Two objects of mass $10 ext{ kg}$ and $20 ext{ kg}$ respectively are connected to the two ends of a rigid

Question

Two objects of mass $10 	ext{ kg}$ and $20 	ext{ kg}$ respectively are connected to the two ends of a rigid rod of length $10 	ext{ m}$ with negligible mass. The distance of the center of mass of the system from the $10 	ext{ kg}$ mass is:

Accepted Answer

Let the position of the $10 	ext{ kg}$ mass be at the origin, so $x_1 = 0$.
The position of the $20 	ext{ kg}$ mass will be at the other end of the rod, so $x_2 = 10 	ext{ m}$.
The position of the center of mass $x_{cm}$ from the origin (which is the $10 	ext{ kg}$ mass) is given by the formula:
$x_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$
Substituting the values:
$x_{cm} = \frac{10(0) + 20(10)}{10 + 20}$
$x_{cm} = \frac{200}{30} = \frac{20}{3} 	ext{ m}$

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