Four bodies have been arranged at the corners of a rectangle shown in figure. Find the centre of mass of the system.
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Let $m_1 = m, m_2 = 2m, m_3 = 3m, m_4= 2m$
Let mass m, be at origin.
$\therefore\text{ For }\text{m}_1;\text{x}_1=0,\text{y}_1=0$
$\text{For }\text{m}_2;\text{x}_2=\text{a}\hat{\text{i}}$
$\text{y}_2=0$
$\text{For }\text{m}_4;\text{x}_4=0,\text{y}-4=\text{b}\hat{\text{j}}$
Coordinates of COM of the system are
$\text{x}=\frac{\text{m}_1\text{x}_1+\text{m}_2\text{x}_2+\text{m}_3\text{x}_3+\text{m}_4\text{x}_4}{\text{m}_1+\text{m}_2+\text{m}_3+\text{m}_4}$
$=\frac{\text{m}\times02\text{m}\times\text{a}\hat{\text{i}}+3\text{m}\times\text{a}\hat{\text{i}}+2\text{m}\times0}{\text{m}+2\text{m}+\text{3m}+2\text{m}}$
$\text{x}=\frac{5\text{ma}\hat{\text{i}}}{8\text{m}}=\frac{5\text{a}\hat{\text{i}}}{8}$
$\text{y}=\frac{\text{m}_1\text{y}_1+\text{m}_2\text{y}_2+\text{m}_3\text{y}_3+\text{m}_4\text{y}_4}{\text{m}_+\text{m}_2+\text{m}_3+\text{m}_4}$
$=\frac{\text{m}\times0+2\text{m}\times0+3\text{m}\times\text{b}\hat{\text{j}}+2\text{m}\times\text{b}\hat{\text{j}}}{\text{m}+2\text{m}+3\text{m}+\text{2m}}$
$=\frac{5\text{mb}\hat{\text{j}}}{8\text{m}}=\frac{5}{8}\text{b}\hat{\text{j}}$
$\therefore$ Centre of mass of system is $\frac{5}{8}(\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}})$
Let mass m, be at origin.
$\therefore\text{ For }\text{m}_1;\text{x}_1=0,\text{y}_1=0$
$\text{For }\text{m}_2;\text{x}_2=\text{a}\hat{\text{i}}$
$\text{y}_2=0$
$\text{For }\text{m}_4;\text{x}_4=0,\text{y}-4=\text{b}\hat{\text{j}}$
Coordinates of COM of the system are
$\text{x}=\frac{\text{m}_1\text{x}_1+\text{m}_2\text{x}_2+\text{m}_3\text{x}_3+\text{m}_4\text{x}_4}{\text{m}_1+\text{m}_2+\text{m}_3+\text{m}_4}$
$=\frac{\text{m}\times02\text{m}\times\text{a}\hat{\text{i}}+3\text{m}\times\text{a}\hat{\text{i}}+2\text{m}\times0}{\text{m}+2\text{m}+\text{3m}+2\text{m}}$
$\text{x}=\frac{5\text{ma}\hat{\text{i}}}{8\text{m}}=\frac{5\text{a}\hat{\text{i}}}{8}$
$\text{y}=\frac{\text{m}_1\text{y}_1+\text{m}_2\text{y}_2+\text{m}_3\text{y}_3+\text{m}_4\text{y}_4}{\text{m}_+\text{m}_2+\text{m}_3+\text{m}_4}$
$=\frac{\text{m}\times0+2\text{m}\times0+3\text{m}\times\text{b}\hat{\text{j}}+2\text{m}\times\text{b}\hat{\text{j}}}{\text{m}+2\text{m}+3\text{m}+\text{2m}}$
$=\frac{5\text{mb}\hat{\text{j}}}{8\text{m}}=\frac{5}{8}\text{b}\hat{\text{j}}$
$\therefore$ Centre of mass of system is $\frac{5}{8}(\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}})$
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