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Question: Three particles of masses 1 kg, \(\frac{3}{2}\) kg , and 2 kg are located at the vertices of an equi...

Three particles of masses 1 kg, 32\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

A

(5a9,2a33)\left( \frac{5a}{9},\frac{2a}{3\sqrt{3}} \right)

B

(2a33,5a9)\left( \frac{2a}{3\sqrt{3}},\frac{5a}{9} \right)

C

(5a9,2a3)\left( \frac{5a}{9},\frac{2a}{\sqrt{3}} \right)

D

(2a3,5a9)\left( \frac{2a}{\sqrt{3}},\frac{5a}{9} \right)

Answer

(5a9,2a33)\left( \frac{5a}{9},\frac{2a}{3\sqrt{3}} \right)

Explanation

Solution

Let the masses 1kg, 32kg\frac{3}{2}kgand 2 kg are located at the vertices A,B and C as shown in figure above. The coordinates of point A,B and C are (0,0), (a,0), (a2,3a2)\left( \frac{a}{2},\frac{\sqrt{3}a}{2} \right)respectively.

The coordinates of centre of mass are

XCM=m1x1+m2x2+m3x3m1+m2+m3=[1×0+32×a+2×a2][1+32+2]=5a9X_{CM} = \frac{m_{1}x_{1} + m_{2}x_{2} + m_{3}x_{3}}{m_{1} + m_{2} + m_{3}} = \frac{\left\lbrack 1 \times 0 + \frac{3}{2} \times a + 2 \times \frac{a}{2} \right\rbrack}{\left\lbrack 1 + \frac{3}{2} + 2 \right\rbrack} = \frac{5a}{9}

YCM=m1y1+m2y2+m3y3m1+m2+m3Y_{CM} = \frac{m_{1}y_{1} + m_{2}y_{2} + m_{3}y_{3}}{m_{1} + m_{2} + m_{3}}

=[1×0+32×0+2×3a2][1+32+2]=23a9=2a33= \frac{\left\lbrack 1 \times 0 + \frac{3}{2} \times 0 + 2 \times \frac{\sqrt{3}a}{2} \right\rbrack}{\left\lbrack 1 + \frac{3}{2} + 2 \right\rbrack} = \frac{2\sqrt{3}a}{9} = \frac{2a}{3\sqrt{3}}