KAHIBARO
Table of Contents
Why the center of mass matters
When several objects move together and interact, their individual motions can be complicated. A very useful idea is to track one special point of the system, called the center of mass. The motion of this point often stays simple even when the parts move in messy ways.
In momentum problems, the center of mass is especially important because its motion is tied directly to the total momentum of the system. This gives a clear way to understand explosions, recoils, and collisions.
Defining the center of mass position
For a system of particles with masses $m_1, m_2, m_3, \dots$ at positions $\vec r_1, \vec r_2, \vec r_3, \dots$, the center of mass position is the mass weighted average of all positions:
$$
\vec R_{\text{cm}} = \frac{\sum_i m_i \vec r_i}{\sum_i m_i}
$$
If the total mass is
$$
M = \sum_i m_i
$$
then we can write
$$
\vec R_{\text{cm}} = \frac{1}{M}\sum_i m_i \vec r_i
$$
This means heavier objects influence the center of mass more strongly than lighter ones.
For two particles on a line, the formula becomes
$$
x_{\text{cm}} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}
$$
If one mass is much larger, the center of mass lies closer to that mass.
Center of mass velocity
To describe center of mass motion, we differentiate the center of mass position with respect to time. Since the masses stay constant in ordinary mechanics,
$$
\vec V_{\text{cm}} = \frac{d\vec R_{\text{cm}}}{dt} = \frac{1}{M}\sum_i m_i \vec v_i
$$
So the center of mass velocity is the total momentum divided by the total mass:
$$
\vec V_{\text{cm}} = \frac{\vec P}{M}
$$
where
$$
\vec P = \sum_i m_i \vec v_i
$$
is the total momentum of the system.
Important relation:
$$
\vec P = M \vec V_{\text{cm}}
$$
This means that the total momentum of a system is exactly the momentum of a single object of mass $M$ moving with the center of mass velocity.
Center of mass acceleration
Differentiating again gives the center of mass acceleration:
$$
\vec A_{\text{cm}} = \frac{d\vec V_{\text{cm}}}{dt}
$$
Using Newton's second law for all particles together, the internal forces cancel in pairs, so only external forces affect center of mass motion. Therefore,
$$
M \vec A_{\text{cm}} = \sum \vec F_{\text{ext}}
$$
or
$$
\vec A_{\text{cm}} = \frac{\sum \vec F_{\text{ext}}}{M}
$$
This is one of the most important results in mechanics. No matter how strongly the parts push on each other, the center of mass responds only to external force.
Center of mass equation of motion:
$$
\sum \vec F_{\text{ext}} = M \vec A_{\text{cm}}
$$
Internal forces do not change the motion of the center of mass.
Connection with conservation of momentum
If the net external force on the system is zero, then
$$
\sum \vec F_{\text{ext}} = 0
$$
so
$$
\vec A_{\text{cm}} = 0
$$
That means the center of mass either stays at rest or moves with constant velocity. Since $\vec P = M\vec V_{\text{cm}}$, the total momentum is then constant.
So conservation of momentum can be stated in center of mass language:
If no net external force acts on a system, the center of mass moves with constant velocity.
For an isolated system,
$$
\sum \vec F_{\text{ext}} = 0 \quad \Rightarrow \quad \vec V_{\text{cm}} = \text{constant}
$$
This is equivalent to conservation of total momentum.
Physical meaning
Imagine two skaters standing on frictionless ice. They push apart. Each skater moves, but if no external horizontal force acts, the center of mass continues exactly as before. If the system was initially at rest, the center of mass remains at the same place.
This can feel surprising, because the objects clearly move. But their motions balance so that the weighted average position behaves simply.
The same idea explains why a firework shell can explode into many fragments while the center of mass still follows the same path the shell would have followed if it had not exploded, as long as external forces are unchanged.
One dimensional examples
Consider two masses, $m_1 = 2\,\text{kg}$ and $m_2 = 3\,\text{kg}$, located at $x_1 = 0\,\text{m}$ and $x_2 = 4\,\text{m}$.
Then
$$
x_{\text{cm}} = \frac{(2)(0) + (3)(4)}{2+3} = \frac{12}{5} = 2.4\,\text{m}
$$
The center of mass lies between the two particles, closer to the larger mass.
Now suppose their velocities are $v_1 = 5\,\text{m/s}$ and $v_2 = -1\,\text{m/s}$. Then total momentum is
$$
P = m_1 v_1 + m_2 v_2 = (2)(5) + (3)(-1) = 10 - 3 = 7\,\text{kg m/s}
$$
The total mass is $5\,\text{kg}$, so
$$
V_{\text{cm}} = \frac{P}{M} = \frac{7}{5} = 1.4\,\text{m/s}
$$
Even though one particle moves left and the other right, the center of mass moves right at $1.4\,\text{m/s}$.
