Physics - Rotation Question with Solution | TestHub

The square frame consists of four identical rods: AB, BC, CD, and DA, each of mass $m$ and length $\ell$.

The moment of inertia of the frame about the x-axis is the sum of the moments of inertia of each rod about the x-axis.

For rod AB, it lies on the x-axis, so its moment of inertia is $I_{AB}=0$.

For rod BC, it is perpendicular to the x-axis with one end on the x-axis, so its moment of inertia is $I_{BC}=\frac{m{\ell }^2}{3}$.

For rod CD, it is parallel to the x-axis at a perpendicular distance $\ell$, so its moment of inertia is $I_{CD}=m{\ell }^2$.

For rod DA, it is perpendicular to the x-axis with one end on the x-axis, so its moment of inertia is $I_{DA}=\frac{m{\ell }^2}{3}$.

The total moment of inertia of the frame about the x-axis is:

$I=I_{AB}+I_{BC}+I_{CD}+I_{DA}$

$I=0+\frac{m{\ell }^2}{3}+m{\ell }^2+\frac{m{\ell }^2}{3}$

$I=\frac{5m{\ell }^2}{3}$

Answer: $\frac{5m{\ell }^2}{3}$