Introduction to Systems of Particles and Rigid Bodies

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Introduction to Systems of Particles and Rigid Bodies

TL;DR

You'll learn how to treat a collection of masses — from a cloud of dust to a spinning wrench — as one object using the center of mass. You'll see why internal forces cancel out, leaving only external forces to control the overall motion. You'll also learn when and why a rigid body needs rotation equations on top of this.

1. The Mental Model

Forget tracking every particle separately — that's hopeless for anything with more than two or three pieces. Instead, find one special point, the center of mass, that moves exactly as if all the mass and all the external force were concentrated there. A rigid body is just a system of particles that can't change shape, so it needs one more idea: rotation about that point. The whole messy system behaves, for translation purposes, like a single point mass at its center of mass.

2. The Core Material

2.1 Center of Mass: Definition and Why It Works

For a system of $N$ particles with masses $m_i$ at positions $\vec{r}_i$, the center of mass (COM) is:

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

where $M = \sum_i m_i$ is total mass. For a continuous object, the sum becomes an integral:

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

Now here's the key derivation that makes this useful. Take the second time-derivative of $\vec{R}_{cm}$:

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

Each $\vec{F}_i$ is the total force on particle $i$, which splits into external forces (gravity, normal forces, applied pushes) and internal forces (particle $i$ pushing or pulling on particle $j$, like spring connections or collisions within the system). By Newton's third law, every internal force comes in an equal-and-opposite pair: particle $i$ pushes on $j$ with force $\vec{f}_{ij}$, and $j$ pushes back on $i$ with $\vec{f}_{ji} = -\vec{f}_{ij}$. When you sum over all particles, every internal pair cancels exactly. What survives is:

$$M\ddot{\vec{R}}_{cm} = \vec{F}_{ext}$$

This is the single most important result in this topic. It says the center of mass obeys Newton's second law as if it were a point particle carrying the entire system's mass, completely ignoring how complicated the internal structure is. A fireworks shell explodes into a hundred fragments — internal forces (the explosion) do wild things to each piece, but the COM keeps sailing along the exact same parabola it was on before the explosion, because the explosion is purely internal.

2.2 Linear Momentum and Angular Momentum of a System

Total linear momentum is $\vec{P} = \sum_i m_i \vec{v}_i = M\vec{V}_{cm}$. Combined with the result above:

$$\frac{d\vec{P}}{dt} = \vec{F}_{ext}$$

If $\vec{F}_{ext} = 0$, momentum is conserved — this is why explosions, collisions, and recoil problems always work through the COM.

Angular momentum needs more care because, unlike force, torque from internal pairs does NOT automatically cancel in general — unless the internal forces are central (act along the line joining each pair, like gravity or electrostatic/contact forces, which is true for almost every system you'll meet). Total angular momentum about a fixed origin $O$:

$$\vec{L}_O = \sum_i \vec{r}_i \times m_i \vec{v}_i$$

Differentiating and using Newton's third law with central internal forces, the internal torques cancel in pairs (because $\vec{f}_{ij}$ lies along $\vec{r}_i - \vec{r}_j$, so the cross products cancel), leaving:

$$\frac{d\vec{L}_O}{dt} = \vec{\tau}_{ext,O}$$

This is the rotational analogue of $\vec{F}=M\vec{a}$. A convenient and very common choice is to compute angular momentum about the COM itself — this works even if the COM is accelerating, giving $\frac{d\vec{L}_{cm}}{dt} = \vec{\tau}_{ext,cm}$, a result that's surprisingly robust and used constantly in rigid body dynamics.

2.3 Why Rigid Bodies Need an Extra Equation

A rigid body is a system of particles where every pairwise distance $|\vec{r}_i - \vec{r}_j|$ stays constant. This constraint means the body has only 6 degrees of freedom in 3D (3 for COM translation, 3 for orientation) no matter how many particles make it up — compare that to $3N$ degrees of freedom for a general system of $N$ free particles.

Because of this, rigid body motion splits cleanly into two independent problems:

1. Translation of the COM: governed by $M\ddot{\vec{R}}_{cm} = \vec{F}_{ext}$ (Section 2.1).
2. Rotation about the COM: governed by $\frac{d\vec{L}_{cm}}{dt} = \vec{\tau}_{ext,cm}$ (Section 2.2), where for a rigid body $\vec{L}_{cm} = I\vec{\omega}$ ($I$ is the moment of inertia, $\vec{\omega}$ is angular velocity).

