**Topic 2B-02: Center of mass**

Two point-like particles of masses m_{1} and m_{2} are attached
to end points Q_{1} and Q_{2} of a massless rod of length L
respectively. We try to balance the rod at a point O as shown in Fig. 1. The rod
forms an angle θ with a horizontal line. The length OQ_{1} is
denoted as x. We need to vary x and θ until the point O is found. Such a
point O is called the center of mass, or the center of gravity of the two
particle system.

The situation can be abstracted as shown in Fig. 2, where position vector
**r**_{1} originates from point O and ends at point Q_{1},
and position vector **r**_{2} originates from point O and ends at
point Q_{2}. Force **F**_{1} originates from point
Q_{1}, points vertically downward, and has a magnitude of
m_{1}*g*. Force **F**_{2} originates from point
Q_{2}, points vertically downward, and has a magnitude of
m_{2}*g*. **F**_{1} and **F**_{2} are
gravitational pulls acting on particles m_{1} and m_{2}
respectively. When the rod is balanced at point O, the rod will not rotate
around point O. The torque **τ**_{1} = **r**_{1} x
**F**_{1} tries to rotate the rod counter-clockwise around O, and the
torque **τ**_{2} = **r**_{2} x **F**_{2}
tries to rotate the rod clockwise around point O. In order to keep the rod
stationary, we must have

**τ**_{1} + **τ**_{2} = 0.

Since **τ**_{1} originates from point O and points vertically
out of the paper, and **τ**_{2} originates from point O and
points vertically into the paper, we must have |**τ**_{1}
= |**τ**_{2}| in order to have **τ**_{1}
+ **τ**_{2} = 0. From Eq. (2) of Topic 2B-01, we have

|**τ**_{1}| =
|**r**_{1}|·|**F**_{1}|·sinθ

= *x*·m_{1}*g*·sinθ ,

and

|**τ**_{2}| =
|**r**_{2}|·|**F**_{2}|·sinθ

= (L - *x*)·m_{2}*g*·sinθ ,

Thus |**τ**_{1}| = |**τ**_{2}|
leads to

m_{1}*x* = m_{2}(L - *x*),

or

*x* = m_{2}L/(m_{1} + m_{2}) .
(1)

This concept of the center of mass can be refined further so that it does not
need to depend on the gravitational pulls. From Eq. (1) we have

|m_{1}**r**_{1}| = m_{1}*x*

=m_{1}m_{2}L / (m_{1} + m_{2}),

and

|m_{2}**r**_{2}| = m_{2}(L - *x*)

=m_{1}m_{2}L / (m_{1} + m_{2}).

Thus |m_{1}**r**_{1}| =
|m_{2}**r**_{2}|. Since
m_{1}**r**_{1} and m_{2}**r**_{2} are
pointing at the opposite directions, we have

m_{1}**r**_{1} + m_{2}**r**_{2} = 0.
(2)

Eq. (2) can be considered as the definition of the center of mass. It says that
first find a point O that satisfies Eq. (2), then point O is the center of mass.
To prove that such a point O is indeed the center of mass as discussed in Eq. (1),
we assume a force m_{1}**f**_{1} applies on point Q_{1},
and a force m_{2}**f**_{2} applies on point Q_{2} as
shown in Fig. 3. We also assume that

|**f**_{1}| = |**f**_{2}| = |**f**|
and **f**_{1} // **f**_{2} // **f**, where **f** is
a mathematical vector without a specific originating point. Torques
**τ**_{1} and **τ**_{2} are defined as

**τ**_{1} = **r**_{1} x (m_{1}**f**_{1})

and

**τ**_{2} = **r**_{2} x (m_{2}**f**_{2})

respectively. Following the second kind of addition rule for torques as discussed in Topic 2B-02,
we have

**τ** = **τ**_{1} + **τ**_{2}

= **r**_{1} x (m_{1}**f**_{1}) +
**r**_{2} x (m_{2}**f**_{2})

= (m_{1}**r**_{1} + m_{2}**r**_{2}) x **f**.

Substituting Eq. (2) into the above equation, we get

**τ** = 0 .

If |**f**_{1}| = |**f**_{2}| = *g*,
and both **f**_{1} and **f**_{2} point vertically downward,
**τ** becomes the torque around point O generated by the gravitational pulls.
**τ** = 0 means that the two particle system under consideration can be
balanced at point O. Thus point O is the center of mass as discussed in Fig. 1.

Though Eq. (2) is an elegant way to define the center of mass, it is not easy to
use. There is a more practical way to compute the center of mass. In Fig. 4 P
is a reference point, like the origin of a coordinate system. Position vectors
**R**_{1} and **R**_{2}, originating from point P, define
points Q_{1} and Q_{2} respectively. Position vector **R**
defines the center of mass point O. Position vectors **r**_{1} and
**r**_{2} are drawn from O to Q_{1} and Q_{2}
respectively. We have

**R**_{1} = **R** + **r**_{1},

and

**R**_{2} = **R** + **r**_{2}.

Thus

m_{1}**R**_{1} + m_{2}**R**_{2} =
m_{1}(**R** + **r**_{1}) + m_{2}(**R** + **r**_{2})

= (m_{1} + m_{2})**R** + (m_{1}**r**_{1} +
m_{2}**r**_{2}).

Since point O is the center of mass and Eq. (2) is satisfied, the above equation
leads to

**R** = (m_{1}**R**_{1} + m_{2}**R**_{2})
/ (m_{1} + m_{2}).
(3)

Eq. (3) allows us to compute the center of mass easily when **R**_{1}
and **R**_{2} are known.

