Four particles of masses m, m, 2m and 2m are placed at the four corners of a square of side a as shown in the figure. The (x, y) coordinates of the centre of mass are

$(\frac{a}{2}, 2 a)$

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$(\frac{a}{2}, a)$

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$(\frac{a}{2}, \frac{2 a}{3})$

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$(a, \frac{a}{3})$

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Solution

The correct option is C
$(\frac{a}{2}, \frac{2 a}{3})$

The (x, y) co-ordinates of the centre of mass are given by
$x = \frac{m_{1} x_{1} + m_{2} x_{2} + m_{3} x_{3} + m_{4} x_{4}}{m_{1} + m_{2} + m_{3} + m_{4}} = \frac{m \times 0 + m \times a + 2 m \times a + 2 m \times 0}{m + m + 2 m + 2 m} = \frac{a}{2}$
and $y = \frac{m_{1} y_{1} + m_{2} y_{2} + m_{3} y_{3} + m_{4} y_{4}}{m_{1} + m_{2} + m_{3} + m_{4}} = \frac{m \times 0 + m \times 0 + 2 m \times a + 2 m \times a}{m + m + 2 m + 2 m} = \frac{2 a}{3}$
Hence, the correct choice is (c).

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The correct option is C (a2,2a3) The (x, y) co-ordinates of the centre of mass are given by x=m1x1+m2x2+m3x3+m4x4m1+m2+m3+m4=m×0+m×a+2m×a+2m×0m+m+2m+2m=a2 and y=m1y1+m2y2+m3y3+m4y4m1+m2+m3+m4=m×0+m×0+2m×a+2m×am+m+2m+2m=2a3 Hence, the correct choice is (c).