Physics - Rotation Question with Solution | TestHub

The moment of inertia of a thin uniform rod about an axis passing through its centre of mass and perpendicular to its length is given by:

$I_{cm}=\frac{M{L}^2}{12}$

The centre of mass is at a distance of $\frac{L}{2}$ from either end. The given axis is at a distance of $\frac{L}{4}$ from one end. The perpendicular distance $h$ between the parallel axes is:

$h=\frac{L}{2}-\frac{L}{4}=\frac{L}{4}$

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

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

$I=\frac{M{L}^2}{12}+M{\left(\frac{L}{4}\right)}^2$

$I=\frac{M{L}^2}{12}+\frac{M{L}^2}{16}$

$I=\frac{4M{L}^2+3M{L}^2}{48}=\frac{7M{L}^2}{48}$

Answer: $\frac{7{\mathrm{ML}}^2}{48}$