A uniform thin rod of length `3 L` is bent at right angles at a distance `L` from one end as shown in Fig. If length of rod is `1.8m`, find the co-ordinates and position vector of the mass of the system.

Explanation

We know that centre of mass of a rod is at its mid-point. The two parts of the given rod can be considered as two point masses located at the mid-points of the two parts of the given rod as shown in Fig.
If `m` is mass of length `L` of the rod, them `m_(1) = m` and `m_(2) = 2m`.
Co-ordinates of `m_(1)` are `x_(1) = L//2, y_(1) = 0`.
Co-ordinates of `m_(2)` are `x_(2) = 0, y_(2) = L`
`x_(cm) = (m_(1)x_(1) + m_(2)x_(2))/(m_(1) + m_(2)) = (m xx L//2 + 2m xx0)/((m + 2m)) = (L)/(6)`
`y_(cm) = (m_(1)y_(1) + m_(2)y_(2))/(m_(1) + m_(2)) = (m xx 0+ 2m xxL)/(3m) = (2L)/(3)`
Total lenght of rod `= 2L + L= 3L = 1.8m`
`:. L = (1.8)/(3) = 0.6m`
`(x_(cm),y_(cm)) = ((L)/(6),(2L)/(3)) = ((0.6)/(6),(2 xx 0.6)/(3))`
Position vector of centre of mass
`oversetrarr(r_(cm)) = hati x_(cm) + hatjy_(cm) = (0.1 hati + 0.4hatj)`