The moment of inertia of a uniform thin rod about a perpendicular axis passing through one end is I₁. The same rod is bent into a ring and its moment of inertia about a diameter is I₂. If I₁/I₂ is (xπ ²)/3, then the value of x will be ____________.

Solution

Physics sir · 4 steps

Same mass, same length: the rod's length becomes the ring's circumference, so R = L/2π.

1. Rod about end. For a uniform rod of mass M and length L, about a perpendicular axis through one end, I₁ = ML²/3.
2. Find the radius. Bending keeps the mass M and makes the length L the circumference. So 2π R = L, which gives R = L/2π.
3. Ring about diameter. About its central perpendicular axis the ring has MR². By the perpendicular axis theorem, two diameters share this equally, so I₂ = MR²/2 = M/2 × L²/4π² = ML²/8π².

   I₁/I₂ = ML²/3 × 8π²/ML² = 8π²/3

4. So. Comparing with xπ²/3 gives x = 8.

The common slip: 4. Using MR² for the ring about a diameter makes I₂ twice too big, giving 4π²/3 and x = 4. That value is for the axis through the centre, perpendicular to the plane.

The idea to keep. When a wire is reshaped, mass and length stay fixed. Write the new size in terms of L first, then use the right axis: a ring about a diameter has MR²/2, not MR².

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