Question

Two thin rods of mass $m$ and length $l$, each are joined to form a $L$ shape as shown. The moment of inertia of rods about an axis passing through free end (O) of a rod and perpendicular to both the rods is:

• A. $\frac{2}{7}m{l}^2$
• B. $\frac{m{l}^2}{6}$
• C. $m{l}^2$
• D. $\frac{5m{l}^2}{3}$

Answer 1

The correct option is D $\frac{5m{l}^2}{3}$

Moment of inertia of rod 1 about O perpendicular to both the rods

$I_1=\frac{m{l}^2}{12}+m{x}^2$ where x is the distance between O and center of mass of rod.

$l=\sqrt{l^2+{\left(\frac{l}{2}\right)}^2}=\sqrt{\frac{{5l}^2}{4}}$

Moment of inertia of rod 2 about axis through O perpendicular to both rods

$I_2=\frac{m{l}^2}{3}$

Total moment of inertia I$=I_1{+I}_2$ -------(1)

$I_1=\frac{m{l}^2}{12}+\frac{5m{l}^2}{4}=\frac{4m{l}^2}{3}$

Thus 1 becomes

$=\frac{4m{l}^2}{3}+\frac{m{l}^2}{3}=\frac{5m{l}^2}{3}$