What the centre of mass is
Until now mechanics treated every object as a point mass. Real bodies have finite size, and a thrown spanner tumbles in a way no single particle does. NCERT §6.1 opens the chapter by noting this inadequacy: an extended body is, in the first place, a system of particles. The centre of mass is the device that rescues the point-mass picture — it is the one point whose motion is as simple as a particle's, even while the body spins and wobbles around it.
Formally, the centre of mass (CM) is the mass-weighted average position of all the particles in a system. It is a mathematical point, not a physical particle; NIOS §7.2 states this plainly — "it is a mathematical tool and there is no physical point as CM." It need not even lie within the material of the body, as a ring or a horseshoe shows. What it depends on is one thing only: how the mass is distributed.
CM of a two-particle system
Start with the simplest case NCERT uses: two particles of masses $m_1$ and $m_2$ on a line taken as the $x$-axis, at distances $x_1$ and $x_2$ from an origin $O$. The centre of mass lies at
$$X = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}.$$
If the two masses are equal, $m_1 = m_2 = m$, this collapses to $X = (x_1 + x_2)/2$ — the CM lies exactly midway between them. When the masses differ, the CM shifts toward the heavier particle. NIOS §7.2 derives the same result from a potential-energy argument and reaches identical algebra.
CM of an n-particle system
For $n$ particles of masses $m_1, m_2, \ldots, m_n$ distributed in space, NCERT generalises the two-particle formula. With total mass $M = \sum m_i$, the three coordinates of the centre of mass are
$$X = \frac{\sum m_i x_i}{M}, \qquad Y = \frac{\sum m_i y_i}{M}, \qquad Z = \frac{\sum m_i z_i}{M}.$$
These three scalar equations combine into one vector statement using position vectors $\vec r_i = x_i\hat i + y_i\hat j + z_i\hat k$:
$$\vec R = \frac{1}{M}\sum m_i \vec r_i.$$
NCERT highlights a useful consequence: if the origin is chosen at the centre of mass itself, then $\sum m_i \vec r_i = 0$. This vanishing-moment property is exactly what later defines the centre of gravity, and it is the algebra behind the cut-out technique below.
Once located, how does the CM move? See Motion of Centre of Mass — the CM obeys $M\vec a_{cm} = \vec F_{ext}$ regardless of internal forces.
Worked example — three masses at the vertices of a triangle
Find the centre of mass of three particles at the vertices of an equilateral triangle of side $0.5\,\text{m}$. The masses are $100\,\text{g}$, $150\,\text{g}$ and $200\,\text{g}$, placed at $O$, $A$ and $B$ respectively.
Set up coordinates. With the axes chosen as in NCERT, $O = (0, 0)$, $A = (0.5, 0)$ and $B = (0.25,\, 0.25\sqrt{3})$. The total mass is $M = 100 + 150 + 200 = 450\,\text{g}$.
Apply the n-particle formula. $$X = \frac{100(0) + 150(0.5) + 200(0.25)}{450} = \frac{125}{450} = \frac{5}{18}\,\text{m}.$$ $$Y = \frac{100(0) + 150(0) + 200(0.25\sqrt{3})}{450} = \frac{50\sqrt{3}}{450} = \frac{\sqrt{3}}{9}\,\text{m}.$$
Read the result. The CM is at $\left(\tfrac{5}{18},\, \tfrac{\sqrt 3}{9}\right)\,\text{m}$. NCERT stresses the punchline: because the masses are unequal, this point is not the geometric centre (centroid) of the triangle. Only for three equal masses does the CM coincide with the centroid.
CM of continuous bodies by symmetry
A rigid body is a system of so many particles that summation becomes integration. NCERT replaces the discrete sums with integrals over mass elements $dm$:
$$X = \frac{1}{M}\int x\,dm, \qquad Y = \frac{1}{M}\int y\,dm, \qquad Z = \frac{1}{M}\int z\,dm.$$
For NEET you almost never evaluate these integrals. NCERT proves, by reflection symmetry, that for a uniform body of regular shape the centre of mass sits at the geometric centre. The argument is clean: for every mass element $dm$ at position $(x, y, z)$ there is an equal element at $(-x, -y, -z)$, so each pair contributes zero to the integral, which therefore vanishes at the centre.
| Uniform body | Location of centre of mass | Inside the body? |
|---|---|---|
| Thin straight rod | Midpoint of the rod | Yes |
| Ring (circular loop) | Centre of the circle | No — empty space |
| Circular disc | Centre of the disc | Yes |
| Solid or hollow sphere | Centre of the sphere | Yes (solid); empty for hollow shell interior |
| Rectangular lamina / cuboid | Intersection of the diagonals (centroid) | Yes |
| Triangular lamina | Centroid (intersection of medians) | Yes |
NCERT Example 6.2 supplies the triangular-lamina result: dividing the lamina into thin strips parallel to a side puts each strip's CM at its midpoint, the locus of which is a median; repeating for all three sides forces the CM onto the centroid where the medians meet. Symmetry, not calculus, does the work.
Worked example — CM of an L-shaped lamina
Find the centre of mass of a uniform L-shaped lamina of mass $3\,\text{kg}$, formed from three squares each of side $1\,\text{m}$.
