Question

A square is made by joining four rods each of mass $m$ and length $L$ each. Its moment of inertia about an axis $PQ,$ on its plane and passing through one of its corners (as shown in the figure) will be:

• A. $6{mL}^2$
• B. $\frac{4}{3}{mL}^2$
• C. $\frac{8}{3}{mL}^2$
• D. $\frac{10}{3}{mL}^2$

Answer 1

The correct option is C $\frac{8}{3}{mL}^2$

We know that for a rod inclined at some angle $\theta$ to the axis on its plane, moment of inertia is given by


$I_1=\frac{m{L}^2}{3}{\text{sin}}^2\theta$, for axis of rotation passing through its one end.
$I_2=\frac{m{L}^2}{12}{\text{sin}}^2\theta$, for axis of rotation passing through its centre of mass.


Moment of inertia of system of rods about $\text{PQ}$ is,
$I=I_{AB}+I_{AC}+I_{BD}+I_{CD}$
$I=2(I_{AB}+I_{BD})\left[\text{due to symmetry}\right]$
$I=2[\frac{m{L}^2}{3}{\text{sin}}^2{45}^{\circ }+(\frac{m{L}^2}{12}{\text{sin}}^2{45}^{\circ }+m{\left(\frac{3L}{2\sqrt{2}}\right)}^2)]$
Since moment of inertia of rods $AB$ and $AC$ about axis $PQ$ will be $\frac{m{L}^2}{3}{\text{sin}}^2{45}^{\circ }$
and moment of inertia of rods $BD$ and $CD$ about axis $PQ$ will be given by,
$I_{BD}=\frac{m{L}^2}{12}{\text{sin}}^2{45}^{\circ }+m{d}^2$, where $d=\frac{3L}{2\sqrt{2}}$
$\Rightarrow I=2[\frac{m{L}^2}{6}+(\frac{m{L}^2}{24}+\frac{9m{L}^2}{8})]$
$\Rightarrow I=2\left[\frac{4m{L}^2}{3}\right]$
$\therefore I=\frac{8m{L}^2}{3}$
Moment of inertia of system of rods shown in figure about the given axis of rotation $\text{PQ}$ will be $\frac{8m{L}^2}{3}$