Question

Moment of inertia of a ring about its diameter is I. Its moment of inertia about axis passing through its centre and normal to its plane is:

(a) 2I

(b) \(\frac{I}{2}\)

(c) \(\frac{3}{2}I\)

(d) I

Answer 1

Given: I_AB = I_CD = I_d = I

I_zz = ?

On applying theorem of perpendicular axes

I_zz' = I_AB + I_CD = I + I = 2I

I_zz' = 2I