Centre of Mass — Discrete and Continuous

Why Centre of Mass is the Hidden Lever in Mechanics

Most mechanics problems become drastically simpler once you compute the centre of mass (CoM). Newton’s second law for a system of particles reduces to a single equation: the net external force equals the total mass times the acceleration of the CoM. Internal forces — collisions, explosions, spring interactions — never affect the CoM motion.

For JEE Main and NEET, centre of mass is a 1- to 2-mark scoring chapter. JEE Advanced uses CoM as the doorway into rigid-body dynamics, and a strong CoM intuition makes rotational mechanics far less painful.

This hub covers both discrete systems (point masses) and continuous systems (rods, plates, hemispheres). We build the formulas from scratch, work through six graded problems, and end with the trick patterns that recur in PYQs.

Key Terms & Definitions

Centre of mass — the unique point where the system’s mass can be considered to act for translational motion. It moves as if all external forces were applied at that point.

Centroid — for a uniform body, the geometric centre coincides with the CoM. For non-uniform bodies, they differ.

Position vector $\vec{r}_i$ — location of the $i$ -th particle relative to a chosen origin.

Linear mass density ( $\lambda$ ) — mass per unit length, used for rods and wires. Surface density ( $\sigma$ ) — mass per unit area, used for plates and shells. Volume density ( $\rho$ ) — mass per unit volume, used for solids.

Discrete Systems

Two-particle system

For two masses $m_1$ and $m_2$ at positions $x_1$ and $x_2$ :

$x_{\text{cm}} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$

The CoM lies on the line joining the two masses, closer to the heavier one. If $m_1 = m_2$ , it sits exactly at the midpoint.

N-particle system

For $N$ particles in 3D:

$\vec{r}_{\text{cm}} = \frac{\sum_{i=1}^{N} m_i \vec{r}_i}{\sum_{i=1}^{N} m_i}$

Apply this component by component: $x_{\text{cm}}, y_{\text{cm}}, z_{\text{cm}}$ are computed independently.

Continuous Systems

For continuous mass distributions, replace the sum with an integral:

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

The trick is choosing the right $dm$ :

Rod: $dm = \lambda\, dx$
Plate: $dm = \sigma\, dA$
Solid: $dm = \rho\, dV$

Famous Continuous Results

Memorise these — they save 5+ minutes per problem.

Body	CoM location
Uniform rod	Midpoint
Uniform triangular plate	Centroid (intersection of medians)
Semicircular wire of radius $R$	$2R/\pi$ from diameter
Semicircular plate of radius $R$	$4R/(3\pi)$ from diameter
Solid hemisphere of radius $R$	$3R/8$ from flat face
Hollow hemisphere of radius $R$	$R/2$ from flat face
Solid cone of height $h$	$h/4$ from base
Hollow cone of height $h$	$h/3$ from base

Solved Examples Graded Easy → Hard

Example 1 (Easy, CBSE-level)

Three particles of masses 1, 2, 3 kg are at $(0, 0)$ , $(1, 0)$ , $(0, 1)$ respectively. Find the CoM.

Solution:

$x_{\text{cm}} = \frac{1 \times 0 + 2 \times 1 + 3 \times 0}{6} = \frac{2}{6} = \frac{1}{3}$

$y_{\text{cm}} = \frac{1 \times 0 + 2 \times 0 + 3 \times 1}{6} = \frac{1}{2}$

CoM at $(1/3, 1/2)$ .

Example 2 (Easy, JEE Main)

A uniform rod of length $L$ and mass $M$ has a small bead of mass $M$ attached to one end. Find the CoM measured from that end.

Solution: Treat the rod as a point mass at its midpoint $L/2$ .

$x_{\text{cm}} = \frac{M \cdot 0 + M \cdot (L/2)}{2M} = \frac{L}{4}$

Example 3 (Medium, JEE Main)

A circular disc of radius $R$ has a small disc of radius $R/2$ removed, with the small disc’s centre at distance $R/2$ from the original centre. Find the new CoM.

Solution: Use the negative-mass trick. Treat the removed piece as a negative mass.

Original disc mass $M$ , area $\pi R^2$ . Removed area $= \pi R^2/4$ , mass $= M/4$ .

Place origin at original centre. The full disc has CoM at origin. The removed disc has CoM at $R/2$ .

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

The CoM shifts a distance $R/6$ opposite to the cut-out.

Example 4 (Medium, NEET)

A man of mass 60 kg stands at one end of a 120 kg, 4 m long boat at rest in still water. He walks to the other end. By how much does the boat shift?

Solution: External force is zero (water is frictionless), so CoM doesn’t move.

Initial: man at $x = 0$ , boat CoM at $x = 2$ . System CoM:

$x_0 = \frac{60 \times 0 + 120 \times 2}{180} = \frac{240}{180} = \frac{4}{3}\,\text{m}$

Let the boat shift by $d$ . Final positions: man at $4 + d$ (he is now at the far end of the shifted boat), boat CoM at $2 + d$ .

$\frac{60(4 + d) + 120(2 + d)}{180} = \frac{4}{3}$

$\frac{240 + 60d + 240 + 120d}{180} = \frac{4}{3} \implies 480 + 180d = 240 \implies d = -\frac{4}{3}\,\text{m}$

The boat shifts $4/3$ m in the opposite direction.

