Consider a badminton racket with length scales as shown in the figure.
If the mass of the linear and circular portions of the badminton racket are same (M) and the mass of the threads are negligible, the moment of inertia of the racket about an axis perpendicular to the handle and in the plane of the ring at, r/2 distance from the end A of the handle will be ................ Mr².

Solution

Physics sir · 4 steps

Split the racket into a rod and a ring, then shift each one to the axis with the parallel axis theorem.

1. Handle as a rod. The handle is a rod of mass M and length 6r. Its centre is 3r from A, so it sits 3r - r/2 = 5r/2 from the axis.
2. Rod about the axis. About its own centre a rod has M(6r)²/12 = 3Mr². Shifting to the axis adds M(5r/2)² = 25/4Mr², so I_(rod) = 37/4Mr².
3. Ring about the axis. The axis lies in the plane of the ring, so about its own centre the ring turns about a diameter: Mr²/2. The ring's centre is 6r + r = 7r from A, which is 13r/2 from the axis. So I_(ring) = Mr²/2 + 169/4Mr² = 171/4Mr².

   I = 37/4Mr² + 171/4Mr² = 208/4Mr² = 52 Mr²

4. So. The moment of inertia of the racket about the given axis is 52 Mr².

The common slip: 40. Measuring the ring's distance to its near edge (11r/2 from the axis) instead of to its centre (13r/2) gives 40, not 52.

The idea to keep. For a composite body, find each part's moment of inertia about its own centre, then shift it with Md². Always measure d to the centre of mass, and check whether the axis is a diameter of the ring.

Answer from the NTA final answer key. Solution by PrepWiser.