Question

Two rods of equal mass $m$ and length $l$ lie along the x axis and y axis with their centres origin. If the moment of inertia of both about the line x=y is given as $\frac{m{l}^2}{x}$. Find $x$.

Answer 1

Line PQ represents $y=x$
MI of AB about PQ which makes an angle ${45}^{\circ }$ with x-axis is two times the MI of AO about PQ.
MI of CD about PQ is same as that of MI of AB about PQ.
Hence, MI of two rods is 4 times the MI of AO about PQ.
$AR\perp PQ$Consider a small length dx of rod AO at a distance x from O.
MI of dx about PQ is $dI=\frac{m}{l}dx(x\sin {45}^{\circ }{)}^2=\frac{m}{2l}\frac{x^3}{3}$
$\therefore I={\int }_0^{\frac{l}{2}}\frac{m}{2l}\frac{x^3}{3}=\frac{m}{2l}\frac{(\frac{l}{2}{)}^3}{3}=\frac{m{l}^2}{48}$
$\therefore$ MI of the two rods is $4\times \frac{m{l}^2}{48}=\frac{m{l}^2}{12}=\frac{m{l}^2}{x}$
$\Rightarrow x=12$