Physics
Rotational Motion
Centre of Mass

Rotational Motion · JEE & NEET Physics

Centre of Mass: notes and previous year questions

The mass-weighted average position: two masses, many particles, solid bodies, holes, curved shapes, and how the centre of mass moves.

35 JEE Main questions (2002–2026)
1 JEE Advanced questions (2009)
12 NEET questions (2002–2026)

Centre of Mass in short

The centre of mass is the mass-weighted average position; for moving as a whole, the body acts like one particle there.
For two masses it lies closer to the heavier one, splitting the gap in the inverse ratio of the masses: m₁r₁ = m₂r₂.
For many particles, each coordinate is its own weighted average.
A uniform body's centre of mass lies on every line of symmetry.

1What is the centre of mass?

Throw a hammer and it spins: the end of the handle loops and wobbles. But one point of it follows a clean parabola, exactly like a thrown ball. That point is the centre of mass (COM).

The centre of mass is the average position of all the mass, where heavier parts count more. For moving as a whole (translation), you can treat the entire body as one particle, with all its mass, sitting at that point.

A light rod with 3 kg at one end and 1 kg at the other tips if you support it in the middle. Slide the support toward the 3 kg end and it balances a quarter of the way along: that balance point is the centre of mass.

2Two particles

Multiply each mass by its position, add, and divide by the total mass:

x_{cm} = \frac{m_1x_1 + m_2x_2}{m_1 + m_2}

With $m_1$ at the origin and $m_2$ a distance $d$ away:

x_{cm} = \frac{m_2 d}{m_1 + m_2}

Its distance from m₁.

The distances from the two masses are $r_1 = \frac{m_2d}{m_1 + m_2}$ and $r_2 = \frac{m_1d}{m_1 + m_2}$ , so

m_1r_1 = m_2r_2

Like a balanced seesaw: the gap splits in the ratio m₂ : m₁.

Equal masses: exactly halfway.
Unequal masses: closer to the heavier one, in the inverse ratio of the masses.

3Many particles

Each coordinate is its own weighted average, with $M = \sum m_i$ the total mass:

x_{cm} = \frac{\sum m_i x_i}{M}, \quad y_{cm} = \frac{\sum m_i y_i}{M}

\vec r_{cm} = \frac{\sum m_i \vec r_i}{M}

The same thing as one vector; add z in three dimensions.

Three equal masses at the corners of a triangle have their centre of mass at the centroid, where the medians meet.

4Solid bodies and symmetry

A real body has its mass spread out. Cut it into tiny pieces of mass $dm$ and add them all up; the sum becomes an integral:

\vec r_{cm} = \frac{1}{M}\int \vec r\, dm

For a uniform rod of length $L$ , $dm = \frac{M}{L}dx$ , so $x_{cm} = \frac{1}{L}\int_0^L x\,dx = \frac{L}{2}$ : the middle.

If a uniform body is symmetric about a line, its centre of mass lies on that line.
With two or more lines of symmetry, it is where they cross.
So a uniform rod, disc, sphere, square plate or ring has its centre of mass at its centre.

For a body made of simple parts, replace each part by a point mass at its own centre of mass, then combine the points.

5Holes and empty space

A body with a hole: think of the full body as the part that is left plus the piece cut out. Equivalently, treat the hole as a negative mass.

The centre of mass need not be inside the material: a ring's is at its empty centre, a boomerang's lies in the gap, a hollow hemisphere's is in the hollow, and a high jumper arching over the bar can have the centre of mass pass under it.

6Curved shapes

These lie on the axis of symmetry. Heights are measured from the centre of the flat edge or base.

Body	Height of the centre of mass	About
Semicircular ring (radius $R$ )	$2R/\pi$	$0.64R$
Semicircular disc	$4R/3\pi$	$0.42R$
Hollow hemisphere	$R/2$	$0.5R$
Solid hemisphere	$3R/8$	$0.375R$
Hollow cone (height $h$ )	$h/3$	$0.33h$
Solid cone	$h/4$	$0.25h$

In each pair the solid body's centre of mass is lower: filling it in adds most mass near the wide base or the flat edge.

