Question

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

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

Answer 1

Answer: (A)

$X_{cm}=\frac{2mL+2mL+\frac{5mL}{2}}{4m}=\frac{13}{8}L$
$Y_{cm}=\frac{2m\times L+m\times \left(\frac{L}{2}\right)+m\times 0}{4m}=\frac{5L}{8}$