Decompose into known centres. The L-shape is three identical squares. Since the lamina is uniform, each square has mass $1\,\text{kg}$, and by symmetry the CM of each square is its geometric centre: $C_1 = (1/2, 1/2)$, $C_2 = (3/2, 1/2)$, $C_3 = (1/2, 3/2)$. Treat each square as a point mass of $1\,\text{kg}$ at its centre.
Combine with the discrete formula. $$X = \frac{1(1/2) + 1(3/2) + 1(1/2)}{3} = \frac{5/2}{3} = \frac{5}{6}\,\text{m},$$ $$Y = \frac{1(1/2) + 1(1/2) + 1(3/2)}{3} = \frac{5/2}{3} = \frac{5}{6}\,\text{m}.$$
Result. The CM lies at $\left(\tfrac{5}{6}, \tfrac{5}{6}\right)\,\text{m}$, on the line of symmetry $OD$ of the L. The method generalises: split any composite body into standard pieces, place a point mass at each piece's centre, and apply $\sum m_i x_i / \sum m_i$.
This decomposition strategy — break a shape into pieces with known centres, then average — is the single most useful CM skill for NEET. The cut-out technique that follows is the same idea with one of the pieces carrying negative mass.
The cut-out (removed-mass) technique
When a piece is removed from a body — a hole punched in a disc, a corner sliced off a sheet — model the removed piece as negative mass. The remaining body equals the full body plus a negative-mass copy of the cut-out. The CM along any axis is then
$$X = \frac{M\,x_{\text{full}} - m\,x_{\text{hole}}}{M - m},$$
where $M$ is the mass of the full body, $m$ the mass of the removed piece, and $x_{\text{full}}, x_{\text{hole}}$ are the geometric centres of the full body and of the removed piece. The minus sign in both numerator and denominator is the whole trick: the removed mass is subtracted, not added.
Worked example — disc with a hole
From a uniform disc of radius $R$, a circular hole of radius $R/2$ is cut out. The centre of the hole is at a distance $R/2$ from the centre of the original disc. Locate the centre of gravity of the resulting flat body.
Mass of the removed piece. Mass scales with area for a uniform disc. The hole's area is $\pi(R/2)^2 = \pi R^2/4$, a quarter of the full disc's $\pi R^2$. So if the full disc has mass $M$, the removed piece has mass $m = M/4$.
Place the centres on an axis through O and the hole. Let the centre $O$ of the full disc be the origin. The full disc's CM is at $x_{\text{full}} = 0$; the removed piece's centre is at $x_{\text{hole}} = R/2$.
Apply the cut-out formula. $$X = \frac{M(0) - \tfrac{M}{4}\left(\tfrac{R}{2}\right)}{M - \tfrac{M}{4}} = \frac{-\,\tfrac{MR}{8}}{\tfrac{3M}{4}} = -\frac{R}{6}.$$
Result. The CM of the remaining body lies at a distance $R/6$ from the centre $O$, on the side away from the hole (the negative sign signals the direction opposite to the cut). Removing mass on one side pulls the CM toward the heavier remaining side.
Centre of mass vs centre of gravity
The centre of mass and the centre of gravity are different concepts that usually coincide. NCERT §6.8.2 defines the centre of gravity (CG) as the point about which the total gravitational torque on the body is zero — the balance point of a cardboard on a pencil tip. The centre of mass, by contrast, depends only on the distribution of mass and "has nothing to do with gravity."
The link is the field $g$. The condition for the CG is $\sum m_i \vec r_i \times \vec g = 0$. When $g$ is uniform across the body it factors out, leaving $\sum m_i \vec r_i = 0$ — which, measured from the CG, is precisely the centre-of-mass condition. So the CG coincides with the CM whenever gravity is uniform over the body, true for every laboratory-sized object. They separate only for a body so extended that $g$ varies appreciably from one part to another.
| Property | Centre of mass | Centre of gravity |
|---|---|---|
| Defined by | Mass distribution alone | Zero net gravitational torque |
| Depends on gravity? | No | Yes |
| Defining condition | $\sum m_i \vec r_i = 0$ about it | $\sum m_i \vec r_i \times \vec g = 0$ about it |
| Exists in gravity-free space? | Yes | Not meaningful |
| When they coincide | Always, provided $g$ is uniform over the body (all NEET-scale objects) | |
Centre of mass in one breath
- CM is the mass-weighted average position: $X = \sum m_i x_i / M$, and likewise for $Y, Z$.
- Two particles: CM lies on the join, closer to the heavier mass, with $m_1 r_1 = m_2 r_2$.
- Uniform regular bodies (rod, ring, disc, sphere) have CM at the geometric centre — by reflection symmetry, no integration.
- The CM can lie outside the material (ring, L-shape, body with a hole).
- Cut-out technique: treat the removed piece as negative mass; $X = (Mx_{\text{full}} - m\,x_{\text{hole}})/(M - m)$.
- For the NCERT disc with an $R/2$ hole at $R/2$, the CM shifts to $R/6$ from centre, away from the hole.
- CM depends on mass only; CG depends on gravity. They coincide whenever $g$ is uniform over the body.