A thin wire of length ' L ' and linear mass density ' m ' is bent into a circular ring (in x - y plane) with centre ' C ' as shown in figure. The moment of inertia of the ring about an axis yy' will be :

Step-by-step solution

Total mass of the ring, M = mL . Since the wire of length L is bent into a circular ring of radius R , its circumference is L . 2 R = L R = L 2 The axis yy' is a tangent to the ring in its plane. The moment of inertia of a ring about its diametric axis is I_ d = 1 2 MR^2 . Using the parallel axis theorem, the moment of inertia of the ring about the tangent yy' is: I_ yy' = I_ d + MR^2 I_ yy' = 1 2 MR^2 + MR^2 = 3 2 MR^2 Substituting the values of M and R : I_ yy' = 3 2 (mL) ( L 2 )^2 I_ yy' = 3 2 mL ( L^2 4 ^2 ) I_