Centre of Mass

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Centre of Mass

TL;DR

You'll learn how to find the single point that represents the average position of a system's mass, and why that point moves as if all external forces act on it alone. You'll calculate centre of mass for discrete particles and continuous bodies. You'll see why this idea lets you ignore messy internal forces entirely.

1. The Mental Model

Picture a system of particles as a cloud of masses scattered in space. The centre of mass is the "balance point" of that cloud — if you held it there, the whole system would balance perfectly. Internal forces (particles pushing or pulling each other) never move this point; only outside forces can. The centre of mass moves exactly like a single particle carrying the system's total mass, driven only by external forces.

2. The Core Material

2.1 Defining the Centre of Mass

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

$$\vec{R}_{cm} = \frac{\sum_{i=1}^{n} m_i \vec{r}_i}{\sum_{i=1}^{n} m_i} = \frac{1}{M}\sum_{i=1}^{n} m_i \vec{r}_i$$

where $M = \sum m_i$ is the total mass. In component form this splits cleanly:

$$x_{cm} = \frac{\sum m_i x_i}{M}, \quad y_{cm} = \frac{\sum m_i y_i}{M}, \quad z_{cm} = \frac{\sum m_i z_i}{M}$$

Notice this is a weighted average, not a simple geometric average. A heavy mass far away pulls the centre of mass toward it much more than a light mass at the same distance. If all masses are equal, the centre of mass reduces to the ordinary centroid (geometric centre).

For a continuous body, sums become integrals. You chop the body into infinitesimal mass elements $dm$ at position $\vec{r}$:

$$\vec{R}_{cm} = \frac{1}{M}\int \vec{r}\, dm$$

To actually compute this, you express $dm$ in terms of density and a spatial variable. For a rod of length $L$ with linear density $\lambda(x)$: $dm = \lambda(x)\,dx$. For a 2D lamina with surface density $\sigma$: $dm = \sigma\, dA$. For a 3D solid with volume density $\rho$: $dm = \rho\, dV$. Symmetry is your best friend here — if a body has a line or plane of symmetry with uniform density, the centre of mass must lie on it. That often kills two of the three coordinates before you integrate anything.

2.2 The Dynamics: Why This Point Is Special

Differentiate the centre of mass position twice with respect to time. First, velocity:

$$\vec{V}_{cm} = \frac{d\vec{R}_{cm}}{dt} = \frac{1}{M}\sum_i m_i \vec{v}_i$$

This says total momentum $\vec{P} = \sum m_i \vec{v}_i = M\vec{V}_{cm}$ — the system's momentum looks exactly like one particle of mass $M$ moving at $\vec{V}_{cm}$.

Now differentiate again to get acceleration, and bring in Newton's second law for each particle. Every particle feels two kinds of force: external forces (gravity, applied pushes, friction from outside) and internal forces (particle $i$ pushing particle $j$, and vice versa). By Newton's third law, every internal force pair is equal and opposite, so when you sum over all particles, all internal forces cancel exactly:

$$M\vec{A}_{cm} = \sum_i m_i \vec{a}_i = \sum_i \vec{F}_i^{ext} + \sum_i \sum_j \vec{F}_{ij}^{internal} = \vec{F}_{ext}^{total} + 0$$

So:

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

This is the big payoff. No matter how complicated the internal interactions are — explosions, collisions, springs, muscles — the centre of mass obeys plain Newtonian mechanics as if it were a point particle feeling only the net external force. A spinning, tumbling wrench thrown through the air has a horribly complex motion for each individual atom, but its centre of mass traces a clean parabola, exactly like a thrown ball.

2.3 Centre of Mass vs. Centre of Gravity vs. Centroid

These three get confused constantly, so pin down the distinctions:

• Centre of mass depends only on mass distribution. It exists regardless of any force field.
• Centre of gravity is the point where gravity's net torque effect acts. In a uniform gravitational field, it coincides exactly with the centre of mass. In a non-uniform field (e.g., a very tall tower where $g$ changes noticeably with height), they can differ slightly.
• Centroid is purely geometric — the "centre" of a shape, assuming uniform density. For a uniform-density object, centroid = centre of mass.

