Question

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

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

Moment of inertia of remaining rod is

$I=\frac{(\frac{M}{2})(\frac{L}{2}{)}^2}{3}$

$\Rightarrow I=\frac{(\frac{M}{2})(\frac{L^2}{4}{)}^2}{3}$

$\Rightarrow I=\frac{M{L}^2}{24}$