It is not difficult to extend the discussions about a two particle system to a
N-particle system. Let particles of masses m_{1}, m_{2},
· · ·, m_{N} located at points Q_{1}, Q_{2},
· · ·, Q_{N} respectively. Let position vectors
**r**_{1}, **r**_{2}, · · ·, **r**_{N}
originate from point O and ends at points

Q_{1}, Q_{2}, · · ·,
Q_{N} respectively. If the equation

m_{1}**r**_{1} + m_{2}**r**_{2} + · ·
· + m_{N}**r**_{N} = 0
(4)

is satisfied, then O is the center of mass of this N-particle system. To demonstrate
the validity of this statement, we imagine that each particle in the system is
connected to point O via a massless rod as shown in Fig. 5. Let forces
m_{1}**f**_{1}, m_{2}**f**_{2}, · ·
·, m_{N}**f**_{N} applied to points Q_{1}, Q_{2},
·, ·, ·, Q_{N} respectively. We assume that |**f**_{1}|
= |**f**_{2}| = · · · = |**f**_{N}| =
|**f**| and **f**_{1} // **f**_{2} // · ·
· // **f**_{N} // **f**, where **f** is a mathematical vector
without a specific originating point. The torque of this system around point O is

**τ** = **r**_{1} x (m_{1}**f**_{1}) +
**r**_{2} x (m_{2}**f**_{2}) + · · · +
**r**_{N} x (m_{N}**f**_{N})

= (m_{1}**r**_{1} + m_{2}**r**_{2} + ·
· · + m_{N}**r**_{N}) x **f**.

Eq. (4) then says that **τ** = 0. If **f** is chosen to point vertically
downward, and |**f**| is set to *g*, then **τ** is the
torque around point O generated by the gravitational pulls of this N-particle system.
**τ** = 0 means that the system as depicted in Fig. 5 can indeed be balanced
at point O. Thus point O can be properly called the center of mass (or center of gravity)
of this system.

The extension of Eq. (3) to the N-particle system is also straightforward. We have

**R** = (m_{1}**R**_{1} + m_{2}**R**_{2} +
· · · + m_{N}**R**_{N}) / (m_{1} +
m_{2} + · · · + m_{N}) ,
(5)

where **R**_{1}, **R**_{2}, · · ·,
**R**_{N} are position vectors that originate from a reference point
P and define points Q_{1}, Q_{2}, · · ·, Q_{N}
respectively, and **R** is the position vector that defines point O.

Eq. (5) is actually more powerful than it appears. We can group any number of
particles under consideration as a subsystem. Let us consider the center of mass
point of particles 1 and 2, and denote the postion vector for this center of
mass point as **R**_{12}. By applying Eq.(5) to the
subsystem of particles 1 and 2, we have

(m_{1} + m_{2})**R**_{
12} = (m_{1R1
+ m2R2).
Substituting this equation into Eq. (5), we get
R = {(m1 + m2)R
12 +
· · · + mNRN}
/ {(m1 +
m2) + · · · + mN}
.
This means that particles 1 and 2 can be replaced by a particle of mass (m1 +
m2) located at the position R
12. This simplifying process can be applied to any group of
particles in the system and as many groups as we want. We will be using this
property later on when it comes to find center of mass points in the
exercises.
}

The number N can be extended to infinity, and the distance among neighboring particles can be reduced to infinitesimally small. Thus the discrete particle system will approach a continuous and extended object. Equations (4) and (5) will also approach a continuous limit. We will not get into details here but rather differ the discussion to upcoming exercises.

There is an important property related to the center of mass and the gravitational
pulls. Let us consider the N-particle system as depicted in Fig. 6. Let forces
**F**_{1}, **F**_{2}, · · · **F**_{N}
apply on points Q_{1}, Q_{2}, · · ·, Q_{N}
respectively. **R**_{1}, **R**_{2}, · · ·,
**R**_{N} are position vectors that originate from a reference point
P and define points Q_{1}, Q_{2}, · · ·, Q_{N}
respectively. Position vector **R** defines the center of mass point O as shown
in Fig. 6. Position vectors OQ_{1}, OQ_{2}, · · ·,
OQ_{N} are denoted as **r**_{1}, **r**_{2}, ·
· ·, **r**_{N} respectively. The torque vector around the
reference point P is

**τ** = **R**_{1} x **F**_{1} +
**R**_{2} x **F**_{2} + · · · +
**R**_{N} x **F**_{N} .

Since **R**_{1} = **r**_{1} + **R**,
**R**_{2} = **r**_{2} + **R**, · · ·,
**R**_{N} = **r**_{N} + **R**, we have, in the
mathematical sense,

**τ** = (**r**_{1} x **F**_{1} +
**r**_{2} x **F**_{2} + · · · +
**r**_{N} x **F**_{N}) + **R** x (**F**_{1} +
**F**_{2} + · · · + **F**_{N}).

The first term at the right-hand side of the above equation is the torque around
the center of mass point O so is denoted as **τ**_{O}. The sum
(**F**_{1} + **F**_{2} + · · · +
**F**_{N}) is a mathematical sum. It means that **F**_{1},
**F**_{2}, · · ·, **F**_{N} are all
displaced without changing their magnitudes and the orientation such that they
all originate from point O and then summed. Let **F**_{1},
**F**_{2}, · · ·, **F**_{N} be
gravitational pulls, that is,

**F**_{1} = m_{1}** g**,

where

Using Eq. (4), we get

Thus

where M = m