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# 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. Askiitians Expert Soumyajit IIT-Kharagpur
28 Points
11 years ago

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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All the best Nitin Vomani !!!

Regards,

Soumyajit Das IIT Kharagpur

4 years ago
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 formula￼In 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
3 years ago
A(0,0),B(1,0),C(1/2,√3/2)
Now as com at x=m1x1+m2x2+m3x3/m1+m2+m3
=1(0)+2(1)+3(1/2)/1+2+3=(2+3/2)6=4+3/12=7/12
com at y=m1y1+m2y2+m3y3/m1+m2+m3
=1(0)+2(0)+3(√3/2)/1+2+3=(3√3/2)/6=3√3/2*6=√3/4
COM=√x*x+y*y=√7/12*7/12+√3/4*√3/4
=√49/144+3/16
=√49+27/144
=√76/144
=2√19/12
=   √19/6 m from A