Centre of Mass of Rod System – Rankers Physics
Topic: Center of Mass , Momentum and Collision
Subtopic: Calculation of Center of Mass

Centre of Mass of Rod System

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:

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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