Motion of the Centre of Mass and Linear Momentum

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Motion of the Centre of Mass and Linear Momentum

TL;DR

The centre of mass of a system moves as though all the mass were concentrated there and all external forces act on it—this is Newton's second law applied to the whole system. Linear momentum is conserved when no external forces act, which is why collisions follow predictable rules. These two ideas are the foundation for analyzing any multi-object system.

1. The Mental Model

Think of a system of particles as a single lump for the purposes of describing overall motion. The centre of mass is that lump's position. External forces push the centre of mass around; internal forces (between particles) don't affect where the centre of mass goes. If you throw a spinning wrench in the air, it tumbles chaotically, but its centre of mass follows a clean parabola—that's this principle at work. The centre of mass is not an illusion; it's the balancing point that obeys Newton's laws on behalf of the entire system.

2. The Core Material

2.1 Defining the Centre of Mass

For a system of $n$ particles with masses $m_1, m_2, \ldots, m_n$ at positions $\vec{r}_1, \vec{r}_2, \ldots, \vec{r}_n$, the centre of mass is located at:

$$\vec{R}_{\text{cm}} = \frac{m_1 \vec{r}_1 + m_2 \vec{r}_2 + \cdots + m_n \vec{r}_n}{m_1 + m_2 + \cdots + m_n} = \frac{1}{M} \sum_{i=1}^{n} m_i \vec{r}_i$$

where $M = \sum m_i$ is the total mass. This is a weighted average of all positions—each mass "pulls" the average toward itself proportionally. If two 1 kg balls are at $x = 0$ and $x = 10$ m, the centre of mass is at $x = 5$ m. If the left ball were 3 kg and the right 1 kg, it shifts to $x = 2.5$ m.

2.2 Velocity and Acceleration of the Centre of Mass

Differentiate the position to get velocity:

$$\vec{V}_{\text{cm}} = \frac{d\vec{R}_{\text{cm}}}{dt} = \frac{1}{M} \sum_{i=1}^{n} m_i \vec{v}_i$$

This is the total momentum $\vec{P}$ divided by total mass. Differentiate again for acceleration:

$$\vec{A}_{\text{cm}} = \frac{d\vec{V}_{\text{cm}}}{dt} = \frac{1}{M} \sum_{i=1}^{n} m_i \vec{a}_i$$

Now apply Newton's second law to each particle: $\vec{F}_i^{\text{ext}} + \sum_{j \neq i} \vec{F}_{ij} = m_i \vec{a}_i$, where $\vec{F}_i^{\text{ext}}$ is the external force on particle $i$ and $\vec{F}_{ij}$ is the internal force from particle $j$. When you sum over all particles, the internal forces cancel by Newton's third law (each $\vec{F}_{ij}$ is paired with $-\vec{F}_{ji}$). You're left with:

$$\vec{F}_{\text{ext}} = M \vec{A}_{\text{cm}}$$

where $\vec{F}_{\text{ext}} = \sum_i \vec{F}_i^{\text{ext}}$ is the total external force. This is the key result: the centre of mass accelerates like a single particle of mass $M$ under the total external force alone. Internal forces are irrelevant to the centre of mass motion.

2.3 Linear Momentum and Conservation

Linear momentum of a particle is $\vec{p} = m\vec{v}$. For a system:

$$\vec{P} = \sum_{i=1}^{n} \vec{p}_i = \sum_{i=1}^{n} m_i \vec{v}_i = M \vec{V}_{\text{cm}}$$

Taking the time derivative:

$$\frac{d\vec{P}}{dt} = M \frac{d\vec{V}_{\text{cm}}}{dt} = M \vec{A}_{\text{cm}} = \vec{F}_{\text{ext}}$$

So momentum changes only due to external forces. If $\vec{F}_{\text{ext}} = 0$, then $\vec{P}$ is constant. This is the law of conservation of linear momentum. In an isolated collision (no friction, no gravity, etc.), momentum before equals momentum after. In a system under gravity alone, horizontal momentum is conserved (gravity is vertical).

graph TD
    A["System of particles<br/>with masses and velocities"] --> B["Calculate centre of mass<br/>position and velocity"]
    B --> C{"External forces<br/>present?"}
    C -->|Yes| D["Centre of mass<br/>accelerates:<br/>F_ext = M·A_cm"]
    C -->|No| E["Centre of mass<br/>moves at constant velocity"]
    D --> F["Internal forces<br/>redistribute momentum<br/>between particles"]
    E --> G["Total linear momentum<br/>is conserved"]
    F --> G

3. Worked Example

Two ice skaters on a frictionless rink: Alice (mass 50 kg) moving at 2 m/s to the right, and Bob (mass 80 kg) initially at rest. They collide and stick together.

