Question

The position vector of the centre of mass $\left(r_{cm}\right)$ of an asymmetric uniform bar of negligible area of cross-section as shown figure is $(aL,bL)$. Find $\frac{a}{b}$.

Answer 1

Answer: 2.6

$X$-coordinate of centre of mass,
$X_{cm}=\frac{2m\times L+m\times 2L+m\times \frac{5L}{2}}{4m}=\frac{13}{8}L$
But $X_{cm}=aL$
$\therefore aL=\frac{13}{8}L$
$\therefore a=\frac{13}{8}$
$Y$-coordinate of centre of mass,
$Y_{cm}=\frac{2m\times L+m\times \left(\frac{L}{2}\right)+m\times 0}{4m}=\frac{5L}{8}$
$Y_{cm}=bL$
$\because bL=\frac{5L}{8}$
$\therefore b=\frac{5}{8}$
$\therefore \frac{a}{b}=\frac{13/8}{5/8}=2.6$