Lesson 6 – Centre of Mass & System of Particles
Stage 2 : Mathematical Definition of Centre of Mass
(Intermediate Physics – Page 6)

In this page, we extend the concept of centre of mass to continuous mass distributions such as rods and laminae. These derivations are important for Intermediate board exams and form the foundation for higher-level problems.

Centre of Mass of a Uniform Rod

Consider a uniform rod of length L and mass M placed along the x-axis. Since the rod is uniform, its mass is evenly distributed along its length.

Let a small element of length dx be taken at a distance x from the origin. The mass of this element is proportional to its length.

Mass per unit length of the rod is:
λ = M / L

Mass of the small element:
dm = λ dx = (M / L) dx

Derivation of Centre of Mass of a Uniform Rod

The position of centre of mass is given by:

x_cm = ( ∫ x dm ) / ( ∫ dm )

Substituting dm = (M / L) dx:

x_cm = ( ∫₀ᴸ x (M/L) dx ) / M

x_cm = (M/L) × ( ∫₀ᴸ x dx ) / M

x_cm = (1/L) × [ x² / 2 ]₀ᴸ = L / 2

Thus, the centre of mass of a uniform rod lies at its midpoint.

Centre of Mass of a Uniform Lamina (Concept)

A lamina is a thin flat sheet with uniform thickness. The centre of mass of a lamina depends on its shape and mass distribution.

For a uniform rectangular lamina, the centre of mass lies at the point where the diagonals intersect.

For a triangular lamina, the centre of mass lies at the intersection of medians.

Centre of Mass of a Triangular Lamina

In a uniform triangular lamina, the centre of mass is located at a distance of 2/3 of the length of each median from the corresponding vertex.

This result is important for direct questions in Intermediate examinations.

Numerical Example (Uniform Rod)

Example:
A uniform rod of length 2 m has mass 4 kg. Find the position of its centre of mass from one end.

Solution:
For a uniform rod, centre of mass lies at midpoint.

x_cm = L / 2 = 2 / 2 = 1 m

Why Integration Is Needed

In continuous bodies, mass is distributed continuously. Therefore, summation used for discrete particles is replaced by integration.

This idea becomes very important in higher classes and competitive examinations.

Intermediate Exam Tips

• Write formula clearly before derivation
• Define symbols properly
• Use neat diagrams for laminae
• Show limits of integration clearly

Stage 2 – Final Recap

✔ COM of continuous bodies uses integration
✔ Uniform rod COM lies at midpoint
✔ Lamina COM depends on symmetry
✔ Integration replaces summation for continuous mass

Transition to Next Stage

In the next stage, we will study the motion of the centre of mass and its relation with external forces, which is extremely important for both Intermediate and IIT-JEE.