Question

Two rods of equal length $\left(L\right)$ and equal mass $\left(M\right)$ are kept along $x$ and $y$ axis respectively such that their centres of mass lie at the origin. The moment of inertia about the line $y=x$, is

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

Answer 1

The correct option is B $\frac{M{L}^2}{12}$


Moment of inertia of a rod of mass $M$ and length $L$ with axis of rotation passing through the center of the rod (perpendicular to the plane) and making angle $\theta$ with the rod:
$I^{\prime }=\frac{M{L}^2}{12}{\sin }^2\theta$

Therefore, moment of inertia of rod $\text{AB}$ and $\text{CD}$ about the line $y=x$ is
$I=2I^{\prime }$
$=2\times \frac{M{L}^2}{12}{\sin }^2{45}^{\circ }$ $(\theta ={45}^{\circ })$
$=\frac{M{L}^2}{12}$