Question

The position vector of the centre of mass $r_{\text{cm}}$ of an asymmetric uniform bar of negligible area of cross - section as shown in figure is:

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

Answer 1

The correct option is B ${\overrightarrow{r}}_{\text{cm}}=\frac{13}{8}L\hat{x}+\frac{5}{8}L\hat{y}$


$x-$ coordinate of centre of mass is,

$X_{\text{cm}}=\frac{2mL+2mL+\frac{5mL}{2}}{4m}=\frac{13}{8}L$

$y-$ coordinate of centre of mass is,

$Y_{\text{cm}}=\frac{2mL+m\left(\frac{L}{2}\right)+m\times 0}{4m}=\frac{5L}{8}$