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

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