Q.Find the centre of mass of three particles at the vertices of an equilateral triangle. The masses of the particles are 100 g, 150 g, and 200 g respectively. Each side of the equilateral triangle is 0.5 m long.

The key idea is to use the definition of the Center of Mass for a system of discrete particles.

1. Set up a coordinate system: Place the 100 g particle at the origin $(0,0)$. The 150 g particle is at $(0.5, 0)$. The 200 g particle is at $(0.5/2, 0.5\sqrt{3}/2) = (0.25, 0.25\sqrt{3})$.
2. Calculate the total mass $M = 100 + 150 + 200 = 450$ g.
3. Apply the formula for the x-coordinate of the center of mass: $X_{CM} = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{M} = \frac{(100)(0) + (150)(0.5) + (200)(0.25)}{450} = \frac{0 + 75 + 50}{450} = \frac{125}{450} = \frac{5}{18}$ m.
4. Apply the formula for the y-coordinate of the center of mass: …

The figure shows an equilateral triangle with its three vertices labelled O, A, and B. Vertex O is placed at the origin $(0,0)$ of the $x$-$y$ coordinate plane. Vertex A lies on the $x$-axis at $(0.5,0)$. Vertex B is at $(0.25,\,0.25\sqrt{3})$, which is the third corner of an equilateral triangle of side length $0.5$ units. A median is drawn from vertex B to the opposite side OA. The centre of mass of the three-particle system is marked inside the triangle at the coordinates $\left(\frac{5}{18},\,\frac{1}{3\sqrt{3}}\right)$. The $x$ and $y$ axes are shown, with the origin at O.

The physical idea is straightforward: when you have discrete masses at known positions, the centre of mass is the weighted average of their positions. Each mass pulls the balance point toward itself in proportion to its mass. In this example, three particles of masses $100\ \text{g}$, $150\ \text{g}$, and $200\ \text{g}$ sit at the vertices. The 200 g mass at B is the heaviest, so the centre of mass lies closer to B than to the lighter vertices — notice that the CM coordinates are not at the triangle’s geometric centre (the centroid), but shifted toward B.

The textbook develops the definition of centre of mass for a system of $n$ particles. For the $x$-coordinate:

and similarly for the $y$-coordinate:

Here $m_i$ is the mass of the $i$-th particle and $(x_i, y_i)$ are its coordinates. The denominator is the total mass $M = \sum m_i$.

Applying these formulas to the figure: total mass $M = 100 + 150 + 200 = 450\ \text{g}$. The $x$-coordinate of the CM is

The $y$-coordinate is

The median drawn from B to side OA is a visual aid — it shows that the CM lies along this median, but not at its midpoint. The CM divides the median in the ratio of the masses at the ends, a property that generalises to any system of two particles.

The center of mass of a system of particles is a crucial concept in physics, representing the average position of all the mass in the system. Imagine trying to balance an object on a single point; that point would be its center of mass. For a system of discrete particles, it's essentially a weighted average of their individual positions, where the 'weight' for each particle is its mass. This means that particles with greater mass have a stronger influence on the location of the center of mass.

To find the center of mass, we first need to define a coordinate system. This allows us to assign specific $(x, y)$ coordinates to each particle. The choice of coordinate system does not affect the physical location of the center of mass, but a smart choice can simplify the calculations significantly.

Here's how we can approach this problem:

1. Set up the Coordinate System and Particle Positions:

   Let the side length of the equilateral triangle be $a = 0.5$ m. We'll place one vertex of the triangle at the origin $(0,0)$ to simplify calculations.

   Let the three particles have masses $m_1 = 100$ g, $m_2 = 150$ g, and $m_3 = 200$ g.

   • Particle 1 ($m_1 = 100$ g): Place it at the origin.

     $P_1 = (x_1, y_1) = (0,0)$

   • Particle 2 ($m_2 = 150$ g): Place it along the positive x-axis.

     $P_2 = (x_2, y_2) = (a, 0) = (0.5, 0)$

   • Particle 3 ($m_3 = 200$ g): The third vertex of an equilateral triangle with one vertex at $(0,0)$ and another at $(a,0)$ will be at $(a/2, a\sqrt{3}/2)$.

     $P_3 = (x_3, y_3) = \left(\frac{0.5}{2}, \frac{0.5\sqrt{3}}{2}\right) = (0.25, 0.25\sqrt{3})$

   Tip

   Choosing one vertex at the origin and another on an axis simplifies the coordinates of those two points, reducing the number of non-zero terms in the center of mass calculation. …

Concept: Centre of Mass of a System of Discrete Particles

Step 1: Set up coordinates for the three vertices

Side of the equilateral triangle $a = 0.5$ m. Place the 100 g particle at the origin, the 150 g particle along the x-axis, and use the equilateral-triangle geometry for the third vertex:

Step 2: Write the centre-of-mass formula

Step 3: Total mass

Step 4: Compute $X_{CM}$ …