Centre of Mass
• The centre of mass of a body or system of
bodies is also called the centre of gravity (for
uniform gravity field) and can be defined as
the point that moves as though all of the
mass were concentrated there and all forces
were applied there.
• The idea of the centre of mass is that it is an
average of the masses factored by each
particle’s distance from a reference point.
• Consider:
• Consider:
• The centre of mass for two objects of mass m1
and m2 is:
m x m x
x com 
1 1
2
m1  m 2
2
• In general: x
com

1
M
n
mx
i
i 1
n
i
M 
m
i 1
i
• In general: x
com

1
M
n
mx
i
n
M 
i
i 1
m
i
i 1
• Further:
y com 
x com 
1
M
n
1
M
m
i
i 1
n
mx
i
i 1
z com 
yi
i
y com 
1
M
1
M
n
mz
i
i 1
n
m
i 1
i
yi
i
z com 
1
M
n
mz
i
i 1
i
• Defining the centre of mass in terms of a
vectors, r  x iˆ  y ˆj  z kˆ
com
• So that,
com
1

rcom 
M
com
n

i 1

m i ri
com
• For solid bodies made of many particles which
can be considered a continuous distribution of
matter,
x com 
1
M
 x dm
y com 
1
M
 y dm
z com 
• where dm are mass elements.
1
M
 z dm
• For solid bodies of uniform density we can
rewrite the previous equations so that,
x com 
1
V
 x dV
y com 
1
V
 y dV
z com 
1
V
 z dV
Example
• Find the location of the centre of mass of the
system of three particles if each particle has a
mass m.
y
(0,L,0)
(0,0,0)
(L,0,0)
x
Example
• Sol:
y com 
x com 
1
M
1
M
y
n
mx
i
i
n
m
i 1
i
yi
(0,L,0)
i 1
z com 
1
M
n
mz
i
i 1
i
(0,0,0)
(L,0,0)
x
Example
• Sol:
y com 
x com 
1
M
y
n
mx
i
i
n
1
M
m
i
(0,L,0)
i 1
z com 
yi
i 1
1
M
1
n
mz
i
i 1
i
2
3
(0,0,0)
(L,0,0)
• Using (0,0,0) as the reference point,
x com 
1
M
n
mx
i
i 1
i

1
3m
 0  m   0  m   L  m 

L
3
x
Example
• Sol:
y com 
x com 
y
n
1
M
mx
i
i
n
1
M
m
i
(0,L,0)
i 1
z com 
yi
i 1
1
M
1
n
mz
i
i 1
i
2
3
(0,0,0)
(L,0,0)
• Using (0,0,0) as the reference point,
x com 
y com 
1
M
1
M
n
mx
i
i

i 1
n
m
i 1
i
yi 
1
3m
1
3m
 0  m   0  m   L  m 
 mL
 0  0 
L
3

L
3
x
Example
• Sol:
y com 
x com 
y
n
1
M
mx
i
i
n
1
m
M
i
(0,L,0)
i 1
z com 
yi
i 1
1
M
1
n
mz
i
i 1
i
2
3
(0,0,0)
(L,0,0)
• Using (0,0,0) as the reference point,
x com 
y com 
1
M
1
M
z com  0
n
mx
i
i

i 1
n
m
i 1
i
yi 
1
3m
1
3m
 0  m   0  m   L  m 
 mL
 0  0 
L
3

L
3
x
• Therefore the centre of mass is
 L L

 , ,0 
 3 3

.
Example: The subtraction method
• A uniform metal plate of radius 2R has a disk
of radius R removed as shown. Find the plate’s
centre of mass.
R
2R
Example: The subtraction method
• Sol: The plate is the composite of two discs of
radius R and 2R. Therefore to find the com we
calculate the com of each disc, subtracting the
mass of the smaller disc.
R
2R
2R
=
-
R
Example: The subtraction method
R
2R
2R
=
-
R
S
P
C
• By symmetry the centre of mass for “C” is at
the centre (0,0) of “P”.
Example: The subtraction method
R
2R
2R
=
-
R
S
P
C
• By symmetry the centre of mass for “C” is at
its centre which also the centre (0,0) of “P”.
• Similarly, the com of “S” is at its centre which
is at (-R,0).
Example: The subtraction method
R
2R
2R
-
=
R
S
P
C=P+S
• The com of the P+S is,
x com 
1
M
n
mx
i
i 1
i
and y com 
1
M
n
m
i 1
i
yi  0
Example: The subtraction method
R
2R
2R
-
=
R
S
P
C=P+S
• The com of the P+S is,
x com 
xPS 
1
M
n
mx
i
i
i 1
m S xS  m P xP
mS  mP
and y com 
1
M
n
m
i 1
i
yi  0
Example: The subtraction method
R
2R
2R
-
=
R
S
P
C=P+S
• The com of the P+S is,
x com 
xPS 
1
M
n
mx
i
i
and y com 
i 1
m S xS  m P xP
mS  mP
0 
1
M
n
m
i
yi  0
i 1
m S xS  m P xP
mS  mP
Example: The subtraction method
R
2R
2R
=
-
R
S
P
 xP   xS
C=P+S
mS
mP
Example: The subtraction method
R
2R
2R
=
-
R
S
P
 xP   xS
C=P+S
mS
mP
However m=density x thickness x area
 xP   xS
 s  t s  As
 p  t p  Ap
Example: The subtraction method
R
2R
2R
=
-
R
S
P
 xP   xS
C=P+S
mS
mP
However m=density x thickness x area
 xP   xS
 s  t s  As
 p  t p  Ap
  xS
As
Ap
  xS
R
2
 2 R    R
2
2
Example: The subtraction method
R
2R
2R
=
-
R
S
P
C=P+S
Substituting for xs
x P    R 
R
2
 2 R    R
2
2

R
3
• Therefore the centre of mass is
 R

,0 

 3

.
Example
• Find the centre of mass for a rod of nonuniform density    0  x / L .
x
dx
Example
• Find the centre of mass for a rod of nonuniform density    0  x / L .
x
dx
x com 
1
M
 x dm
Where, dm   dx
Example
• Find the centre of mass for a rod of nonuniform density    0  x / L .
x
dx
x com 
 x com 
1
M
 x dm
1
 dm
 x dm
Where, dm   dx

1
  dx
 x  dx
Example
• Find the centre of mass for a rod of nonuniform density    0  x / L .
x
dx
x com 
 x com 
1
M
 x dm
1
 dm
 x dm
Where, dm   dx

1
  dx
 x  dx

1

0
x / Ldx
L
 x 
0
0
x / L  dx
Example
• Find the centre of mass for a rod of nonuniform density    0  x / L .
x
dx
x com 
1
 0 / L  xdx
L
  0 / L  x dx
2
0
Example
• Find the centre of mass for a rod of nonuniform density    0  x / L .
x
dx
x com 
1
 0 / L  xdx
L
  0 / L  x dx 
1
2
0
x
2
2
0
L

x
3
3
0
L
Example
• Find the centre of mass for a rod of nonuniform density    0  x / L .
x
dx
x com 
L
1
 0 / L  xdx
3
 x com 
L /3
2
L /2
  0 / L  x dx 
1
2

2L
3
0
x
2
2
0
L

x
3
3
0
L