You genuinely need both equations to predict what happens — translation alone tells you nothing about spin, and vice versa. A classic example: push a hockey puck off-center. The COM equation tells you it accelerates in the direction you pushed, full stop, regardless of where on the puck you pushed. The torque equation tells you separately that it also starts spinning, because your push created a torque about the COM. These two motions are decoupled and happen simultaneously.

flowchart TD
    A["System of N particles"] --> B{"Rigid constraint?<br/>distances fixed"}
    B -- No --> C["General system:<br/>track COM only via<br/>sum of external forces"]
    B -- Yes --> D["Rigid body: 6 DOF"]
    D --> E["Translation:<br/>M·a_cm = F_ext"]
    D --> F["Rotation about COM:<br/>dL_cm/dt = τ_ext,cm"]
    E --> G["COM follows<br/>Newton's 2nd law"]
    F --> H["Spin/orientation<br/>via I and ω"]

3. Worked Example

Setup: A uniform rod of mass $M=2\,\text{kg}$ and length $L=1\,\text{m}$ lies at rest on frictionless ice. You strike it with a horizontal impulse $J = 4\,\text{N·s}$, applied perpendicular to the rod at a point $0.3\,\text{m}$ from its center. Find the velocity of the COM immediately after, and the angular velocity of the rod.

Step 1 — Translation (COM motion). The impulse changes total momentum regardless of where it was applied:

$$J = M v_{cm} \implies v_{cm} = \frac{J}{M} = \frac{4}{2} = 2\ \text{m/s}$$

The entire rod's COM moves at 2 m/s in the direction of the strike. Notice: it does not matter that you hit it off-center — the COM speed only depends on total impulse and total mass. This is the direct consequence of $M\ddot{\vec{R}}_{cm}=\vec{F}_{ext}$ integrated over time.

Step 2 — Rotation (angular momentum about COM). The impulsive torque about the COM is:

$$\tau \,\Delta t = J \cdot d = 4 \times 0.3 = 1.2\ \text{N·m·s}$$

This equals the change in angular momentum about the COM: $\Delta L_{cm} = I_{cm}\,\omega$.

For a uniform rod about its center, $I_{cm} = \frac{1}{12}ML^2 = \frac{1}{12}(2)(1)^2 = \frac{1}{6}\ \text{kg·m}^2$.

$$\omega = \frac{\Delta L_{cm}}{I_{cm}} = \frac{1.2}{1/6} = 7.2\ \text{rad/s}$$

Result: Right after the strike, the rod's center moves at 2 m/s in a straight line, while the rod simultaneously spins at 7.2 rad/s about that moving center. Both motions proceed independently from that instant on — the COM keeps moving at constant velocity (no more forces act), and the rod keeps spinning at constant $\omega$ (no more torques act), tracing out a path where the rod's ends swing around a center line that travels in a straight line. This is exactly the translation/rotation split from Section 2.3 in action.

4. Key Takeaways

4.1 Most Important Concepts

4.1 Most Important Concepts

• Center of mass formula: $\vec{R}_{cm} = \frac{1}{M}\sum_i m_i \vec{r}_i$ — the mass-weighted average position of a system.
• Internal forces always cancel in pairs: Newton's third law guarantees $\sum \vec{f}_{ij} + \vec{f}_{ji} = 0$, so only external forces move the COM.
• $M\ddot{\vec{R}}_{cm} = \vec{F}_{ext}$: the COM obeys Newton's second law regardless of internal chaos (explosions, collisions, deformation).
• Momentum conservation: if $\vec{F}_{ext}=0$, total momentum $\vec{P}$ stays constant — the foundation of all collision problems.
• $\frac{d\vec{L}_{cm}}{dt} = \vec{\tau}_{ext,cm}$: angular momentum about the COM changes only due to external torque, valid even if the COM accelerates.
• Rigid bodies have exactly 6 degrees of freedom: 3 translational (COM position) + 3 rotational (orientation), independent of particle count.
• Translation and rotation are decoupled: you can solve the COM motion and the spin motion as two separate problems and combine them.

4.2 Common Misconceptions

• "The center of mass is always inside the object." False — a boomerang or donut has its COM in empty space.
• "Where you apply a force doesn't matter." It matters enormously for rotation (it determines torque), but not at all for the COM's translational motion — only the magnitude and direction of the net external force matter there.
• "Internal forces don't affect anything." They don't affect the COM's trajectory or total momentum, but they absolutely can change the shape, spin, and internal energy of the system (think of a diver tucking mid-air).
• "Angular momentum conservation needs zero net force." It actually needs zero net torque about the chosen point — these are independent conditions; you can have force but no torque, or torque but no force.

4.3 Compare & Contrast

5. Now Try It

Problem: Two skaters, A (60 kg) and B (40 kg), stand together at rest on frictionless ice, holding a compressed spring between them. When released, the spring pushes them apart. Skater A ends up moving at 2 m/s. (a) Find skater B's speed. (b) Describe what happens to the system's center of mass throughout this process, and explain why using the physics from Section 2.1.

Work it through using momentum conservation ($\vec{F}_{ext}=0$ since ice is frictionless and gravity/normal cancel vertically) and the COM argument from the worked example's logic.