Four particles of masses $m_{1} = 2 m, m_{2} = 4 m, m_{3} = m$ and $m_{4}$ are placed at four corners of a square. What should be the value of $m_{4}$ , so that the centre of mass of all the four particles are exactly at the centre of the square?

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Solution

For centre of mass to be at centre of square the X and Y coordinates of centre of mass should be zero.

$X_{c m} = \frac{m_{1} x_{1} + m_{2} x_{2} + m_{3} x_{3} + m_{4} x_{4}}{m_{1} + m_{2} + m_{3} + m_{4}}$

$(2 m_{1}) (- x) + (4 m_{2}) (x) + (m_{3}) (x) + (m_{4}) (- x) = 0$

$m_{4} = 3 m$

$Y_{c m} = \frac{m_{1} y_{1} + m_{2} y_{2} + m_{3} y_{3} + m_{4} y_{4}}{m_{1} + m_{2} + m_{3} + m_{4}}$

$(2 m_{1}) (- x) + (4 m_{2}) (- x) + (m_{3}) (x) + (m_{4}) (x) = 0$

$m_{4} = 5 m$

So we get different value of $m_{4}$ for both $X$ and $Y$ co ordinate of centre of mass to be zero, which is not possible.

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