Find the centre of mass of a uniform (a) half-disc, (b) quarter-disc.

NCERT Exemplar · Q23

Q. Find the centre of mass of a uniform

(a) half-disc,

(b) quarter-disc.

Karnataka PUCLong· 5mImportance★★★★★est

91% · 52/57 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 uniform laminae with symmetry, the center of mass lies on the symmetry axis; integration in polar coordinates gives $\bar{y} = \frac{4R}{3\pi}$ from the straight edge for a half-disc and $\bar{x} = \bar{y} = \frac{4R}{3\pi}$ from the respective straight edges for a quarter-disc.

Why the center of mass shifts inward

The center of mass of a body is the weighted average position of all its mass elements. For a uniform lamina—one with constant density—the center of mass coincides with the geometric centroid. A full disc has its center of mass at the geometric center by symmetry. When you remove half or three-quarters of the disc, you break that symmetry, and the center of mass shifts toward the remaining material.

The key insight: symmetry pins down some coordinates immediately. A half-disc is symmetric about the diameter that forms its straight edge, so the center of mass must lie on that axis. We need only find how far it sits from the edge. A quarter-disc has two perpendicular symmetry axes (the two radii), and by symmetry $\bar{x} = \bar{y}$ .

Both problems reduce to a single integral in polar coordinates, where the area element $dA = r \, dr \, d\theta$ and the position of each element is $(r \cos \theta, r \sin \theta)$ .

(a) Half-disc

Consider a half-disc of radius $R$ lying in the upper half-plane, with its straight edge along the $x$ -axis and center at the origin.

Symmetry argument: The half-disc is symmetric about the $y$ -axis, so $\bar{x} = 0$ . We need only find $\bar{y}$ .
Set up the integral: The $y$ -coordinate of the center of mass is

\bar{y} = \frac{1}{A} \iint y \, dA,

where $A = \frac{1}{2} \pi R^2$ is the area of the half-disc.

Polar coordinates: In polar coordinates, $y = r \sin \theta$ and $dA = r \, dr \, d\theta$ . The half-disc is described by $0 \le r \le R$ and $0 \le \theta \le \pi$ . Thus

\bar{y} = \frac{1}{\frac{1}{2} \pi R^2} \int_0^\pi \int_0^R (r \sin \theta) \cdot r \, dr \, d\theta = \frac{2}{\pi R^2} \int_0^\pi \sin \theta \, d\theta \int_0^R r^2 \, dr.

Evaluate the radial integral:

\int_0^R r^2 \, dr = \frac{R^3}{3}.

Evaluate the angular integral:

\int_0^\pi \sin \theta \, d\theta = \left[ -\cos \theta \right]_0^\pi = -(-1 - 1) = 2.

Combine:

\bar{y} = \frac{2}{\pi R^2} \cdot 2 \cdot \frac{R^3}{3} = \frac{4 R^3}{3 \pi R^2} = \frac{4R}{3\pi}.

\text{Half-disc: } \quad \bar{x} = 0, \quad \bar{y} = \frac{4R}{3\pi} \approx 0.424 R.

The center of mass lies on the axis of symmetry, about $42\%$ of the radius from the straight edge.

(b) Quarter-disc

Now consider a quarter-disc of radius $R$ in the first quadrant, with straight edges along the positive $x$ - and $y$ -axes.

Symmetry argument: The quarter-disc is symmetric under reflection across the line $y = x$ (swapping $x$ and $y$ ). Therefore $\bar{x} = \bar{y}$ . We need only compute one coordinate.
Set up the integral for $\bar{x}$ :

\bar{x} = \frac{1}{A} \iint x \, dA,

where $A = \frac{1}{4} \pi R^2$ .

Polar coordinates: Here $x = r \cos \theta$ , and the quarter-disc is $0 \le r \le R$ , $0 \le \theta \le \frac{\pi}{2}$ . Thus …

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.

KCET 2025Set D-41 markMCQ
$\mu = \cot\theta$
›Reveal solutionSolution
Resolve the three string tensions at the junction O, get the horizontal pull on the floor block, then set that pull equal to the limiting friction $\mu N$ with $N = Mg$ .