Before collision:
- Alice: $m_A = 50$ kg, $v_A = 2$ m/s
- Bob: $m_B = 80$ kg, $v_B = 0$ m/s
- Total momentum: $P = 50 \times 2 + 80 \times 0 = 100$ kg·m/s

After collision (they move together):
- Combined mass: $M = 50 + 80 = 130$ kg
- Common velocity: $v_{\text{final}} = ?$

Apply conservation of momentum (no external horizontal forces):
$$P_{\text{before}} = P_{\text{after}}$$
$$100 = 130 \times v_{\text{final}}$$
$$v_{\text{final}} = \frac{100}{130} \approx 0.77 \text{ m/s}$$

The centre of mass of the system moves at $V_{\text{cm}} = \frac{P}{M} = \frac{100}{130} \approx 0.77$ m/s throughout—before, during, and after the collision. Internally, momentum is shuffled between Alice and Bob, but the centre of mass trajectory is unaffected by the collision because there are no external horizontal forces.

4. Key Takeaways

4.1 Most Important Concepts

1. Centre of mass is a weighted average position. Each mass contributes proportionally; denser regions pull it closer.

2. External forces alone determine centre of mass motion. Internal forces shuffle momentum around but don't change the total.

3. $\vec{F}_{\text{ext}} = M\vec{A}_{\text{cm}}$ is Newton's second law for the whole system. Apply it to find how the system's centre of mass accelerates.

4. Momentum is conserved when external forces sum to zero. This holds in collisions, explosions, and any isolated system.

5. The centre of mass velocity equals total momentum divided by total mass. $\vec{V}_{\text{cm}} = \vec{P} / M$, always.

6. Internal forces come in action-reaction pairs and cancel. This is why they don't appear in the centre of mass equation.

7. Momentum is a vector. Conservation applies component by component—horizontal and vertical momentum are independent.

4.2 Common Misconceptions

• Misconception: "The centre of mass is always at the geometric middle of the system."
  Truth: It's a mass-weighted average. A 10 kg ball next to a 1 kg ball puts the centre of mass much closer to the heavy one.

• Misconception: "If a system is spinning, momentum isn't conserved."
  Truth: Linear momentum is still conserved if external forces sum to zero. Rotation doesn't change this; it's described by angular momentum, which is separate.

• Misconception: "In a collision, kinetic energy is always conserved."
  Truth: Momentum is always conserved in an isolated system. Kinetic energy is only conserved in elastic collisions; inelastic collisions lose energy to heat, sound, deformation, etc.

• Misconception: "Internal forces can move the centre of mass."
  Truth: Internal forces redistribute motion among particles but cannot change the centre of mass velocity without external forces.

4.3 Compare & Contrast

5. Now Try It

Problem: A cannon (mass 1000 kg) at rest on a frictionless surface fires a cannonball (mass 10 kg) horizontally at 100 m/s to the right.

(a) Calculate the recoil velocity of the cannon (to the left).
(b) Verify that the centre of mass remains at rest after firing.
(c) If the cannon were on a ramp inclined at 30°, would momentum still be conserved? Explain why or why not.

Work through it step-by-step:
- Write the momentum before firing.
- Apply conservation to find the cannon's final velocity.
- Calculate $\vec{V}_{\text{cm}}$ before and after to confirm it doesn't change.
- For part (c), identify all external forces and decide whether they have a component along the ramp.

Success looks like: You find the cannon recoils at 1 m/s to the left, the centre of mass velocity is zero both before and after, and you explain that on a ramp, gravity has a component parallel to the surface, so momentum along the ramp is not conserved—but momentum perpendicular to the ramp is.

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