The centre of mass (CM) of a rigid body is the unique point where the entire mass of the body can be considered concentrated for analysing its translational motion. NCERT Class 11 Physics Chapter 6, page 95 defines it as the mass-weighted average position of all constituent particles.
The formula. For a system of n particles:
R_cm = (Σ m_i r_i) / (Σ m_i)
For a continuous rigid body, the sum becomes an integral: R_cm = (1/M) ∫ r dm, where M is the total mass.
The trap that costs marks. A common confusion is defaulting to the geometric centre — assuming the CM is always at the midpoint of the body. This is true ONLY for uniform, symmetric bodies. When the mass distribution is non-uniform, or when a rigid body is composed of parts with different densities, the CM shifts toward the heavier region. For two particles of unequal mass on a rod, the CM sits closer to the heavier particle: its distance from mass m₁ is m₂L/(m₁ + m₂), not L/2.
Key ideas for NEET.
Watch out: when a problem says "uniform rigid body," that guarantees the geometric centre IS the CM. If "uniform" is absent, check the mass distribution before assuming symmetry.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
The centre of mass of a uniform solid sphere lies:
Answer: B. For a uniform symmetric body, the centre of mass coincides with the geometric centre by symmetry (NCERT Class 11 Physics Chapter 6, page 95).
Why A is wrong: A is wrong because the CM of a solid sphere is not on the surface — the symmetric mass distribution places it at the interior geometric centre.
Why C is wrong: C is wrong because the CM lies outside the body only for certain shapes like a ring or hollow hemisphere, not a solid sphere.
Why D is wrong: D is wrong because there is no R/2 offset for a uniform solid sphere — the mass-weighted average position is exactly the geometric centre.
The centre of mass of a uniform ring of radius R is located:
Answer: D. By symmetry, the CM of a uniform ring is at its geometric centre — a point where no material exists (NCERT Class 11 Physics Chapter 6, page 95).
Why A is wrong: A is wrong because the CM is not at any particular point on the ring — the symmetric mass distribution averages to the geometric centre.
Why B is wrong: B is wrong because the CM lies in the plane of the ring, not along the axis at a distance R.
Why C is wrong: C is wrong because the symmetry of the ring places the CM exactly at the centre, not at a radial offset of R/2.
Which of the following statements about the centre of mass is correct?
Answer: C. By definition, the CM is the mass-weighted average position that represents the body for translational dynamics. External forces act as if applied at the CM (NCERT Class 11 Physics Chapter 6, page 95).
Why A is wrong: A is wrong because the CM can lie outside the body — for example, a uniform ring has its CM at the centre where no material is present.
Why B is wrong: B is wrong because the CM coincides with the geometric centre only for uniform, symmetric bodies. Non-uniform mass distributions shift the CM.
Why D is wrong: D is wrong because the position of the CM relative to the body is independent of coordinate origin — choosing a different origin shifts all coordinates equally, leaving the mass-weighted average unchanged relative to the body.
Two particles of masses 2 kg and 3 kg are placed at the ends of a 1.0 m long massless rod. The distance of the centre of mass from the 2 kg mass is:
Answer: A. Distance from m₁ = m₂ × L / (m₁ + m₂) = 3 × 1.0 / (2 + 3) = 0.60 m. The CM is closer to the heavier 3 kg mass (NCERT Class 11 Physics Chapter 6, page 95).
Why B is wrong: B is wrong because 0.40 m = m₁L/(m₁ + m₂) gives the distance from the 3 kg mass, not from the 2 kg mass. This reversal is a common sign error.
Why C is wrong: C is wrong because 0.50 m assumes equal masses (L/2) — this is the trap of defaulting to the midpoint regardless of mass ratio (trap: equal-distance default).
Why D is wrong: D is wrong because 0.67 m = 2/3 of L, which does not match the correct mass-weighted formula. This likely results from an incorrect ratio.
Three particles of equal mass m are placed at the vertices of an equilateral triangle of side a. The centre of mass is located at:
Answer: D. For equal masses at the vertices of any triangle, R_cm = (m·r₁ + m·r₂ + m·r₃)/(3m) = (r₁ + r₂ + r₃)/3, which is the centroid (NCERT Class 11 Physics Chapter 6, page 95).
Why A is wrong: A is wrong because placing three equal masses symmetrically means no single vertex is preferred — the CM is the average of all three positions, which is the centroid.
Why B is wrong: B is wrong because the midpoint of a side is equidistant from only two of the three vertices. The third mass pulls the CM away from any side's midpoint to the centroid.
Why C is wrong: C is wrong because for any set of masses placed at the vertices of a triangle, the CM lies inside (or on the boundary of) the triangle, never outside.
A uniform circular disc of mass M and radius R has a smaller disc of radius R/2 removed from it such that the smaller disc's edge touches the centre of the larger disc. The centre of mass of the remaining portion is at a distance from the centre of the original disc equal to:
Answer: A. The removed disc has mass M/4 (mass ∝ area; area ratio = (R/2)² / R² = 1/4) and its centre is at R/2 from the original centre. Using the negative-mass method: x_cm = (M·0 − (M/4)·(R/2)) / (M − M/4) = (−MR/8) / (3M/4) = −R/6. The magnitude is R/6, directed opposite to the removed portion.
