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Obtain the general expression of centre of mass for distributed n particles of system in three dimension. |
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Answer» Solution :Suppose the coordinates of n particles of masses `m_(1),m_(2),…,m_(n)` are `(x_(1),y_(1),z_(1)),(x_(2),y_(2),z_(2)),…,(x_(n),y_(n),z_(n))` respectively. `therefore` The position of centre of mass, `(x,y,Z)=(m_(1)(x_(1),y_(1),z_(1))+m_(2)(x_(2),y_(2),z_(2))+...+m_(n)(x_(n),y_(n),z_(n)))/(m_(1)+m_(2)+...+m_(n))` `=(Sigmam_(i)(x_(i),y_(i),z_(i)))/(Sigmam_(i))` where `i=1,2,3,....,n` OR X-coordinate, Y-coordinate and Z-coordinate of centre of mass of a system, `X=(m_(1)x_(1)+m_(2)x_(2)+...+m_(n)x_(n))/(m_(1)+m_(2)+...+m_(n))=(Sigmam_(i)x_(i))/(M)` `Y=(m_(1)y_(1)+m_(2)y_(2)+...+m_(n)y_(n))/(m_(1)+m_(2)+...+m_(n))=(Sigmam_(i)y_(i))/(M)` `Z=(m_(1)z_(1)+m_(2)z_(2)+...+m_(n)z_(n))/(m_(1)+m_(2)+...+m_(n))=(Sigmam_(i)z_(i))/(M)` where `i=1,2,3,...,n` and `m_(1)+m_(2)+.....+m_(n)=Sigmam_(i)=M` is the total mass of system. If we want to find the position vector of centre of mass in i then position vector of `VECR=x_(i)hati+y_(i)hatj+z_(i)hatk` and the position vector of centre of mass `vecR=(Xhati+Yhatj+Zhatk)` `therefore vecR=(Sigmam_(i)vecr_(i))/(M)` If the origin of frame of reference (coordinate system) TAKEN as centre of mass of a system then for the system of particles `Sigmam_(i)vec(r_(i))=0` |
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