$(n-1)$ equal point masses each of mass $m$ are placed at the vertices of a regular $n$-polygon. The vacant vertex has a position vector $\vec{a}$ with respect to the centre of the polygon. Find the position vector of centre of mass.

(n-1) equal point masses each of mass m are placed at the vertices of a regular n-polygon. The vacant vertex h

NCERT Exemplar · Q22

(n-1)

Sikkim CbseShort· 3mImportance★★★★★est

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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 problem asks for the position vector of the center of mass of $(n-1)$ equal point masses placed at the vertices of a regular $n$ -polygon, with one vertex vacant. By considering the complete polygon and subtracting the effect of the missing mass, we find the position vector of the center of mass to be $\boxed{\frac{-\vec{a}}{n-1}}$ .

To find the position vector of the center of mass for a system of particles, we use the principle that the center of mass represents the average position of the total mass of the system. When dealing with a regular polygon and missing masses, a powerful technique is to consider the complete system first and then account for the missing part.

Here's the intuition:

A regular $n$ -polygon with equal point masses at all $n$ vertices has its center of mass exactly at its geometric center. This is due to the symmetry of the arrangement. If we place the center of the polygon at the origin of our coordinate system, the position vector of the center of mass for the complete system would be $\vec{0}$ .

Now, imagine this complete system. It consists of two parts:

The $(n-1)$ masses that are actually present.
A single mass $m$ that would be at the vacant vertex.

The center of mass of the complete system is the weighted average of the center of mass of these two parts. Since we know the center of mass of the complete system (it's the origin), we can use this relationship to find the center of mass of the $(n-1)$ present masses.

Define the coordinate system and components:

Let the center of the regular $n$ -polygon be the origin $\vec{O} = \vec{0}$ .

Each of the $(n-1)$ point masses has mass $m$ .

The vacant vertex has a position vector $\vec{a}$ with respect to the center of the polygon. This means if a mass $m$ were placed at this vacant vertex, its position vector would be $\vec{a}$ .
Consider the complete system:

Imagine a hypothetical system where all $n$ vertices of the regular polygon have an equal point mass $m$ . Due to the symmetry of a regular polygon, the center of mass of this complete system would coincide with the geometric center of the polygon, which we have set as the origin.

So, the position vector of the center of mass of the complete system, $\vec{R}_{CM, \text{total}}$ , is $\vec{0}$ .
Apply the Center of Mass formula:

The general formula for the position vector of the center of mass of a system of particles is:

$\vec{R}_{CM} = \frac{\sum_{i=1}^{N} m_i \vec{r}_i}{\sum_{i=1}^{N} m_i}$

We can think of the complete system (total mass $nm$ ) as being composed of two subsystems:
- The $(n-1)$ masses that are actually present. Let their total mass be $M_1 = (n-1)m$ , and their center of mass be $\vec{R}_{CM}$ . This is what we need to find. …

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Q. $(n-1)$ equal point masses each of mass $m$ are placed at the vertices of a regular $n$ -polygon. The vacant vertex has a position vector $\vec{a}$ with respect to the centre of the polygon. Find the position vector of centre of mass.

Concept understanding — Center of Mass

What is the Center of Mass?

Why does this matter?

The precise definition

Breaking it down with an example

Key properties to remember

Concept: Build the Missing Mass Back In, Using the Full Polygon's Known Centre of Mass

Q.(n−1)(n-1)(n−1) equal point masses each of mass mmm are placed at the vertices of a regular nnn-polygon. The vacant vertex has a position vector a⃗\vec{a}a with respect to the centre of the polygon. Find the position vector of centre of mass.

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Q. $(n-1)$ equal point masses each of mass $m$ are placed at the vertices of a regular $n$ -polygon. The vacant vertex has a position vector $\vec{a}$ with respect to the centre of the polygon. Find the position vector of centre of mass.

Unlock everything free for 14 days