Question

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

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

Answer 1

Answer: (C)

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