Centre of mass — how to find for discrete and continuous bodies

Question

How do we find the centre of mass (COM) for a system of discrete particles and for continuous bodies (rods, semicircles, hemispheres)?

Solution — Step by Step

For $n$ particles with masses $m_1, m_2, \ldots, m_n$ at positions $x_1, x_2, \ldots, x_n$ :

x_{CM} = \frac{m_1 x_1 + m_2 x_2 + \cdots + m_n x_n}{m_1 + m_2 + \cdots + m_n} = \frac{\sum m_i x_i}{\sum m_i}

Similarly for $y_{CM}$ and $z_{CM}$ .

Example: Two particles of mass 2 kg at $x = 0$ and 3 kg at $x = 5$ m:

x_{CM} = \frac{2(0) + 3(5)}{2 + 3} = \frac{15}{5} = 3 \text{ m}

The COM is closer to the heavier particle — always.

Replace the sum with an integral. For a body with linear mass density $\lambda(x)$ :

x_{CM} = \frac{\int x \, dm}{\int dm}

where $dm = \lambda \, dx$ for a rod, $dm = \sigma \, dA$ for a surface, or $dm = \rho \, dV$ for a volume.

Example — Uniform rod of length L:

x_{CM} = \frac{\int_0^L x \cdot \frac{M}{L} dx}{M} = \frac{\frac{M}{L} \cdot \frac{L^2}{2}}{M} = \frac{L}{2}

No surprise — the COM of a uniform rod is at its midpoint.

Body	COM location
Uniform rod	Centre ( $L/2$ from either end)
Semicircular ring (radius $R$ )	$\frac{2R}{\pi}$ from centre
Semicircular disc (radius $R$ )	$\frac{4R}{3\pi}$ from centre
Hemisphere (solid, radius $R$ )	$\frac{3R}{8}$ from flat face
Hemisphere (hollow, radius $R$ )	$\frac{R}{2}$ from flat face
Triangular lamina	At centroid ( $\frac{1}{3}$ of median)
Cone (solid, height $H$ )	$\frac{H}{4}$ from base
Cone (hollow, height $H$ )	$\frac{H}{3}$ from base

For a body with a hole (like a disc with a circular cavity), use:

x_{CM} = \frac{M_{full} \cdot x_{full} - M_{removed} \cdot x_{removed}}{M_{full} - M_{removed}}

This avoids complicated integration. Treat the removed part as having negative mass.

flowchart TD
    A["Find COM"] --> B{"Type of system?"}
    B -->|"Discrete particles"| C["Use sum formula: sum mi xi / sum mi"]
    B -->|"Continuous body"| D{"Shape known?"}
    D -->|"Standard shape"| E["Use memorised result from table"]
    D -->|"Non-standard"| F["Integrate: integral of x dm / integral of dm"]
    B -->|"Body with cavity"| G["Use subtraction: full body minus removed part"]

Why This Works

The centre of mass is the mass-weighted average position. For discrete systems, it is a simple weighted average. For continuous bodies, the sum becomes an integral because we are adding up infinitely many infinitesimally small mass elements.

The subtraction trick works because of superposition: a disc with a hole equals a full disc minus the removed piece. Since COM calculation is linear (weighted sum), we can subtract the removed part’s contribution.

Alternative Method

For complex geometries, use symmetry first. If a body has an axis of symmetry, the COM lies on that axis. This reduces a 3D problem to 1D. For a uniform body with two axes of symmetry, the COM is at the intersection — no calculation needed.

Common Mistake

When using the subtraction method, students often subtract positions instead of using the formula correctly. You must write $x_{CM} = \frac{M_{full} \cdot x_{full} - M_{cavity} \cdot x_{cavity}}{M_{full} - M_{cavity}}$ . A common error is to write $x_{full} - x_{cavity}$ directly, which is dimensionally and physically wrong. The masses act as weights in the average — you cannot skip them. JEE Main 2023 had a disc-with-cavity problem where this was the primary source of wrong answers.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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