One of the most powerful results of classical mechanics is that the center of mass of a system behaves like a single particle with mass equal to the total mass and subject only to external forces. This section proves this principle and demonstrates its applications.
Momentum of the center of mass
Total momentum of the system can be expressed as:
\[ \mathbf{P} = \sum_i \mathbf{p}_i = \sum_i m_i \mathbf{v}_i \]
From the definition of center of mass:
\[ \mathbf{r}_{\text{cm}} = \frac{\sum_i m_i \mathbf{r}_i}{M} \]
Taking the time derivative:
\[ \mathbf{v}_{\text{cm}} = \frac{d\mathbf{r}_{\text{cm}}}{dt} = \frac{\sum_i m_i \mathbf{v}_i}{M} \]
Therefore:
\[ M\mathbf{v}_{\text{cm}} = \sum_i m_i \mathbf{v}_i = \mathbf{P} \]
Result: The total momentum of a system equals total mass times center-of-mass velocity.
Newton's second law for center of mass
The center-of-mass acceleration is:
\[ \mathbf{a}_{\text{cm}} = \frac{d\mathbf{v}_{\text{cm}}}{dt} = \frac{\sum_i m_i \mathbf{a}_i}{M} \]
Applying Newton's second law to each particle \( i \):
\[ \mathbf{F}_{i,\text{net}} = m_i \mathbf{a}_i \]
Summing over all particles:
\[ \sum_i \mathbf{F}_{i,\text{net}} = \sum_i m_i \mathbf{a}_i = M\mathbf{a}_{\text{cm}} \]
Internal forces (between particles) appear as action-reaction pairs and cancel. Only external forces survive:
\[ \mathbf{F}_{\text{ext}} = M\mathbf{a}_{\text{cm}} \]
Equivalently:
\[ \mathbf{F}_{\text{ext}} = \frac{d\mathbf{P}}{dt} \]
This is Newton's second law for the center of mass.
Physical meaning
The center of mass of a system acts like a fictitious particle:
- Mass: total mass \( M \) of the system
- Force: only external forces (internal forces are ignored)
- Acceleration: \( \mathbf{a}_{\text{cm}} = \mathbf{F}_{\text{ext}} / M \)
Internal forces may cause the system to rotate or deform, but they do not change the translational motion of the center of mass.
Example 1: Two objects connected by a spring
Two objects placed horizontally with only a compressed spring between them:
- Object A: mass \( m_A = 2 \) kg
- Object B: mass \( m_B = 3 \) kg
- Spring is released
Question: What happens to the center of mass?
Answer: The spring forces (A pushes B, B pushes A) are internal forces. If the system is isolated (no external forces), then \( \mathbf{F}_{\text{ext}} = 0 \). Therefore:
\[ \mathbf{a}_{\text{cm}} = 0 \]
The center of mass remains at constant velocity (or at rest). The two objects move in opposite directions, but the center of mass does not move.
Example 2: Ball thrown in Earth's gravity field
A person throws a ball at an angle. Ignore air resistance.
Question: What is the trajectory of the center of mass?
Answer: The only external force is weight:
\[ \mathbf{F}_{\text{ext}} = M\mathbf{g} \]
(where \( \mathbf{g} = -9.8\hat j \) m/s²)
Therefore:
\[ \mathbf{a}_{\text{cm}} = \mathbf{g} \]
The center of mass accelerates exactly like a single particle under gravity. Its path is a parabola — the same as in Chapter 4! This holds even if the ball spins.
Example 3: Binary star system
Two stars:
- Star A: mass \( M_A = 2 M_{\odot} \) (two solar masses)
- Star B: mass \( M_B = 3 M_{\odot} \)
They orbit each other in circular paths.
Question: Where is the center of mass and what does it do?
Answer: If isolated (no nearby star), only internal forces (mutual gravity) act. Therefore:
\[ \mathbf{F}_{\text{ext}} = 0 \implies \mathbf{v}_{\text{cm}} = \text{const} \]
The center of mass moves with constant velocity (or is at rest). In the frame where the center of mass is stationary, both stars orbit in circles around it.
Conservation of momentum (§9.3 preview)
Since \( \mathbf{F}_{\text{ext}} = \frac{d\mathbf{P}}{dt} \), if \( \mathbf{F}_{\text{ext}} = 0 \) (isolated system), then:
\[ \mathbf{P} = \text{const} \]
Total momentum is conserved. This is one of the most fundamental principles in physics, following directly from Newton's second law for the center of mass.
What you should be able to do
- Express total momentum as \( M\mathbf{v}_{\text{cm}} \)
- Apply Newton's second law to center of mass: \( \mathbf{F}_{\text{ext}} = M\mathbf{a}_{\text{cm}} \)
- Recognize that internal forces do not affect center-of-mass motion
- See that \( \mathbf{F}_{\text{ext}} = 0 \Rightarrow \mathbf{P} = \text{constant} \)
Preview of §9.3
The center of mass is crucial not just for translational motion. For rigid bodies that rotate (like a spinning wheel), we must also understand motion about the center of mass. This introduces torque and angular momentum.
📚 See also: Halliday Vol 1, Ch 9, §9.2.
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