Q 11-06-228JEE MainJEE Main 2019 (12 Jan, Shift 1)Medium
The position vector of the centre of mass $\vec{r}_{cm}$ of an asymmetric uniform bar of negligible area of cross-section as shown in figure is:
Answer: (A) $\vec{r}_{cm} = \dfrac{13}{8}L\hat{x} + \dfrac58L\hat{y}$
Treat each straight part as a point mass at its midpoint:
- $2m$ at $(L, L)$
- $m$ at $(2L, L/2)$
- $m$ at $(2.5L, 0)$
$$x_{cm} = \frac{2m(L) + m(2L) + m(2.5L)}{4m} = \frac{13}{8}L,\qquad y_{cm} = \frac{2m(L) + m(L/2) + 0}{4m} = \frac58L$$
Solution by Sreeraj P, M.Sc Physics