Question

Moment of inertia of a thin rod of mass m and length l about at axis passing through a point ${\displaystyle \frac{l}{4}}$ from one and perpendicular to the rod is:
$A.{\displaystyle \frac{m{l}^2}{12}}$
$B.{\displaystyle \frac{m{l}^2}{13}}$
$C.{\displaystyle \frac{7m{l}^2}{48}}$
$D.{\displaystyle \frac{m{l}^2}{9}}$

Answer 1

Hint: Use the theorem of parallel axes. Substitute the values in the formula and get a moment of inertia of a thin rod of mass m and length l about at the axis passing through a point ${\displaystyle \frac{l}{4}}$ from one and perpendicular to the rod.

Formula used:
$I=I_{CM}+m{d}^2$

Complete answer:

Given: $d={\displaystyle \frac{l}{4}}$
According to the theorem of parallel axes,
$I=I_{CM}+m{d}^2$ …(1)
where, I: Moment of inertia of thin rod
           $I_{CM}$: Moment of Inertia at center of mass
But, we know $I_{CM}={\displaystyle \frac{1}{12}}m{l}^2$
Therefore, substituting the values in the equation. (1) we get,
$I={\displaystyle \frac{1}{12}}m{l}^2+m{\left({\displaystyle \frac{l}{4}}\right)}^2$
$\therefore I={\displaystyle \frac{m{l}^2}{12}}+{\displaystyle \frac{m{l}^2}{16}}$
$\therefore I={\displaystyle \frac{7m{l}^2}{48}}$
Therefore, Moment of inertia of a thin rod of mass m and length l about at axis passing through a point ${\displaystyle \frac{l}{4}}$ from one and perpendicular to the rod is $I={\displaystyle \frac{7m{l}^2}{48}}$.

So, the correct answer is “Option C”.

Note:
For a uniform rod with negligible thickness, the moment of inertia about its center of mass is $I_{CM}={\displaystyle \frac{1}{12}}m{l}^2$. And the moment of inertia about the end of the rod is $I_{end}={\displaystyle \frac{1}{3}}m{l}^2$.