Motion of the Centre of Mass and Linear Momentum

<<>>

Motion of the Centre of Mass and Linear Momentum

TL;DR

The centre of mass moves as if all the system's mass and all external forces acted on a single point — internal forces cancel out completely. This single idea turns messy multi-body problems (explosions, collisions, rockets) into simple single-particle calculations. You'll learn to locate the centre of mass, track its motion, and use momentum conservation when no external force acts.

1. The Mental Model

Forget tracking every particle in a system — track one imaginary point instead. That point, the centre of mass, obeys Newton's second law using only the external forces, no matter how chaotic the internal motion is. Internal forces (particles pushing each other, pieces of an exploding shell interacting) always come in Newton's-third-law pairs and cancel in the total. A system's messy internal dance never changes where its centre of mass is heading.

2. The Core Material

2.1 Locating and Moving the Centre of Mass

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

$$\vec{R}_{cm} = \frac{\sum_i m_i \vec{r}_i}{\sum_i m_i} = \frac{\sum_i m_i \vec{r}_i}{M}$$

where $M$ is total mass. For a continuous object, the sum becomes an integral: $\vec{R}_{cm} = \frac{1}{M}\int \vec{r}\, dm$.

Differentiate once with respect to time and you get the velocity of the centre of mass:

$$\vec{V}_{cm} = \frac{\sum_i m_i \vec{v}_i}{M}$$

Notice something important: $M\vec{V}_{cm} = \sum_i m_i \vec{v}_i = \sum_i \vec{p}_i = \vec{P}_{total}$. The total linear momentum of a system is exactly the total mass times the velocity of the centre of mass. This is not an approximation — it's an identity that follows directly from the definition.

Differentiate again to get acceleration, then multiply by $M$:

$$M\vec{A}_{cm} = \sum_i m_i \vec{a}_i = \sum_i \vec{F}_i$$

Now split each force on particle $i$ into external forces (from outside the system) and internal forces (from other particles in the system). By Newton's third law, every internal force has an equal-and-opposite partner, so when you sum over all particles, every internal force cancels with its pair. Only external forces survive:

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

This is the big result. The centre of mass of any system — rigid body, cloud of gas, exploding firework, two people on a frictionless cart pushing each other — moves exactly like a single particle of mass $M$ acted on only by the net external force. Internal explosions, collisions, muscle forces, spring releases — none of it matters for $\vec{A}_{cm}$.

2.2 Conservation of Linear Momentum

Since $\vec{F}_{ext} = \frac{d\vec{P}}{dt}$, if the net external force on a system is zero, then $\vec{P}_{total}$ is constant — this is conservation of linear momentum, and it holds componentwise. You can have $F_{ext,x} = 0$ while $F_{ext,y} \neq 0$ (e.g., gravity acts vertically but nothing acts horizontally), and momentum will be conserved in $x$ only.

This is why conservation of momentum works beautifully for explosions and collisions: during the brief instant of interaction, internal forces (the explosive charge, the colliding surfaces) are huge compared to external forces like gravity or friction, so external impulse is negligible over that short time, and momentum just before equals momentum just after.

flowchart TD
    A["System of n particles, each with internal forces between them"] --> B{"Sum all forces: internal forces cancel in pairs (Newton's 3rd law)"}
    B --> C["Net force on system = F_external only"]
    C --> D["M * A_cm = F_external"]
    D --> E{"Is F_external = 0?"}
    E -->|Yes| F["P_total = M * V_cm = constant (momentum conserved)"]
    E -->|No| G["V_cm changes according to F_ext, like a single particle of mass M"]

2.3 Why Internal Forces Truly Vanish

It's worth seeing why this cancellation is so complete, not just accepting it. Consider two particles exerting forces on each other: $\vec{F}_{12}$ (force of particle 2 on particle 1) and $\vec{F}_{21}$ (force of particle 1 on particle 2). Newton's third law says $\vec{F}_{12} = -\vec{F}_{21}$. When you write Newton's second law for each particle and add them:

$$m_1\vec{a}_1 + m_2\vec{a}_2 = \vec{F}_{1,ext} + \vec{F}_{2,ext} + \vec{F}_{12} + \vec{F}_{21} = \vec{F}_{1,ext} + \vec{F}_{2,ext}$$

The internal pair cancels exactly regardless of what kind of force it is — gravitational, electrical, a spring, a collision, friction between the two surfaces. This generalizes to any number of particles: every internal force has a partner, and the sum over all pairs is zero. This is also the deep reason rockets work — the rocket and the expelled gas are an internal action-reaction pair; total momentum of rocket + exhaust gas is conserved (ignoring gravity/drag), even though the rocket itself keeps speeding up.

A subtlety worth flagging: momentum conservation applies to the system, not to individual parts. A single particle within the system can gain or lose momentum freely, as long as another particle's momentum changes to compensate.

3. Worked Example

Setup: A 60 kg person stands at rest at one end of a stationary 90 kg boat of length 3 m, floating on still water (ignore water resistance/friction so there's no horizontal external force). The person walks to the other end of the boat. How far does the boat move, and where does the centre of mass end up?

