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nithin vommi Grade: 11
        Three particles of masses 1.0kg,2.0kg and 3.0kg are placed at the corners A,B and C respectively of an equilateral triangle ABC of edge 1 m.Locate the centre of mass of the system. 
8 years ago

Answers : (4)

Askiitians Expert Soumyajit IIT-Kharagpur
28 Points

Dear Nitin Vomani,

Ans:- Let the trangle ABC  is placed in such a way in the co ord plane so that the mid point of the side BC is the origin

point A=1kg

point B=2kg

point C=3kg

hence the X co ord is=[(2×(-1/2)+(3×1/2)+(1×0)]/(2+3+1)=1/12

The Y co ord is=[(2×0)+(3×0)+(1×√3/2)]/(1+2+3)=√3/12

Hence the COM is At=( 1/12  , √3/12)

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Askiitians Experts
Soumyajit Das IIT Kharagpur

8 years ago
8 Points
Sir the answer given by u does not match the answer given in the book the Xcm is right but the Ycm is given as root 3 /4 
3 years ago
prateek sonker
19 Points
										To find the center of mass:Fix one vertex of triangle as origin and give coordinates to other 2 vertices.apply the center of mass formulaIn the figure shown above A(0, 0) , B (1,0) and C(1/2, sqrt(3)/2)Xcm = M1X1+M2X2+M3X3/(M1+M2+M3) = 1(0)+2(1)+3(1/2)/(1+2+3) = 7/12Ycm = M1Y1+M2Y2+M3Y3/(M1+M2+M3) = 1(0)+2(0)+3(sqrt(3)/2)/(1+2+3) = sqrt(3)/4
one year ago
39 Points
Now as com at x=m1x1+m2x2+m3x3/m1+m2+m3 
com at y=m1y1+m2y2+m3y3/m1+m2+m3
          =   √19/6 m from A
5 months ago
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