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Mathematics

Distance between Two Points on the Same Coordinate Axes

Four particle...

Question

Four particles of masses $1 k g$ , $2 k g,$ $3 k g$ and $4 k g$ are placed at the four vertices $A$ , $B$ , $C$ and $D$ , respectively of a square of side $1 m$ . Find the position of centre of mass of the particles.

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Solution

Assuming $D$ as the origin. $D C$ as x-axis and $D A$ as y-axis we have
$m_{1} = 1 k g, (x_{1}, y_{1}) = (0, 1 m)$
$m_{2} = 2 k g, (x_{2}, y_{2}) = (1 m, 1 m)$
$m_{3} = 3 k g, (x_{3}, y_{3}) = (1 m, 0)$
$m_{4} = 4 k g, (x_{4}, y_{4}) = (0, 0)$
Coordination of their $C M$ are calculated as follows.
$x_{C M} = \frac{m_{1} x_{1} + m_{2} x_{2} + m_{3} x_{3} + m_{4} x_{4}}{m_{1} + m_{2} + m_{3} + m_{4}}$
$= \frac{(1) (0) + 2 (1) + 3 (1) + 4 (0)}{1 + 2 + 3 + 4}$
$\frac{5}{10} = \frac{1}{2} m = 0.5 m$
Similarly, $y_{C M} = \frac{m_{1} y_{1} + m_{2} y_{2} + m_{3} y_{3} + m_{4} y_{4}}{m_{1} + m_{2} + m_{3} + m_{4}}$
$= \frac{(1) (1) + 2 (1) + 3 (0) + 4 (0)}{1 + 2 + 3 + 4}$
$= \frac{3}{10} m = 0.3 m$
$∴ (x_{C M}, y C M) = (0.5 m, 0.3 m)$

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