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Most textbooks use "centre of mass" and "centre of gravity" almost interchangeably, and in most problems that is fine. But they are different ideas, and examiners sometimes ask exactly when they part company. This post gives the two definitions, shows why they normally sit at the same point, works an example where they do not, and lists the confusions that cost marks.
What is the difference between the two?
The NCERT chapter on systems of particles and rotational motion draws the line clearly:
The centre of mass has nothing to do with gravity. It depends only on the distribution of mass in the body.
The centre of gravity (CG) is the point about which the total gravitational torque on the body is zero. Support the body there and it balances.
The CBSE Physics curriculum for 2026-27 covers the centre of mass of a rigid body and the equilibrium of rigid bodies in Class 11, and the JEE (Main) 2026 syllabus includes the centre of mass of a rigid body. The CG section is part of the same NCERT chapter.
Why do they usually coincide?
NCERT's argument takes three lines. Measure positions rᵢ from the CG. Zero gravitational torque means Σ rᵢ × mᵢg = 0. If g is the same for every particle, it comes out of the sum, leaving Σ mᵢrᵢ = 0. That is exactly the condition for the origin to be the centre of mass. So in uniform gravity, the CG and the centre of mass are the same point.
The textbook adds the other half: if the body is so extended that g varies from part to part, the two will not coincide.
Worked example 1: a loaded metre rule
A uniform 1 m rule of mass 200 g has a 100 g mass fixed at its 80 cm mark (sample numbers). Where does it balance?
Centre of mass: (200 × 50 + 100 × 80) / (200 + 100) = (10,000 + 8,000) / 300 = 60 cm.
Over a 1 m rule, g does not change measurably, so the CG is also at 60 cm and a knife edge there balances it. This is how the metre-scale and principle-of-moments activities work.
You can find the CG of a flat irregular shape the same way NCERT describes: hang it from one point and mark the vertical through that point, then hang it from another. The lines cross at the CG. For a small body, that is also its centre of mass.
Worked example 2: when does g change enough to matter?
Take a uniform vertical tower of height L = 1 km standing on the ground (a hypothetical, made-up case). Its centre of mass is at 500 m.
g falls with height. The NCERT chapter on gravitation gives, for heights much smaller than Earth's radius, g(h) ≈ g(1 − 2h/R), with R = 6400 km.
Each slice of the tower is pulled a little less the higher it is, so the point where the weight effectively acts sits slightly below the middle. Weighting each height by the local g and doing the integral to first order gives:
Shift = L²/(6R) = (1000 m)² / (6 × 6.4 × 10⁶ m) = 0.026 m
So the CG is about 2.6 cm below the centre of mass. A full numerical calculation with the exact 1/(R + h)² law gives the same 2.6 cm. For a 100 m building the same formula gives about 0.26 mm, and for anything on a lab bench the difference is far too small to measure.
Where else do they differ?
Gravity's direction also changes over large distances, because it always points towards Earth's centre. For an object spread over a large distance, the pulls on its parts are not parallel. Then there is in general no single point that works as a CG for every orientation, while the centre of mass is still perfectly well defined.
Common mistakes and how to avoid them
"The CG is always at the geometric centre." Only for a uniform, symmetric body. Example 1 has its CG at 60 cm, not 50 cm.
"The centre of mass must lie inside the body." For a ring or a horseshoe it lies in empty space.
Mixing the definitions. Centre of mass is about mass alone; CG is about where gravity's torque balances. If a question puts the body in space with no gravity, only the centre of mass exists.
Toppling questions. A body on a base tips when the vertical line through its CG passes outside the base. This uses the CG, which for everyday objects you can find as the centre of mass.
For more practice finding centres of mass and using them to simplify motion, see our post on centre of mass problems. For balance with several forces and torques, see torque and equilibrium problems with ladders.
A practice routine
Write both definitions from memory, then NCERT's three-line argument for why they coincide.
Rework example 1 with the 100 g mass at 30 cm, and predict the balance point before calculating.
Recompute the tower shift for L = 500 m and check that it is a quarter of the 1 km value.
Find the CG of a cardboard cut-out at home by the hanging method.
If torque and equilibrium still feel shaky, the Fermi Concept Library covers every concept in physics, chemistry, biology, mathematics and computing.
Try step 4 today: cut out an irregular shape, hang it from two corners, and balance it on a pencil tip where the lines cross.
