Centre of Mass

PhysicsSystem Of Particles And Rotational MotionFor NEET aspirants

The centre of mass (CoM) of a system of particles is the single point where the entire mass of the system can be considered concentrated for describing its translational motion. Under an external force, the CoM moves exactly as a single particle of the same total mass would move under that force. For a system of $n$ particles with masses $m_{i}$ at position vectors $r_{i}$ , the CoM position is $R_{c m} = \frac{1}{M} \sum m_{i} r_{i}$ , where $M = \sum m_{i}$ . For continuous bodies, the sum becomes an integral. This concept is fundamental to JEE and NEET mechanics, especially rotational motion and collisions.

Key Formulas - Quick Reference

Position of CoM (n particles): $R_{c m} = \frac{\sum m _{i} r _{i}}{\sum m _{i}}$
CoM coordinates: $X_{c m} = \frac{\sum m _{i} x _{i}}{M}$ , $Y_{c m} = \frac{\sum m _{i} y _{i}}{M}$ , $Z_{c m} = \frac{\sum m _{i} z _{i}}{M}$
Two-particle CoM (on line): $x_{c m} = \frac{m _{1} x _{1} + m _{2} x _{2}}{m _{1} + m _{2}}$
Continuous body: $R_{c m} = \frac{1}{M} \int r d m$
Velocity of CoM: $v_{c m} = \frac{\sum m _{i} v _{i}}{M}$
Acceleration of CoM: $a_{c m} = \frac{\sum m _{i} a _{i}}{M} = \frac{F _{e x t}}{M}$
Total momentum: $P = M v_{c m}$

1. What is Centre of Mass?

Every physical system of particles or extended body has associated with it a single point whose motion completely characterises the translational motion of the whole system. When external forces act on the system, this point moves as if the entire mass of the system were concentrated there and all the external forces were applied at this point. This point is called the centre of mass.

Definition: The centre of mass of a system of

n

point particles is that point about which the mass-weighted sum of position vectors vanishes. Equivalently, it is the mass-weighted average position of all the particles.

The centre of mass is a purely mathematical point. It need not coincide with any actual particle of the system, and it may even lie outside the body (for example, the CoM of a ring lies at its geometric centre, in empty space).

2. CoM of a System of Point Particles

Consider $n$ particles of masses $m_{1}, m_{2}, \dots, m_{n}$ located at position vectors $r_{1}, r_{2}, \dots, r_{n}$ from a chosen origin. The position vector of the centre of mass is

R_{c m} = \frac{m _{1} r _{1} + m _{2} r _{2} + \dots + m _{n} r _{n}}{m _{1} + m _{2} + \dots + m _{n}} = \frac{1}{M} i = 1 \sum n m_{i} r_{i}

where $M = \sum m_{i}$ is the total mass of the system. Writing $r_{i} = x_{i} \hat{i} + y_{i} \hat{j} + z_{i} \hat{k}$ , the Cartesian coordinates of the CoM are

X_{c m} = \frac{\sum m _{i} x _{i}}{M}, Y_{c m} = \frac{\sum m _{i} y _{i}}{M}, Z_{c m} = \frac{\sum m _{i} z _{i}}{M}

Two-particle system

For two particles of masses $m_{1}$ and $m_{2}$ separated by distance $d$ , the CoM lies on the line joining them, at distances

r_{1} = \frac{m _{2} d}{m _{1} + m _{2}} from m_{1}, r_{2} = \frac{m _{1} d}{m _{1} + m _{2}} from m_{2}

Notice that $m_{1} r_{1} = m_{2} r_{2}$ : the CoM is closer to the heavier particle. This is the fundamental balance relation.

Figure 1: Two-particle system with centre of mass at CM. The balance relation is

m_{1} r_{1} = m_{2} r_{2}

Solved Example 1

Three equal masses

m

are placed at the vertices of an equilateral triangle with side

a

: at

(0, 0)

(a, 0)

, and

(a /2, 3 a /2)

. Find the centre of mass.

