The centre of mass (CM) of a two-particle system is the mass-weighted average position of the two particles. NCERT Class 11 Physics Chapter 7 (System of Particles and Rotational Motion), page 2, defines it as the point where the entire mass of the system can be considered concentrated for describing translational motion.
The trap that costs marks: When two unequal masses sit on a rod, many aspirants default to placing the CM at the midpoint (L/2). This is correct only when the masses are equal. For unequal masses, the CM shifts toward the heavier particle.
The formula. For two particles of masses m₁ and m₂ separated by distance L, with the origin at m₁:
x_cm = (m₁ × 0 + m₂ × L) / (m₁ + m₂) = m₂L / (m₁ + m₂)
Equivalently, the distance of the CM from m₁ is d₁ = m₂L/(m₁ + m₂), and from m₂ is d₂ = m₁L/(m₁ + m₂). Notice the cross-relationship: the distance from each mass is proportional to the other mass.
Key property: m₁d₁ = m₂d₂. The CM is the balance point — the torques about it are equal and opposite.
NEET context. This topic appears as a direct-application question (estimated ~0.4 questions per year, medium weight in the chapter). The standard format: two masses on a massless rod, find distance of CM from one end. The distractor that catches aspirants is L/2 — the equal-distribution default.
Watch out: Always check which end the question asks the distance from. "Distance of CM from the heavier mass" gives a smaller number than "distance from the lighter mass." Misreading which mass the answer refers to is an easy −1.
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 system of particles is defined as the point where:
For a two-particle system with masses m₁ and m₂ separated by distance L, the position of the centre of mass from m₁ is:
In a two-particle system, the distances d₁ and d₂ of the centre of mass from masses m₁ and m₂ respectively satisfy:
Two particles of masses 3 kg and 5 kg are placed at the ends of a 2.0 m long massless rod. The distance of the centre of mass from the 3 kg mass is:
Two particles of masses 2 kg and 6 kg are separated by 4.0 m. How far is the centre of mass from the 6 kg particle?
Two particles of masses m and 3m are joined by a rigid massless rod of length L. The centre of mass divides the rod in the ratio (from mass m to mass 3m):
Two particles of masses 4 kg and 8 kg are placed at coordinates (1.0, 0) m and (4.0, 0) m respectively on the x-axis. The x-coordinate of the centre of mass is:
Two particles of masses 1.0 kg and 4.0 kg are separated by 1.00 m. If the 1.0 kg mass is moved 0.50 m closer to the 4.0 kg mass (separation becomes 0.50 m), how far does the centre of mass shift?
Given
- m₁ = 2.0 kg - m₂ = 3.0 kg - L = 0.50 m (separation)
Required
Distance of centre of mass from m₁ (the 2.0 kg particle).
Concept
The centre of mass of a two-particle system lies at the mass-weighted average position. Placing the origin at m₁, the CM position equals m₂L/(m₁ + m₂). The CM is closer to the heavier particle (NCERT Class 11 Physics Chapter 7, page 2).
Formula
d₁ = m₂ × L / (m₁ + m₂)
Substitution
d₁ = 3.0 × 0.50 / (2.0 + 3.0)
Calculation
d₁ = 1.50 / 5.0 = 0.30 m
Final answer
The centre of mass is **0.30 m** from the 2.0 kg particle (and 0.20 m from the 3.0 kg particle). Note on significant figures: the masses and length are each given to 2 significant figures. The integers in the formula (the addition and division) are exact counting operations and do not limit significant figures. The answer 0.30 m retains 2 significant figures.
Common trap
Answering 0.25 m (= L/2) by assuming equal distribution. The mass ratio is 2 : 3, so the CM is not at the midpoint. Always apply the formula — the midpoint shortcut works only when masses are equal.
Similar NEET-style question
"A system consists of two point masses, 5.0 kg and 15.0 kg, separated by 1.00 m. Find the distance of the centre of mass from the 15.0 kg mass." [Answer: d₂ = 5.0 × 1.00/(5.0 + 15.0) = 0.25 m. Note: from the *heavier* mass, so the answer is the smaller distance.] ---
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)
Test yourself on this topic with real past-paper questions:
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