Does the centre of mass of a body necessarily lie within the material of the body?

No. The centre of mass is a mathematical point and may lie in empty space outside the material, as seen in a uniform ring, hollow cylinder, or hollow sphere where the CM lies at the geometric center where no matter exists.

Can internal forces change the state of motion of the centre of mass?

No. By Newton's third law, internal forces always cancel in action-reaction pairs (sum of F_int = 0). Only net external forces (F_ext) can produce acceleration of the centre of mass (A_cm = F_ext / M). For example, when an artillery shell explodes in mid-air, the trajectory of its centre of mass remains an unaltered parabola until fragments hit the ground.

What is the centre of mass location for a uniform semicircular wire and a uniform semicircular disc?

For a uniform semicircular wire (ring) of radius R, the CM is located at y = 2R/pi along the axis of symmetry. For a uniform semicircular disc of radius R, the CM is located at y = 4R/(3pi) from the center.

§ 6.1NCERT Class 11 · Physics · Chapter 6

Centre of Mass & Motion of Centre of Mass

QCC Notes · Class 11 Physics NCERT-aligned Updated 26 Sep 2026 11 min read

A real physical object is not a solitary mathematical point; it is an extended assembly of billions of interacting particles. The Centre of Mass (CM) is the unique point at which the entire mass of the system may be assumed to be concentrated and where all external forces may be considered to act for the analysis of translational motion.

1. Mathematical Formulation of Centre of Mass

A. Two-Particle Discrete System

Consider two point masses m₁ and m₂ situated along the x-axis at coordinates x₁ and x₂. The position of their centre of mass X_cm is given by:

X_cm = (m₁x₁ + m₂x₂) / (m₁ + m₂)

If the origin is placed at mass m₁ (so x₁ = 0 and x₂ = d), the distance of the CM from m₁ is:

r₁ = [m₂ / (m₁ + m₂)] d and r₂ = [m₁ / (m₁ + m₂)] d

This reveals the fundamental inverse mass law: m₁r₁ = m₂r₂. The centre of mass divides the line joining two particles in the inverse ratio of their masses, lying closer to the heavier particle.

B. System of N Particles in 3D Space

For n particles with masses m₁, m₂, ..., m_n and position vectors r₁, r₂, ..., r_n:

R_cm = (∑ m_i r_i) / M = (1/M) ∑ m_i r_i where M = ∑ m_i

In Cartesian components:

X_cm = (m₁x₁ + m₂x₂ + ... + m_nx_n) / M
Y_cm = (m₁y₁ + m₂y₂ + ... + m_ny_n) / M
Z_cm = (m₁z₁ + m₂z₂ + ... + m_nz_n) / M

2. Continuous Mass Distributions & Standard Geometries

For an extended rigid body with continuous mass distribution, the summation is replaced by integration over infinitesimal mass elements dm:

X_cm = (1/M) ∫ x dm, Y_cm = (1/M) ∫ y dm, Z_cm = (1/M) ∫ z dm

Standard Symmetrical Body	Mass Distribution / Element	Location of Centre of Mass (from base or center)
Uniform Straight Rod (Length L)	Linear density λ = M/L	x_cm = L / 2 (midpoint)
Semicircular Wire (Ring) of Radius R	Linear wire along upper half	y_cm = 2R / π ≈ 0.637 R
Semicircular Disc of Radius R	Surface density σ = 2M/(πR²)	y_cm = 4R / (3π) ≈ 0.424 R
Hemispherical Hollow Shell of Radius R	Hollow surface	y_cm = R / 2 = 0.5 R
Solid Hemisphere of Radius R	Uniform volume density	y_cm = 3R / 8 = 0.375 R
Solid Uniform Right Circular Cone (Height h)	Volume integration from base	y_cm = h / 4 (from flat base)
Hollow Right Circular Cone (Height h)	Surface integration from base	y_cm = h / 3 (from flat base)

Negative Mass Technique for Cavity Problems

When a portion of mass m_removed is scooped out from an original symmetrical body of mass M_total:

X_rem = (M_total X_orig − m_removed x_cavity) / (M_total − m_removed)

This elegant subtractive principle converts asymmetric cavity configurations into standard two-body problems with negative mass.

