Define center of mass.

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Q. Define center of mass.

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Concept understanding — Center of Mass

What is the Center of Mass?

Imagine you pick up a broom by the handle and try to balance it horizontally on one finger. You instinctively slide your finger along the handle until the broom stays level. That point — the one where the broom doesn't tip — is its center of mass.

Now think about throwing a cricket bat. It spins and wobbles in the air, but there is one point on the bat that follows a smooth, parabolic path, as if all the bat's mass were concentrated there. That point is also the center of mass.

The core idea is simple: the center of mass is the average position of all the mass in an object. It's the point where you could imagine the entire mass of the object being concentrated, and the object would behave the same way under the influence of external forces.

Why does this matter?

When you push an object at its center of mass, it moves in a straight line without rotating. Push it anywhere else, and it will both move and spin. This is why:

A car's stability depends on where its center of mass is (lower = safer).
A tightrope walker holds a long pole — moving the pole shifts their combined center of mass back over the rope.
In projectile motion, the center of mass of a system (like an exploding firework) continues along the original parabolic path, even though the fragments scatter.

The precise definition

For a system of particles, the center of mass is the weighted average of their positions, where the weight is the mass of each particle.

\vec{R}_{\text{CM}} = \frac{m_1 \vec{r}_1 + m_2 \vec{r}_2 + \cdots + m_n \vec{r}_n}{m_1 + m_2 + \cdots + m_n} = \frac{\sum m_i \vec{r}_i}{\sum m_i}

Here:

$\vec{R}_{\text{CM}}$ is the position vector of the center of mass
$m_i$ is the mass of the $i$ -th particle
$\vec{r}_i$ is the position vector of that particle

For a continuous object (like a rod or a sphere), the sum becomes an integral:

\vec{R}_{\text{CM}} = \frac{1}{M} \int \vec{r} \, dm

where $M$ is the total mass and $dm$ is an infinitesimal mass element.

Breaking it down with an example

Take two masses on a light rod: $m_1 = 2 \text{ kg}$ at $x = 0$ , and $m_2 = 3 \text{ kg}$ at $x = 5 \text{ m}$ .

The center of mass is:

x_{\text{CM}} = \frac{(2)(0) + (3)(5)}{2 + 3} = \frac{0 + 15}{5} = 3 \text{ m}

So the center of mass is at $x = 3 \text{ m}$ , closer to the heavier mass. That makes intuitive sense — the heavier mass "pulls" the average toward itself.

Note

The center of mass does not have to be inside the object. A ring or a hollow sphere has its center of mass at the geometric center, which is empty space.

Key properties to remember

External forces only — Internal forces (like collisions between parts of the system) do not affect the motion of the center of mass. Only external forces can change its velocity.
If no external force acts, the center of mass moves with constant velocity (or stays at rest). This is the law of conservation of momentum applied to the whole system.
For symmetric objects with uniform density, the center of mass coincides with the geometric center. For irregular shapes, it shifts toward the region with more mass.
In a uniform gravitational field, the center of mass and the center of gravity are the same point. (They differ only if gravity varies significantly across the object — not something you'll see in school problems.)

A final intuition

Think of the center of mass as the balance point of an object. If you could place a tiny, invisible support exactly at that point, the object would be perfectly balanced in any orientation. Every piece of mass on one side is exactly counterbalanced by the pieces on the other side.

That's why, when you jump off a boat, the boat moves backward — your center of mass and the boat's center of mass shift relative to each other, but the center of mass of the whole system (you + boat) stays put (if no external horizontal force acts). This is the heart of why the center of mass concept is so powerful: it lets you treat a complicated, spinning, wobbling object as a single point for many problems.

Looking up "Center of Mass: definition, formula & real-world examples" is a good habit before an exam, and it is worth knowing that Center of Mass is a core, NCERT-aligned topic from the System of Particles and Rotational Motion portion of the Class 11 Physics curriculum, and it is tested regularly in CBSE board exams as well as in JEE Main and NEET. Cross-checking this explanation against the relevant NCERT Physics chapter and solving a few past-year questions will round out your preparation.

Step 1. State the definition. The center of mass is the point at which the entire mass of a body (or a system of particles) may be taken to be concentrated for the purpose of describing the body's overall translational motion.