Two dimensional motion
In two dimensions, the center of mass has separate coordinates:
$$
x_{\text{cm}} = \frac{\sum_i m_i x_i}{M}, \qquad
y_{\text{cm}} = \frac{\sum_i m_i y_i}{M}
$$
and similarly in three dimensions.
The velocity components are
$$
V_{\text{cm},x} = \frac{\sum_i m_i v_{i,x}}{M}, \qquad
V_{\text{cm},y} = \frac{\sum_i m_i v_{i,y}}{M}
$$
So center of mass motion can be studied component by component, just like momentum.
Internal motion versus overall motion
A system can have two kinds of motion at the same time. The whole system can move through space, and the parts can also move relative to one another.
For example, two masses connected by a spring may oscillate back and forth while the entire pair drifts to the right. The oscillation is internal motion. The drift of the center of mass is the overall motion.
Momentum conservation controls the overall motion, not necessarily the internal motion.
Example, explosion of a system initially at rest
Suppose a stationary object of mass $10\,\text{kg}$ explodes into two pieces, one of mass $4\,\text{kg}$ and the other of mass $6\,\text{kg}$. If the $4\,\text{kg}$ piece moves to the right at $12\,\text{m/s}$, find the velocity of the other piece.
Initially the total momentum is zero, so the center of mass is at rest and must remain at rest if external force is negligible.
Thus final momentum must also be zero:
$$
(4)(12) + (6)v_2 = 0
$$
$$
48 + 6v_2 = 0
$$
$$
v_2 = -8\,\text{m/s}
$$
The second piece moves left. The center of mass remains at rest.
Example, walking inside a boat
A person stands on a boat floating in still water. If the person walks forward, the boat moves backward. This happens because the total horizontal momentum stays constant if water resistance is neglected.
Although the person and boat move relative to each other, the center of mass of the person plus boat system stays fixed horizontally if it started at rest. The backward motion of the boat balances the forward motion of the person.
This is a direct application of center of mass motion.
Continuous objects
For an extended object, the center of mass can be found by imagining it as made of many tiny mass pieces. The discrete sum becomes an integral:
$$
\vec R_{\text{cm}} = \frac{1}{M}\int \vec r \, dm
$$
For many symmetric objects of uniform density, the center of mass lies at the geometric center. For irregular objects or nonuniform density, it may not.
The details of calculating center of mass for continuous objects belong to a separate topic, but the motion rule stays the same:
$$
\sum \vec F_{\text{ext}} = M \vec A_{\text{cm}}
$$
A useful viewpoint for collisions
In a collision, the individual forces during contact can be large, but they are internal to the system if both colliding objects are included. Therefore they do not affect the motion of the center of mass of the two object system.
If external forces are negligible during the short collision time, then the center of mass moves with constant velocity before, during, and after the collision.
This is why momentum conservation is such a powerful collision tool.
Common situations
| Situation | Net external force | Center of mass motion |
|---|---|---|
| Isolated system at rest | $0$ | Stays at rest |
| Isolated system moving | $0$ | Moves with constant velocity |
| System under constant external force | Constant | Accelerates uniformly |
| Explosion in midair, neglecting air resistance | Gravity only | Follows ordinary projectile path |
| Collision on frictionless surface | Approximately $0$ horizontally | Horizontal center of mass velocity stays constant |
Visual picture
The center of mass lies closer to the larger mass.
Key ideas to remember
The center of mass is the mass weighted average position of a system. Its velocity is the total momentum divided by total mass. Its acceleration depends only on the net external force. Internal forces can change the motions of the parts, but they cannot change the motion of the center of mass.
Essential formulas for center of mass motion:
$$
\vec R_{\text{cm}} = \frac{1}{M}\sum_i m_i \vec r_i
$$
$$
\vec V_{\text{cm}} = \frac{1}{M}\sum_i m_i \vec v_i = \frac{\vec P}{M}
$$
$$
\sum \vec F_{\text{ext}} = M \vec A_{\text{cm}}
$$
If $\sum \vec F_{\text{ext}} = 0$, then
$$
\vec P = \text{constant}, \qquad \vec V_{\text{cm}} = \text{constant}
$$
Understanding center of mass motion turns a many body problem into something much simpler, the motion of one effective point that represents the whole system.