Step 1 — Identify the forces at the junction O.

Three strings meet at $O$ , so $O$ is a massless point in equilibrium under three tensions:

$T_1$ — along the horizontal string that runs to the block on the floor (it pulls $O$ towards that block, i.e. horizontally to the left).

$T_2$ — along the inclined string going up to the wall at angle $\theta$ above the horizontal (it pulls $O$ up and to the right).

$T_3$ — along the vertical string, pulling $O$ straight down. The hanging block of mass $M$ is in equilibrium, so

$T_3 = Mg$
Step 2 — Equilibrium of the junction O.

Vertical direction: only the inclined string has an upward component, and it must hold up the hanging weight.
$T_2\sin\theta = T_3 = Mg \qquad \Rightarrow \qquad T_2 = \frac{Mg}{\sin\theta}$
Horizontal direction: the inclined string's horizontal component is balanced by the horizontal string.
$T_1 = T_2\cos\theta = \frac{Mg}{\sin\theta}\cdot\cos\theta = Mg\cot\theta$
Step 3 — Equilibrium of the block on the floor.

The horizontal string pulls this block towards $O$ with force $T_1 = Mg\cot\theta$ . The only force that can oppose it is static friction from the floor. The string is horizontal, so it adds no vertical component and the normal reaction is just the block's own weight:

$N = Mg$ …
Start your 14-day free trial to see the full solution →
KCET 2025Set D-41 markMCQ
$1$
›Reveal solutionSolution
Set up coordinates with A at the origin and AB along the x-axis, read off the vertex positions of the equilateral triangle, then apply the weighted-average centre-of-mass formula separately to $x$ and $y$ .

Step 1 — The concept and formula

The centre of mass of a system of discrete point masses is the mass-weighted average position:
$x_{cm} = \frac{\sum m_i x_i}{\sum m_i}, \qquad y_{cm} = \frac{\sum m_i y_i}{\sum m_i}$
It is a weighted average because each particle "pulls" the balance point toward itself in proportion to its mass. Note the answer depends entirely on our choice of origin — and the question fixes that for us: vertex A is the origin.

Step 2 — Set up the coordinates of the triangle

Take A at the origin and let side AB lie along the $+x$ axis. For an equilateral triangle of side $a = 1$ m:

$\mathbf{A} = (0,\, 0)$ — carries $m_A = 1$ kg

$\mathbf{B} = (a,\, 0) = (1,\, 0)$ — carries $m_B = 2$ kg

$\mathbf{C} = \left(\dfrac{a}{2},\ \dfrac{\sqrt{3}}{2}a\right) = \left(\dfrac{1}{2},\ \dfrac{\sqrt{3}}{2}\right)$ — carries $m_C = 3$ kg

Why C sits there: C is equidistant from A and B, so it lies above the midpoint of AB, at $x = a/2$ . Its height is found from Pythagoras on the half-triangle:
$h = \sqrt{a^2 - \left(\tfrac{a}{2}\right)^2} = \sqrt{a^2 - \tfrac{a^2}{4}} = \frac{\sqrt{3}}{2}a$
Step 3 — Total mass
$M = m_A + m_B + m_C = 1 + 2 + 3 = 6~\text{kg}$
Step 4 — The $x$ -coordinate
$x_{cm} = \frac{m_A x_A + m_B x_B + m_C x_C}{M} = \frac{(1)(0) + (2)(1) + (3)\left(\tfrac12\right)}{6}$ $x_{cm} = \frac{0 + 2 + 1.5}{6} = \frac{3.5}{6} = \frac{7}{12}~\text{m}$
Step 5 — The $y$ -coordinate …
Start your 14-day free trial to see the full solution →
COMEDK 2025Set 2025-A1 markMCQ
$Q. The separation between C and O atoms in CO is 0.12 nm . The distance of C atom from the centre of mass is (A) 0.1 nm (B) 0.05 nm (C) 0.03 nm (D) 0.07 nm$
›Reveal solutionSolution
The center of mass of a diatomic molecule lies closer to the heavier atom. For CO, with atomic masses 12 u (C) and 16 u (O) and bond length 0.12 nm, the distance from C to the center of mass is 0.07 nm, so option (D) is correct.

