Question

The position vectors of three particles of mass `m_(1) = 1kg, m_(2) = 2kg and m_(3) = 3kg` are `r_(1) = ((hat(i) + 4hat(j) +hat(k))`m, `r_(2) = ((hat(i)+(hat(j)+hat(k))`m and `r_(3) = (2hat(i) - (hat(j) -(hat(2k))`m, respectively. Find the position vector of their center of mass.

Answer 1

The position vector of CM of the three particles will be given by
`r_(CM) = (m_(1)r_(1) + m_(2)r_(2) + m_(3)r_(3))/(m_(1)+ m_(2) +m_(3))`
substituting, the values, we get
`r_(cm) = (1(i+4j+k) + 2(i+j+k)+3(2i-j-2k))/(1+2+3)`
`=(9i + 3j -3k)/6 rArr r_(cm) =1/2(3i + j -k)m`