Where $2R/\pi$ comes from. A bit of the half ring at angle $\theta$ has mass $dm = \frac{M}{\pi}d\theta$ and height $R\sin\theta$ . So

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

7How the centre of mass moves

Let the position formula change with time. Its rate of change gives the velocity of the centre of mass, and the rate of change of that gives its acceleration:

\vec v_{cm} = \frac{\sum m_i\vec v_i}{M} = \frac{\vec P_{total}}{M}

\vec a_{cm} = \frac{\sum m_i\vec a_i}{M} = \frac{\vec F_{ext}}{M}

\vec F_{ext} = M\vec a_{cm}

Newton's second law for a whole system.

Why only external forces? Forces between the parts come in third-law pairs, equal and opposite. Adding all the forces, every pair cancels. So pushes and pulls inside a system can never change how its centre of mass moves.

8No outside force

With no net external force, $\vec v_{cm}$ stays constant. If everything starts at rest, the centre of mass never moves, so each body moves a distance inversely proportional to its mass:

m_1 d_1 = m_2 d_2

ExampleSkaters and a boat

Skaters of 30 kg and 60 kg, 6 m apart, pull on a rope: the light one moves 4 m, the heavy one 2 m, and they meet at the centre of mass. With 40 kg and 120 kg, 8 m apart, they meet 6 m from the light skater.

A person of mass $m$ walks the length $L$ of a boat of mass $M$ . The boat moves back $d$ and the person moves $L - d$ over the water. The centre of mass is fixed, so $m(L - d) = Md$ :

d = \frac{mL}{m + M}

60 kg on a 140 kg, 4 m boat: $d = 240/200 = 1.2$ m back, and the person moves 2.8 m forward. 50 kg walking 6 m on a 250 kg raft: $d = 300/300 = 1$ m.

Two masses $m$ and $2m$ , a distance $d$ apart, pulled together by their own attraction, meet at the centre of mass, $\frac{2d}{3}$ from $m$ .

9Explosions

An explosion is driven by internal forces. From outside only gravity acts, before and after, so the centre of mass keeps to the original path (while all the pieces are still in the air).

ExampleShell bursting at the top

Fired at 20 m/s at 60°, with $g = 10$ m/s². Range $R = \frac{u^2\sin 2\theta}{g} = \frac{400 \times \sqrt3/2}{10} = 20\sqrt3 \approx 34.6$ m. At the top the speed is $20\cos 60° = 10$ m/s.

It bursts into two equal pieces, and one retraces its path back to the launch point (velocity −10 m/s). Momentum: $2m \times 10 = m(-10) + mv_2$ , so $v_2 = 30$ m/s. Both land together; their midpoint lands at $R$ , so the other piece lands at $2R = 40\sqrt3 \approx 69.3$ m.

If instead one piece stops dead, the other needs $v_2 = 20$ m/s for the centre of mass to keep the old path.

A shell moving at $v$ bursts into two equal pieces, one going forward at $3v$ : $2mv = 3mv + mv_2$ , so $v_2 = -v$ , backward at $v$ .

10Centre of mass and energy

The gravitational potential energy of a whole body is its mass times $g$ times the height of its centre of mass: $U = Mgh_{cm}$ .

Summary

Key ideas

The centre of mass is the mass-weighted average position; for moving as a whole, the body acts like one particle there.
For two masses it lies closer to the heavier one, splitting the gap in the inverse ratio of the masses: m₁r₁ = m₂r₂.
For many particles, each coordinate is its own weighted average.
A uniform body's centre of mass lies on every line of symmetry.
Replace each part of a body by a point mass at its own centre; treat a hole as a negative mass.
The centre of mass can be in empty space: a ring, a boomerang, a hollow hemisphere.
Centre of gravity and centre of mass coincide when gravity is uniform.
Solid hemispheres and cones have lower centres of mass than hollow ones.
v_cm = P/M and F_ext = M a_cm: internal forces cancel in pairs and cannot move the centre of mass.
With no outside force and everything at rest, the centre of mass stays put, so lighter bodies move farther.
After an explosion, the centre of mass keeps to the old path while the pieces are in the air.
Gravitational potential energy of a body is M g h_cm.

