Physics - Rotation Question with Solution | TestHub

The moment of inertia of a uniform rod of mass $M$ and length $L$ about an axis passing through its centre of mass and perpendicular to its length is $I_{cm}=\frac{M{L}^2}{12}$.

The new axis passes through the point $(0,\frac{2L}{3})$ and is parallel to the $z$-axis. The perpendicular distance $d$ between the parallel axes is $d=\frac{2L}{3}$.

Using the parallel axis theorem, the moment of inertia about the new axis is given by:

$I=I_{cm}+M{d}^2$

$I=\frac{M{L}^2}{12}+M{\left(\frac{2L}{3}\right)}^2$

$I=\frac{M{L}^2}{12}+\frac{4M{L}^2}{9}$

$I=\frac{3M{L}^2+16M{L}^2}{36}$

$I=\frac{19}{36}M{L}^2$

Answer: $\frac{19}{36}M{L}^2$