Example 5 (Medium, JEE Advanced)

Find the CoM of a uniform semicircular wire of radius $R$ .

Solution: Place the diameter along the x-axis, with origin at centre. By symmetry, $x_{\text{cm}} = 0$ . For $y_{\text{cm}}$ :

Element at angle $\theta$ : position $(R\cos\theta, R\sin\theta)$ , length $R\,d\theta$ , mass $dm = \lambda R\,d\theta$ where $\lambda = M/(\pi R)$ .

$y_{\text{cm}} = \frac{1}{M}\int_0^\pi R\sin\theta \cdot \lambda R\,d\theta = \frac{\lambda R^2}{M}\int_0^\pi \sin\theta\,d\theta$

$y_{\text{cm}} = \frac{\lambda R^2}{M} \times 2 = \frac{2R}{\pi}$

Example 6 (Hard, JEE Advanced)

A uniform solid hemisphere of mass $M$ and radius $R$ rests on its flat face. A small block of mass $m$ is placed at the apex. Find the location of the system CoM measured from the flat face.

Solution: Hemisphere CoM at $3R/8$ from flat face. Block at $R$ .

$y_{\text{cm}} = \frac{M(3R/8) + mR}{M + m} = \frac{3MR/8 + mR}{M + m}$

For $m = M$ : $y_{\text{cm}} = (3R/8 + R)/2 = 11R/16$ .

Exam-Specific Tips

JEE Main — Discrete CoM problems and the negative-mass trick are the most common. Memorise the 8 standard continuous results.

JEE Advanced — Boat-on-water and conservation-of-CoM problems are the favourites. Always start by asking “is the net external force zero?” If yes, CoM is fixed.

NEET — Pure formula application. Master the discrete formula and the named results table.

When a body has a removed piece, negative mass beats integration. Use the formula:

$x_{\text{cm}} = \frac{M_{\text{full}} x_{\text{full}} - M_{\text{removed}} x_{\text{removed}}}{M_{\text{full}} - M_{\text{removed}}}$

For symmetric bodies, identify the symmetry axis first — the CoM must lie on it. This often reduces a 2D problem to a 1D integral.

Common Mistakes to Avoid

Mistake 1: Forgetting that internal forces (collisions, explosions) cannot move the CoM. If the system starts at rest with no external force, the CoM stays put forever.

Mistake 2: Confusing the formulas for hemispheres. Solid hemisphere: $3R/8$ . Hollow hemisphere: $R/2$ . They differ because density is concentrated differently.

Mistake 3: Plugging mass densities into discrete formulas, or using point masses where integration is needed. Always classify the problem first.

Mistake 4: Wrong sign in the negative-mass trick. The removed mass enters with a minus in both numerator and denominator.

Mistake 5: Treating CoM velocity as zero when external forces are present. Friction, gravity, and normal forces all change CoM motion.

Practice Questions

Q1. Particles of mass 2, 3, 5 kg sit at positions 0, 2, 4 m on the x-axis. Find CoM.

$x_{\text{cm}} = (0 + 6 + 20)/10 = 2.6$ m.

Q2. A uniform L-shaped wire has two arms of equal length $L$ at right angles. Find CoM.

Each arm has CoM at midpoint, mass $M/2$ . CoM of arm 1: $(L/2, 0)$ . CoM of arm 2: $(0, L/2)$ . System: $(L/4, L/4)$ .

Q3. A man of 70 kg jumps from a 30 kg boat. How far does the boat recoil if the man moves 4 m?

Conservation: $70 \times 4 = 30 \times d \implies d = 28/3 \approx 9.33$ m.

Q4. Find the CoM of a uniform semicircular plate of radius $R$ .

By symmetry $x_{\text{cm}} = 0$ . Standard result: $y_{\text{cm}} = 4R/(3\pi)$ .

Q5. A uniform solid sphere of radius $R$ has a spherical cavity of radius $R/2$ tangent to its surface. Find the new CoM.

Removed sphere has mass $M/8$ , centre at $R/2$ from original centre. New CoM: $-(M/8)(R/2) / (7M/8) = -R/14$ .

FAQs

Why is the CoM useful even when an object is rotating? Newton’s second law for the CoM is exact regardless of rotation: $\vec{F}_{\text{ext}} = M\vec{a}_{\text{cm}}$ . Rotation adds extra equations but doesn’t break this one.

Does the CoM lie inside the body? Not always. A boomerang or a hollow ring has CoM in empty space.

Why is the solid-hemisphere CoM at $3R/8$ , not $R/2$ ? Because mass is concentrated near the flat base — wider cross-sections at the bottom outweigh the top.

Can the CoM accelerate without any particle accelerating? No. CoM acceleration is the mass-weighted average of particle accelerations. If the average is nonzero, at least one particle is accelerating.

Is centroid the same as CoM? Yes for uniform bodies, no for non-uniform. Geometry textbooks compute centroids; physics requires the mass-weighted version.

How does CoM relate to torque? For uniform gravity, the gravitational torque on a body is the same as if all the weight were applied at the CoM. That is why we balance objects on their CoM.