Solution:

Using $X_{c m} = \frac{\sum m _{i} x _{i}}{\sum m _{i}}$ :

X_{c m} = \frac{m ( 0 ) + m ( a ) + m ( a /2 )}{3 m} = \frac{a}{2}

Y_{c m} = \frac{m ( 0 ) + m ( 0 ) + m ( 3 a /2 )}{3 m} = \frac{a}{2 3}

So the CoM is at $(\frac{a}{2}, \frac{a}{2 3})$ , which is the centroid of the triangle. This makes sense: for equal masses at vertices, the CoM is always at the geometric centroid.

3. CoM of Continuous Bodies

For a body with a continuous mass distribution, we replace the summation with an integral. If $d m$ is a mass element at position $r$ , then

R_{c m} = \frac{1}{M} \int r d m

X_{c m} = \frac{1}{M} \int x d m, Y_{c m} = \frac{1}{M} \int y d m, Z_{c m} = \frac{1}{M} \int z d m

The integral is taken over the entire body. Choose the mass element $d m$ to be a strip, ring, or shell whose position is easy to describe, and use the mass per unit length ( $λ$ ), area ( $σ$ ), or volume ( $ρ$ ) to express $d m$ in terms of the coordinate.

Derivation: Uniform rod

Consider a uniform rod of mass $M$ and length $L$ with one end at the origin, lying along the x-axis. Linear mass density $λ = M / L$ , so $d m = λ d x = (M / L) d x$ .

X_{c m} = \frac{1}{M} \int_{0}^{L} x (\frac{M}{L}) d x = \frac{1}{L} \cdot \frac{L ^{2}}{2} = \frac{L}{2}

The CoM of a uniform rod lies at its midpoint, as expected from symmetry.

Standard CoM formulas (memorise for JEE/NEET)

The following table gives centre-of-mass locations for common uniform bodies. All positions are measured from the natural symmetry point (centre, apex, or diameter).

Body	Location of CoM
Uniform rod, length $L$	Midpoint (at $L /2$ from either end)
Uniform circular ring, radius $R$	At geometric centre
Uniform circular disc, radius $R$	At geometric centre
Semicircular ring (wire), radius $R$	$\frac{2 R}{π}$ from centre, along axis of symmetry
Semicircular disc (plate), radius $R$	$\frac{4 R}{3 π}$ from centre, along axis of symmetry
Hollow hemisphere, radius $R$	$\frac{R}{2}$ from centre, along axis
Solid hemisphere, radius $R$	$\frac{3 R}{8}$ from centre, along axis
Solid cone, height $h$	$\frac{h}{4}$ from base (or $\frac{3 h}{4}$ from apex)
Hollow cone, height $h$	$\frac{h}{3}$ from base (or $\frac{2 h}{3}$ from apex)
Triangular plate (uniform)	At centroid, $\frac{h}{3}$ from any base

Note: For any body with a plane of symmetry, the CoM lies on that plane. For a body with two or more planes of symmetry, the CoM lies on their intersection. This shortcut saves a great deal of integration in exam problems.

Solved Example 2

A semicircular disc of radius

R

and mass

M

is placed with its diameter along the x-axis. Find the y-coordinate of the CoM.

Solution:

Representative ring element: semicircular strip of radius r, thickness dr. Integrating over r from 0 to R sweeps the whole disc.

By symmetry $X_{c m} = 0$ . For $Y_{c m}$ , take a thin strip of radius $r$ and thickness $d r$ ; this strip is a semicircular ring of mass $d m = σ \cdot π r d r$ where $σ = \frac{M}{π R ^{2} /2} = \frac{2 M}{π R ^{2}}$ . The CoM of a semicircular ring of radius $r$ is at $y = \frac{2 r}{π}$ .

Y_{c m} = \frac{1}{M} \int_{0}^{R} \frac{2 r}{π} \cdot σ π r d r = \frac{2 σ}{M} \int_{0}^{R} r^{2} d r = \frac{2 σ R ^{3}}{3 M} = \frac{2}{3 M} \cdot \frac{2 M}{π R ^{2}} \cdot R^{3} = \frac{4 R}{3 π}

So $Y_{c m} = \frac{4 R}{3 π}$ , matching the standard formula.