3. Kinematics and Dynamics of Centre of Mass

A. Velocity and Acceleration of Centre of Mass

Differentiating the position vector R_cm with respect to time:

V_cm = dR_cm/dt = (∑ m_i v_i) / M = P_total / M
A_cm = dV_cm/dt = (∑ m_i a_i) / M = (∑ F_i) / M

B. Newton's Second Law for a System of Particles

Forces acting on particles consist of internal forces (mutual interactions) and external forces (environment):

∑ F_i = ∑ F_ext + ∑ F_int

Because internal forces obey Newton's Third Law (equal and opposite in collinear pairs, F_ij = −F_ji), the vector sum of all internal forces is identically zero: ∑ F_int = 0.

F_ext = M A_cm = dP_total / dt

The centre of mass moves as if the entire mass of the system were concentrated at that point and all external forces were applied directly to it.

JEE TRAP: Internal forces (springs between blocks, chemical explosions, muscular contractions, gravitational attraction between binary stars) can alter the velocities and kinetic energies of individual particles, but they can NEVER accelerate or deflect the Centre of Mass!

C. Conservation of Linear Momentum of a System

When the net external force on a system is zero (F_ext = 0):

A_cm = 0 ⇒ V_cm = constant ⇒ P_total = constant

If the centre of mass was initially at rest (V_cm = 0), it remains permanently at rest in the inertial frame:

ΔR_cm = (∑ m_i Δr_i) / M = 0 ⇒ ∑ m_i Δr_i = 0

This yields the famous zero-external-force displacement relation: m₁ Δx₁ + m₂ Δx₂ = 0 (e.g., person walking on a floating frictionless boat).

4. Visualising Centre of Mass Shift and Conservation

Figure 6.1 — Projectile Mid-Air Explosion vs Center of Mass Trajectory

                  Height
                    ^              [Explosion occurs here!]
                    |                     * (Fragments fly in all directions)
                    |                    /                     |                   /                       |                  /     \  <-- Real pieces scatter
                    |                 /   .                       |                /  .   .                      |               / .       .                     |              /.           .\  <-- Parabola of CM is PRESERVED!
                    +-------------+---------------+-------------------------> Distance (x)
                                Launch         Unbroken Parabola Land
      

Exam Read-out: Gravity is the only external force acting before and after the explosion. Because the internal blast force generates zero net force on the system, the Centre of Mass continues along the identical parabolic trajectory as if no explosion had occurred.

Step-by-Step Solved Numerical Examples

Solved Example 1: Circular Disc with Circular Hole (Cavity Problem)

Problem: From a uniform circular disc of radius R and center at origin O(0,0), a smaller circular disc of radius r = R/2 is removed. The center of the hole lies at (R/2, 0). Find the coordinates of the centre of mass of the remaining portion.

Centre of Mass: (−R/6, 0)

Step 1: Determine Mass Ratio:
Let σ be the uniform mass surface density.
Mass of full original disc: M₁ = σ · πR²
Mass of removed portion: M₂ = σ · π(R/2)² = σ · πR²/4 = M₁/4
Remaining mass: M_rem = M₁ − M₂ = 3M₁/4.

Step 2: Apply the Negative Mass Formula:
Center of original disc: x₁ = 0
Center of removed cavity: x₂ = R/2
X_cm = (M₁x₁ − M₂x₂) / (M₁ − M₂)
X_cm = [M₁(0) − (M₁/4)(R/2)] / [3M₁/4] = [−M₁ R / 8] / [3M₁/4] = −(R/8) × (4/3) = −R/6.
By symmetry across the x-axis, Y_cm = 0. Therefore, the new CM is at (−R/6, 0).

Solved Example 2: Man Walking on a Frictionless Boat

Problem: A man of mass m = 50 kg stands at one end of a boat of mass M = 150 kg and length L = 4 m floating stationary on still, frictionless water. The man walks to the other end of the boat. Find the displacement of the boat relative to the water.

Displacement of Boat: 1.0 m (in opposite direction)

Step 1: Check External Forces:
Horizontal external force F_ext,x = 0 (water is frictionless). Thus, the centre of mass of the (man + boat) system has zero horizontal displacement: ΔX_cm = 0.

