internal
The center-of-mass potential energy is what we have been dealing
with throughout this semester. We have neglected the energy which
can be stored inside an object due to chemical, nuclear, and other
interactions between the atoms and molecules which make up a real
block, pulley, or other macroscopic object. The later chapters in this
book cover a more complete discussion of energy leading to the fields
of Statistical Mechanics and Thermal Physics (also called
Thermodynamics). Due to time constraints and our belief in a less is
more approach to physics education, we don’t cover 20+ chapters in
PHYS1224 like some universities so. Thus, we will limit our
discussion in this subject to the time available. You will be required to
take advanced courses in these subjects if you are an engineer or
physicists. The important thing to remember is that a strong
understanding of classical mechanics is necessary for success in a
wide range of related subjects like thermodynamics and not just for
solving blocks moving on planes.
In thermodynamics, we usually separate quantities based upon
macroscopic motion (center-of-mass) and microscopic motion (with
respect to the center of mass). Here are several important terms of
thermodynamics and their origins in mechanics:
1) Microscopic kinetic energy is called heat, Q.
2) Temperature, T, is a measure of the average kinetic energy of an
atom or molecule. The correct temperature scale in physics
problems is the Kelvin scale as you can’t have negative
temperatures as they imply imaginary speeds!!
3) The total energy of a system can be written as
E  K cm  U cm  K Internal U Internal
The change in the center-of-mass energy can be accounted by work
done on the system by external forces, W. Thus, we can write the
energy equation as
ΔE  W  K Internal final  K Internal Initial  U Internal final  U Internal Initial 
ΔE  W  ΔQ  ΔU
This equation basically says that the total energy of a system can
be changed by doing work on the system, adding heat energy or
increasing the microscopic internal energy of the system.