Obtain an expression for the position vector of centre of mass of a system of n particles in one dimension.

Obtain an expression for the position vector of centre of mass of a sy

1.	Obtain an expression for the position vector of centre of mass of a system of n particles in one dimension.
Answer» Solution :According to as shown in figure. Let the distances of the two particles be `x_(1)andx_(2)` respectively from ORIGIN O. Let `m_(1) and m_(2)` be respectively the masses of the two particles on X-axis. The centre of mass of the system is that point C which is at a DISTANCE X from O, then `X=(m_(1)x_(1)+m_(2)x_(2))/(m_(1)+m_(2))` [X is denoted by `r_(cm)` also] where X is the mass-weighted mean of `x_(1)andx_(2)`. If the two particles have the same mass `m_(1)=m_(2)=m`, then `X=(mx_(1)+mx_(2))/(m+m)=(x_(1)+x_(2))/(2)` Hence, for two particles of EQUAL mass the centre of mass lies exactly midway between them. Similarly, n particles of masses `m_(1),m_(2),m_(3),...,m_(n)` their respectively, distances `x_(1),x_(2),x_(3),...,x_(n)` from origin on X-axis then centre of mass of system of particles is given by, `X=(m_(1)x_(1)+m_(2)x_(2)+m_(3)x_(3)+...m_(n)x_(n))/(m_(1)+m_(2)+m_(3)+...m_(n))` `=(Sigmam_(i)x_(i))/(Sigmam_(i))` But `Sigmam_(i)=m_(1)+m_(2)+m_(3),...,m_(n)` `Sigma=m` = TOTAL mass of system `therefore (Sigmam_(i)x_(i))/(M)` (where `i=1,2,3,...,n`)