three particles each of mass 5g are placed at vertices of an equilateral triangle of side 60cm. find;distance of its center of mass from any of its vertex
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Given :
Three masses each of 5g side=a=60cm=60/100=0.6m
To find :
Distance of its centre of mass from any of its vertex.
Solution:
From symmetry, we can conclude that centre of mass is located at the centre of circumcircle of triangle.
so distane of its centre of mass is nothing but radius of its cirumcircle
R=√3/ 3 xa
=√3/ 3 x 0.6
=1.73/3 x 0.6
=0.576 x 0. 6=0.3456m
Three masses each of 5g side=a=60cm=60/100=0.6m
To find :
Distance of its centre of mass from any of its vertex.
Solution:
From symmetry, we can conclude that centre of mass is located at the centre of circumcircle of triangle.
so distane of its centre of mass is nothing but radius of its cirumcircle
R=√3/ 3 xa
=√3/ 3 x 0.6
=1.73/3 x 0.6
=0.576 x 0. 6=0.3456m
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