Question

$I_1$ is moment of inertia of a thin circular ring about its own axis. The ring is cut at a point then it is unfolded into a straight rod. If $I_2$ is moment of inertia of the rod about an axis perpendicular to the length of rod and passing through its centre, then the ratio of $I_1$ to $I_2$ is:

• A. $3:\pi$
• B. $3:{\pi }^2$
• C. $4:\pi$
• D. $4:{\pi }^2$

Answer 1

The correct option is B $3:{\pi }^2$

Let mass and radius of ring be $m$ and $r$.

So $I_1=m{r}^2$

Now when cutted and unfolded into rod, the length of the rod is $l=2\pi r$

So $I_2=m{l}^2/12=m\times 4{\pi }^2{r}^2/12={\pi }^{2}m{r}^2/3$

So, $I_1:I_2=3:{\pi }^2$