The density of a non-uniform rod of length 1m is given by $\rho(x) = a(1+bx^2)$ where a and b are constants and $0 \le x \le 1$. The centre of mass of the rod will be at (a) $\dfrac{3(2+b)}{4(3+b)}$ (b) $\dfrac{4(2+b)}{3(3+b)}$ (c) $\dfrac{3(3+b)}{4(2+b)}$ (d) $\dfrac{4(3+b)}{3(2+b)}$

The density of a non-uniform rod of length 1m is given by ρ(x) = a(1+bx²) where a and b are constants and 0 ≤

NCERT Exemplar · Q7

\rho(x) = a(1+bx^2)

(a)

\dfrac{3(2+b)}{4(3+b)}

(b)

\dfrac{4(2+b)}{3(3+b)}

(c)

\dfrac{3(3+b)}{4(2+b)}

(d)

\dfrac{4(3+b)}{3(2+b)}

Kerala DhseMCQ· 1mImportance★★★★★est

63% · 36/57 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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. …

Start your 14-day free trial to unlock the full solution →

For a non-uniform rod, the center of mass is the mass-weighted average position. Integrate $x \cdot dm$ and divide by the total mass; with $\rho(x) = a(1+bx^2)$ , this yields $x_{cm} = \dfrac{3(2+b)}{4(3+b)}$ .

Why the center of mass requires integration

The center of mass is the point where all the mass can be considered concentrated for translational motion. For a uniform rod it sits at the geometric center, but when density varies with position we must account for how mass is distributed.

Think of the rod as a collection of infinitesimal pieces. Each piece at position $x$ contributes to "pulling" the center of mass toward itself, weighted by how much mass it contains. Mathematically, the center of mass is:

x_{cm} = \frac{\int x \, dm}{\int dm} = \frac{\text{moment of mass}}{\text{total mass}}

The denominator is simply the total mass $M$ , while the numerator sums up each position multiplied by the mass at that position.

Step-by-step calculation

1. Express the mass element

For a rod with linear density $\rho(x)$ , a small segment of length $dx$ at position $x$ has mass:

dm = \rho(x) \, dx = a(1 + bx^2) \, dx

2. Find the total mass

Integrate from $x = 0$ to $x = 1$ :

M = \int_0^1 a(1 + bx^2) \, dx = a \int_0^1 (1 + bx^2) \, dx

M = a \left[ x + b \cdot \frac{x^3}{3} \right]_0^1 = a \left( 1 + \frac{b}{3} \right) = a \cdot \frac{3 + b}{3}

3. Calculate the moment about the origin

The numerator of $x_{cm}$ requires:

$\int_0^1 x \, dm = \int_0^1 x \cdot a(1 + bx^2) \, dx = a \int_0^1 (x + bx^3) \, dx$ …

Start your 14-day free trial to unlock the full solution →

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

KEAM 2026Set eng-2026-04184 marksMCQ
$Q. Two masses 2 kg and 6 kg lie on the x-axis at distances of 3 m and 6 m, respectively, from the origin. The distances of the centre of mass of the system from the origin and from the 2 kg mass are in the ratio (A) 3 : 7 (B) 1 : 4 (C) 2 : 1 (D) 7 : 3 (E) 3 : 1$
›Reveal solutionSolution

The centre of mass is at $5.25$ m; from the origin it is $5.25$ m and from the $2$ kg mass $2.25$ m, a ratio $7:3$ .

The centre of mass lies at

$x_{cm}=\dfrac{2(3)+6(6)}{2+6}=\dfrac{6+36}{8}=5.25\ \text{m}.$ …
Start your 14-day free trial to see the full solution →
KEAM 2026Set eng-2026-04224 marksMCQ
$Q. The objects and the corresponding positions of centre of mass are given below. The FALSE one is (A) Uniform rod : middle point of the rod (B) Circular ring : centre of the ring (C) Triangular lamina : point of intersection of altitudes (D) Cylinder : middle point on it axis (E) Cubical box : intersection of diagonals$
›Reveal solutionSolution
[!TLDR]

A triangular lamina's centre of mass is its centroid (intersection of medians), so option (C), which names the intersection of altitudes, is false.

Concept

The centre of mass of a symmetric or uniform body lies at its geometric centre (NCERT/CBSE Class-11 System of Particles). For a triangular lamina this is the centroid.

Solution

Check each:

(A) Uniform rod $\to$ midpoint — correct.

(B) Circular ring $\to$ centre — correct. …
Start your 14-day free trial to see the full solution →
KEAM 2026Set pha-2026-0420F4 marksMCQ
$Gm^2$
›Reveal solutionSolution

No external force → $a_{cm}=0$ .

