Three particles of masses 1,kg,frac{3}{2},kg, | vedclass

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(A) Let the masses $m_1 = 1\,kg$,$m_2 = \frac{3}{2}\,kg$,and $m_3 = 2\,kg$ be located at the vertices $A(0, 0)$,$B(a, 0)$,and $C\left(\frac{a}{2}, \fr

Vedclass · Accepted Answer

$\left( \frac{5a}{9}, \frac{2a}{3\sqrt{3}} \right)$

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(A) Let the masses $m_1 = 1\,kg$,$m_2 = \frac{3}{2}\,kg$,and $m_3 = 2\,kg$ be located at the vertices $A(0, 0)$,$B(a, 0)$,and $C\left(\frac{a}{2}, \frac{\sqrt{3}a}{2}\right)$ respectively.The $x$-coordinate of the centre of mass is given by:$X_{CM} = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{m_1 + m_2 + m_3} = \frac{1(0) + \frac{3}{2}(a) + 2(\frac{a}{2})}{1 + \frac{3}{2} + 2} = \frac{\frac{3a}{2} + a}{\frac{2+3+4}{2}} = \frac{\frac{5a}{2}}{\frac{9}{2}} = \frac{5a}{9}$.The $y$-coordinate of the centre of mass is given by:$Y_{CM} = \frac{m_1 y_1 + m_2 y_2 + m_3 y_3}{m_1 + m_2 + m_3} = \frac{1(0) + \frac{3}{2}(0) + 2(\frac{\sqrt{3}a}{2})}{1 + \frac{3}{2} + 2} = \frac{\sqrt{3}a}{\frac{9}{2}} = \frac{2\sqrt{3}a}{9} = \frac{2a}{3\sqrt{3}}$.Thus,the coordinates of the centre of mass are $\left( \frac{5a}{9}, \frac{2a}{3\sqrt{3}} \right)$.

Three particles of masses $1\,kg$,$\frac{3}{2}\,kg$,and $2\,kg$ are located at the vertices of an equilateral triangle of side $a$. The $x, y$ coordinates of the centre of mass are

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