For almost all problems you'll face (object sizes much smaller than Earth's radius), centre of mass and centre of gravity are interchangeable.

flowchart TD
    A["System of particles with internal forces"] --> B["Sum forces over all particles"]
    B --> C["Internal forces cancel in pairs (Newton's 3rd Law)"]
    C --> D["Only external forces survive"]
    D --> E["F_ext = M * A_cm"]
    E --> F["CM moves like one point particle of mass M"]

3. Worked Example

Problem: Three particles lie in the xy-plane: $m_1 = 2\,\text{kg}$ at $(0, 0)$, $m_2 = 3\,\text{kg}$ at $(4, 0)$, and $m_3 = 5\,\text{kg}$ at $(2, 6)$ (distances in metres). Find the centre of mass. Then, if a fourth, external force of $20\,\text{N}$ acts on this system along the $+x$ direction (ignore gravity for this part), find the instantaneous acceleration of the centre of mass.

Step 1 — Total mass:
$$M = 2 + 3 + 5 = 10\,\text{kg}$$

Step 2 — x-coordinate of CM:
$$x_{cm} = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{M} = \frac{(2)(0) + (3)(4) + (5)(2)}{10} = \frac{0 + 12 + 10}{10} = 2.2\,\text{m}$$

Step 3 — y-coordinate of CM:
$$y_{cm} = \frac{(2)(0) + (3)(0) + (5)(6)}{10} = \frac{0 + 0 + 30}{10} = 3.0\,\text{m}$$

So $\vec{R}_{cm} = (2.2, 3.0)\,\text{m}$. Notice this point doesn't coincide with any actual particle — that's normal.

Step 4 — Acceleration of CM under external force:
Using $\vec{F}_{ext} = M\vec{A}_{cm}$:
$$A_{cm} = \frac{F_{ext}}{M} = \frac{20\,\text{N}}{10\,\text{kg}} = 2\,\text{m/s}^2 \text{ along } +x$$

This is true no matter what internal forces (springs, collisions between the three masses) are doing simultaneously. The individual particles could be accelerating wildly in different directions due to internal interactions, but the average motion — the CM — accelerates at a clean $2\,\text{m/s}^2$ in the direction of the net external push. This is the entire power of the concept: it decouples the "overall system motion" from the "internal mess."

4. Key Takeaways

4.1 Most Important Concepts

• CM is a mass-weighted average position: $\vec{R}_{cm} = \frac{1}{M}\sum m_i \vec{r}_i$, not a simple geometric average.
• Internal forces always cancel when summed over the whole system, due to Newton's third law pairing.
• $\vec{F}_{ext} = M\vec{A}_{cm}$ is the master equation: the CM obeys single-particle dynamics regardless of internal complexity.
• Momentum of system = $M\vec{V}_{cm}$: total momentum is fully captured by tracking just the CM's velocity.
• Symmetry simplifies integrals: a uniform body's CM lies on any axis or plane of symmetry — use this before integrating.
• CM need not be inside the body: think of a thin ring or an L-shaped bracket — the CM can sit in empty space.
• CM, centre of gravity, and centroid coincide under uniform density and uniform gravitational field, but are conceptually distinct.

4.2 Common Misconceptions

• "Centre of mass is always at the geometric centre." False — only true for uniform-density, symmetric objects. An L-shaped bracket or a system of unequal masses has its CM shifted toward the heavier/denser region, often lying outside the material entirely.

• "Internal forces don't matter at all." They matter enormously for each particle's individual motion, but they cancel out only when you sum forces over the whole system to find the CM's motion. Internal forces can still change the system's shape, rotation, and internal energy.

• "If the CM isn't accelerating, nothing is moving." False — individual particles can be moving and accelerating in complicated ways (think of an exploding firework) while the CM moves at constant velocity or stays still, as long as net external force is zero.

• "Centre of mass and centre of gravity are always the same thing." They coincide only when the gravitational field is uniform across the object. For objects comparable in size to planetary distances, they can differ slightly.

4.3 Compare & Contrast

5. Now Try It

Exercise: A uniform thin rod of length $L = 6\,\text{m}$ and mass $M = 9\,\text{kg}$ lies along the x-axis from $x=0$ to $x=6$. A small point mass $m = 3\,\text{kg}$ is attached at $x = 6\,\text{m}$ (the far end of the rod). Find the x-coordinate of the centre of mass of the rod+mass system.

Work it in two steps:
1. Find the rod's own CM alone (use symmetry — no integral needed).
2. Treat the rod (as one lump at its own CM) and the point mass as a two-particle system, and apply the discrete CM formula to combine them.

Success looks like: you get $x_{cm} = 4.5\,\text{m}$, and you can explain in one sentence why replacing the rod with a single point mass at its own CM is valid for this calculation.

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