Three particles of masses 1kg,2kg and 3kg are situated at the corners of an equilateral triangle of side b. The (x,y) coordinates respectively for the centre of mass of the system of particles will be:
A
[7b12,√3b12]
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B
[3√3b12,7b12]
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C
[7b12,3√3b12]
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D
[7b12,3√3b4]
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Solution
The correct option is C[7b12,3√3b12] The coordinates of points A, B and C are A→(0,0) B→(b,0) C→(b2,bsin60∘)=(b2,√3b2)
Applying the formula for coordinates of COM of system: xCM=m1x1+m2x2+m3x3m1+m2+m3 xCM=(1×0)+(2×b)+(3×b2)1+2+3=7b12 Simillarly, yCM=m1y1+m2y2+m3y3m1+m2+m3 yCM=(1×0)+(2×0)+(3×√3b2)1+2+3=3√3b12 So, the coordinates of the centre of mass are [7b12,3√3b12]