Question

Three particles of mass 1 kg, 2 kg, and 3 kg are placed at the vertices A, B and C respectively of an equilateral triangle ABC of side 1 m. The centre of mass of the system from vertices A (located at origin) is

• A. \( \left( \frac{7}{12}, 0 \right) \)
• B. \( (0, 0) \)
• C. \( \left( \frac{7}{12}, \frac{3\sqrt{3}}{12} \right) \)
• D. \( \left( \frac{9}{12}, \frac{3\sqrt{3}}{12} \right) \)

Hint:

When calculating the center of mass, always consider the mass-weighted averages of the positions of the objects.

Answer 1

The Correct Option is C

The center of mass \( \text{CM} \) of a system of particles is given by the formula: \[ x_{\text{cm}} = \frac{\sum m_i x_i}{\sum m_i}, \quad y_{\text{cm}} = \frac{\sum m_i y_i}{\sum m_i} \] Where \( m_i \) are the masses of the particles, and \( (x_i, y_i) \) are the coordinates of the particles. Given the coordinates of the three particles placed at vertices A, B, and C of an equilateral triangle of side 1 m, we assign the coordinates of the points: - \( A = (0, 0) \) - \( B = (1, 0) \) - \( C = \left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right) \) Now, we can calculate the center of mass: 1. x-coordinate: \[ x_{\text{cm}} = \frac{1 \times 0 + 2 \times 1 + 3 \times \frac{1}{2}}{1 + 2 + 3} = \frac{0 + 2 + 1.5}{6} = \frac{3.5}{6} = \frac{7}{12} \] 2. y-coordinate: \[ y_{\text{cm}} = \frac{1 \times 0 + 2 \times 0 + 3 \times \frac{\sqrt{3}}{2}}{1 + 2 + 3} = \frac{0 + 0 + \frac{3\sqrt{3}}{2}}{6} = \frac{\frac{3\sqrt{3}}{2}}{6} = \frac{3\sqrt{3}}{12} \] Thus, the center of mass is at \( \left( \frac{7}{12}, \frac{3\sqrt{3}}{12} \right) \).