4. Motion of the Centre of Mass

Differentiating $M R_{c m} = \sum m_{i} r_{i}$ with respect to time gives the velocity of the CoM:

v_{c m} = \frac{d R _{c m}}{d t} = \frac{\sum m _{i} v _{i}}{M} = \frac{P _{t o t a l}}{M}

Thus the total linear momentum of a system equals the total mass times the velocity of the CoM: $P = M v_{c m}$ . Differentiating again gives the acceleration of the CoM:

a_{c m} = \frac{\sum m _{i} a _{i}}{M} = \frac{F _{n e t, e x t}}{M}

By Newton's third law, internal forces between particles cancel in pairs, so only the net external force enters. This is a remarkable result:

Key result: The centre of mass of a system moves as if all the mass were concentrated at that point and all the external forces acted directly on it. Internal forces do not affect the motion of the CoM.

Because of this, when a shell explodes in mid-air, its centre of mass continues to follow the original parabolic trajectory - the fragments may fly apart, but their weighted average position obeys projectile motion under gravity alone.

Solved Example 3

Two particles A (mass 1 kg) and B (mass 2 kg) are projected simultaneously from the same vertical line, 90 m apart. A is on the ground moving upward with

u_{A} = 200

m/s; B is at height 90 m moving downward with

u_{B} = 50

m/s. Find the maximum height attained by the CoM. Take

g = 10

m/s².

Solution:

Initial position of CoM (measuring height from ground):

y_{c m, 0} = \frac{( 1 ) ( 0 ) + ( 2 ) ( 90 )}{1 + 2} = 60 m

Initial velocity of CoM (upward positive):

u_{c m} = \frac{( 1 ) ( 200 ) + ( 2 ) ( - 50 )}{3} = \frac{100}{3} m/s

Only gravity acts externally, so $a_{c m} = - g = - 10$ m/s² throughout. Using $v^{2} = u^{2} - 2 g h$ with $v = 0$ at maximum height above the starting CoM position:

h = \frac{u _{c m}^{2}}{2 g} = \frac{( 100/3 ) ^{2}}{20} = \frac{10000/9}{20} = 55.55 m

Maximum height of CoM above ground $= 60 + 55.55 = 115.55$ m.

5. CoM of a Body with a Cavity

To find the CoM of a body from which a portion has been removed (a cavity), treat the removed portion as a negative mass. If the original body has mass $M$ with CoM at $r_{M}$ and the removed portion has mass $m$ with CoM at $r_{m}$ , then the CoM of the remaining body is

R_{c m} = \frac{M r _{M} - m r _{m}}{M - m}

This trick avoids doing a fresh integration over an awkwardly-shaped body.

Solved Example 4

A uniform disc of radius

R

has a circular hole of radius

R /2

cut from it. The centre of the hole is at distance

R /2

from the centre of the disc. Find the CoM of the remaining portion.

Solution:

Disc of radius R centred at O; hole of radius R/2 centred at C. The CoM of the remaining portion lies on the line OC, on the side opposite the hole.

Let the original disc have mass $M$ centred at the origin. The removed disc has area $π (R /2)^{2} = π R^{2} /4$ , one-quarter that of the full disc, so its mass is $m = M /4$ . Its CoM is at $x = R /2$ .

Using the negative-mass method along the x-axis:

X_{c m} = \frac{M ( 0 ) - ( M /4 ) ( R /2 )}{M - M /4} = \frac{- M R /8}{3 M /4} = - \frac{R}{6}

The CoM lies on the axis through both centres, at distance $R /6$ from the original centre, on the side opposite to the hole.