Step 2. Give the formula. For a system of point masses $m_i$ at position vectors $\vec r_i$ , the center of mass is located at

\vec r_{CM}=\frac{\sum_i m_i\vec r_i}{M},\qquad M=\sum_i m_i.

Step 3. Note why it matters. However complicated the true motion of a body's individual particles (tumbling, deforming, exploding), its center of mass alone always moves exactly as a single point particle of the same total mass would move, under the net external force — e.g. a thrown, spinning bat has only its center of mass tracing a clean parabola.

✓Final answer

\vec r_{CM}=\dfrac{\sum m_i\vec r_i}{M}

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 22 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$Q. In the absence of an external force, velocity of the center of mass -$
(a) Remains constant
(b) Decreases
(c) is Zero
(d) Increases
›Reveal solutionSolution

In the absence of an external force, the velocity of the centre of mass of a system remains constant (Newton's first law applied to the system as a whole).

For a system of particles, the equation of motion of the centre of mass is F_ext = M (d v_cm/dt), where M is the total mass and F_ext is the net external force. This follows because all internal forces (between particles of the system) occur in Newton's-third-law pairs and cancel out when summed over the whole system. If F_ext = 0, then d v_cm/dt = 0, meaning the centre of mass moves with constant velocity (it neither speeds up nor slows down nor changes direction) — even if the individual particles of the system are accelerating due to internal forces.

✓Final answer
$The correct option is (a) Remains constant.$
CBSE 2026Set ANNUAL1 markMCQ
$Q. There are two objects of masses 1 kg and 2 kg located at (1, 2) and (-1, 3) respectively. The coordinates of the centre of mass are:$
(a) (2, -1)
(b) (8/3, -1/3)
(c) (-1/3, 8/3)
(d) None of the above
›Reveal solutionSolution

$x_{cm}=\dfrac{m_1x_1+m_2x_2}{m_1+m_2}$ , $y_{cm}=\dfrac{m_1y_1+m_2y_2}{m_1+m_2}$ give $\left(-\tfrac13,\tfrac83\right)$ .

For a system of point masses, the centre of mass is:

$x_{cm}=\frac{m_1x_1+m_2x_2}{m_1+m_2}, \qquad y_{cm}=\frac{m_1y_1+m_2y_2}{m_1+m_2}$

With $m_1=1\text{ kg}$ at $(1,2)$ and $m_2=2\text{ kg}$ at $(-1,3)$ :

$x_{cm}=\frac{1(1)+2(-1)}{1+2}=\frac{1-2}{3}=-\frac13$

$y_{cm}=\frac{1(2)+2(3)}{1+2}=\frac{2+6}{3}=\frac83$

So the centre of mass is at $\left(-\dfrac13,\ \dfrac83\right)$ .

✓Final answer
$\left(-\dfrac13,\ \dfrac83\right)$
CBSE 2026Set ANNUAL1 markMCQ
$Q. If a uniform rod has length l, then its centre of mass will be located at$
(a) l/2
(b) 3l/4
(c) l/3
(d) none of these
›Reveal solutionSolution

A uniform rod's centre of mass is at its midpoint, l/2. Answer (A).

For a uniform rod the linear mass density is constant, so the mass is symmetric about the middle of the rod. By symmetry the centre of mass lies at the geometric centre, a distance l/2 from each end.

✓Final answer
$(A) l/2.$
CBSE 2026Set ANNUAL1 markMCQ
$Q. If the centre of gravity of a body coincides with its centre of mass, then the gravitational fields acting on different parts of the body are$
(a) zero
(b) equal
(c) different
(d) none of these
›Reveal solutionSolution

Centre of gravity coincides with centre of mass when the gravitational field is uniform (equal on all parts). Answer (B).

The centre of mass depends only on the mass distribution. The centre of gravity is the point where the total weight appears to act, which depends on how g varies over the body.

When the gravitational field is uniform, g is the same for every element of the body, and the two points coincide. This happens for ordinary-sized bodies in a uniform field. Hence the field acting on different parts must be equal.

✓Final answer
$(B) equal.$
CBSE 2026Set ANN1 markMCQ
$Q. For which of the following shapes, the centre of mass does not lie on the body?$
(a) Pencil
(b) Dice
(c) Shotput
(d) Bangle
›Reveal solutionSolution
A bangle's centre of mass lies at its geometric centre - empty space - so the centre of mass does not lie on the body.

