Q.A uniform square plate lies in the x-y plane with its centre at the origin (the z-axis passing through the centre, perpendicular to the plate). A small irregular piece Q, originally located in the second quadrant (x<0, y>0) of the plate, is cut out and re-glued at the centre (origin) of the plate, leaving a hole at its original position. In which quadrant of the x-y plane does the centre of mass of the resulting plate now lie?
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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.
RCM=m1+m2+⋯+mnm1r1+m2r2+⋯+mnrn=∑mi∑miri
Here:
- RCM is the position vector of the center of mass
- mi is the mass of the i-th particle
- ri is the position vector of that particle
For a continuous object (like a rod or a sphere), the sum becomes an integral:
RCM=M1∫rdm
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: m1=2 kg at x=0, and m2=3 kg at x=5 m.
The center of mass is:
xCM=2+3(2)(0)+(3)(5)=50+15=3 m
So the center of mass is at x=3 m, closer to the heavier mass. That makes intuitive sense — the heavier mass "pulls" the average toward itself.
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
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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. …
Removing mass from the second quadrant shifts the centre of mass toward the diagonally opposite (fourth) quadrant; re-gluing the piece at the centre does not shift it back. So the CM ends up in quadrant IV. …
A hole in the second quadrant makes the remaining material's centre of mass move to the opposite quadrant (IV). Gluing the removed piece at the origin adds mass at r=0, which cannot shift the CM. Hence the final centre of mass is in quadrant IV.
Concept
For a composite (or a body with a hole), rcm=∑mi∑miri; a hole is treated as negative mass.
Steps
Let the full plate have mass M (CM at origin), and let the piece Q have mass μ located at rQ in quadrant II (so xQ<0, yQ>0).
- After removing Q, the plate-with-hole has CM at
rhole=M−μM(0)−μrQ=M−μ−μrQ.
Since −rQ has x>0, y<0, this lies in quadrant IV. …
Concept: Centre of Mass of a Body With a Hole, Then Mass Added Elsewhere
Step 1: Effect of removing piece Q (originally in quadrant II)
rhole=M−μM(0)−μrQ=M−μ−μrQ
Since rQ has x<0,y>0, −rQ has x>0,y<0 — quadrant IV.
Step 2: Effect of gluing Q back at the origin …
- KEAM 2026Set eng-2026-04184 marksMCQQ.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
xcm=2+62(3)+6(6)=86+36=5.25 m. …
- KEAM 2026Set eng-2026-04224 marksMCQQ.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 → midpoint — correct.
- (B) Circular ring → centre — correct. …
- KEAM 2026Set pha-2026-0420F4 marksMCQQ.Two particles of masses m and 2m kept 1 m apart are attracted to each other by gravitational force. The acceleration of their centre of mass is (G = gravitational constant) (A) Gm (B) 2Gm (C) 3Gm (D) Gm2 (E) zero
›Reveal solutionSolution
No external force → acm=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:
Mtotalacm=Fexternal=0. …
- KEAM 2025Set eng-2025-04234 marksMCQQ.Two objects of masses 1 kg and 2 kg are moving towards each other with accelerations 2 ms−2 and 3 ms−2 respectively on a smooth horizontal surface. The acceleration of centre of mass of the system is (A) (34)ms−2 in the direction of acceleration of 2 kg mass (B) (32)ms−2 in the direction of acceleration of 1 kg mass (C) (32)ms−2 in the direction of acceleration of 2 kg mass (D) (34)ms−2 in the direction of acceleration of 1 kg mass (E) zero
›Reveal solutionSolution
The centre of mass accelerates at 34ms−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, acm=∑m∑F. 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: F1=1×2=2N.
- Force on 2 kg mass: F2=2×3=6N, opposite in direction.
- Net force =6−2=4N in the direction of the 2 kg mass's acceleration. …
- KEAM 2025Set eng-2025-04274 marksMCQQ.The centre of mass of a thin uniform rod of length L lies at a distance (from one end) (A) 32L (B) 43L (C) 2L (D) 3L (E) 4L
›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 λ, the centre-of-mass position measured from one end is …
- KEAM 2025Set pha-2025-0429F4 marksMCQQ.Three particles of equal mass lie at distances of 1cm, 2cm and 3cm from the origin. The distance of their centre of mass from the origin is (A) 2cm (B) 1cm (C) 2.5cm (D) 3cm (E) 6cm
›Reveal solutionSolution
The centre of mass of equal masses is the arithmetic mean of their positions: (1+2+3)/3=2cm.
For particles of equal mass m at positions xi, the centre of mass is
xcm=3mmx1+mx2+mx3=3x1+x2+x3. …
- KEAM 2023Set eng-2023-P1-A14 marksMCQQ.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. …
- KEAM 2021Set eng-2021-P1-A14 marksMCQQ.The X and Y coordinates of the three particles of masses m, 2m and 3m are respectively (0,0), (1,0) and (−2,0). The X-coordinate of the centre of mass of the system is (A) 31 (B) 32 (C) −31 (D) −32 (E) 61
›Reveal solutionSolution
The x-coordinate of the centre of mass is −32.
Concept and Intuition
The centre of mass along x is the mass-weighted average of the particle positions, xcm=∑mi∑mixi.
Step-by-Step Solution
- Numerator: m(0)+2m(1)+3m(−2)=0+2m−6m=−4m.
- Total mass: m+2m+3m=6m.
- xcm=6m−4m=−32. …
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