The masses and positions (in rectangular coor | vedclass

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(D) The position vector of the center of mass $(COM)$ of a system of particles is given by:$r_{COM} = \frac{m_1 r_1 + m_2 r_2 + m_3 r_3 + m_4 r_4}{m_1

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$-0.4 a \hat{j}$

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(D) The position vector of the center of mass $(COM)$ of a system of particles is given by:$r_{COM} = \frac{m_1 r_1 + m_2 r_2 + m_3 r_3 + m_4 r_4}{m_1 + m_2 + m_3 + m_4}$Given masses and positions:$m_1 = 1 \ kg, r_1 = (a \hat{i} + a \hat{j})$$m_2 = 2 \ kg, r_2 = (-a \hat{i} + a \hat{j})$$m_3 = 3 \ kg, r_3 = (-a \hat{i} - a \hat{j})$$m_4 = 4 \ kg, r_4 = (a \hat{i} - a \hat{j})$Total mass $M = 1 + 2 + 3 + 4 = 10 \ kg$$r_{COM} = \frac{1(a \hat{i} + a \hat{j}) + 2(-a \hat{i} + a \hat{j}) + 3(-a \hat{i} - a \hat{j}) + 4(a \hat{i} - a \hat{j})}{10}$$r_{COM} = \frac{(a - 2a - 3a + 4a) \hat{i} + (a + 2a - 3a - 4a) \hat{j}}{10}$$r_{COM} = \frac{0 \hat{i} - 4a \hat{j}}{10} = -0.4 a \hat{j}$

The masses and positions (in rectangular coordinates) of four particles are as follows: $1 \ kg$ at $(a, a)$,$2 \ kg$ at $(-a, a)$,$3 \ kg$ at $(-a, -a)$,and $4 \ kg$ at $(a, -a)$. The position vector of the centre of mass of the system of four particles is:

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