Four particles of masses $1 kg, 2 kg, 3 kg$ and $4 kg$ are placed at the four vertices $A,B,C$ and $D$ of a square of side $1 m$ as shown in figure. Find the position of centre of mass of the particles.

(0.3, 0.5) m

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(0.1, 0.2) m

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(0.5, 0.3) m

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(0.5, 0.5) m

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Solution

The correct option is C $(0.5, 0.3) m$

As shown in diagram, $D$ is the origin, so assuming $DC$ as $x -$ axis and $DA$ as $y -$ axis.

From such orientation, we get

$m_{1} = 1 kg, (x_{1}, y_{1}) = (0, 1) m$
$m_{2} = 2 kg, (x_{2}, y_{2}) = (1, 1) m$
$m_{3} = 3 kg, (x_{3}, y_{3}) = (1, 0) m$
$m_{4} = 4 kg, (x_{4}, y_{4}) = (0, 0)$

So, the $x -$ coordinates of the COM is given by

$x_{C O M} = \frac{m_{1} x_{1} + m_{2} x_{2} + m_{3} x_{3} + m_{4} x_{4}}{m_{1} + m_{2} + m_{3} + m_{4}}$

$x_{C O M} = \frac{(1) (0) + 2 (1) + 3 (1) + 4 (0)}{1 + 2 + 3 + 4}$

$x_{C O M} = \frac{5}{10} = 0.5 m$

Similarly,

$y_{C O M} = \frac{m_{1} y_{1} + m_{2} y_{2} + m_{3} y_{3} + m_{4} y_{4}}{m_{1} + m_{2} + m_{3} + m_{4}}$

$y_{C O M} = \frac{(1) (1) + 2 (1) + 3 (0) + 4 (0)}{1 + 2 + 3 + 4}$

$y_{C O M} = \frac{3}{10} = 0.3 m$

$∴ (x_{C O M}, y_{C O M}) = (0.5, 0.3) m$

Therefore, option $(C)$ is correct.

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