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three particles of masses 1kg,2kg,3kg,are placed at three vertices A,B and C of an equilataral triangle of edge 1m.what is the distance between center of mass and A
Hi Rajat,
Take A to be the origin (0,0). Let side of eq triangle be "a".
B is (a,0) and C is (a/2,√3a/2)
Now Co-ord of centre of mass C of three particles m1,m2,m3 placed at (x1,y1) (x2,y2) (x3,y3) is:
[(m1x1+m2x2+m3x3)/(m1+m2+m3),(m1y1+m2y2+m3y3)/(m1+m2+m3)]
Using this get the co-ord of Centre of mass in the above triangle.
Then use distance formula to get distance between A and Centre of mass.
Centre of mass is (7a/12,3√3a/12) where a=1(edge of triangle).
So now distance = (√19)/6.
Hope that helps.
All the best.
Regards,
Ashwin (IIT Madras).
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