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?
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.
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.
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. …
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. …
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2026Set ANNUAL2 marks
Q.A uniform circular plate of radius (R) has a smaller circular portion of radius (R/2) cut out from its edge. Determine the position of the center of mass of the remaining portion of the plate with respect to the center of the original plate.
›Reveal solutionSolution
The center of mass of the remaining portion shifts by R/6 from the center of the original plate, on the side opposite to the cut-out.
Let the original uniform circular plate have radius R, mass M, and surface mass density σ = M/(πR²), centered at the origin O.
A smaller circular portion of radius R/2 is cut out from the edge — meaning its own edge is tangent to the boundary of the original plate, so its center lies at a distance R/2 from O (call this point along the +x axis, at x = R/2).
Mass of the original full plate: m₁ = σπR² = M (at position x = 0).
Mass of the cut-out piece: m₂ = σπ(R/2)² = σπR²/4 = M/4 (at position x = R/2).
The center of mass of the (full plate) can be thought of as: [full plate] = [remaining portion] + [cut-out piece]. Since the full plate's COM is at the origin:
Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2025Set ANNUAL2 marks
Q.Prove that the center of mass of a rod having uniformly distributed mass lies in the middle of the rod.
›Reveal solutionSolution
Integrating xdm over a rod with uniform linear mass density gives xcm=L/2 — the middle of the rod.
Consider a thin uniform rod of total mass M and length L, lying along the x-axis from x=0 to x=L. Because the mass is distributed uniformly, the linear mass density (mass per unit length) is constant:
λ=LM
Consider a small element of the rod of length dx at position x; its mass is dm=λdx.
The x-coordinate of the centre of mass is defined as: