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

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Question

Let the masses $1 \mathrm{kg}, \frac{3}{2} \mathrm{kg}$ and $2 \mathrm{kg}$ are

located at the vertices $A, B$ and $C$ as shown

in figure above.

Vedclass · Accepted Answer

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

Vedclass · Answer

Let the masses $1 \mathrm{kg}, \frac{3}{2} \mathrm{kg}$ and $2 \mathrm{kg}$ are

located at the vertices $A, B$ and $C$ as shown

in figure above. The coordinates of points

$\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are $(0,0),(\mathrm{a}, 0),\left(\frac{\mathrm{a}}{2}, \frac{\sqrt{3} \mathrm{a}}{2}ight)$

respectively.

The coordinates of centre of mass are

$\mathrm{X}_{\mathrm{CM}}=\frac{\mathrm{m}_{1} \mathrm{x}_{1}+\mathrm{m}_{2} \mathrm{x}_{2}+\mathrm{m}_{3} \mathrm{x}_{3}}{\mathrm{m}_{1}+\mathrm{m}_{2}+\mathrm{m}_{3}}=\frac{\left[1 	imes 0+\frac{3}{2} 	imes \mathrm{a}+2 	imes \frac{\mathrm{a}}{2}ight]}{\left[1+\frac{3}{2}+2ight]}$

$=\frac{5 a}{9}$

$\mathrm{Y}_{\mathrm{CM}}=\frac{\mathrm{m}_{1} \mathrm{y}_{1}+\mathrm{m}_{2} \mathrm{y}_{2}+\mathrm{m}_{3} \mathrm{y}_{3}}{\mathrm{m}_{1}+\mathrm{m}_{2}+\mathrm{m}_{3}}$

$=\frac{\left[1 	imes 0+\frac{3}{2} 	imes 0+2 	imes \frac{\sqrt{3} \mathrm{a}}{2}ight]}{\left[1+\frac{3}{2}+2ight]}=\frac{2 \sqrt{3} \mathrm{a}}{9}=\frac{2 \mathrm{a}}{3 \sqrt{3}}$

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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