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

Question

Find the centre of mass of a system of three particles: 2 kg at $(0, 0)$ , 3 kg at $(4, 0)$ , and 5 kg at $(0, 3)$ . Also explain how to find the COM of a semicircular disc of radius $R$ .

(JEE Main / CBSE Class 11 pattern)

Solution — Step by Step

flowchart TD
    A["Find COM"] --> B{"System type?"}
    B -->|"Discrete particles"| C["Use summation formula\nxcm = Σmᵢxᵢ / Σmᵢ"]
    B -->|"Continuous body\n(uniform)"| D{"Standard shape?"}
    D -->|Yes| E["Use known results\n(memorise for rod, disc, etc.)"]
    D -->|No| F["Use integration\nxcm = ∫x dm / ∫dm"]
    B -->|"Body with cavity"| G["Subtraction method\nTreat cavity as\nnegative mass"]

x_{\text{cm}} = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{m_1 + m_2 + m_3} = \frac{2(0) + 3(4) + 5(0)}{2+3+5} = \frac{12}{10} = \mathbf{1.2}

y_{\text{cm}} = \frac{m_1 y_1 + m_2 y_2 + m_3 y_3}{m_1 + m_2 + m_3} = \frac{2(0) + 3(0) + 5(3)}{10} = \frac{15}{10} = \mathbf{1.5}

The centre of mass is at $\mathbf{(1.2, 1.5)}$ .

By symmetry, $x_{\text{cm}} = 0$ (taking the diameter as the x-axis, the flat edge).

For $y_{\text{cm}}$ , we integrate using thin semicircular strips or use the known result:

y_{\text{cm}} = \frac{4R}{3\pi}

This result comes from integrating $y \cdot dm$ over the disc area, where $dm = \frac{2M}{\pi R^2} \cdot y \, dx \, dy$ in appropriate coordinates. The derivation uses polar coordinates and gives $\frac{4R}{3\pi} \approx 0.424R$ above the flat edge.

Body	COM position (from the symmetric point)
Uniform rod	Centre of the rod
Semicircular ring	$\frac{2R}{\pi}$ from centre
Semicircular disc	$\frac{4R}{3\pi}$ from centre
Hemispherical shell	$\frac{R}{2}$ from centre
Solid hemisphere	$\frac{3R}{8}$ from centre
Triangular lamina	$\frac{h}{3}$ from base
Cone (solid)	$\frac{h}{4}$ from base

Why This Works

The centre of mass is the mass-weighted average position. For discrete particles, it is a simple weighted sum. For continuous bodies, the sum becomes an integral. The physical meaning: if you support the body at its COM, it balances perfectly — no net torque due to gravity.

The subtraction method for cavities works because COM is additive: if a full disc has COM at the centre, and we remove a smaller disc (treated as negative mass), the COM of the remaining body shifts away from the cavity.

Alternative Method — Subtraction for Bodies with Cavities

For a disc of radius $R$ with a circular hole of radius $R/2$ cut from one edge:

Total mass $M$ , cavity mass $m = M/4$ (area ratio), cavity COM at distance $R/2$ from centre.

x_{\text{cm}} = \frac{M \cdot 0 - m \cdot (R/2)}{M - m} = \frac{-\frac{M}{4} \cdot \frac{R}{2}}{M - \frac{M}{4}} = \frac{-MR/8}{3M/4} = -\frac{R}{6}

The COM shifts $R/6$ away from the cavity.

For JEE, the subtraction method is tested almost every year. The trick: treat the cavity as a body with negative mass and apply the two-body COM formula. This avoids integration entirely and solves complex problems in under 2 minutes.

Common Mistake

When using the subtraction method, students often forget to use the remaining mass in the denominator. The denominator is $(M - m)$ , not $M$ . Also, the sign of the cavity mass must be negative. If you write positive mass for the cavity, the COM shifts toward the cavity instead of away from it — physically absurd.

Question

Solution — Step by Step

Why This Works

Alternative Method — Subtraction for Bodies with Cavities

Common Mistake

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