Question

Two rods $\text{OA}$ and $\text{OB}$ of equal lengths and masses are lying in the $xy$ plane as shown in figure. Let $I_x,I_y$ and $I_z$ be the moments of inertia of both the rods about $x,y$ and $z$ axis respectively. Then,

• A. $I_x>I_y$
• B. $I_x=I_y$
• C. $I_x<I_y$
• D. $I_x+I_y=I_z$

Answer 1

Answer: (B), (D)

The correct options are
B $I_x=I_y$
D $I_x+I_y=I_z$

Let the mass and length of both rods be $m$ and $l$ respectively.
Moment of inertia of both rods $\text{OA}$ and $\text{OB}$ about $x$ - axis:
$I_x=2\left[\frac{m{l}^2}{3}{\sin }^2\theta \right]$
$[\text{where}\theta ={45}^{\circ }]$
$\Rightarrow I_x=2\left[\frac{m{l}^2}{3}{\sin }^2{45}^{\circ }\right]=\frac{m{l}^3}{3}$
Moment of inertia of both rods $\text{OA}$ and $\text{OB}$ about $y$ - axis:
$I_y=2\left[\frac{m{l}^2}{3}{\cos }^2\theta \right]$
$[\because \theta ={45}^o]$
$I_y=2\left[\frac{m{l}^3}{3}{\cos }^2{45}^{\circ }\right]=\frac{m{l}^3}{3}$
Moment of inertia of both rods $\text{OA}$ and $\text{OB}$ about $z$ - axis :
$I_z=2\left[\frac{m{l}^2}{3}\right]=\frac{2}{3}m{l}^2$

So, $I_x=I_y=\frac{m{l}^2}{3}$ and $I_z=\frac{2}{3}m{l}^2$
$\therefore I_x=I_y<I_z$
Option (b) and (d) are correct.