The moment of inertia of a thin rod of mass $M$ and length $L$ about an axis passing through a point at a…
The moment of inertia of a thin rod of mass $M$ and length $L$ about an axis passing through a point at a distance $\frac{L}{4}$ from its centre and perpendicular to its length is
$\frac{M L^3}{48}$
$\frac{M L^2}{48}$
$\frac{M L^2}{12}$
$\frac{7 M L^2}{48}$
Solution
Let, $M$ be the mass of the thin rod and $L$ be the length of thin rod.
Moment of inertia of thin rod about its centre of mass, $I_{\mathrm{COM}}=\frac{1}{12} M L^2$
By parallel axis theorem,
Moment of inertia about an axis = Moment of inertia about centre of mass + Mass $\times$ (distance from axis $)^2$
$
\begin{aligned}
\Rightarrow \quad I & =I_{\mathrm{coM}}+M(d)^2 \\
& =\frac{1}{12} M L^2+M\left(\frac{L}{4}\right)^2 \quad\left(\text { Given, } d=\frac{L}{4}\right) \\
& =\left(\frac{4+3}{48}\right) M L^2=\frac{7}{48} M L^2
\end{aligned}
$