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\times0}{4m} = \frac{5L}{8} $