Why B is wrong: B is wrong because R/2 is the distance of the removed disc's centre from the original centre — not the CM shift of the remaining body. Using R/2 as the answer skips the mass-weighting calculation entirely.
Why C is wrong: C is wrong because R/3 uses the correct mass fraction 1/4 but places the removed disc's centre at R instead of R/2: (M/4)·R/(3M/4) = R/3.
Why D is wrong: D is wrong because R/4 comes from placing the removed disc's centre at R instead of R/2 and dividing by M: (M/4)·R / M = R/4. The removed disc's centre is at R/2, and the remaining mass is 3M/4.
Particles of masses 1 kg, 2 kg, and 3 kg are placed at positions (0, 0), (1, 0), and (0, 2) respectively (coordinates in metres). The position of the centre of mass is:
Answer: B. x_cm = (1×0 + 2×1 + 3×0)/(1+2+3) = 2/6 = 1/3 m. y_cm = (1×0 + 2×0 + 3×2)/(6) = 6/6 = 1 m. So the CM is at (1/3, 1) m.
Why A is wrong: A is wrong because x_cm = 2/3 would require the 2 kg mass to contribute 4/6 to the x-coordinate, which means an arithmetic error — the correct contribution is 2×1/6 = 2/6 = 1/3.
Why C is wrong: C is wrong because x = 1/2 would need the x-moment 2 × 1 = 2 kg·m divided by 4 kg, but the total mass is 6 kg, so x_cm = 2/6 = 1/3 (trap: equal-distance default).
Why D is wrong: D is wrong because (1, 1/3) swaps the x and y coordinates of the CM — a common bookkeeping error when handling 2D problems.
A uniform L-shaped lamina is made of two identical thin rectangular strips, each of mass m and length L, joined perpendicularly at one end. Taking the junction as the origin with one strip along the x-axis and the other along the y-axis, the coordinates of the centre of mass of the L-shaped lamina are:
Answer: C. The CM of the horizontal strip is at (L/2, 0) and the CM of the vertical strip is at (0, L/2). Both have mass m. x_cm = (m·L/2 + m·0)/(2m) = L/4. y_cm = (m·0 + m·L/2)/(2m) = L/4. So CM = (L/4, L/4).
Why A is wrong: A is wrong because (L/2, L/2) treats each strip's CM as if it alone determines the system's CM. But two equal-mass strips at (L/2, 0) and (0, L/2) average to (L/4, L/4), not (L/2, L/2) (trap: equal-distance default — forgetting to average).
Why B is wrong: B is wrong because (L/4, L/2) breaks the symmetry of two identical strips — the x and y components must be equal by the identical-mass and symmetric-geometry argument.
Why D is wrong: D is wrong because (L/2, L/4) also breaks the symmetry. If the strips are identical, x_cm must equal y_cm.
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Pattern: Two-particle CM on a rigid rod (PYQ 2022 pattern).
Given
Two particles: m₁ = 4.0 kg at position x₁ = 0, and m₂ = 6.0 kg at position x₂ = 2.0 m, connected by a massless rigid rod.
Required
Position of the centre of mass of the system.
Concept
The centre of mass is the mass-weighted average position. For a two-particle system, the CM lies along the line joining them, closer to the heavier particle.
Formula
x_cm = (m₁ x₁ + m₂ x₂) / (m₁ + m₂)
Substitution
x_cm = (4.0 × 0 + 6.0 × 2.0) / (4.0 + 6.0) = (0 + 12.0) / 10.0
Calculation
x_cm = 12.0 / 10.0 = 1.2 m
Final answer
The centre of mass is at x = 1.2 m from the 4.0 kg mass, i.e., 0.8 m from the 6.0 kg mass — closer to the heavier particle as expected.
Common trap
Answering 1.0 m (the midpoint L/2 = 2.0/2 = 1.0 m) by assuming equal mass weighting. The midpoint answer ignores the 4:6 mass ratio entirely.
Similar NEET-style question
Two atoms in a diatomic molecule have masses 14 u and 16 u, separated by 1.2 × 10⁻¹⁰ m. Find the distance of the CM from the heavier atom.
Answer sketch: Distance from m₂ (16 u) = m₁ L / (m₁ + m₂) = 14 × 1.2 × 10⁻¹⁰ / 30 = 5.6 × 10⁻¹¹ m.
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The centre of mass of a system of particles is the unique point where the entire mass of the system may be considered to be concentrated for purposes of describing translational motion. For a system of n particles: R_cm = (Σ m_i r_i) / (Σ m_i).
-- NCERT Class 11 Physics, Ch. 6, p. 96The position of the centre of mass equals the mass-weighted average of particle positions. For continuous bodies use integral form.
| Symbol | Quantity | SI Unit |
|---|---|---|
| R_cm | CoM position | m |
| m_i | mass of i-th particle | kg |
| r_i | position of i-th particle | m |
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