The moment of inertia of a uniform thin rod about a perpendicular axis passing through one end is I 1 . The same rod is bent into a ring and its moment of inertia about a diameter is I 2 . If I 1 I 2 is x π 2 3 , then the value of x will be ______.

Step-by-step solution

Moment of inertia of rod rotating about an axis passing through one end is I 1 = m l 2 3           ⋯ 1 Moment of inertia of ring rotating about one of its diameters is I 2 = m R 2 2 . Since the rod is bent into a ring, the circumference of the ring and the length of the rod are equal, l = 2 π R Then, I 2 = m R 2 2 = m l 2 8 π 2 Thus, the ratio I 1 I 2 = 8 π 2 3 , so x = 8 .