The moment of inertia of a ring about an axis perpendicular to its plane and passing through its center is…
The moment of inertia of a ring about an axis perpendicular to its plane and passing through its center is $4 \mathrm{~kg} \mathrm{~m}^2$. Its moment of inertia about the tangent in the plane is
$6 \mathrm{~kg} \mathrm{~m}^2$
$8 \mathrm{~kg} \mathrm{~m}^2$
$4 \mathrm{~kg} \mathrm{~m}^2$
$2 \mathrm{~kg} \mathrm{~m}^2$
Solution
Ring has moment of inertia $M R^2$ about the symmetric central axis. Using perpendicular axis theorem one can get moment of inertia about the planar diagonal as $\frac{M R^2}{2}$
Now, using to the parallel axis theorem, we shift by a distance R to be at the tangential position:
$I_t=\frac{1}{2} M R^2+M R^2=\frac{3}{2} M R^2$
Given $M R^2=4 \mathrm{kgm}^2$ :
$I_t=\frac{3}{2} \times 4 \mathrm{kgm}^2=6 \mathrm{kgm}^2$