Question

$I$ is moment of inertia of a thin rod of uniform thickness about an axis perpendicular to its length and passing through its centre. The rod at one of two sides of axis is cut and removed. The moment of inertia of remaining rod about same axis is :

• A. $\frac{I}{2}$
• B. $\frac{I}{3}$
• C. $\frac{I}{4}$
• D. $\frac{I}{6}$

Answer 1

The correct option is A $\frac{I}{2}$

Let initial mass and length be $m$ and $l$.

So MI about an axis perpendicular to its length and passing through its center is $I=m{l}^2/12$

After cutting into half, mass is $m/2$ and length is $l/2$

So MI about the same axis is $\left(m/2\right)\times (l/2{)}^2/3=m{l}^2/24=I/2$