Question

Moment of inertia of the system of two rods having same linear mass density $\lambda$, about the axis $AB$ as shown in the figure is

• A. $\frac{\lambda L^3}{4\sqrt{3}}$
• B. $\frac{\lambda L^3}{12\sqrt{2}}$
• C. $\frac{\lambda L^3}{4}$
• D. $\frac{\lambda L^3}{6\sqrt{2}}$

Answer 1

The correct option is D $\frac{\lambda L^3}{6\sqrt{2}}$

Let us take a small element of mass $dm$ at a distance $x$ from end $A$ along rod $AC$ and the distance of the element is $r$ from axis $AB$.


$L_0=\frac{\left(\frac{L}{2}\right)}{\sin {45}^{\circ }}=\frac{L}{\sqrt{2}}$
Moment of inertia of mass $dm$ about $AB$
$dI=dm{r}^2$
$\Rightarrow dI=\lambda dx(x\sin {45}^{\circ }{)}^2$
[$\because \lambda =\frac{dm}{dx}$ & $r=x\sin {45}^{\circ }$ from diagram]

Putting the limit of $x=0\text{to}x=\frac{L}{\sqrt{2}}$ for rod $AC$ $\&$ integrating both sides
$\Rightarrow \underset{0}{\overset{I}{\int }}dI=\frac{\lambda }{2}\underset{0}{\overset{L\sqrt{2}}{\int }}{x}^{2}dx$
$\Rightarrow I=\frac{\lambda }{2}{\left[\frac{x^3}{3}\right]}_0^{L/\sqrt{2}}$
$\Rightarrow I=\frac{\lambda }{6}\left[\frac{L^3}{2\sqrt{2}}\right]$
$\therefore I=\frac{\lambda L^3}{12\sqrt{2}}$ for rod $AC$.

For rod $BC$, moment of inertia about axis $AB$ will be equal to $I=\frac{\lambda L^3}{12\sqrt{2}}$, due to symmetry.
$\Rightarrow$Moment of inertia of the system of two rods about axis $AB$ will be,
$I^{\prime }=I+I=2I$
$\therefore I^{\prime }=2\times \frac{\lambda L^3}{12\sqrt{2}}=\frac{\lambda L^3}{6\sqrt{2}}$