The centre of gravity of a body on the earth coincides with its centre of mass for a 'small' object whereas fo

NCERT Exemplar · Q14

Q. The centre of gravity of a body on the earth coincides with its centre of mass for a 'small' object whereas for an 'extended' object it may not. What is the qualitative meaning of 'small' and 'extended' in this regard? For which of the following the two coincides? A building, a pond, a lake, a mountain?

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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. …

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The centre of gravity (COG) and centre of mass (COM) coincide only when Earth's gravitational field $g$ is essentially uniform across the body. 'Small' means the object's vertical extent is small enough that $g$ barely changes from its top to its bottom; 'extended' means that variation becomes non-negligible. Of the four examples, a building and a pond count as small enough for COG and COM to coincide; a lake and a mountain are extended enough that they don't.

Why COG and COM can differ

The centre of mass is a purely geometric point — the mass-weighted average position, $\vec{R}_{COM}=\dfrac{\sum m_i\vec{r}_i}{\sum m_i}$ . The centre of gravity is the point where the total weight effectively acts: $\vec{R}_{COG}=\dfrac{\sum m_i\, g(\vec{r}_i)\,\vec{r}_i}{\sum m_i\, g(\vec{r}_i)}$ . If $g$ is the same at every point of the body, the $g$ 's cancel and $\vec{R}_{COG}=\vec{R}_{COM}$ . If $g$ varies noticeably across the body, the two points separate slightly, with COG shifting toward the region of stronger gravity (i.e. closer to Earth's surface).

How much does $g$ actually vary with height?

Near Earth's surface, $g$ decreases with height $h$ as

g(h)\approx g_0\left(1-\frac{2h}{R_E}\right), \qquad R_E\approx 6400\text{ km}

so the fractional change in $g$ over a height $h$ is roughly $\dfrac{2h}{R_E}$ .

Applying this to the four examples

Building (height of order $10^2$ m): fractional change $\sim \dfrac{2\times100}{6.4\times10^6}\approx 3\times10^{-5}$ — negligible. COG and COM coincide.
Pond (depth of a few metres): the fractional change is smaller still — negligible. COG and COM coincide. …

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Why COG and COM can differ

How much does ggg actually vary with height?

Applying this to the four examples

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How much does $g$ actually vary with height?

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