Step 1 — No external horizontal force, so $V_{cm}$ stays zero.
Nothing pushes the person-boat system sideways, so the centre of mass doesn't move horizontally at all throughout the whole process. It started at rest, so $V_{cm} = 0$ always, meaning the position of the centre of mass is also fixed in the water frame.

Step 2 — Compute initial centre of mass position.
Let the boat occupy $x = 0$ (far end, boat's centre) to $x = 3$ m along its length, with the person starting at $x = 0$ end. Treat the boat's own mass as concentrated at its centre, $x_{boat} = 1.5$ m. Person starts at $x_{person} = 0$.

$$x_{cm} = \frac{m_{person} \cdot x_{person} + m_{boat} \cdot x_{boat}}{m_{person}+m_{boat}} = \frac{60(0) + 90(1.5)}{150} = \frac{135}{150} = 0.9 \text{ m}$$

Step 3 — After the walk, person is at the other end.
Let the boat shift by displacement $d$ (boat moves relative to water). The person's new position relative to the boat is $x = 3$ m from the start, so relative to water the person is now at $(3 + d)$, and the boat's centre is now at $(1.5 + d)$.

Since $x_{cm}$ must stay at 0.9 m (no external force moved it):

$$0.9 = \frac{60(3+d) + 90(1.5+d)}{150}$$

$$0.9 \times 150 = 180 + 60d + 135 + 90d$$

$$135 = 315 + 150d$$

$$d = \frac{135 - 315}{150} = \frac{-180}{150} = -1.2 \text{ m}$$

Interpretation: The boat moves 1.2 m in the direction opposite to the person's walk. The person walked 3 m along the boat, but relative to the water they only moved $3 + (-1.2) = 1.8$ m. The system's centre of mass never budged — it's exactly where it started, at $x_{cm} = 0.9$ m, confirming our governing idea: internal forces (person's feet pushing on boat) rearranged the parts but could not shift $\vec{R}_{cm}$ because there was no external horizontal force.

4. Key Takeaways

4.1 Most Important Concepts

• Centre of mass formula: $\vec{R}_{cm} = \frac{\sum m_i \vec{r}_i}{M}$ — a mass-weighted average position, not a geometric centre (unless mass is uniformly distributed and symmetric).
• $M\vec{A}_{cm} = \vec{F}_{ext}$: The centre of mass obeys Newton's second law using only external forces — internal forces are irrelevant to its motion.- Momentum–centre of mass link: $\vec{P}_{total} = M\vec{V}_{cm}$ is an exact identity, true at every instant, not just when momentum is conserved.
• Conservation condition: If $\vec{F}_{ext} = 0$ (or zero in a particular direction), $\vec{P}_{total}$ stays constant — this can hold componentwise even when it doesn't hold overall.
• Internal forces cancel in pairs: Every internal force has a Newton's-third-law partner, so their sum over the whole system is always zero, regardless of the force type.
• Momentum conservation is a system property: Individual particles can gain or lose momentum freely; only the total is protected when $\vec{F}_{ext}=0$.
• Explosions/collisions approximation: During very short interaction times, internal forces dominate over external ones (gravity, friction), so momentum is conserved across the interaction even if not conserved over longer timescales.

4.2 Common Misconceptions

• "The centre of mass is always at the geometric centre." False — it's the mass-weighted average; an irregular or non-uniform object has its centre of mass shifted toward the denser side.
• "If the total momentum is zero, nothing is moving." False — individual particles can have large velocities in opposite directions that cancel; the boat-and-person example shows both objects move while $\vec{P}_{total}=0$ throughout.
• "Internal forces don't affect anything." They affect individual particles' motion and shape/configuration of the system enormously — they just don't affect the centre of mass's trajectory or the total momentum.
• "Momentum conservation requires no forces at all." It only requires zero net external force (or along a specific direction); internal forces, and even some external forces that happen to cancel, are fine.

4.3 Compare & Contrast

5. Now Try It

Problem: A 2 kg object at rest explodes into three fragments on a frictionless horizontal surface (ignore gravity's vertical effect — no horizontal external force). Fragment A (0.5 kg) flies off at 6 m/s along the +x axis. Fragment B (0.5 kg) flies off at 4 m/s along the +y axis. Find the velocity (magnitude and direction) of fragment C (1 kg).

Work it by setting total initial momentum to zero, writing $\vec{P}_A + \vec{P}_B + \vec{P}_C = 0$, solving for $\vec{P}_C$ component-wise, then dividing by C's mass to get its velocity vector. State the final answer as a magnitude in m/s and an angle measured from the +x axis.

Success looks like: you get $\vec{v}_C = (-3, -2)$ m/s, magnitude $\sqrt{13} \approx 3.61$ m/s, directed into the third quadrant at about $213.7°$ from +x (or equivalently $\tan^{-1}(2/3)\approx33.7°$ below the −x axis).

<<>>