Common Mistakes to Avoid

Watch out

Confusing centre of mass with centre of gravity. They coincide in a uniform gravitational field, but the CoM depends only on mass distribution while the centre of gravity depends on how gravity varies. For JEE/NEET problems, they can usually be treated as identical.
Forgetting to use the negative-mass trick for cavity problems and instead attempting a direct integration over the awkward remaining shape.
Using $m_{1} r_{1} = m_{2} r_{2}$ without recognising it requires distances $r_{1}, r_{2}$ measured from the CoM itself, not from an arbitrary origin.
Assuming the CoM must lie inside the body. For a ring, a semicircular wire, or an L-shaped plate, the CoM lies outside the material.
Applying internal forces (like the force between two colliding balls) when writing $F_{e x t} = M a_{c m}$ . Only external forces determine CoM motion.
Confusing signs in the semicircular disc formula: $4 R / (3 π)$ is measured from the diameter (flat edge), not from the curved edge.

Frequently Asked Questions

Q1. Where does the centre of mass of a uniform triangular plate lie?

At the centroid of the triangle - the intersection of its three medians. This is at distance $h /3$ from each base, where $h$ is the corresponding altitude. For a right triangle with legs along the axes at $(0, 0)$ , $(a, 0)$ , $(0, b)$ , the centroid is at $(a /3, b /3)$ .

Q2. Why does the centre of mass of a ring lie at its centre, in empty space?

By symmetry: for every mass element on one side of the centre there is an identical element on the diametrically opposite side. Their contributions to the position vector cancel exactly, leaving the CoM at the geometric centre. The CoM is a mathematical point and does not need to coincide with any material.

Q3. What is the difference between centre of mass and centre of gravity?

The centre of mass depends only on the mass distribution of the body. The centre of gravity is the point where the resultant gravitational force acts. In a uniform gravitational field (which is an excellent approximation for any object smaller than a country), the two coincide. For very large bodies where $g$ varies across the body, they differ slightly. For JEE and NEET, treat them as the same.

Q4. How does the CoM move when a projectile explodes in mid-flight?

The CoM continues to follow the original parabolic path as if the explosion never happened. The explosion produces only internal forces, which cannot change the motion of the CoM. Only gravity (external) acts on the system, so the CoM's trajectory is unchanged - though the individual fragments follow their own paths.

Q5. Can the centre of mass of a system be at rest while the individual particles are moving?

Yes. If the total momentum $P = M v_{c m} = 0$ , then $v_{c m} = 0$ even though individual particles have non-zero velocities. Example: two equal-mass particles moving with equal and opposite velocities have their CoM at rest at the midpoint.

Q6. How do I choose the mass element $d m$ when integrating for CoM?

Pick a strip, ring, or shell whose position along the required axis is a single value (or nearly so), and whose mass can be expressed in terms of that coordinate. For a rod, use a strip $d x$ ; for a disc's CoM along the axis, use a ring of radius $r$ ; for a hemisphere, use a disc at height $y$ . Then $d m = σ \cdot d A$ or $ρ \cdot d V$ where $σ, ρ$ are the density.

Q7. For a system of particles moving under only mutual gravitational attraction, how does the CoM move?

With constant velocity in a straight line (or stays at rest). Since gravity between the particles is internal, there is no external force on the system, so $a_{c m} = 0$ . This is why the CoM of a binary star system moves in a straight line while the two stars orbit around it.

Previous year questions on Centre of Mass

11 questions from past papers, each with a step-by-step solution.

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1. What is Centre of Mass?

2. CoM of a System of Point Particles

Two-particle system

3. CoM of Continuous Bodies

Derivation: Uniform rod

Standard CoM formulas (memorise for JEE/NEET)

4. Motion of the Centre of Mass

5. CoM of a Body with a Cavity

Common Mistakes to Avoid

Frequently Asked Questions

Q1. Where does the centre of mass of a uniform triangular plate lie?

Q2. Why does the centre of mass of a ring lie at its centre, in empty space?

Q3. What is the difference between centre of mass and centre of gravity?

Q4. How does the CoM move when a projectile explodes in mid-flight?

Q5. Can the centre of mass of a system be at rest while the individual particles are moving?

Q6. How do I choose the mass element dm when integrating for CoM?

Q7. For a system of particles moving under only mutual gravitational attraction, how does the CoM move?

Previous year questions on Centre of Mass

Ready to master System Of Particles And Rotational Motion?

Q6. How do I choose the mass element $d m$ when integrating for CoM?