Four particles of masses m, 2m, 3m and 4m are | vedclass

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(B) Let the masses be placed at the vertices of the parallelogram. Let mass $m$ be at $(0, 0)$. Since the side length is $a$ and one angle is $60^o$,t

Vedclass · Accepted Answer

$\left( 0.95a, \frac{\sqrt{3}}{4}a \right)$

Vedclass · Answer

(B) Let the masses be placed at the vertices of the parallelogram. Let mass $m$ be at $(0, 0)$. Since the side length is $a$ and one angle is $60^o$,the mass $4m$ is at $(a, 0)$. The mass $2m$ is at $(a \cos 60^o, a \sin 60^o) = (a/2, \sqrt{3}a/2)$. The mass $3m$ is at $(a + a/2, \sqrt{3}a/2) = (3a/2, \sqrt{3}a/2)$.The $x$-coordinate of the centre of mass is $x_{cm} = \frac{m(0) + 4m(a) + 2m(a/2) + 3m(3a/2)}{m + 2m + 3m + 4m} = \frac{0 + 4ma + ma + 4.5ma}{10m} = \frac{9.5ma}{10m} = 0.95a$.The $y$-coordinate of the centre of mass is $y_{cm} = \frac{m(0) + 4m(0) + 2m(\sqrt{3}a/2) + 3m(\sqrt{3}a/2)}{10m} = \frac{5\sqrt{3}ma/2}{10m} = \frac{\sqrt{3}a}{4}$.Thus,the centre of mass is at $(0.95a, \frac{\sqrt{3}}{4}a)$.

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