Four particles, each of mass 1 kg. are placed at the corners of a square of side 12 cm. if mass $m_{3}$ is removed , find the shift in the centre of mass of the system (see fig.) is $x \hat{i} + y \hat{j}$ then find $x + y$

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Detailed Solution

As shown in Example 2, the $x$ and y coordinates of the original system are $x_{C M} = 6 c m a n d y_{C M} = 6 c m$
If mass $m_{3}$ is removed, the $x$ and $y$ coordinates of the centre of mass of the remaining system are

$x_{C M} = \frac{1 \times 0 \times 1 \times 12 \times 1 \times 0}{1 + 1 + 1} = 4 c m$
$y_{C M} = \frac{1 \times 0 \times 1 \times 0 \times 1 \times 12}{1 + 1 + 1} = 4 c m$

$r_{1} = 6 \hat{i} + 6 \hat{j}$

$\begin{array}{l} r_{2} = 4 \hat{i} + 4 \hat{j} \\ \begin{aligned} Δ r = r_{2} - r_{1} & = (6 \hat{i} + 6 \hat{j}) - (4 \hat{i} + 4 \hat{j}) \\ = 2 \hat{i} + 2 \hat{j} \end{aligned} \end{array}$