JEE Mains Physics · System of Particles and Rotational Motion

Centre of Mass and Its Motion

The centre of mass is the mass-weighted average position of a system; it moves as if all the mass sat there and every external force acted there.

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Why this matters

Twenty-two PYQs, seventeen of them multiple choice, and four from 2026. Eleven ask where the centre of mass is: point masses, a bent rod, a plate with a hole cut out, a rod or plate whose density varies. Eleven ask how it moves: the shape of its path, keeping it fixed while masses shift, explosions and recoil.

Concept 1 of 2: Locating the centre of mass

The centre of mass is an average position in which heavy parts count more. Find it one coordinate at a time. Look for symmetry first: a symmetric body has its centre of mass on the line of symmetry, which fixes one coordinate for free. A hole is easiest to handle as a piece of negative mass.

Definition

Point masses: $x_{cm} = \dfrac{\sum m_i x_i}{\sum m_i}$ , and the same for y and z.
A continuous body: $x_{cm} = \dfrac{\int x\,dm}{\int dm}$ . For a rod with $\lambda = \lambda_0(1 - x^{2}/L^{2})$ , $\int_0^L x\lambda\,dx = \lambda_0L^{2}/4$ and $\int_0^L \lambda\,dx = 2\lambda_0L/3$ , so $x_{cm} = 3L/8$ .
A bent rod: treat each straight piece as a point mass at its own midpoint, with mass in proportion to its length.
Standard results: semicircular ring $2R/\pi$ from the centre; semicircular disc $4R/3\pi$ ; both on the line of symmetry.
A cut-out: full body minus the hole, $x_{cm} = \dfrac{Mx_1 - mx_2}{M - m}$ . For a uniform plate, mass is in proportion to AREA, so a hole of half the radius has a quarter of the mass.

Centre of mass

x_{cm} = \frac{\sum m_i x_i}{\sum m_i} \qquad x_{cm} = \frac{\int x\,dm}{\int dm} \qquad x_{cm} = \frac{Mx_1 - mx_2}{M - m}

Worked example

A uniform square plate of side 6 cm lies with one corner at the origin and its sides along the axes. A square of side 3 cm is cut from the far corner (the corner at (6, 6)). Find the centre of mass of the rest.

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The same idea in a real exam question:

JEE Mains · 2025 · 23 January 2025 · Q20Moderate

Example 1 · System of Particles and Rotational Motion · Centre of Mass and Its Motion

Consider a circular disc of radius 20 cm with centre located at the origin. A circular hole of a radius 5 cm is cut from this disc in such a way that the edge of the hole touches the edge of the disc. The distance of centre of mass of residual or remaining disc from the origin will be-

Averaging positions without the masses

The centre of mass of 1 kg at x = 0 and 3 kg at x = 4 m is at 3 m, not at the midpoint 2 m. Each position must be multiplied by its mass before dividing by the total mass.

A hole's mass goes with its area

A circular hole of half the disc's radius removes a quarter of the mass, not half. For a uniform solid, mass goes with volume, so half the radius removes an eighth.

Semicircular ring and semicircular disc differ

A ring has all its mass on the rim, so its centre of mass is further out: 2R/π ≈ 0.64R. A disc has mass spread inwards: 4R/3π ≈ 0.42R.

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Concept 2 of 2: Motion of the centre of mass, explosions and recoil

Internal forces cancel in pairs, so only external forces move the centre of mass. With no external force it keeps its velocity: a body at rest that explodes leaves its centre of mass where it was. Its path is a straight line when its velocity and acceleration point the same way, and a parabola when they do not.