The key idea is that the center of mass divides the bond in inverse proportion to the masses. The heavier atom (oxygen) is closer to the center of mass, so the lighter carbon atom is farther from it. We can find the exact distance by setting the torques (mass × distance) about the center of mass equal.

Define the system and variables

Let the carbon atom have mass $m_C = 12$ u (atomic mass units) and the oxygen atom have mass $m_O = 16$ u. The bond length (distance between nuclei) is $d = 0.12$ nm. Let $x$ be the distance from the carbon atom to the center of mass. Then the distance from oxygen to the center of mass is $d - x$ .

Apply the center of mass condition

The center of mass is the point where the weighted distances balance:

$m_C \cdot x = m_O \cdot (d - x)$
This says the “mass × distance” on one side equals that on the other.

Solve for $x$

$12x = 16(0.12 - x)$ $12x = 1.92 - 16x$ $12x + 16x = 1.92$ $28x = 1.92$ $x = \frac{1.92}{28} = 0.06857\ \text{nm} \approx 0.07\ \text{nm}$

Interpret the result …
Start your 14-day free trial to see the full solution →
COMEDK 2025Set 2025-E1 markMCQ
$Q. Select the correct statement from the following: The position of the centre of mass of a system : (A) Remains the same in any translatory motion (B) Depend on the choice of coordinates (C) Does not depend on the shape and size of the system (D) Depends on the distribution of its mass.$
›Reveal solutionSolution
The centre of mass is defined as the weighted average position of all mass in a system, so it inherently depends on how mass is distributed. The correct statement is (D).

The centre of mass (COM) is a conceptual point where the entire mass of a system can be considered to be concentrated for many mechanical purposes. To decide which statement is correct, we need to recall its definition and properties.

Definition: For a system of particles, the position vector of the centre of mass is
$\mathbf{R}_{\text{COM}} = \frac{\sum m_i \mathbf{r}_i}{\sum m_i}$
where $m_i$ are the masses and $\mathbf{r}_i$ their positions. This formula makes clear that the COM depends on both the masses and their locations — i.e., the mass distribution.

Now evaluate each option:

Option (A): “Remains the same in any translatory motion.”

If the whole system undergoes pure translation (all particles move by the same displacement), the COM also moves by that displacement. So its position changes relative to a fixed origin. This statement is false unless “remains the same” means relative to the system itself — but the wording “in any translatory motion” implies motion relative to an external frame, so it’s incorrect.

Option (B): “Depends on the choice of coordinates.”

The numerical coordinates of the COM do change if you shift the origin or rotate axes, but the physical point in space is independent of coordinate choice. The statement is ambiguous; usually “depends on the choice of coordinates” is considered false because the COM is a physical property, not a coordinate artifact.

Option (C): “Does not depend on the shape and size of the system.” …
Start your 14-day free trial to see the full solution →
COMEDK 2025Set 2025-M1 markMCQ
$R$
›Reveal solutionSolution

Treating the hole as negative mass, the centre of mass shifts by $\dfrac{m_{\text{hole}}\,d}{m_{\text{remaining}}}=\dfrac{\pi R^2\cdot R}{3\pi R^2}=\dfrac{R}{3}$ away from the hole — option (A).

Concept. The centre of mass of a body with a piece removed is found by modelling the removed piece as a body of negative mass located at its own centre, then combining it with the full body.

Step 1 — Geometry and masses.

Original disc radius $2R \Rightarrow$ area $=\pi(2R)^2 = 4\pi R^2$ .

The smaller disc has the larger disc's radius $2R$ as its diameter, so its radius is $R$ and area $=\pi R^2$ . Its centre lies at distance $R$ from the centre $O$ .

