Question:

Centre of mass of the given system of particles will be at

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Center of mass shifts towards the heavier mass. Use symmetry and CM formulas when masses differ.
Updated On: Apr 23, 2025
  • OA
  • OB
  • OC
  • OD
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The Correct Option is B

Solution and Explanation

Let’s take the square of side \( a \), with O at the center. Coordinates: - A(0, a), mass = \(2m\) - B(a, a), mass = \(4m\) - C(a, 0), mass = \(2m\) - D(0, 0), mass = \(2m\) Using the center of mass formula: \[ x_{cm} = \frac{\sum m_i x_i}{\sum m_i} = \frac{2(0) + 4(a) + 2(a) + 2(0)}{2 + 4 + 2 + 2} = \frac{6a}{10} = 0.6a \] \[ y_{cm} = \frac{2(a) + 4(a) + 2(0) + 2(0)}{10} = \frac{6a}{10} = 0.6a \] So, the center of mass lies along OB, as it lies closer to the heavier mass (4m at B).
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