Consider the meter stick.
It is an extended object. In other words we can’t simply treat it as a particle.
For example if we twirl it, each piece moves in a different radius circle with
different speeds!
Imagine that this meter stick is moving as well as spinning.
Now the motion of each piece of the meter stick is even more complicated!
However, there is one point on the meter stick that moves just like a simple particle!
This point is also special because it is the point on which the meter stick would
balance.
In fact, if we threw the meter stick through the air, this point would act just like a
projectile, it would follow a parabolic path.
Hence, this point acts as if all of the mass of the object is concentrated at that point.
Center of Mass – the point in an object or system of particles which acts as a
simple particle
– the point at which all of the mass of an object or system of
particles can be thought to be concentrated
– the point that moves as if all of the external forces were
applied at that point
– the balancing point
Rotating Projectile
Consider two masses m1 and m2 connected by a thin massless rod.
m1
cm
m2
How do the masses of the
two objects compare?
Since the center of mass is
closer to m2,
m2 > m1!
Consider two masses m1 and m2 connected by a thin massless rod.

FN
m1

Fg1
Since the system is balanced
it is in static equilibrium!

0
1   2  0
Fg1r1  Fg 2 r2  0
m2

Fg 2
Consider two masses m1 and m2 connected by a thin massless rod.
y
Since the system is balanced
it is in static equilibrium!

xcm
m1

x1
cm

x2
m2

0
x
1   2  0
Fg1r1  Fg 2 r2  0

FN
m1

Fg1

m2

Fg 2
Consider two masses m1 and m2 connected by a thin massless rod.
y

xcm
m1

x1
cm

x2
m2
x
Fg1r1  Fg 2 r2  0


 
m1g  xcm  x1   m2 g  x2  xcm   0




m1xcm  m1x1  m2 x2  m2 xcm  0




m1xcm  m2 xcm  m1x1  m2 x2



m1  m2 xcm  m1x1  m2 x2
Consider two masses m1 and m2 connected by a thin massless rod.
y

xcm

xcm
m1

x1
cm

x2
m2
x


m1 x1  m2 x2

m1  m2
What if the masses were both m?


mx1  mx2

xcm 
mm
 
x1  x2

xcm 
2
In a more general system of n particles, the position of the center of mass would be

xcm



m1x1  m2 x2      mn xn

m1  m2      mn
1 n 

xcm 
 mi xi
M i 1
where M  total system mass  m1  m2      mn
In 3 dimensions the center of mass would have coordinates given by
1 n 

xcm   mi xi
M i 1
1 n 

ycm   mi yi
M i 1
1 n 

zcm   mi zi
M i 1
Remember that the position can be described in 3 dimensions as
  ˆ ˆ  ˆ
r  xi  yj  z k
Hence the center of mass can be written as
1 n 

rcm 
 mi ri
M i 1
Example 1
Two objects one of mass 10. kg and one of mass 5.0 kg,
have their centers separated by 3.0 m. Where is their center of mass?
m1
m2
3.0 m
1 2 

xcm   mi xi
M i 1


m1x1  m2 x2

xcm 
m1  m2

Since x1  0
5.0 kg 3.0 m 

xcm 
10. kg  5.0 kg

xcm  1.0 m
Example 2
Four particles of mass 1.0 kg, 2.0 kg, 3.0 kg, and 4.0 kg are arranged as
shown below. What is the center of mass of the system?
m4  4m
4.0 kg
m3  3m
5.0 m
2.0 m
1.0 kg
m1  m
Line of symmetry
3.0 kg
1 4 

xcm   mi xi
M i 1




m1x1  m2 x2  m3 x3  m4 x4

xcm 
m1  m2  m3  m4
2.0 kg
 
m2  2m
Since x1  x4  0


2mx  3mx

xcm 
1m  2m  3m  4m
1

xcm  x
2

xcm  2.5 m
Example 2
Four particles of mass 1.0 kg, 2.0 kg, 3.0 kg, and 4.0 kg are arranged as
shown below. What is the center of mass of the system?
m4  4m
m3  3m
3.0 kg
1 4 

ycm 
 mi yi
M i 1




2.0 m
m1 y1  m2 y2  m3 y3  m4 y4

ycm 
m1  m2  m3  m4
1.0 kg
2.0 kg
 
m1  m
m2  2m
Since y1  y2  0


3my  4my

ycm 

ˆ
ˆ
rcm  2.5 m i  1.4 m  j
1m  2m  3m  4m
4.0 kg
5.0 m
7 

ycm  y
10

ycm  1.4 m
As in the last example, the center of mass need not lie within an object.
Example 3
A uniform sheet of plywood has a mass of 20.0 kg. If the plywood is
shaped as in the diagram, what are the coordinates of its center of mass?
m
60. cm
30. cm
3m
45,15
90. cm
75,45
Method 1
60. cm
1 2 

xcm   mi xi
M i 1


m1x1  m2 x2

xcm 
m1  m2
3m45  m75

xcm 
3m  m

xcm  52.5 cm
Example 3
A uniform sheet of plywood has a mass of 20.0 kg. If the plywood is
shaped as in the diagram, what are the coordinates of its center of mass?
m
60. cm
30. cm
75,45
Method 1
60. cm
3m
45,15
90. cm

rcm  52.5 cmiˆ  22.5 cm ˆj
1 2 

ycm 
 mi yi
M i 1


m1 y1  m2 y2

ycm 
m1  m2
3m15  m45

ycm 
3m  m

ycm  22.5 cm
Example 3
A uniform sheet of plywood has a mass of 20.0 kg. If the plywood is
shaped as in the diagram, what are the coordinates of its center of mass?
m
75,30
60. cm
60. cm
m
30. cm
Method 2
30,15
90. cm
30 cm  75 cm

xcm 
2

xcm  52.5 cm
Since the pieces are the same
mass, the center of mass of the
entire sheet must be exactly half
the way between the individual
center of masses.
15 cm  30 cm

ycm 
2

ycm  22.5 cm

rcm  52.5 cmiˆ  22.5 cm ˆj