Question

If $I_1$ the moment of inertia of a thin rod about an axis perpendicular to its length and passing through its centre of mass and $I_2$ the moment of inertia of the ring formed by the same rod about an axis passing through the centre of mass of the ring and perpendicular to the planet of the ring. Then find the ratio $\frac{I_1}{I_2}$

Answer 1

$I_1=\frac{M{L}^2}{12}$

find the radius when ring is bent in the form of circle

$L=2\pi r$

implies that $r=\frac{L}{2\pi }$

$MI$ of a ring about its central axis is

$I_2=M{r}^2$

$=M{\left(\frac{L}{2\pi }\right)}^2$

$I_1:I_2=\frac{\frac{M{L}^2}{12}}{m{\left(\frac{L}{2\pi }\right)}^2}$

$={\pi }^2:3$