Four particles of mass $m_{1} = 2,, m_{2} = 4 m, m_{3} = m$ 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

X-coordinate of CM is at origin

$h e n c e, 0 = (\frac{2 m (- a) + 4 m \times a + m \times a + m_{4} (- a)}{2 m + 4 m + m + m_{4}})$
$o r 0 = 3 m a - m_{4} a$
$∴ m_{4} = 3 m$

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Four particles of mass m1=2,,m2=4m,m3=m and m4 are placed at four corners of a square. What should be the value of M4 , so that the centre of mass of all the four particles are exactly at the centre of the square?

X-coordinate of CM is at originhence,0=(2m(−a)+4m×a+m×a+m4(−a)2m+4m+m+m4)or0=3ma−m4a∴m4=3m

Four particles of mass $m_{1} = 2,, m_{2} = 4 m, m_{3} = m$ 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?

X-coordinate of CM is at origin

$h e n c e, 0 = (\frac{2 m (- a) + 4 m \times a + m \times a + m_{4} (- a)}{2 m + 4 m + m + m_{4}})$
$o r 0 = 3 m a - m_{4} a$
$∴ m_{4} = 3 m$