Centre Of Mass

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Two point masses of 2 kg and 6 kg are fixed at the two ends of a light rigid rod of length 1 m. At what distance from the 2 kg mass does the centre of mass of the system lie?

RankGuru · Accepted Answer

The centre of mass of a system of point particles is the mass-weighted average of their positions, defined by $x_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$. Because the connecting rod is light, only the two end masses contribute to the calculation. Take the 2 kg mass at the origin and the 6 kg mass at $x = 1	ext{ m}$. Then $x_{cm} = \frac{2(0) + 6(1)}{2 + 6} = \frac{6}{8} = 0.75	ext{ m}$ measured from the 2 kg end. The value 0.25 m is the distance from the heavier 6 kg mass rather than the lighter one, so it answers a different question. The value 0.50 m would only be correct if the two masses were equal, which they are not. The value 0.60 m comes from incorrectly weighting by length rather than by mass. This is the standard NCERT treatment of the centre of mass of a two-particle system. As a sanity check, the centre of mass must always lie closer to the heavier mass, and 0.75 m is indeed nearer the 6 kg end, confirming the result.

Centre Of Mass

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