Questions

# Three identical rods each of length L and mass m are joined to form a U-shape as shown in figure .The distance of center of mass of the combined system from the corner  'O' is

a
$\frac{\mathrm{L}}{2}$
b
$\frac{\mathrm{L}}{3}$
c
$\frac{\sqrt{8}\mathrm{L}}{6}$
d
$\frac{\sqrt{13}\mathrm{L}}{6}$

detailed solution

Correct option is D

${\mathrm{x}}_{\mathrm{cm}}=\frac{{\mathrm{m}}_{1}{\mathrm{x}}_{1}+{\mathrm{m}}_{2}{\mathrm{x}}_{2}+{\mathrm{m}}_{3}{\mathrm{x}}_{3}}{{\mathrm{m}}_{1}+{\mathrm{m}}_{2}+{\mathrm{m}}_{3}}$

$=\frac{0×\mathrm{m}×\frac{\mathrm{L}}{2}×\mathrm{m}×\mathrm{L}}{\mathrm{m}+\mathrm{m}+\mathrm{m}}$

$=\frac{\mathrm{m}\left(\frac{3\mathrm{L}}{2}\right)}{3\mathrm{m}}=\frac{3\mathrm{L}}{6}=\frac{\mathrm{L}}{2}$

${\mathrm{y}}_{\mathrm{cm}}=\frac{{\mathrm{m}}_{1}{\mathrm{y}}_{1}+{\mathrm{m}}_{2}{\mathrm{y}}_{2}+{\mathrm{my}}_{3}}{{\mathrm{m}}_{1}+{\mathrm{m}}_{2}+{\mathrm{m}}_{3}}$

$=\left(\frac{\mathrm{m}×\frac{\mathrm{L}}{2}+\mathrm{m}×0+\mathrm{m}×\frac{\mathrm{L}}{2}}{3\mathrm{m}}\right)$

$=\frac{\frac{\mathrm{mL}}{2}+\frac{\mathrm{mL}}{2}}{3\mathrm{m}}=\frac{\frac{2\mathrm{mL}}{2}}{3\mathrm{m}}=\frac{\mathrm{L}}{3}$

${\mathrm{r}}_{\mathrm{cm}}={\mathrm{x}}_{\mathrm{cm}}\mathrm{i}+{\mathrm{y}}_{\mathrm{cm}}\mathrm{j}$

$=\frac{\mathrm{L}}{2}\mathrm{i}+\frac{\mathrm{L}}{3}\mathrm{j}$

$=\sqrt{{\left(\frac{\mathrm{L}}{2}\right)}^{2}+{\left(\frac{\mathrm{L}}{3}\right)}^{2}}$

$=\sqrt{\frac{{\mathrm{L}}^{2}}{4}+\frac{{\mathrm{L}}^{2}}{9}}=\sqrt{\frac{9{\mathrm{L}}^{2}+4{\mathrm{L}}^{2}}{36}}$

$=\sqrt{\frac{13{\mathrm{L}}^{2}}{36}}=\frac{\sqrt{13}\mathrm{L}}{6}$