Question

The moment of inertia of rod of mass $M$ length $I$ about an axis perpendicular to it through one end is

• A. $ \frac{Ml^{2}}{12} $
• B. $ \frac{Ml^{2}}{2} $
• C. $ \frac{Ml^{2}}{3} $
• D. $ \frac{Ml^{2}}{4} $

Answer 1

Answer: (C)

For the rod of mass $M$ and length $l$ ,
$I = M ^{2} / 12 .$ Using the parallel axes theorem,
$I '= I + Ma ^{2}$ with $a =1 / 2$ we get,
$I'= M \frac{ l ^{2}}{12}+ M \left(\frac{1}{2}\right)^{2}=\frac{ Ml ^{2}}{3}$
We can check this independently since $I$ is half the moment of inertia of a rod of mass $2 M$ and length $2l$ about its midpoint,
$I' =2 M \cdot \frac{4 l ^{2}}{12} \times \frac{1}{2}=\frac{ Ml ^{2}}{3}$