Taking surface mass density $\sigma$ :
$m_{\text{full}} = 4\pi R^2\sigma,\qquad m_{\text{hole}} = \pi R^2\sigma,\qquad m_{\text{remaining}} = 3\pi R^2\sigma.$
Step 2 — Set up coordinates.

Put $O$ at the origin and the hole's centre at $(R,0)$ . The full disc's centre of mass is at the origin, and it equals the mass-weighted average of the remaining part and the removed part: …
Start your 14-day free trial to see the full solution →
COMEDK 2024Set 2024-M1 markMCQ
$Q. Which of the following statement is true regarding the centre of mass of a system? (A) The centre of mass depends on the size and shape but does not depend on the distribution of mass of the body. (B) The centre of mass depends on the coordinate system. (C) The centre of mass of a system depends on the size and shape of the body but independent of the co-ordinate system (D) The centre of mass of a body always lies inside the body.$
›Reveal solutionSolution
The centre of mass is defined by the weighted average of positions of all mass elements; it depends on mass distribution and shape, but not on the choice of coordinate system. The correct option is (C).

The centre of mass (COM) is a point that represents the average location of the total mass of a system. The key idea is that it is a physical property of the mass distribution itself — not a mathematical artifact of where we place our axes. So any statement that says the COM depends on the coordinate system is false, because shifting the origin just shifts the COM coordinates by the same amount; the point in space remains the same relative to the body.

Let’s examine each option carefully.

Option (A): “The centre of mass depends on the size and shape but does not depend on the distribution of mass of the body.” This is false. The COM formula is

$\mathbf{R}_{\text{COM}} = \frac{\sum m_i \mathbf{r}_i}{\sum m_i}$
for discrete masses, or
$\mathbf{R}_{\text{COM}} = \frac{1}{M} \int \mathbf{r} \, dm$
for continuous bodies. Both explicitly involve how mass is arranged (the distribution). For example, a uniform rod and a rod with a heavy end have the same size and shape but different COM positions. So distribution matters.

Option (B): “The centre of mass depends on the coordinate system.”

This is a classic trap. If you change the coordinate system, the coordinates of the COM change, but the physical point in space does not. The COM is a property of the system, not of the axes. So this statement is false.

Option (C): “The centre of mass of a system depends on the size and shape of the body but independent of the co-ordinate system.” …
Start your 14-day free trial to see the full solution →
COMEDK 2023Set 2023-M1 markMCQ
$m_1$
›Reveal solutionSolution

The definition of centre of mass gives $\dfrac{m_1x_1+m_2x_2}{m_1+m_2}$ .

For two particles of masses $m_1,m_2$ at coordinates $x_1,x_2$ from the origin, the centre of mass is

$x_{cm}=\frac{\sum m_i x_i}{\sum m_i}=\frac{m_1 x_1+m_2 x_2}{m_1+m_2}.$ …
Start your 14-day free trial to see the full solution →
COMEDK 2022Set 20221 markMCQ
${R \over {(4\pi - 2)}}$
›Reveal solutionSolution

Distance from the centre = R / (4pi - 2).

Concept: centre of mass of a body with a portion removed (negative-mass method).

The square is cut out with a RADIUS of the disc as its diagonal, so:

diagonal d = R -> side a = R/sqrt(2)

area of square = a^2 = R^2/2

The square's centre lies at the midpoint of that radius, i.e. at distance R/2 from the disc's centre.

Take surface density sigma = 1 (it cancels).

Whole disc: area = pi R^2, centre at x = 0

Removed square: area = R^2/2, centre at x = R/2

Centre of mass of the remaining portion (measured from the disc centre, on the side opposite the cut): …
Start your 14-day free trial to see the full solution →
KCET 2021Set B-21 markMCQ
$8\,\mathrm{kg}$
›Reveal solutionSolution
The net gravitational force on the 2 kg mass at the centroid is the vector sum of pulls from the two 8 kg masses at A and B. To cancel this resultant, the 4 kg mass must be placed on the extension of CG beyond G, at a distance $PG = 1/\sqrt{2}$  m from G. The correct option is (B).

