Centre of Mass of Rod System – Rankers Physics

Calculation of Center of Mass: Practice Problem & Solution

Two objects of mass $10\text{ kg}$ and $20\text{ kg}$ respectively are connected to the two ends of a rigid rod of length $10\text{ m}$ with negligible mass. The distance of the centre of mass of the system from the $10\text{ kg}$ mass is : (2022)
$5\text{ m}$
$\frac{10}{3}\text{ m}$
$\frac{20}{3}\text{ m}$
$10\text{ m}$

Solution Explained:

To solve this problem, we apply the core principles of Calculation of Center of Mass. Understanding the underlying formula is key to arriving at the correct answer below:

Centre of mass formula is $x_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$. Taking $10\text{ kg}$ at origin and $20\text{ kg}$ at $10\text{ m}$, we get $x_{cm} = \frac{10(0) + 20(10)}{10+20} = \frac{20}{3}\text{ m}$. Option (c) is correct.

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