The centre of mass of a body need not lie within the material of the body; it depends on how the mass is distributed. For symmetric bodies it lies at the geometric centre.

Pencil: centre of mass lies on the pencil (its midpoint).

Dice (cube): centre of mass at the solid centre, inside the body.

Shotput (solid sphere): centre of mass at its centre, inside the body.

Bangle (ring): centre of mass at the centre of the ring, which is hollow - so it lies off the material of the body.

✓Final answer
$(d) Bangle.$
CBSE 2025Set ANNUAL1 markMCQ
$Q. Where is the centre of mass of a uniform rod located?$
(a) At one end of the rod
(b) At the midpoint of the rod
(c) At the top of the rod
(d) At the bottom of the rod
›Reveal solutionSolution

For any uniformly (symmetrically) shaped and uniformly dense body like a straight rod, the centre of mass lies at the geometric centre -- the midpoint.

Centre of mass, x_cm = (Integral of x dm) / (Integral of dm), over the length of the rod.

For a rod of uniform linear mass density lying along the x-axis from 0 to L, by symmetry every mass element at position x has a mirror-image element at (L - x) with equal mass. These pairs balance exactly at the midpoint, x = L/2.

✓Final answer
$(b) At the midpoint of the rod.$
CBSE 2025Set ANNUAL1 markMCQ
$Q. Case study: The 'centre of mass' of a body is a point at which whole of the mass of the body is supposed to be concentrated. The centre of mass of a two body system is- X_cm = (m1 x1 + m2 x2) / (m1 + m2) where symbols have their usual meaning. The velocity of centre of mass is V_cm = dX_cm / dt.$
(1) Which of the following statement is false?
(a) Centre of mass of body may lie outside the body.
(b) Centre of mass of body may lie on the body.
(c) Centre of mass of two bodies of different masses lie at the centre of line joining the masses.
(d) Centre of mass of a ring lies at a point where actually no mass of the body exists.
›Reveal solutionSolution
Statement (c) is the false one — the COM of two unequal masses lies closer to the heavier mass, not at the geometric midpoint.

The centre of mass of a two-particle system lying on a line, with masses $m_1$ and $m_2$ at positions $x_1$ and $x_2$ , is given by:

$X_{cm} = \dfrac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$

Check each statement:

'Centre of mass of body may lie outside the body' — TRUE. Example: a ring, a horseshoe, or a boomerang — the COM lies in the hollow/empty region, outside the actual material of the body.

'Centre of mass of body may lie on the body' — TRUE. For a uniform, solid, symmetric body like a sphere or cube, the COM lies at the geometric centre, which is a point within the body's material.

'Centre of mass of two bodies of different masses lie at the centre of line joining the masses' — FALSE. If $m_1 \\ne m_2$ , the weighted-average formula $X_{cm} = \dfrac{m_1x_1+m_2x_2}{m_1+m_2}$ does NOT give the geometric midpoint of the line joining them — it is pulled closer to whichever mass is larger. The COM coincides with the midpoint ONLY in the special case that $m_1 = m_2$ .

'Centre of mass of a ring lies at a point where actually no mass of the body exists' — TRUE. By symmetry, a uniform ring's COM is at its geometric centre, which is empty space (no material of the ring is actually there).

✓Final answer
$The correct option is (c), since it is the false statement — the centre of mass of two different masses is closer to the heavier mass, not at the exact midpoint of the line joining them.$
CBSE 2025Set ANNUAL1 markMCQ
$Q. (2) If two particle of equal mass are placed at a distance 'd' apart, the centre of mass is located$
(a) At one of the particles.
(b) At the midpoint of the distance between them.
(c) Outside the line joining the two particles.
(d) None of above
›Reveal solutionSolution

Two equal masses have their centre of mass exactly at the midpoint of the line joining them.

Let the two particles, each of mass m, be placed at positions $x_1 = 0$ and $x_2 = d$ on a line. Using the centre-of-mass formula:

$X_{cm} = \dfrac{m_1 x_1 + m_2 x_2}{m_1 + m_2} = \dfrac{m(0) + m(d)}{m+m} = \dfrac{md}{2m} = \dfrac{d}{2}$

Since $X_{cm} = d/2$ , the centre of mass lies exactly halfway between the two particles — this makes sense because with equal masses, neither particle 'pulls' the COM toward itself more than the other, so by symmetry it sits precisely in the middle.

