Four particles of masses 1,kg, 2 ,kg, 3 ,kg a | vedclass

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(B) Based on the coordinate system shown in the figure:- Particle at $D$ (origin) has mass $m_D = 4\,kg$ at $(0, 0)$.- Particle at $C$ has mass $m_C =

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$(0.5 \,m, 0.3 \,m)$

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(B) Based on the coordinate system shown in the figure:- Particle at $D$ (origin) has mass $m_D = 4\,kg$ at $(0, 0)$.- Particle at $C$ has mass $m_C = 3\,kg$ at $(1, 0)$.- Particle at $B$ has mass $m_B = 2\,kg$ at $(1, 1)$.- Particle at $A$ has mass $m_A = 1\,kg$ at $(0, 1)$.The $x$-coordinate of the centre of mass is:$X_{cm} = \frac{m_A x_A + m_B x_B + m_C x_C + m_D x_D}{m_A + m_B + m_C + m_D} = \frac{1(0) + 2(1) + 3(1) + 4(0)}{1 + 2 + 3 + 4} = \frac{5}{10} = 0.5\,m$.The $y$-coordinate of the centre of mass is:$Y_{cm} = \frac{m_A y_A + m_B y_B + m_C y_C + m_D y_D}{m_A + m_B + m_C + m_D} = \frac{1(1) + 2(1) + 3(0) + 4(0)}{1 + 2 + 3 + 4} = \frac{3}{10} = 0.3\,m$.Thus,the coordinates of the centre of mass are $(0.5\,m, 0.3\,m)$.

Four particles of masses $1\,kg, 2 \,kg, 3 \,kg$ and $4\, kg$ are placed at the four vertices $A, B, C$ and $D$ of a square of side $1\, m$ as shown in the figure. The coordinates of the centre of mass of the particles are:

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