Question

A thin rod of mass $M$ and length $L$ is rotating about an axis perpendicular to the length of the rod and passing through its centre. If half of the total length rod that lies at one side of axis is cut and removed then the moment of inertia of remaining rod is:

• A. $\frac{M{L}^2}{12}$
• B. $\frac{M{L}^2}{18}$
• C. $\frac{M{L}^2}{24}$
• D. $\frac{M{L}^2}{36}$

Answer 1

Answer: (C)

$\frac{M{L}^2}{24}$

$I=\frac{M{L}^2}{12}$
If a half rod is removed. mass becomes $\frac{M}{2}$ and the length of the rod becomes $\frac{L}{2}$
$I^{\prime }=\frac{1}{3}\left(\frac{M}{2}\right){\left(\frac{L}{2}\right)}^2$
$=\frac{1}{24}M{L}^2$