✓Final answer
$The correct option is (b) At the midpoint of the distance between them.$
CBSE 2025Set ANNUAL1 markMCQ
$Q. (3) The motion of centre of mass of a system is determined by$
(a) External forces only
(b) Internal forces only
(c) Both internal and external forces
(d) Number of particles in the system
›Reveal solutionSolution

Only external forces determine the motion of the centre of mass; internal forces always cancel by Newton's third law.

For a system of particles, the total mass M times the acceleration of the centre of mass equals the sum of ALL forces acting on all the particles:

$M a_{cm} = \sum F_{external} + \sum F_{internal}$

By Newton's third law, every internal force (a particle of the system exerting a force on another particle of the same system) comes as an action-reaction pair that is equal in magnitude and opposite in direction. When summed over the whole system, all such internal force pairs cancel exactly:

$\sum F_{internal} = 0$

So the equation of motion for the centre of mass simplifies to:

$M a_{cm} = \sum F_{external}$

This shows that however complicated the internal forces (collisions, explosions, mutual attraction, etc.) between the particles of a system might be, they can never change the motion of the system's centre of mass — only a net force from OUTSIDE the system can accelerate the centre of mass.

✓Final answer
$The correct option is (a) External forces only.$
CBSE 2025Set ANNUAL1 markMCQ
$Q. (4) A body moving in a straight line explodes in 20 pieces. The path of centre of mass will be$
(a) Parabolic
(b) Straight line
(c) Circular
(d) Hyperbolic
›Reveal solutionSolution

The COM keeps moving along the original straight-line path, since explosion forces are purely internal.

As derived from Newton's laws applied to a system of particles, the acceleration of the centre of mass of a system is determined ONLY by the net EXTERNAL force acting on it; internal forces (however violent) always cancel in pairs by Newton's third law and cannot change the COM's motion.

When a body moving in a straight line explodes into 20 pieces, the forces of the explosion (which fling the pieces apart) are entirely internal to the system made up of all 20 pieces together. No new external force is introduced by the explosion itself (only gravity, if present, would be external — and the question is framed purely on the explosion's effect).

Therefore, the centre of mass of the 20 pieces (taken together) is completely unaffected by the explosion — it continues to move exactly as the original single body was moving before it exploded, i.e., along the same straight line, with the same velocity it had at the moment of explosion.

✓Final answer
$The correct option is (b) Straight line.$
CBSE 2024Set ANNUAL1 markMCQ
$Q. Out of two particles of same mass, one is stationary and acceleration of another is a⃗. The acceleration of centre of mass of the system will be (A) zero (B) a⃗/2 (C) a⃗ (D) 2a⃗$
›Reveal solutionSolution

Centre-of-mass acceleration of the two-particle system is $\vec a/2$ .

For a system of particles, $\vec a_{cm}=\dfrac{\sum m_i\vec a_i}{\sum m_i}$ . Here both particles have equal mass $m$ : one is stationary ( $\vec a_1=0$ ) and the other has acceleration $\vec a$ .

$\vec a_{cm}=\dfrac{m(0)+m(\vec a)}{m+m}=\dfrac{m\vec a}{2m}=\dfrac{\vec a}{2}$ .

✓Final answer
$\vec a/2$
CBSE 2024Set ANNUAL1 markMCQ
$Q. If a stationary firecracker explodes into a number of particles, then the centre of mass will$
(a) move in vertical direction
(b) move in concentric circles
(c) remain stationary
(d) move in parabolic path.
›Reveal solutionSolution

Internal explosive forces cannot move the centre of mass; only an external force can, so it stays at rest.

The forces that blow the firecracker into fragments are internal forces (action-reaction pairs between the fragments) — by Newton's third law they always occur in equal and opposite pairs and cancel out when summed over the whole system. The motion of the centre of mass of a system is governed only by the net external force acting on it: $M\vec{a}_{cm} = \vec{F}_{ext}$ . Since gravity is the only external force here (assume negligible over the short duration, and the cracker was stationary before exploding), the centre of mass has no net external force to change its state of rest, so it continues to remain exactly where it was — stationary — even as the fragments fly apart in all directions.

✓Final answer
$The correct option is (c) remain stationary.$

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