Question

A thin uniform rod is of mass $\text{M}$ and length $\text{L}$. Find the radius of gyration for rotation about an axis passing through a point at a distance of $\frac{L}{4}$ from the centre and perpendicular to the rod is $\sqrt{\frac{n}{48}}L$.Then $n$ is

Answer 1

As we know,
MOI of the rod about axis passing through its centre and perpendicular to the rod $=\frac{M{L}^2}{12}$
By parallel axis theorem, MOI about new axis:-
$=\frac{M{L}^2}{12}+M{\left(\frac{L}{4}\right)}^2=M{K}^2$
$\Rightarrow \frac{L^2}{12}+\frac{L^2}{16}=K^2$
or $K=\sqrt{\frac{7}{48}}L$
So, $n=7$