Question

Find the position vector of the centre of mass ${\overrightarrow{r}}_{\text{com}}$ of an asymmetric uniform bar of negligible area of cross-section, shown in the figure given below.

• A. ${\overrightarrow{r}}_{\text{com}}=\frac{13}{8}L\hat{i}+\frac{5}{8}L\hat{j}$
• B. ${\overrightarrow{r}}_{\text{com}}=\frac{11}{8}L\hat{i}+\frac{3}{8}L\hat{j}$
• C. ${\overrightarrow{r}}_{\text{com}}=\frac{3}{8}L\hat{i}+\frac{11}{8}L\hat{j}$
• D. ${\overrightarrow{r}}_{\text{com}}=\frac{5}{8}L\hat{i}+\frac{13}{8}L\hat{j}$

Answer 1

The correct option is A ${\overrightarrow{r}}_{\text{com}}=\frac{13}{8}L\hat{i}+\frac{5}{8}L\hat{j}$

The given bar can be considered as a combination of $3$ individual rods, whose coordinates of centres of mass have been shown in the figure.


The position vector of the centre of mass of the combination of bodies is given by,
${\overrightarrow{r}}_{\text{com}}=\frac{m_1\overrightarrow{r_1}+m_2\overrightarrow{r_2}+m_3\overrightarrow{r_3}}{m_1+m_2+m_3}$
$\Rightarrow {\overrightarrow{r}}_{\text{com}}=\frac{2m\times (L\hat{i}+L\hat{j})+m(2L\hat{i}+\frac{L}{2}\hat{j})+m(\frac{5}{2}L\hat{i}+0\hat{j})}{2m+m+m}$
$\Rightarrow {\overrightarrow{r}}_{\text{com}}=\frac{\frac{13}{2}mL\hat{i}+\frac{5}{2}mL\hat{j}}{4m}$
$\therefore {\overrightarrow{r}}_{\text{com}}=\frac{13}{8}L\hat{i}+\frac{5}{8}L\hat{j}$