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 2 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:
The centre of mass of a uniform ring of radius R is located:
Which of the following statements about the centre of mass is correct?
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:
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:
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:
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:
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:
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. ---
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. 2For a particle: L = r x p. For a rigid body about its rotation axis: L = I omega. Vector quantity.
| Symbol | Quantity | SI Unit |
|---|---|---|
| L | angular momentum | kg*m^2/s |
| I | moment of inertia | kg*m^2 |
| omega | angular velocity | rad/s |
The 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 |
Standard moments of inertia about the symmetry axis. For other axes use parallel/perpendicular axes theorems.
| Symbol | Quantity | SI Unit |
|---|---|---|
| M | mass | kg |
| R | radius | m |
| L | length | m |
| I | moment of inertia | kg*m^2 |
Moment of inertia about any axis = moment about parallel axis through CM + Md^2.
| Symbol | Quantity | SI Unit |
|---|---|---|
| I | MOI about given axis | kg*m^2 |
| I_cm | MOI about parallel CM axis | kg*m^2 |
| M | total mass | kg |
| d | perpendicular distance | m |
For planar lamina: MOI about axis perpendicular to plane = sum of MOI about two perpendicular in-plane axes through same point.
| Symbol | Quantity | SI Unit |
|---|---|---|
| I_z | MOI perp to plane | kg*m^2 |
| I_x, I_y | MOI in plane | kg*m^2 |
Rotational analogues of linear kinematic equations under constant angular acceleration.
| Symbol | Quantity | SI Unit |
|---|---|---|
| omega | angular velocity | rad/s |
| alpha | angular acceleration | rad/s^2 |
| theta | angular displacement | rad |
| t | time | s |
Energy of rotation about an axis. Adds to translational KE for rolling bodies.
| Symbol | Quantity | SI Unit |
|---|---|---|
| I | moment of inertia | kg*m^2 |
| omega | angular velocity | rad/s |
Cross product of position vector and force vector. Magnitude r F sin(theta).
| Symbol | Quantity | SI Unit |
|---|---|---|
| tau | torque | N*m |
| r | position from pivot | m |
| F | force | N |
| theta | angle between r and F | rad |
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Category: Overthinking
Student answers L/2 for two-particle CoM regardless of mass ratio.
Question gives two masses on rigid rod and asks for CoM distance.
R_cm from m1 = m2*L/(m1+m2). Heavier mass pulls CoM closer to it.
Category: Similar Terms
Student conserves rotational KE when angular momentum is conserved (or vice versa). When I changes, L = Iω is conserved but KE = ½Iω² is NOT (it depends on I and ω together).
Question describes a body whose moment of inertia changes (skater pulling arms in, star collapsing).
L conservation requires zero external torque. KE conservation requires no work done — different criteria. When I changes via internal forces, L conserved, ω increases, KE increases.
Category: Similar Terms
Student confuses 2/5 (solid sphere) with 2/3 (hollow sphere) or 1/2 (disc) with 1 (ring).
Question gives a specific geometry and asks for I or radius of gyration.
Memorise: solid sphere 2/5, hollow sphere 2/3, disc/cylinder 1/2, ring/hoop 1, rod-centre 1/12, rod-end 1/3.
Category: Unit Conversion
Student plugs rpm directly into formulas requiring rad/s. 1 rpm = 2π/60 rad/s.
Question gives ω in rpm and asks for kinematic quantities in SI units.
Convert: ω(rad/s) = (2π/60) × rpm. Always check units before substituting.
Root cause: concept gap
L = Iω is conserved when external torque is zero. KE = ½Iω² is NOT conserved when I changes (since ω changes too). When skater pulls arms in, L conserved, ω increases, KE increases (work done by muscles).
Root cause: concept gap
Memorise standard moments of inertia. Solid sphere has more mass near axis (smaller MOI = 2MR²/5); hollow sphere has all mass at radius R (larger MOI = 2MR²/3).
Root cause: unit error
Convert: ω(rad/s) = (2π/60) × ω(rpm). For example, 1200 rpm = 1200 × 2π/60 = 125.66 rad/s.
The angular acceleration of a body, moving along the circumference of a circle, is
forgets conversion rpm to rad s
Treats rpm as rad/s without 2*pi/60 conversion
uses equal distribution
Default to L/2 regardless of mass ratio
uses energy conservation instead
Confusing L conservation with KE conservation
swap solid hollow coefficients
Confusing 2/5 (solid sphere) with 2/3 (hollow sphere)
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