Step 2: Set up Relative Displacements:
Let the boat shift backward by distance x relative to the shore.
Displacement of boat relative to water: Δx_boat = −x
Displacement of man relative to boat: +L
Displacement of man relative to water: Δx_man = L − x

Step 3: Apply Center of Mass Conservation:
m · Δx_man + M · Δx_boat = 0
50(4 − x) + 150(−x) = 0
200 − 50x − 150x = 0 ⇒ 200x = 200 ⇒ x = 1.0 m.
The boat moves 1.0 m backward relative to the still water.

5. Solved NEET & JEE Practice Questions

Q1. Three point masses 1 kg, 2 kg, and 3 kg are placed at the vertices of an equilateral triangle of side 1 m. If the 1 kg mass is at the origin (0,0) and the 2 kg mass is at (1, 0), the coordinates of the centre of mass are:

A. (7/12, √3/4)
B. (5/12, √3/2)
C. (1/2, √3/6)
D. (7/6, √3/2)

Answer: A

Explanation: Coordinates of the vertices: (0,0) for 1 kg, (1,0) for 2 kg, and (0.5, √3/2) for 3 kg.
Total mass M = 1 + 2 + 3 = 6 kg.
X_cm = [1(0) + 2(1) + 3(0.5)] / 6 = (2 + 1.5)/6 = 3.5/6 = 7/12 m.
Y_cm = [1(0) + 2(0) + 3(√3/2)] / 6 = (3√3 / 2) / 6 = √3/4 m.

Q2. A bomb at rest explodes into three fragments of masses in the ratio 1 : 1 : 2. The two equal fragments fly off mutually perpendicular to each other with a speed of 30 m/s each. What is the velocity of the third, heavier fragment?

A. 15 m/s
B. 15√2 m/s
C. 30√2 m/s
D. 45 m/s

Answer: B

Explanation: Let masses be m, m, and 2m. Initial momentum P_i = 0.
First fragment: p₁ = m(30 î)
Second fragment: p₂ = m(30 ĵ)
Resultant momentum of the two equal pieces: p₁₂ = 30m(î + ĵ), magnitude = √(30² + 30²) m = 30√2 m.
By conservation of linear momentum: p₃ = −p₁₂ ⇒ (2m) v₃ = 30√2 m ⇒ v₃ = (30√2) / 2 = 15√2 m/s in the direction opposite to the resultant.

Q3. The distance of the centre of mass of a uniform solid hemisphere of radius R from its flat planar base is:

A. R / 2
B. 2R / π
C. 3R / 8
D. 4R / (3π)

Answer: C

Explanation: By standard volume integration (dm = ρ π (R² − y²) dy), the centre of mass of a uniform solid hemisphere lies at y_cm = 3R / 8 from the flat circular base along the axis of symmetry. For a hemispherical shell, it is R/2.

Q4. A shell fired from a cannon follows a parabolic path. In flight, it explodes into fragments. The centre of mass of all the fragments combined will:

A. Move vertically downwards
B. Continue along the original parabolic path
C. Move horizontally forward
D. Come instantaneously to rest

Answer: B

Explanation: The explosive forces are purely internal forces. The only external force acting on the system is gravity (F_ext = M g downward). Since F_ext is unchanged, the acceleration of the centre of mass remains a_cm = g downward, and its centre of mass follows the original parabolic trajectory until fragments strike the ground.

6. Frequently Asked Questions (FAQs)

Why does the centre of mass depend only on mass distribution and not on the choice of coordinate axes?

The centre of mass is an intrinsic physical point fixed relative to the body's physical geometry. While the numerical values of its (x, y, z) coordinates change when the coordinate origin is shifted or rotated, the physical location of the point in space relative to the particles remains completely invariant.

When does the Centre of Mass (CM) differ from the Centre of Gravity (CG)?

The Centre of Mass (CM) is where the mass distribution is balanced (∑ m_i r_i = 0), whereas the Centre of Gravity (CG) is the point where the net gravitational torque vanishes (∑ m_i g_i × r_i = 0). For ordinary-sized objects on Earth, gravitational field g is uniform, so CM and CG coincide exactly. For exceptionally massive or tall structures (like a mountain or satellite orbiting near Earth where g decreases with altitude), the gravitational field varies across the body, causing CG to lie slightly lower than CM.

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