The moment of inertia of a ring about an axis perpendicular to its plane and passing through its center is 4 ~kg ~m ^2 . Its moment of inertia about the tangent in the plane is

Step-by-step 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 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= 1 2 M R^2+M R^2= 3 2 M R^2 Given M R^2=4 kgm ^2 : I_t= 3 2 4 kgm ^2=6 kgm ^2