Question

A thin uniform rod of mass $M$ and length $L$ has its moment of inertia $I_1$, about its perpendicular bisector. The rod is bend in the form of a semicircular arc. Now its moment of inertia through the centre of the semi circular arc and perpendicular to its plane is $I_2$. The ratio of $I_1:I_2$ will be ________ :

• A. < 1
• B. > 1
• C. = 1
• D. can't be said

Answer 1

The correct option is A < 1

For a rod we have moment of inertia as $\frac{M{L}^2}{12}$. When it is bent the radius of curvature will be given by $\frac{L}{\pi }$.
Now as the moment of inertia of a ring about its center is $M{R}^2$, thus using parallel axis theorem moment of inertia of ring about a tangential axis is $2M{R}^2$. Thus for half of ring would be $M{R}^2$.
For the given bent semicircular arc we will have moment of inertia as $\frac{M{L}^2}{{\pi }^2}=\frac{M{L}^2}{9.85}$ which is greater than moment of inertia of the rod. Thus $I_1:I_2<1$