Short on time? This page is the recap for the calculus-based course (separate from Physics 1, which uses sums only). For explanations, diagrams and worked examples, use the full study guide.
Recap
- A system is what you put inside a boundary; the rest is the environment. Say it in words.
- A system’s behaviour comes from the interactions between its parts. Mass or energy may cross the boundary (an open system).
- Parts can move differently from the whole. Internal structure matters for some questions, and it can change when outside conditions change (water surging in a braking tank).
- If inner details do not matter, model the system as one object at its center of mass.
- A uniform symmetric object has its center of mass on its lines of symmetry. The center of mass can be in empty space.
Key relationships
| Relationship | Units | When to use |
|---|---|---|
| x_cm = Σmᵢxᵢ / Σmᵢ (same for y, z) | m | particles, or uniform pieces placed at their own centers |
| x_cm = ∫x dm / ∫dm | m | continuous object |
| λ = dm/dx, so dm = λ dx | kg/m | rod or other thin object |
| M = ∫λ dx | kg | rod with varying density |
| M = ∫σ dA; M = ∫ρ dV | kg | plate (σ in kg/m²); solid (ρ in kg/m³) |
| Hole: include it as a piece of negative mass | — | composite shapes |
Assumptions behind the numbers
- Thin rods are treated as one-dimensional; plates have negligible thickness.
- “Uniform” means constant density, so symmetry places the center of mass.
- Coefficients in a density function carry units: in λ = 0.50 + 1.5x, the 1.5 is in kg/m².
Mistakes to avoid
- Dividing by the number of particles instead of the total mass.
- Losing the sign of a negative coordinate.
- Using λ × length when λ varies. Integrate λ dx.
- Stopping at ∫x dm. That has unit kg·m; divide by M.
- Assuming half the mass lies on each side of the center of mass.
- Insisting the center of mass is on the object. Rings and brackets say otherwise.
- Treating every system as a point, even when its internal structure decides the answer.
Quick self-check
- A 1.0 kg particle is at (0, 0) and a 3.0 kg particle at (0.40 m, 0.20 m). Where is the center of mass? ((0.30 m, 0.15 m))
- A 2.0 m rod has λ = 3.0x (kg/m, x in m from one end). Find M and x_cm. (M = 6.0 kg; x_cm = 4/3 m ≈ 1.3 m)
- Where is the center of mass of a uniform thin ring? (At its center, where there is no material.)
Next: practice questions.