Three particles of masses 1kg, 2kg and 3 kg are situated at the corners of an equilateral triangle of side b. The coordinates of the centre of mass are :

To find the coordinates of the center of mass for three particles located at the corners of an equilateral triangle, we can follow a systematic approach. The center of mass is essentially the weighted average of the positions of all the particles, taking into account their masses. Let's break this down step by step.

Setting Up the Problem

Assume the vertices of the equilateral triangle are positioned in a coordinate system as follows:

• Particle 1 (1 kg) at coordinates (0, 0)
• Particle 2 (2 kg) at coordinates (b, 0)
• Particle 3 (3 kg) at coordinates (b/2, (√3/2)b)

Finding the Coordinates of the Center of Mass

The formula for the center of mass (CM) of a system of particles is given by:

• X_cm = (Σ(m_i * x_i)) / Σm_i
• Y_cm = (Σ(m_i * y_i)) / Σm_i

Where:

• m_i = mass of each particle
• (x_i, y_i) = coordinates of each particle

Calculating the Components

Now, let's calculate the total mass (M) of the system:

M = 1 kg + 2 kg + 3 kg = 6 kg

Next, we compute the X coordinate of the center of mass:

X_cm = (1 kg * 0 + 2 kg * b + 3 kg * (b/2)) / 6 kg

X_cm = (0 + 2b + (3b/2)) / 6

X_cm = (2b + 1.5b) / 6 = (3.5b) / 6 = (7b) / 12

Now, for the Y coordinate:

Y_cm = (1 kg * 0 + 2 kg * 0 + 3 kg * (√3/2)b) / 6 kg

Y_cm = (0 + 0 + (3√3/2)b) / 6

Y_cm = (3√3/2)b / 6 = (√3/4)b

Final Coordinates of the Center of Mass

Thus, the coordinates of the center of mass for the three particles are:

X_cm = (7b)/12

Y_cm = (√3/4)b

This result shows that the center of mass is located at a position that balances the influence of all three masses in the system, taking into account their respective weights and distances from the origin. This concept is crucial in physics as it helps in analyzing systems of particles in various applications, from engineering to astrophysics.