The mutual gravitational attraction between the two particles is an internal force of the system; the forces on the two masses are equal and opposite. By Newton's laws, the centre of mass accelerates only under a net external force:

$M_{total}\,a_{cm} = F_{external} = 0.$ …
Start your 14-day free trial to see the full solution →
KEAM 2025Set eng-2025-04234 marksMCQ
$ms^{-2}$
›Reveal solutionSolution
The centre of mass accelerates at $\tfrac{4}{3}\,ms^{-2}$ in the direction of the 2 kg mass's acceleration.

Concept and Intuition

The acceleration of the centre of mass equals the net external force divided by total mass, $a_{cm} = \frac{\sum \vec{F}}{\sum m}$ . The two forces are the products $ma$ of each body, directed oppositely since the bodies move toward each other.

Step-by-Step Solution

Force on 1 kg mass: $F_1 = 1\times 2 = 2\,N$ .

Force on 2 kg mass: $F_2 = 2\times 3 = 6\,N$ , opposite in direction.

Net force $= 6 - 2 = 4\,N$ in the direction of the 2 kg mass's acceleration. …
Start your 14-day free trial to see the full solution →
KEAM 2025Set eng-2025-04274 marksMCQ
$\frac{2L}{3}$
›Reveal solutionSolution

A thin uniform rod has uniform linear density, so its centre of mass is at the geometric centre, a distance $L/2$ from one end.

For a rod of length $L$ with uniform mass per unit length $\lambda$ , the centre-of-mass position measured from one end is …
Start your 14-day free trial to see the full solution →
KEAM 2025Set pha-2025-0429F4 marksMCQ
$1\,cm$
›Reveal solutionSolution

The centre of mass of equal masses is the arithmetic mean of their positions: $(1+2+3)/3 = 2\,\mathrm{cm}$ .

For particles of equal mass $m$ at positions $x_i$ , the centre of mass is

$x_{cm} = \frac{m x_1 + m x_2 + m x_3}{3m} = \frac{x_1+x_2+x_3}{3}.$ …
Start your 14-day free trial to see the full solution →
KEAM 2023Set eng-2023-P1-A14 marksMCQ
$Q. A uniform thin rod of mass 3 kg has a length of 1 m. If a point mass of 1 kg is attached to it at a distance of 40 cm from its center, the center of mass shifts by a distance of: (A) 2.5 cm (B) 5 cm (C) 8 cm (D) 10 cm (E) 20 cm$
›Reveal solutionSolution
Adding the 1 kg mass shifts the centre of mass by 10 cm.

Concept and Intuition

The uniform rod's centre of mass is at its geometric centre. Placing a point mass off-centre pulls the combined centre of mass toward it by an amount set by the mass-weighted average of positions.

Step-by-Step Solution

Take the rod centre as origin; rod mass 3 kg at 0, point mass 1 kg at 0.40 m. …
Start your 14-day free trial to see the full solution →
KEAM 2021Set eng-2021-P1-A14 marksMCQ
$m$
›Reveal solutionSolution
The $x$ -coordinate of the centre of mass is $-\dfrac{2}{3}$ .

Concept and Intuition

The centre of mass along $x$ is the mass-weighted average of the particle positions, $x_{cm}=\dfrac{\sum m_i x_i}{\sum m_i}$ .

Step-by-Step Solution

Numerator: $m(0)+2m(1)+3m(-2)=0+2m-6m=-4m$ .

Total mass: $m+2m+3m=6m$ .

$x_{cm}=\dfrac{-4m}{6m}=-\dfrac{2}{3}$ . …
Start your 14-day free trial to see the full solution →

←Previous questionQ6 — A uniform square plate lies in the x-y plane with its centre at the or…→Next questionQ8 — A Merry-go-round, made of a ring-like platform of radius R and mass M,…

Q.The density of a non-uniform rod of length 1m is given by $\rho(x) = a(1+bx^2)$ where a and b are constants and $0 \le x \le 1$ . The centre of mass of the rod will be at

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

Why the center of mass requires integration

Step-by-step calculation

Concept: Centre of Mass of a Continuous Body via Integration

Concept

Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Q.The density of a non-uniform rod of length 1m is given by ρ(x)=a(1+bx2)\rho(x) = a(1+bx^2)ρ(x)=a(1+bx2) where a and b are constants and 0≤x≤10 \le x \le 10≤x≤1. The centre of mass of the rod will be at

Why the center of mass requires integration

Step-by-step calculation

🎓Unlock everything free for 14 days

Q.The density of a non-uniform rod of length 1m is given by $\rho(x) = a(1+bx^2)$ where a and b are constants and $0 \le x \le 1$ . The centre of mass of the rod will be at

Unlock everything free for 14 days