The center of mass of a system of particles does not depend upon, (a) position of particles (b) relative dista

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Q. The center of mass of a system of particles does not depend upon,

(a) position of particles

(b) relative distance between particles

(d) force acting on particle

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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. Write down the defining formula for the center of mass.

For a system of $n$ particles with masses $m_1, m_2, \ldots, m_n$ at position vectors $\vec r_1, \vec r_2, \ldots, \vec r_n$ , the center of mass is

\vec R_{CM} = \dfrac{\sum_i m_i \vec r_i}{\sum_i m_i}.

Step 2. Identify exactly which quantities appear on the right-hand side.

Only two kinds of input appear: the masses $m_i$ of the particles, and their position vectors $\vec r_i$ (which fix both the individual positions and, once you take differences, the relative distances between particles). Nowhere in this formula does any force appear — the definition is purely kinematic/geometric, built before any dynamics is even considered.

Step 3. Check each option against the formula.

(a) position of particles — appears directly as $\vec r_i$ , so the CM does depend on this. Not the answer.
(b) relative distance between particles — determined by the set of $\vec r_i$ 's (moving particles relative to each other shifts $\vec R_{CM}$ unless compensated), so it does depend on this. Not the answer.
(c) masses of particles — appears directly as $m_i$ , the weighting factor itself. Not the answer.
(d) force acting on particle — never enters the defining sum at all. A force determines how the CM moves (via $\vec F_{ext} = M\vec a_{CM}$ ), but the location/definition of the CM itself is force-independent.

Step 4. Conclude.

The center of mass depends on positions, relative distances, and masses — but not on any force acting on the particles.

✓Final answer

(d) force acting on particle

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