Centre of Mass

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Centre of Mass

TL;DR

The centre of mass is the single point that moves as if all a system's mass were concentrated there. For a uniform object it's at the geometric centre; for scattered masses you find it by averaging their positions, weighted by how heavy each one is. This point obeys Newton's laws regardless of how the individual parts rotate or collide.

1. The Mental Model

Imagine you're holding a ruler horizontally by one finger. The ruler balances at a specific spot—that's roughly its centre of mass. If you replaced the entire ruler with a point mass sitting at that spot, it would behave identically under gravity. Now generalise: any system of particles or extended object has one such magical point where all the mass effectively "lives" for the purposes of translation. It's where the system's centre of gravity sits, and crucially, it moves according to $\vec{F}_{\text{net}} = M_{\text{total}} \vec{a}_{\text{cm}}$ as if nothing inside were rotating or deforming.

2. The Core Material

What Centre of Mass Actually Is

The centre of mass is a weighted average of position. If you have $n$ 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 position is:

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

where $M = \sum m_i$ is the total mass. Each position gets multiplied by how heavy that bit is, then you divide by the total. Heavy bits pull the average toward themselves. For a continuous object, you replace the sum with an integral:

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

The centre of mass is just the average location of the "stuff", weighted by mass.

Why This Point Matters: Newton's Second Law for Systems

Here's the payoff. If you sum up Newton's second law for every particle in your system, internal forces cancel (they come in equal-and-opposite pairs between any two particles), and you get:

$$\vec{F}_{\text{ext, net}} = M \vec{a}_{\text{cm}}$$

The net external force on the system equals the total mass times the acceleration of the centre of mass. This means you can ignore all the internal friction, collisions, and deformations—the centre of mass moves as if it's a single point mass under only the external forces. A firework exploding in mid-air: the centre of mass follows a parabola the entire time, even though the fragments scatter wildly. That parabola depends only on gravity; the explosion is irrelevant.

Finding Centre of Mass in Practice

For uniform objects: symmetry does the work. A uniform rod's centre of mass is at its midpoint. A uniform sphere's is at its centre. Any uniform object's centre of mass sits at its geometric centre.

For non-uniform or composite objects: break the system into simpler pieces, find their centres of mass, then apply the weighted-average formula to those pieces as if they were point masses. Example: an L-shaped steel bracket made from two rectangular pieces—find the CM of each rectangle, then treat those two CMs as point masses at the locations you found.

Coordinate-by-coordinate: you can find $x_{\text{cm}}$, $y_{\text{cm}}$, and $z_{\text{cm}}$ separately using the same formula applied to just the $x$-, $y$-, or $z$-coordinates.

graph TD
    A["System of particles<br/>or extended object"] --> B{"Is it uniform<br/>and symmetric?"}
    B -->|Yes| C["CM at geometric centre"]
    B -->|No| D{"Continuous or<br/>discrete?"}
    D -->|Discrete| E["Use sum formula:<br/>r_cm = Σ(m_i × r_i) / Σm_i"]
    D -->|Continuous| F["Use integral formula:<br/>r_cm = ∫r dm / M"]
    E --> G["Plug in coordinates<br/>for each particle"]
    F --> H["Set up density function ρ(r)<br/>and integrate"]
    G --> I["Centre of mass found"]
    H --> I
    C --> I

Velocity and Momentum of Centre of Mass

Differentiate $\vec{r}_{\text{cm}}$ with respect to time and you get the centre-of-mass velocity:

$$\vec{v}_{\text{cm}} = \frac{1}{M} \sum_{i} m_i \vec{v}_i = \frac{\vec{p}_{\text{total}}}{M}$$

where $\vec{p}_{\text{total}}$ is the total momentum of the system. Rearranging: the total momentum equals the total mass times the centre-of-mass velocity. This is a clean way to think about collisions: the CM velocity doesn't change unless an external force acts. Internal collisions rearrange the pieces but leave $\vec{p}_{\text{total}}$ untouched (if there's no friction with the ground or other external forces).

3. Worked Example

A 2 kg mass sits at position $(0, 0)$ m. A 3 kg mass sits at $(4, 0)$ m. Where is the centre of mass?

Using the formula for $x_{\text{cm}}$ and $y_{\text{cm}}$ separately:

$$x_{\text{cm}} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} = \frac{(2)(0) + (3)(4)}{2 + 3} = \frac{12}{5} = 2.4 \text{ m}$$

$$y_{\text{cm}} = \frac{m_1 y_1 + m_2 y_2}{m_1 + m_2} = \frac{(2)(0) + (3)(0)}{2 + 3} = 0 \text{ m}$$

The centre of mass is at $(2.4, 0)$ m. Notice it's closer to the 3 kg mass (at $x = 4$) than to the 2 kg mass (at $x = 0$)—the heavier object pulls it toward itself. If a net external force of 5 N acts horizontally on this two-particle system, the CM accelerates at $a_{\text{cm}} = F / M = 5 / 5 = 1 \text{ m/s}^2$, regardless of how the two masses interact with each other internally.

4. Key Takeaways

4.1 Most Important Concepts

1. Centre of mass is a weighted average of position. Every point is weighted by its mass; heavy bits dominate the average.

2. The CM moves under external forces alone. Internal forces (friction between parts, collisions within the system) never affect $\vec{a}_{\text{cm}}$.

3. For uniform symmetric objects, CM is at the geometric centre. This saves calculation for rods, spheres, cubes, etc.

4. Total momentum = total mass × CM velocity. This is the bridge between particle dynamics and centre-of-mass dynamics.

5. CM follows Newton's second law. $\vec{F}_{\text{ext}} = M \vec{a}_{\text{cm}}$ works just like for a point particle, making complex systems tractable.

6. Use the sum for discrete systems, integrals for continuous ones. The formula adapts to your problem; the concept stays the same.

7. Symmetry breaks the problem into pieces. Composite objects can be solved by finding CM of each piece, then treating those CMs as point masses.

4.2 Common Misconceptions

4.3 Compare & Contrast

5. Now Try It

Problem: A uniform rod of mass 4 kg and length 2 m lies along the $x$-axis from $x = 0$ to $x = 2$ m. A 2 kg point mass is attached at $x = 3$ m. Find the centre of mass of the system.

Your job:
1. Treat the rod as a single object with mass 4 kg at its geometric centre ($x = 1$ m).
2. Use the two-body formula (rod as one "particle", attached mass as the other).
3. Calculate $x_{\text{cm}}$.
4. Check: is your answer between 1 m and 3 m? Does it sit closer to the heavy rod than to the light mass?

Hint: The rod's CM is at $x = 1$ m (midpoint of a uniform rod). Now apply:
$$x_{\text{cm}} = \frac{(4)(1) + (2)(3)}{4 + 2}$$

Success looks like: You find $x_{\text{cm}} = \frac{10}{6} = 1.67$ m, and you can explain why this point is closer to the rod (1 m) than to the mass (3 m)—because the rod is twice as heavy.

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