Question

What is the moment of a rod of mass M, length l about an axis perpendicular to it through one end?

Answer 1

Hint: We know that moment of inertia of mass M, length l about the axis passing through its centre can be given by the formula, $I={\displaystyle \frac{1}{12}}M{L}^2$. Now, to find the moment of inertia along one of the ends of the rod, we will use parallel axis theorem which can be given as, $I^{\prime }=I+m{a}^2$, and we will find the answer.

Formula used: $I={\displaystyle \frac{1}{12}}M{L}^2$, $I^{\prime }=I+m{a}^2$

Complete step by step answer:
In question we are asked to find the moment of inertia of a rod of mass M and length l, along the axis perpendicular to its end, so first of all we will draw a figure for our simplicity,

Now, the moment of inertia of rod along the axis passing through its centre can be given by the formula,
$I={\displaystyle \frac{1}{12}}M{L}^2$
Where, M is mass of rod and l is length of rod
Now, to find the moment of inertia along axis perpendicular to one end of rod can be given by parallel axis theorem, which can be given as,
$I^{\prime }=I+m{a}^2$
Where, m is mass of rod, a is distance of axis from centre and I is moment of inertia along centre axis.
So, from it can be said that axis passing through one of the ends is at distance ${\displaystyle \frac{l}{2}}$, from centre, so on substituting this in expression we will get,
$I^{\prime }={\displaystyle \frac{1}{12}}M{L}^2+M{\left({\displaystyle \frac{L}{2}}\right)}^2$
$\Rightarrow I^{\prime }={\displaystyle \frac{1}{12}}M{L}^2+M{\displaystyle \frac{L^2}{4}}={\displaystyle \frac{M{L}^2+3M{L}^2}{12}}$
$\Rightarrow I^{\prime }={\displaystyle \frac{M{L}^2(1+3)}{12}}={\displaystyle \frac{4M{L}^2}{12}}={\displaystyle \frac{M{L}^2}{3}}$
Thus, the moment of inertia of rod along the axis perpendicular to one of its ends is ${\displaystyle \frac{M{L}^2}{3}}$.

Note: One can also calculate this by considering the moment of inertia as half of over all moment of inertia at one of the ends, mass becomes double as everything is concentrated at one point and length is considered as double, and by applying the formula of moment of inertia of rod passing through centre of one end, so the expression will be, $I^{\prime }={\displaystyle \frac{\left(2M\right){\left(2L\right)}^2}{12}}\times {\displaystyle \frac{1}{2}}={\displaystyle \frac{M{L}^2}{3}}$. So, it can be considered as an alternate method of solving such problems.