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Center of Mass as an Average Point

Four particle...

Question

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

A

m

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B

2 m

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C

4 m

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D

6 m

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Solution

The correct option is C $4 m$

Considering a square with side $a$ and assigning origin $(0, 0)$ at mass $m_{1}$ .

The $x$ and $y$ coordinates of all masses can be written as :
$m_{1} = 2 m, (0, 0)$
$m_{2} = 4 m, (a, 0)$
$m_{3} = m, (a, a)$
$m_{4} = m_{4}, (0, a)$

$∵$ $COM$ of system of particles should be at the centre of the square. Hence coordinates of $COM$ is represented by $(x_{COM}, y_{COM}) = (\frac{a}{2}, \frac{a}{2}) . . . . (i)$

Applying the formula and equating the $x$ and $y$ coordinates of $COM$ from Eq. $(i)$

$x_{COM} = \frac{m_{1} x_{1} + m_{2} x_{2} + m_{3} x_{3} + m_{4} x_{4}}{m_{1} + m_{2} + m_{3} + m_{4}}$

$\Rightarrow \frac{a}{2} = \frac{0 + 4 m a + m a + 0}{2 m + 4 m + m + m_{4}} = \frac{5 m a}{(7 m + m_{4})} . . . (i i)$

Simillarly, $y_{COM} = \frac{m_{1} y_{1} + m_{2} y_{2} + m_{3} y_{3} + m_{4} y_{4}}{m_{1} + m_{2} + m_{3} + m_{4}}$

$\Rightarrow \frac{a}{2} = \frac{0 + 0 + m a + m_{4} a}{7 m + m_{4}} = \frac{(m + m_{4}) a}{7 m + m_{4}} . . . . . (i i i)$

Equating Eq. $(i i)$ & $(i i i)$ :
$5 m a = (m + m_{4}) a$
$\Rightarrow m_{4} = 5 m - m = 4 m$
$∴ m_{4} = 4 m$

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