Every equation

Two particles: $x_{cm} = \frac{m_1x_1 + m_2x_2}{m_1 + m_2}$
From m₁, distance d apart: $x_{cm} = \frac{m_2 d}{m_1 + m_2}$
Moment balance: $m_1r_1 = m_2r_2$
Many particles: $x_{cm} = \frac{\sum m_ix_i}{M}$
As a vector: $\vec r_{cm} = \frac{\sum m_i\vec r_i}{M}$
Continuous body: $\vec r_{cm} = \frac{1}{M}\int \vec r\,dm$
Uniform rod: $x_{cm} = L/2$
L-shape, from the corner: $\left(\tfrac{L}{4}, \tfrac{L}{4}\right),\ \tfrac{L}{2\sqrt2}$
Disc with hole R/2 at R/2: $x = -R/6$
Semicircular ring: $2R/\pi$
Semicircular disc: $4R/3\pi$
Hollow hemisphere: $R/2$
Solid hemisphere: $3R/8$
Hollow cone: $h/3$
Solid cone: $h/4$
Velocity of the COM: $\vec v_{cm} = \vec P_{total}/M$
Acceleration of the COM: $\vec a_{cm} = \vec F_{ext}/M$
Distances moved, COM fixed: $m_1d_1 = m_2d_2$
Person on a boat: $d = \frac{mL}{m + M}$
Potential energy: $U = Mgh_{cm}$
Chain leaving a table: $v = \sqrt{g(L^2 - b^2)/L}$

Previous year questions with solutions

Real JEE and NEET questions on centre of mass. Try each one before you open the solution.

Q1JEE Main 2026One correct option

Given below are two statements :

Statement I : For a mechanical system of many particles total kinetic energy is the sum of kinetic energies of all the particles.

Statement II : The total kinetic energy can be the sum of kinetic energy of the center of mass w.r.t to the origin and the kinetic energy of all the particles w.r.t. the center of mass as the reference.

In the light of the above statements, choose the correct answer from the options given below :

ABoth Statement I and Statement II are false
BStatement I is false but Statement II is true
CStatement I is true but Statement II is false
DBoth Statement I and Statement II are true

Show answer and solution

Answer: Option D

Statement I is the definition: the kinetic energy of a system is the sum of the kinetic energies of its particles, and since none of them can be negative, nothing cancels. Statement II is the split used in the last question: $K = \tfrac{1}{2}Mv_{\mathrm{cm}}^{2} + \sum \tfrac{1}{2}mu^{2}$ , the kinetic energy of the centre of mass about the origin plus the kinetic energies of the particles relative to the centre of mass. The cross terms vanish because the momenta measured relative to the centre of mass add up to zero. Both statements are true — D. The trap is to think II contradicts I; it is the same total, sorted into the part that goes with the centre and the part that does not.

Q2NEET 2023One correct option

Two particles A and B initially at rest, move towards each other under mutual force of attraction. At an instance when the speed of A is v and speed of B is 3v, the speed of centre of mass is :

A2v
Bzero
Cv
D4v

Show answer and solution

Answer: Option B

Only the mutual attraction acts — an internal force — and the pair starts from rest, so the centre of mass stays at rest: its speed is zero at every instant. The speeds $v$ and $3v$ reveal the masses (A is three times as heavy as B, since $m_{A}v = m_{B}(3v)$ ) but not the centre's speed. $2v$ is the plain average of the speeds, $4v$ adds them, and $v$ is A's own speed — none of them is the centre, which never moved.

Q3JEE Main 2025One correct option

Consider a circular disc of radius 20 cm with centre located at the origin. A circular hole of 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

A1.5 cm
B2.0 cm
C0.5 cm
D1.0 cm

Show answer and solution

Answer: Option D

Mass goes with area, so with $\sigma\pi$ cancelling the disc counts as $20^{2} = 400$ and the hole as $5^{2} = 25$ . The hole touches the edge from inside, so its centre is $20 - 5 = 15\ \mathrm{cm}$ from the origin. Then $x_{cm} = \dfrac{400 \times 0 - 25 \times 15}{400 - 25} = -\dfrac{375}{375} = -1\ \mathrm{cm}$ : $1.0\ \mathrm{cm}$ from the origin, on the side away from the hole. The traps are putting the hole's centre on the edge itself, at $20\ \mathrm{cm}$ , which gives $1.33\ \mathrm{cm}$ , and dividing by $400$ instead of $375$ , which gives $0.94\ \mathrm{cm}$ — close to the right option for the wrong reason.