The key idea is that gravitational forces are vectors. The two 8 kg masses at A and B pull the 2 kg mass at the centroid G along the lines GA and GB. Because the triangle is equilateral and G is the centroid, GA = GB = GC = 1 m, and the angle between GA and GB is 120°. The resultant of these two equal forces lies along the median from C (i.e., along CG), pointing from G toward C. To cancel it, the 4 kg mass must be placed on the same line CG, but on the opposite side of G, so that its pull on the 2 kg mass is exactly opposite and equal in magnitude to that resultant.

Force from one 8 kg mass The gravitational force on the 2 kg mass due to a single 8 kg mass at distance 1 m is

$F = G \frac{8 \times 2}{1^2} = 16G.$
So each 8 kg mass exerts a force of magnitude $16G$ along GA and GB respectively.

Resultant of the two forces The angle between GA and GB is 120° (since the centroid is equidistant from all vertices and the triangle is equilateral). The magnitude of the resultant of two equal forces $F$ with angle $\theta$ is

$R = \sqrt{F^2 + F^2 + 2F^2\cos\theta} = F\sqrt{2(1+\cos\theta)}.$
With $\cos 120^\circ = -1/2$ ,
$R = 16G \sqrt{2\left(1 - \frac12\right)} = 16G \sqrt{2 \times \frac12} = 16G \sqrt{1} = 16G.$
So the resultant force on the 2 kg mass from the two 8 kg masses has magnitude $16G$ and points along the median from C toward C (since the two pulls are symmetric about CG).

Cancelling the resultant with the 4 kg mass Let the 4 kg mass be placed at a point P on the line CG, on the side opposite to C (i.e., beyond G away from C). The force it exerts on the 2 kg mass is $F_4 = G \frac{4 \times 2}{r^2} = \frac{8G}{r^2},$ …
Start your 14-day free trial to see the full solution →
COMEDK 2021Set 20211 markMCQ
$Q. Centre of mass of the given system of particles will be at (A) OA (B) OB (C) OC (D) OD$
›Reveal solutionSolution

So the CM is at (0.2, 0.2), i.e. displaced from O along the direction (1, 1) - straight towards corner B, the heavier 4m corner. It lies on OB.

Concept: centre of mass of discrete particles, R_cm = (sum m_i r_i)/(sum m_i).

From the figure, take O (the centre of the square, where the diagonals meet) as the origin, with the square of half-diagonal 1 unit. Corners:

A (top-left) = (-1, +1), mass 2m

B (top-right) = (+1, +1), mass 4m

C (bottom-right) = (+1, -1), mass 2m

D (bottom-left) = (-1, -1), mass 2m

Total mass = 10m.

x_cm = [2(-1) + 4(+1) + 2(+1) + 2(-1)]m / 10m = (-2 + 4 + 2 - 2)/10 = +0.2 …
Start your 14-day free trial to see the full solution →
KCET 2020Set A-11 markMCQ
$x$
›Reveal solutionSolution

Use the perpendicular-axis theorem to get $I_z$ from $I_x$ and $I_y$ , then convert to a radius of gyration with $I = Mk^2$ .

Step 1 — Why the perpendicular-axis theorem applies.

The theorem states that for a planar (lamina) body, the moment of inertia about an axis $z$ perpendicular to the plane equals the sum of the moments of inertia about two mutually perpendicular axes $x$ and $y$ lying in the plane and intersecting the $z$ -axis:
$I_z = I_x + I_y$
A thin uniform rectangular plate is exactly such a lamina, and the figure shows $x$ , $y$ in the plane meeting at $O$ with $z$ perpendicular through $O$ — every condition of the theorem is met.

Step 2 — Compute $I_z$ .
$I_z = I_x + I_y = 0.2 + 0.3 = 0.5 \;\text{kg m}^2$
Step 3 — Definition of the radius of gyration.

The radius of gyration $k$ about an axis is the distance from that axis at which the whole mass could be concentrated to give the same moment of inertia:

$I = Mk^{2} \quad\Longrightarrow\quad k = \sqrt{\frac{I}{M}}$ …
Start your 14-day free trial to see the full solution →