Center of Mass Formula: x = Σmx/Σm, List and Solved Examples

The center of mass formula for two particles on a line is $x_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$ , and for any number of particles it is $x_{cm} = \frac{\sum m_i x_i}{\sum m_i}$ . It is a weighted average of positions, where each position is weighted by its mass, so the point always lies closer to the heavier particle.

Indian books spell it centre of mass, and the topic opens the Class 11 chapter on systems of particles. The idea is powerful: however complicated a body or a group of bodies, you can treat its overall motion as if all the mass sat at this one point and every external force acted there. The same point is usually the center of gravity as well.

Quick facts

Two particles
$x_{cm} = \frac{m_1x_1 + m_2x_2}{m_1+m_2}$
Many particles
$x_{cm} = \frac{\sum m_i x_i}{M}$ , same for y and z
Continuous body
$x_{cm} = \frac{1}{M}\int x\,dm$
SI unit
metre (m), since it is a position
Motion
$M\vec{a}_{cm} = \vec{F}_{ext}$
Velocity
$\vec{v}_{cm} = \frac{\sum m_i \vec{v}_i}{M}$

How the center of mass formula weighs each particle

Look at the two particle form. If the masses are equal, the result is the midpoint, $\frac{x_1 + x_2}{2}$ . If $m_2$ is much heavier, the point slides toward $x_2$ . The distances from the center of mass to each particle scale opposite to the masses: $\frac{r_1}{r_2} = \frac{m_2}{m_1}$ .

In two or three dimensions, apply the formula once for each axis: $x_{cm} = \frac{\sum m_i x_i}{M}$ , $y_{cm} = \frac{\sum m_i y_i}{M}$ , $z_{cm} = \frac{\sum m_i z_i}{M}$ , where $M = \sum m_i$ . You may put the origin anywhere; the point itself does not move, only its coordinates change.

Worked examples

Example 1: two blocks on a line

A 2 kg block sits at x = 0 and a 3 kg block at x = 5 m. Where is the center of mass?

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

It is 3 m from the lighter block and 2 m from the heavier one, the inverse of the 2 : 3 mass ratio.

Example 2: three particles in a plane

Masses of 1 kg, 2 kg and 3 kg sit at (0, 0), (4 m, 0) and (0, 6 m). Find the center of mass.

$x_{cm} = \frac{1(0) + 2(4) + 3(0)}{6} = \frac{8}{6} \approx 1.33\ \text{m}$

$y_{cm} = \frac{1(0) + 2(0) + 3(6)}{6} = 3\ \text{m}$ . The point is (1.33 m, 3 m).

Example 3: Earth and Moon

Earth (5.97 × 10²⁴ kg) and the Moon (7.35 × 10²² kg) are 3.84 × 10⁸ m apart. How far from Earth's centre is their center of mass?

Put Earth at x = 0: $x_{cm} = \frac{(7.35 \times 10^{22})(3.84 \times 10^8)}{5.97 \times 10^{24} + 7.35 \times 10^{22}} \approx 4.67 \times 10^6\ \text{m}$ .

That is about 4670 km, less than Earth's radius of 6370 km, so the point lies inside Earth.

Centre of mass formula list for Class 11, a quick formula sheet

Location of the center of mass of uniform bodies
Body	Center of mass
Uniform rod of length L	At its midpoint, L/2 from either end
Semicircular ring (thin arc) of radius R	$\frac{2R}{\pi}$ from the centre, on the symmetry axis
Semicircular disc of radius R	$\frac{4R}{3\pi}$ from the centre
Solid hemisphere of radius R	$\frac{3R}{8}$ from the flat face
Hollow hemispherical shell of radius R	$\frac{R}{2}$ from the flat face
Solid cone of height h	$\frac{h}{4}$ above the base
Triangular plate	At the centroid, where the medians meet

Every row comes from $x_{cm} = \frac{1}{M}\int x\,dm$ . For a body with a piece cut out, treat the hole as a negative mass: $x_{cm} = \frac{M_1 x_1 - M_2 x_2}{M_1 - M_2}$ .

Motion of the center of mass

Differentiate the position formula and you get $\vec{v}_{cm} = \frac{\sum m_i\vec{v}_i}{M}$ , so the total momentum is $M\vec{v}_{cm}$ . Differentiate again: $M\vec{a}_{cm} = \vec{F}_{ext}$ . Internal forces cancel in pairs by Newton's third law and never move the center of mass.

A shell that bursts in mid air is the classic test. The fragments scatter, but their center of mass carries on along the original parabola until the first piece lands. Likewise, if no external force acts, the center of mass stays at rest or moves at constant velocity, which is the root of conservation of momentum.

Links with rotation

The center of mass is the reference point for the parallel axis theorem in the moment of inertia page, and a free body that spins in flight rotates about it. Torques about this point decide whether a body tips. The rotation hub collects the remaining results of this unit.

The center of mass need not lie inside the material. For a ring or a boomerang it sits in empty space.

Common slips

Forgetting signs: a particle on the negative x side contributes a negative $m_i x_i$ .
Dividing by the number of particles instead of the total mass.
Measuring positions from different origins for different particles.
Using the hollow hemisphere result for a solid one, or the reverse.

Frequently asked questions

What is the formula for center of mass?

For particles, xcm = Σmᵢxᵢ / Σmᵢ, and the same form gives y and z. For two masses this is (m₁x₁ + m₂x₂) / (m₁ + m₂).

Is centre of mass the same as center of mass?

Yes. Centre is the spelling used in Indian and British books, center in US books. The formula and meaning are identical.

Can the center of mass lie outside a body?

Yes. For a ring, a hollow sphere or an L shaped plate, the point lies in the empty space the body surrounds.

What is the difference between center of mass and center of gravity?

The center of mass depends only on how mass is spread. The center of gravity is where the total weight acts. In a uniform gravitational field, as near Earth's surface, the two coincide.

Which formulas for center of mass should a Class 11 student memorise?

The particle formula, the velocity form, M times a(cm) = F(external), and the positions for a rod, semicircular ring, semicircular disc, solid and hollow hemispheres and a cone.