Practice questions, easy to hard

Three questions from the centre of mass practice ladder: one easy, one medium, one hard.

Q4Numerical answer

Divide changes in velocity by the time taken, just as positions gave velocities, and the centre's acceleration appears: $\vec{a}_{\mathrm{cm}} = \dfrac{m_{1}\vec{a}_{1} + m_{2}\vec{a}_{2} + \dots}{M}$ , with signs — or $\hat{i}$ and $\hat{j}$ components — kept as before.

On a smooth floor a $1\ \mathrm{kg}$ block accelerates at $6\ \mathrm{m/s^{2}}$ to the right while a $2\ \mathrm{kg}$ block accelerates at $1.5\ \mathrm{m/s^{2}}$ to the left. Taking right as positive, what is the acceleration of their centre of mass, in $\mathrm{m/s^{2}}$ ?

Show answer and solution

Answer: 1 m/s²

$a_{\mathrm{cm}} = \dfrac{1 \times 6 + 2 \times (-1.5)}{1 + 2} = \dfrac{6 - 3}{3} = 1\ \mathrm{m/s^{2}}$ , to the right. Dropping the sign gives $\dfrac{6 + 3}{3} = 3$ , and ignoring the masses gives $\dfrac{6 - 1.5}{2} = 2.25$ . The recipe is the one used for positions and velocities, unchanged: weight each by its mass, keep directions, divide by the total mass.

Q5Numerical answer

The commonest layout is three masses at the corners of a right-angled triangle. Put the origin at the right angle and the two perpendicular sides along the axes: then every corner has at least one coordinate equal to zero, and most terms drop out.

A $2\ \mathrm{kg}$ mass sits at the right angle, a $3\ \mathrm{kg}$ mass $4\ \mathrm{m}$ from it along the $x$ -axis, and a $5\ \mathrm{kg}$ mass $2\ \mathrm{m}$ from it along the $y$ -axis. How far, in m, is the centre of mass from the $y$ -axis (that is, what is $x_{\mathrm{cm}}$ )?

Show answer and solution

Answer: 1.2 m

Only the $3\ \mathrm{kg}$ mass is off the $y$ -axis, so $x_{\mathrm{cm}} = \dfrac{2 \times 0 + 3 \times 4 + 5 \times 0}{2 + 3 + 5} = \dfrac{12}{10} = 1.2\ \mathrm{m}$ . The trap is leaving the $5\ \mathrm{kg}$ mass out of the bottom line because it sits at $x = 0$ : that gives $\dfrac{12}{5} = 2.4\ \mathrm{m}$ . A mass with zero $x$ adds nothing on top but still weighs in underneath. For completeness, $y_{\mathrm{cm}} = \dfrac{5 \times 2}{10} = 1\ \mathrm{m}$ .

Q6Numerical answer

Both particles can move at once. Each adds its own $m\Delta x$ on top, with a sign: $+$ one way along the line, $-$ the other.

A $2\ \mathrm{kg}$ block moves $6\ \mathrm{cm}$ to the right while a $3\ \mathrm{kg}$ block on the same line moves $2\ \mathrm{cm}$ to the left. How far, in cm, does the centre of mass move to the right?

Show answer and solution

Answer: 1.2 cm

Right is $+$ : $\Delta x_{\mathrm{cm}} = \dfrac{2 \times 6 + 3 \times (-2)}{2 + 3} = \dfrac{12 - 6}{5} = 1.2\ \mathrm{cm}$ to the right. The trap is dropping the sign, which gives $\dfrac{12 + 6}{5} = 3.6\ \mathrm{cm}$ . The left move pulls the centre back, so the two terms partly cancel. Had the $3\ \mathrm{kg}$ block moved $4\ \mathrm{cm}$ left, $3 \times 4$ would have cancelled $2 \times 6$ exactly, and the centre would not have moved at all.