Definition

$\vec v_{cm} = \dfrac{\sum m_i\vec v_i}{M}$ , $\vec a_{cm} = \dfrac{\sum m_i\vec a_i}{M}$ , and $M\vec a_{cm} = \vec F_{ext}$ .
No external force: $\vec v_{cm}$ is constant. Path of the centre of mass: straight if $\vec v_{cm} \parallel \vec a_{cm}$ (or $\vec v_{cm} = 0$ ); a parabola if a constant $\vec a_{cm}$ is at an angle to $\vec v_{cm}$ .
Keeping the centre of mass fixed: $m_1\Delta x_1 + m_2\Delta x_2 = 0$ .
Atwood machine: the blocks move in opposite directions, so $a_{cm} = \dfrac{m_1a - m_2a}{m_1 + m_2} = \left(\dfrac{m_1 - m_2}{m_1 + m_2}\right)^{2}g$ , downward.
Explosion from rest: the momenta are equal and opposite, so speeds go inversely as masses. With $K = p^{2}/2m$ , $K_A/K_B = m_B/m_A$ : the lighter piece carries more energy. With three pieces, the third balances the vector sum of the other two momenta.
Recoil: a gun of mass M firing a bullet m at v recoils at $mv/M$ . A machine gun firing n bullets per second needs a force $nmv$ to hold it.
Kinetic energy of a system $= \tfrac{1}{2}Mv_{cm}^{2}$ + kinetic energy relative to the centre of mass.

Velocity of the centre of mass, and Newton's second law for a system

\vec v_{cm} = \frac{\sum m_i\vec v_i}{\sum m_i} \qquad M\vec a_{cm} = \vec F_{ext}

Worked example

A 1 kg body moves with velocity

3\hat i

m/s and no acceleration. A 2 kg body moves with velocity

3\hat j

m/s and acceleration

2\hat i

m/s². Find the velocity and acceleration of their centre of mass, and the shape of its path.

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The same idea in a real exam question:

JEE Mains · 2026 · 6 Apr 2026 Shift 2 · Q7Moderate

Example 2 · System of Particles and Rotational Motion · Centre of Mass and Its Motion

Two identical bodies A and B of equal masses have initial velocities

{\overrightarrow{v}}_{1}= 4\widehat{i}m/s

and

{\overrightarrow{v}}_{2}= 4\widehat{j}m/s

respectively. The body A has acceleration

{\overrightarrow{a}}_{1}= 6\widehat{i}+ 6\widehat{j}m/s^{2}

while the acceleration of the other body

B

is zero. The centre of mass of the two bodies moves in

\_\_\_\_

path.

Equal momenta, not equal speeds

After a body at rest explodes into two pieces, the pieces have equal and opposite MOMENTA. The lighter piece is faster and carries more kinetic energy, in the inverse ratio of the masses.

Signs in an Atwood machine

One block goes up while the other comes down, so their accelerations enter a_cm with opposite signs. Adding them as if both moved down gives a_cm = a, which is too large.

An accelerating centre of mass need not curve

A constant acceleration gives a straight path when the velocity of the centre of mass is along it, or zero. Only an acceleration at an angle to the velocity gives a parabola.

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Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Locating the centre of mass
Centre of mass
$x_{cm} = \frac{\sum m_i x_i}{\sum m_i} \qquad x_{cm} = \frac{\int x\,dm}{\int dm} \qquad x_{cm} = \frac{Mx_1 - mx_2}{M - m}$
Motion of the centre of mass, explosions and recoil
Velocity of the centre of mass, and Newton's second law for a system
$\vec v_{cm} = \frac{\sum m_i\vec v_i}{\sum m_i} \qquad M\vec a_{cm} = \vec F_{ext}$

Watch out for (6)

Averaging positions without the masses
→ Locating the centre of mass
A hole's mass goes with its area
→ Locating the centre of mass
Semicircular ring and semicircular disc differ
→ Locating the centre of mass
Equal momenta, not equal speeds
→ Motion of the centre of mass, explosions and recoil
Signs in an Atwood machine
→ Motion of the centre of mass, explosions and recoil
An accelerating centre of mass need not curve
→ Motion of the centre of mass, explosions and recoil

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